text:
⇑ back to "U.S."
Jehovah's Witnesses Paying Millions in California Sex Abuse Case
4:00PM EST 11/4/2014	 Marty Graham
Jose Lopez was 7-years-old when he was molested during a Bible study. (Fox 2 News)
A Californian judge has ordered the Jehovah's Witnesses to pay $13.5 million to a man who was sexually abused in the 1980s by a church leader, attorney for the victim said on Friday.
Church elders had assigned a man to work with Jose Lopez on Bible studies, even though they knew he had admitted to molesting another boy in 1982, because they felt he was "repentant," Lopez's attorney Irwin Zalkin said.
Lopez, who filed the civil suit in 2010, first went to the Spanish-speaking Kingdom Hall church in San Diego in 1985 at the age of seven, according to the complaint.
Mario Moreno, a lawyer for the church—whose parent group is the Watchtower Bible and Tract Society of New York—told the San Diego Union-Tribune newspaper that the church will appeal Superior Court Judge Joan Lewis' judgment.
The trial judge's decision is a drastic action for any judge to take given the circumstances of this case," the statement in the paper said. "Jehovah's Witnesses abhor child abuse and strive to protect children from such acts."
The U.S.-based Jehovah's Witnesses number about 8 million worldwide and are known for their foreign ministries as well as their door-to-door campaigns.
Join us on our podcast each weekday for an interesting story, well told, from Charisma News. Listen at charismapodcastnetwork.com.
Related topics: Christianity | Church
Nine Days in Heaven


By any objective standard, Prime Minister Benjamin Netanyahu's trip to Washington last week was a stunning success. US President Donald Trump and first lady Melania Trump greeted the premier and his wife, Sara, as old friends. The substantive talks Netanyahu held with Trump on Iran and a host of other key issues were extraordinarily positive.
Then of course, there was the rock-star welcome Netanyahu received at the AIPAC conference.
If Israelis were astounded by the royal treatment Netanyahu received during his visit to India last year, his visit to Washington made clear that what happened in India was no fluke. He is quite clearly one of the most well-regarded statesmen in the world.
Then again, most Israelis could be excused for having little idea either that Netanyahu was treated like a king in India or that he was treated like a second coming of Winston Churchill in the US.
The Israeli media barely covered his trip to India. As for his trip to the US, the media presented everything Netanyahu said and did in the context of the police's obsessive-compulsive criminal probes of Netanyahu.
The political crisis over the haredi draft law, which this week brought the Knesset to the brink of dissolution and Israel to a new general election, was entirely the result of the joint efforts of police investigators and the media to delegitimize Netanyahu as a leader and criminalize him as a person.
Several ministers from parties in Netanyahu's coalition, together with the media, presented Netanyahu's handling of the crisis as self-serving.
Netanyahu, the media and his coalition partners alleged, was putting his personal interest above the national interest. Netanyahu, they insisted, was allowing a minor crisis (ostensibly about whether or not a bill would pass that sets out draft quotas for haredi youths before the annual budget bill went through the Knesset), to become a major crisis that would bring down the government. He was doing so, they said, because he wanted an early election to strengthen his position vis-a-vis the police investigators.
In other words they said, Netanyahu wanted to plunge Israel into political uncertainty for months on end, at the cost of billions of shekels, just to win another election.
Unlike the egotistical premier, the ministers and media said, Netanyahu's coalition partners worked around the clock to solve the crisis without him and force him to accept the compromise they hammered out. Unlike Netanyahu, they sneered, they put the national interest first.
"We know the public doesn't want an election.
And we're serving the public," they declared. The media delightedly agreed.
In truth, those denunciations of Netanyahu were little more than spin.
In reality, the public has an interest in renewing the government's mandate. Working together, the police and the media have used the probes against Netanyahu to destabilize the government and delegitimize its power.
Like their borderline delusional coverage of Netanyahu's trip to the US, the media present every move made by every government minister from every party in the coalition in the context of the police probes and the prospects for an early election.
In this framework, everything the government does is suspect. The fact that the police have demonstrated no credible proof of their claims that Netanyahu accepted bribes in any of their leaks or official statements regarding any of their multiple probes is of no consequence. As far as the major radio, television and print media are concerned, Netanyahu is a crook.
And since the media portray Netanyahu as a crook, everything he says and does and everything his ministers say and do is presented against the backdrop of that specious, unsupported and certainly unproven conclusion.
Under the circumstances, the need for a new mandate is self-evident. The public is the source of the government's power. The police and media insist that the mandate the government received is no longer legitimate. So the public has to express its views on the government in one way or another.
In other words, to the extent Netanyahu wanted to disband the Knesset and announce an election this week, he was right to feel the way he did.
He wasn't being egotistical. The public's interest is harmed by the delegitimization of its government.
Broadly speaking there are three ways the public can make its position known.
First, its elected representatives can assert it.
Netanyahu's coalition members could issue a declaration supporting him. The 67 members of Knesset and the ministers from the parties in the governing coalition can put out a declaration announcing that as they do in all of their endeavors, with everything related to the police probes of Netanyahu, they intend to respect the rule of law.
Since under Basic Law: The Government, a prime minister is only expected to resign from office if he receives a final judgment convicting him of committing crimes, the coalition members could assert that in conformance with the rule of law, they expect Netanyahu to serve out his term and will support him through the end of the term.
A statement along these lines from the ministers and lawmakers in the coalition would constitute a renewed mandate for Netanyahu and the government to govern. It would send the message that in Israel the public chooses its leader.
The police and even the attorney-general cannot replace the public.
For whatever reason, to date, no such declaration has been produced. Which left Netanyahu with two other options this week.
First, he could have brought down his government, disbanded the Knesset and called an election in June. Since the public has an interest in having a legitimate government and since the media and the police are calling the legitimacy of the government into question, going to an election would serve the public's interest. Certainly, elections are the most direct way to find out if the public agrees with the police and the media commentators and thinks Netanyahu should go, or if the public wants him to stay on and continue leading the country.
While elections are the most direct way to get a mandate, the commentators and ministers are right that elections are an extreme step. They cost a lot of money. What's more, war can break out at any moment with Hezbollah-controlled Lebanon. Israel would be worse off if an interim government was leading it in time of war.
And this brings us to the final way a government can renew its mandate. It's what happened this week.
Two things happened this week.
First, news organizations had pollsters for the big news companies ask the public a few key questions.
The pollsters asked whether the public wanted an early election. Seventy percent said no. If the public thought the police and media were right, and that Netanyahu should be thrown out of office, the polls would have received opposite results. Everyone would have wanted an election.
The public was then asked whom it supported for prime minister. Netanyahu out-polled his closest rival – media and police favorite Yesh Atid Party leader Yair Lapid – 3 to 1. If the public agreed with the media and the police, which both support Lapid, the former television talk show host, then the results would have reflected that agreement.
Finally, the pollsters asked which party the public intended to vote for in the next election. The Likud under Netanyahu gained three Knesset seats. Yesh Atid lost two.
The implications, again, are self-evident. The public wants Netanyahu to keep leading the country.
But a poll taken on any given day isn't a mandate.
Which brings us to the second thing that happened this week. Both the opposition parties and every coalition member other than the Likud made clear that they completely opposed an election.
This rare unanimity demonstrated clearly that the political world is certain the public supports Netanyahu and wants him to continue in office.
Taken together, the polls and the wall-to-wall opposition to an election among coalition and opposition parties alike provided Netanyahu and the government with a renewed mandate to govern.
This then brings us back to the investigations that triggered the coalition crisis and fomented Netanyahu's need for a new mandate. From the police leaks over the past couple of weeks it is fairly clear that investigators have hit the wall.
They recommended that Netanyahu be indicted for accepting bribes in Cases 1000 and 2000 and the public yawned and rolled its eyes. The attorney-general sent them back to get real evidence.
Case 3000 is so weak the police can't even figure out how to begin investigating Netanyahu for anything.
As for Case 4000, the latest leaks suggest the police have nothing whatsoever to accuse Netanyahu of having done. The notion that he would give preferential treatment to telecommunications giant Bezeq because Bezeq's owner Shaul Elovitch had Walla website reporters write a couple of nice articles about him and his wife is absurd on its face. And this week the police leaked their new smoking gun: Netanyahu allegedly tried to convince his billionaire friends to start an Israeli version of Fox News.
To which his voters reply: Good for him! That latest leak indicates two things. First, again, the police have no proof that Netanyahu committed any wrongdoing. And second, the police is completely cut off from the public. Netanyahu has run for prime minister four times and won every race by running against the media. If the public didn't share his conviction that the media are impossibly biased, he would either have lost the elections or run on a different platform. The fact that the police think that the public will turn against Netanyahu because he tried to get fair media coverage is ridiculous.
But who knows? Maybe the police will pull a rabbit out a hat and prove something truly terrible.
In the meantime, Netanyahu received a new mandate to lead the country from the public this week. So whatever happens with the investigations, he and his ministers can credibly and comfortably lead the country until November 2019.
Caroline B. Glick is the Director of the Israeli Security Project at the David Horowitz Freedom Center in Los Angeles and the senior contributing editor of the Jerusalem Post, where this article first appeared.


Do you like to scrapbook? Make your own greeting cards? Decorate bulletin boards?
If you said yes to any of the above, then you will love the Cricut Expression cutting machine!
I have had my Provo Craft Cricut Expression machine for several years now… and I LOVE it! It has taken my paper crafting to a whole new level of professionalism and fun.
Use the sticky cutting mat to adhere your selected paper.
Choose your image on the keypad.
See it on the display screen.
Then, sit back and watch the Cricut machine automatically move the mat back and forth as it precisely cuts the image.
The Original Cricut uses a 6″x12″ cutting mat.
The Cricut Expression can use 12″x12″ or 12″x24″ cutting mats.
You can use paper the same size as your mat or small scraps, depending on the size you want your image to be.
The size is chosen using a dial on the side of the Cricut.
The machine will tell you if you choose an image that is too large for your paper.
Each Cricut cartridge includes a plastic overlay to place on the machine's keypad, plus an instruction booklet. The plastic overlay will indicate key selections for that cartridge only.
By using special keys called "creative features," every regular key will cut multiple things. The booklet will show you exactly what the options are for each key.
There are 2 types of cartridges: shape and font.
Whew – that's a lot, right?!
Each Cricut cartridge has different "creative features," which can allow literally hundreds of different cuts from just one cartridge. Since you can use paper of any size and color, the possibilities are endless!
There are additional features on the Cricut Expression that provide even more options — such as quantity, flip, portrait, fit-to page, and more.
See if the price includes any Cricut cartridges or tools. I have seen them advertised several ways (from nothing included to your choice of 2 or more cartridges). I got mine on eBay and it included 2 cartridges and 2 cutting mats.
Cricut tools are usually sold separately. A Cricut cutting mat is required, because it has a thin layer of tacky adhesive which gently grips materials and holds them flat to prevent tearing or shifting during trimming. You will also need a Cricut spatula to help remove cuts from the mat.
I tried using mine without a spatula, but couldn't do it. I even tried using a real spatula from my kitchen (a tiny one, not much bigger than the Cricut spatula), but no luck.
When you cut out a word (such as "HAPPY") the word will be 5 individual letters, not one connected word. This can be a real challenge, especially with small letters. I have a pair of craft tweezers that I use for this, and I typically use a fat font (like Roly Poly) because it is easier to work with. Really small, thin fonts can be very difficult to pull off the sticky mat without tearing them (truly impossible without that Cricut spatula). Many cartridges have "phrases" as an option, which will be cut as one continuous shape. There is Cricut software available called Design Studio which is available. It allows you to "weld" letters and shapes together to form one word.
The Cricut sticky mats don't last forever. Fortunately, there are lots of tips and videos on how to make the mat "re-sticky" using products like Aleene's Tack It Over & Over or repositionable adhesive sprays. I have successfully done that to one of mine. Be careful with a new or newly sticky mat and thin paper; it might tear as you pull it off. Even some cardstock will be tricky to remove from very sticky mats.
Speaking of Cricut tips and hints, they are everywhere! So don't worry if you get your new Cricut and can't figure out how to do what you want. They have a great online community full of people who are willing to share their expertise and ideas. Plus, there are tons of "how-to" Cricut videos posted by users online.
There are over 100 cartridges available for both Cricut machines! But you won't need that many. I got by for about a year with just 2 cartridges — one with fonts and one with shapes. Now I have 8. There are still so many things I can cut with my 8 Cricut cartridges that I'll never get to them all!
If, like me, you're always watching for bargains and you regularly shop at Hobby Lobby, Michael's or JoAnn Fabrics, then you are probably already familiar with their weekly coupons. Unfortunately, they all state that the coupon cannot be used on any Cricut products. Those stores do occasionally have good sales on Cricut items though.
Which Cricut Machine Is Best?
If you're considering a Cricut and think you want the original, smaller Cricut Personal Cutter, I would encourage you to reconsider and get the Cricut Expression instead!
With the original Cricut, you will have to purchase the special 6″x12″ paper or always cut your own paper to that size.
The Cricut Expression can use paper commonly used in scrapbooking: 8.5″x11″ or 12×12″.
These next 2 videos might help you see some of the similarities and differences between the original Cricut Personal Cutter and the Cricut Expression machine.
Cricut vs. Silhouette: Which One Is Better?
Whether you're a teacher, homeschooling mom, or just an avid crafter, the Cricut machine is a tool that you will use over and over again.
Regardless of which Cricut you ultimately choose, you can bet that you'll get years of enjoyment from it!


text:
Of course, one of the most important steps in the process of selling products is setting prices. One of benefits of using Amazon to sell goods online is that you can see what other merchants are charging for similar items. If you want to make sure your copy of The Scarlet Letter sells over the other ones listed, set the price with shipping lower than your competitors. You can also specify whether you want to handle the shipping yourself or allow Amazon to do so.
Prime started as a two-day shipping membership for devoted Amazon shoppers, and it didn't add any additional benefits until 2011. However, since then, Prime has grown into a subscription service with more than 100 million paid users worldwide, while the service itself has evolved to include additional perks over the years, including a Prime credit card now with 5 percent cash back. (Amazon also operates Amazon Pay for purchasing online goods elsewhere with your Amazon account, and the Amazon Cash service for translating cash into store credit using a barcode, although neither are restricted to Prime users.)
Perhaps the most prominent Prime perk, however, is access to Amazon Prime Video. The video on-demand service started in 2006 as Amazon Unboxed, but was rebranded in 2008 and integrated into the Prime service three years later, where it became a huge selling point for Amazon's annual subscription. It now boasts thousands of free TV shows, films, and games, all accessible on pretty much every screen available.
AWS started way back in 2000 as a way to help other retailers manage e-commerce operations, but it soon expanded into much more when key project members managed to convince Bezos that improving and evolving Amazon's own infrastructure may hold the key to a new business model. In 2006, the product as we know it today launched into public availability and proved to be a pioneer for the entire cloud computing industry, offering cloud storage, hosting, and a suite of other tools for managing entire digital infrastructures in remote data centers. The division now pulls in roughly $6 billion every quarter and continues to grow at breakneck pace. It earned $17.5 billion in revenue in all of 2017 and regularly outperforms the company's entire North American retail division in terms of profit.
The Amazon sales rank (ASR) provides an indication of the popularity of a product sold on any Amazon locale. It is a relative indicator of popularity that is updated hourly. Effectively, it is a "best sellers list" for the millions of products stocked by Amazon.[134] While the ASR has no direct effect on the sales of a product, it is used by Amazon to determine which products to include in its bestsellers lists.[134] Products that appear in these lists enjoy additional exposure on the Amazon website and this may lead to an increase in sales. In particular, products that experience large jumps (up or down) in their sales ranks may be included within Amazon's lists of "movers and shakers"; such a listing provides additional exposure that might lead to an increase in sales.[135] For competitive reasons, Amazon does not release actual sales figures to the public. However, Amazon has now begun to release point of sale data via the Nielsen BookScan service to verified authors.[136] While the ASR has been the source of much speculation by publishers, manufacturers, and marketers, Amazon itself does not release the details of its sales rank calculation algorithm. Some companies have analyzed Amazon sales data to generate sales estimates based on the ASR,[137] though Amazon states:
Shelfari was a social cataloging website for books. Shelfari users built virtual bookshelves of the titles which they owned or had read and they could rate, review, tag and discuss their books. Users could also create groups that other members could join, create discussions and talk about books, or other topics. Recommendations could be sent to friends on the site for what books to read. Amazon bought the company in August 2008.[100] Shelfari continued to function as an independent book social network within the Amazon until January 2016, when Amazon announced that it would be merging Shelfari with Goodreads and closing down Shelfari.[102][103]
Before you can sell your goods on Amazon, you need to create a store through Seller Central. As a seller, you have the choice of creating either an individual or professional account. While individuals can list their goods for free, selling as a professional costs $39.99 a month plus additional fees based on the value of the final sale. Individuals typically pay $0.99 per item sold in addition to variable fees based on an item's category.
This service isn't just for packaged goods, either. Many stores offer shoppers the opportunity to order fresh produce, meats, dairy, and other groceries online and get these items delivered directly to their homes. In addition, many products that aren't necessarily found easily in stores can be found online — sometimes for a much cheaper price. You might even be able to take advantage of online rewards or cash-back offers, too.
With all the reasons to celebrate and get ready for dress season, we're coming in full swing with spring dresses for prom, graduation dresses, and crowd pleasing wedding guest dresses. The wildest time of the year is soon to follow with festival season! We'll be dressing down in stunning swimwear & rocking our hottest outfits to feel that festival fire under the desert sun. See you there, babe!
Amazon's transformation into the world's more pervasive retail operation wouldn't be complete unless the company began a seemingly counterintuitive push from online to offline. Starting with its brick-and-mortar bookstores in 2015 — first in Seattle and now in Chicago and New York City — Amazon established its intent to compete on all fronts with its retail competitors.

redpajama_set_name: RedPajamaCommonCrawl


text:
Q: Different o/p - Lists I tried to make a program that gives me errors, but actually I can't put inputs.
What I want to do is something that receives a list of marks, from 0 to 20, and a minimum passing grade min_grade, and returns the number of entries in the grades list with a value greater than or equal to min_grade, and another function with the media_approved:
The output I want will be something like this:
approved([11.2, 13.5, 7.6, 4.4, 12.2, 12.1, 18.2, 9.4, 18.3, 12.33], 9.5)
6
approved([7.5, 3.3, 6.6, 6.4], 9.5)
0
average_passed([11.2, 13.5, 7.6, 4.4, 12.2, 12.1, 18.2, 9.4, 18.3, 12.3], 9.5)
13.9166
average_passed([7.5, 3.3, 6.6, 6.4], 9.5)
0

The program I made is wrong, because it only calculates the average, it is badly programmed and does not have a part of what I said earlier.
Then goes the code:
def calc_average(s, scores):
    scores = []

c = float(input())

def average(grades, c):
    return sum(scores) / len(scores)

s = [11.2, 13.5, 7.6, 4.4, 12.2, 12.1, 18.2, 9.4, 18.3, 12.3]
print(average(s))

Can anyone help me, to change my code so that it is the same as the output I put earlier?
Thank you for your help.

A: If you just want to calculate number of values greater than the given i/p. One of the potential solutions is:

def fn(input_list, target):
    approved = []

    for i in range(len(input_list)):
        if input_list[i] >= target:
            approved.append(input_list[i])

    return len(approved)

This surely solves the first problem. I am not sure what media_approved means. Could you please elaborate on the same.
Thank you, I hope this helps.

A: def averaged_passed(s, c):
  scores = []
  for i in range(len(s)):
     if(s[i]>=c):
       scores.append(s[i])
  try:
    return sum(scores)/len(scores)
  except:
    return('No one passed')

def approved(s, c):
  count =0
  for i in range(len(s)):
     if(s[i]>=c):
       count=count+1
  return count
c = float(input())

s = [7, 3, 6, 6]
print(pass(s,c))

A: From what I understand, you want two functions:

*

*approved to return the number of marks that are greater than or equal to min_grade.

*average_passed to return the average of all passed marks.

def approved(grades, min_grade):
    return len([grade for grade in grades if grade >= min_grade])

def average_passed(grades, min_grade):
    passed = [grade for grade in grades if grade >= min_grade]
    return sum(passed)/len(passed) if passed else 0

Output:
approved([11.2, 13.5, 7.6, 4.4, 12.2, 12.1, 18.2, 9.4, 18.3, 12.33], 9.5)
7
approved([7.5, 3.3, 6.6, 6.4], 9.5)
0
average_passed([11.2, 13.5, 7.6, 4.4, 12.2, 12.1, 18.2, 9.4, 18.3, 12.3], 9.5)
13.971428571428572
average_passed([7.5, 3.3, 6.6, 6.4], 9.5)
0
If you are beginner, you might want to refer to Python list comprehension and condensed if-else statement to understand the code.
For example,
return sum(passed)/len(passed) if passed else 0 is equivalent to
if passed: # returns True if the list is not empty, else False
    return sum(passed)/len(passed) # calc average if list is not empty
else:
    return 0 # return zero if list is empty

redpajama_set_name: RedPajamaStackExchange


text:
\section{Introduction}
An important tool in the study of the group $GL(n,{\mathbb{Z}})$ is provided by the geometric realization of the partially ordered set (poset) of proper direct summands of ${\mathbb{Z}}^n$. The natural inclusion ${\mathbb{Z}}^n\to {\mathbb{Q}}^n$ gives a one-to-one correspondence between proper direct summands of ${\mathbb{Z}}^n$ and proper subspaces of ${\mathbb{Q}}^n$, so that this poset is isomorphic to the spherical building $X_n$ for $GL(n,{\mathbb{Q}})$.  The term ``spherical" comes from the  Solomon-Tits theorem \cite{Sol}, which says that $X_n$ has the homotopy type of a bouquet of spheres:

\begin{theorem*} [Solomon-Tits Theorem]  {\sl The geometric realization of the poset of proper subspaces of an $n$-dimensional vector space has the homotopy type of a bouquet of spheres of dimension $n-2$. }
\end{theorem*}

The building $X_n$ encodes the structure of parabolic subgroups of $GL(n,{\mathbb{Q}})$: they are the stabilizers of simplices.  $X_n$ also parametrizes the Borel-Serre boundary of the homogeneous space for $GL(n,{\mathbb{R}})$.  The top-dimensional homology $H_{n-2}(X_n)$ is the Steinberg module $I_n$ for $GL(n,{\mathbb{Q}})$, and is a dualizing module for the homology of $GL(n,{\mathbb{Z}})$, i.e. for all coefficient modules $M$ there are isomorphisms
$$H^i(GL(n,{\mathbb{Z}});M)\to H_{d-i}(GL(n,{\mathbb{Z}});M\otimes I_n),$$
where $d=n(n-1)/2$ is the virtual cohomological dimension  of $GL(n,{\mathbb{Z}})$.

If one replaces $GL(n,{\mathbb{Z}})$ by the group $Aut(F_n)$ of automorphisms of the free group of rank $n$, the natural analog $FC_n$ of $X_n$ is the geometric realization of the poset of proper free factors of $F_n$. The abelianization map $F_n\to {\bf {\mathbb{Z}}}^n$ induces a map from $FC_n$ to the poset of summands of ${\mathbb{Z}}^n$.  In this paper we prove the analog of the Solomon-Tits theorem for $FC_n$:

\begin{theorem}   The geometric realization of the poset of proper free factors of $F_n$
has the homotopy type of a bouquet of spheres of dimension $n-2$.
\end{theorem}

By analogy, we call the top homology $H_{n-2}(FC_n)$ the {\it Steinberg module} for $Aut(F_n)$.  This leaves open some intriguing questions.  It has recently been shown that $Aut(F_n)$ is a virtual duality group~\cite{BF}; does the Steinberg module act as a dualizing module?  [This is answered in the negative for $n=5$ in the preprint \cite{HMNP} posted in 2022.]  There is an analog, called Autre space, of the homogeneous space for $GL(n,{\mathbb{Z}})$ and the Borel-Serre boundary; what is the relation  between this and the ``building" of free factors?

In \cite{Qui}, Quillen developed tools for studying the homotopy type of the geometric realization $|X|$ of a poset $X$.  Given an order-preserving map $f\colon X\to Y$ (a ``poset map"), there is a spectral sequence relating the homology of $|X|$, the homology of $|Y|$, and the homology of the ``fibers" $|f/y|$,  where
$$f/y=\{x\in X|f(x)\leq y\}$$
with the induced poset structure.

To understand $FC_n$ then, one might try to apply Quillen's theory using the poset map $FC_n\to X_n$.  However, it seems to be difficult to understand the fibers of this map.  Instead, we proceed by modeling the poset of free factors topologically, as the poset $B_n$ of simplices of a certain subcomplex of the ``sphere complex" $S(M)$ studied in \cite{Hat}. There is a natural poset map from $B_n$ to $FC_n$; we compute the homotopy type of $B_n$ and of the Quillen fibers of the  poset map, and apply Quillen's spectral sequence to obtain the result.

\section{Sphere systems}

Let $M$ be the compact 3-manifold obtained by taking a connected sum of $n$ copies of $S^1\times S^2$ and removing the interior of a closed ball.  A {\it sphere system} in $M$ is a non-empty finite set of disjointly embedded 2-spheres in the interior of $M$, no two of which are isotopic, and none of which bounds a ball or is isotopic to the boundary sphere of $M$.  The complex $S(M)$ of sphere systems in $M$ is defined to be the simplicial complex whose $k$-simplices are isotopy classes of sphere systems with $k+1$ spheres.

Fix a basepoint $p$ on $\partial M$. The fundamental group $\pi_1(M,p)$ is isomorphic to $F_n$.  Any automorphism of $F_n$ can be realized by a homeomorphism of $M$ fixing $\partial M$.  A theorem of Laudenbach \cite{Lau} implies that such a homeomorphism inducing the identity on $\pi_1(M,p)$ acts trivially on isotopy classes of sphere systems, so that in fact $Aut(F_n)$ acts on $S(M)$.

For $H$ a subset of $\pi_1(M,p)$, define $S_H$ to be the subcomplex of $S(M)$ consisting of isotopy classes of sphere systems $S$ such that $\pi_1(M-S,p)\supseteq H$.  When $H$ is trivial, $S_H$ is $S(M)$, and in this case the following result was proved in \cite{Hat}.

\begin{theorem}\label{SH} The complex $S_H$ is contractible for each $n\geq 1$.
\end{theorem}

The proof will be a variant of the proof in \cite{Hat}, using the following fact.

\begin{lemma}\label{disjoint}  Any two simplices in $S_H$ can be represented by sphere systems $\Sigma$ and $S$ such that every element of $H$ is representable by a loop disjoint from both $\Sigma$ and $S$.
\end{lemma}

\begin{proof}  Enlarge $\Sigma$ to a maximal
sphere system $\Sigma'$, so the components of $M - \Sigma'$ are three-punctured spheres.  By Proposition~1.1 of \cite{Hat} we may isotope $S$ to be in normal form with respect to $\Sigma'$.  This means that $S$ intersects each component of $M - \Sigma'$ in a collection of surfaces, each having at most one boundary circle on each of the three punctures; and if one of these surfaces is a disk then it separates the two punctures not containing its boundary.

We can represent a given element of $H$ by a loop $\gamma_0$ based at $p$, such that $\gamma_0$ is disjoint from $S$ and transverse to $\Sigma'$.  The points of intersection of $\gamma_0$ with $\Sigma^\prime$ divide $\gamma_0$ into a finite set of arcs, each entirely contained in one component of $M-\Sigma^\prime$.  Suppose one of these arcs $\alpha$, in a component $P$ of $M-\Sigma^\prime$,  has both endpoints on the same boundary sphere $\sigma$ of $P$.  Since the map $\pi_0(\sigma - (S\cap \sigma)) \to \pi_0(P -(S\cap P))$ is injective (an easy consequence of normal form), there is an arc $\alpha'$ in $\sigma - (S\cap \sigma)$ with $\partial \alpha' = \partial\alpha$. Since $P$ is simply-connected, $\alpha$ is homotopic to $\alpha'$ fixing endpoints. This homotopy gives a homotopy of $\gamma_0$ eliminating the two points of $\partial\alpha$ from $\gamma_0 \cap \Sigma'$, without introducing any intersection points with $S$.  After repeating this operation a finite number of times, we may assume there are no remaining arcs of $\gamma_0-(\gamma_0\cap\Sigma^\prime)$ of the specified sort.

Now consider a homotopy $F \colon I \times I \to M$ of $\gamma_0$ to a loop $\gamma_1$ disjoint from $\Sigma$. Make $F$ transverse to $\Sigma^\prime$ and look at $F^{-1}(\Sigma')$.  This consists of a collection of disjoint arcs and circles.  These do not meet the left and right edges of $I\times I$ since these edges map to the basepoint $p$.

We claim that every arc component of $F^{-1}(\Sigma^\prime)$ with one endpoint on $I\times\{0\}$ must have its other endpoint on $I\times\{1\}$.  If not,  choose an ``edgemost" arc with both endpoints on $I\times \{0\}$, i.e. an arc such that the interval of $I\times\{0\}$ bounded by the endpoints contains no other point of $F^{-1}(\Sigma^\prime)$.  Then $\gamma_0$ maps this interval to an arc $\alpha$ in $M-S$ which is entirely contained in one component $P$ of $M-\Sigma^\prime$ and has both endpoints on the same boundary sphere of $P$, contradicting our
assumption that all such arcs have been eliminated.

Since the loop $\gamma_1$ is disjoint from $\Sigma$, it follows that $\gamma_0$ must be disjoint from $\Sigma$, and by construction $\gamma_0$ was disjoint from $S$.
\end{proof}

\begin{proof}[Proof of Theorem \ref{SH}]
Following the method in \cite{Hat}, a contraction of $S_H$ can be constructed by performing a sequence of surgeries on an arbitrary system $S$ in $S_H$ to eliminate its intersections with a fixed system $\Sigma$ in $S_H$, after first putting $S$ into normal form with respect to a maximal system $\Sigma'$ containing $\Sigma$.  In \cite{Hat} the system $\Sigma$ itself was maximal but for the present proof $\Sigma$ must be in $S_H$ so it cannot be maximal if $H$ is nontrivial.  We can choose $\Sigma$ to be a single sphere defining a vertex of $S_H$ for example.  Once $S$ has been surgered to be disjoint from $\Sigma$ it will lie in the star of $S$ which is contractible so this will finish the proof.  (Alternatively, we could use the simpler contraction technique of \cite{HaVo}, which reverses the roles of $S$ and $\Sigma$.)

Each surgery on $S$ is obtained by taking a circle of intersection of $S$ and $\Sigma$ which is innermost on $\Sigma$ among the remaining circles of $S\cap \Sigma$, bounding a disk $D$ in $\Sigma$ with $D\cap S=\partial D$, then taking the two spheres obtained by attaching parallel copies of $D$ to parallel copies of the two disks of $S-\partial D$. By the lemma, elements of $H$ are representable by loops disjoint from $\Sigma$ and $S$, so these loops remain disjoint from sphere systems obtained by surgering $S$ along $\Sigma$ because such surgery produces spheres lying in a neighborhood of $S\cup\Sigma$.

In order to ensure a continuous retraction the surgery process is made canonical by  performing surgery on all innermost spheres at once.  As a result, surgery can produce trivial spheres bounding  balls in $M$ or   balls punctured by the sphere $\partial M$.   At the end of the surgery process we discard all trivial spheres in the resulting sphere system, and it must be checked that at least one nontrivial sphere remains. To check this we note that the end result could be achieved by doing a single surgery at a time and then renormalizing, so it suffices to  show that a single surgery cannot produce two trivial spheres.

Suppose, to the contrary, that a nontrivial sphere $s$ is surgered to produced two trivial spheres $s'$ and $s''$. The spheres $s\cup s'\cup s''$ form the boundary of a three-punctured sphere $P$ in $M$.  If $s'$ or $s''$, say $s'$, bounds a ball or punctured ball on the same side of $s'$ as $P$ then $P$ will be contained in this ball or punctured ball, hence so will $s$, contradicting the nontriviality of $s$. Thus $s'$ and $s''$ both bound balls or punctured balls on the opposite side from $P$. They cannot both bound punctured balls since $M$ has only one puncture, so one of them bounds a ball. This forces the other one to be isotopic to $s$, but this is not possible since $s$ is nontrivial while $s'$ and $s''$ are trivial.
\end{proof}

For an inductive argument in the next theorem we will need a generalization of the preceding theorem to manifolds with more than one boundary sphere. Let $M_k$ be the manifold  obtained from the connected sum of $n$ copies of $S^1 \times S^2$ by deleting the interiors of $k$ disjoint closed balls rather than just a single ball.  Choose the basepoint $p$ on one of the spheres in $\partial M$. For $H\subseteq \pi_1(M_k,p)$, define $S_H(M_k)$ to be the complex of isotopy classes of sphere systems $S$ in $M_k$ no two of which are isotopic and none of which bounds a ball or is isotopic to a sphere of $\partial M_k$, and such that $\pi_1(M-S,p)\supseteq H$.

\begin{lemma}\label{SkH}
For $n\geq 1$ and $k\geq 1$ the complex $S_H(M_{k+1})$ deformation retracts onto a subcomplex isomorphic to $S_H(M_k)$.
\end{lemma}

\begin{proof}
When $H$ is trivial this is Lemma 2.2 in \cite{Hat} and the
following proof extends the proof there to the general case after one small refinement to take $H$ into account.  Let the spheres of $\partial M_{k+1}$ be $\partial_0, \cdots,\partial_k$ with $p\in \partial_0$.  Call a vertex of $S_H(M_{k+1})$ {\it special\/} if it splits off a three-punctured sphere from $M_{k+1}$ having $\partial_k$ as one of its boundary components.  Let $S'_H(M_{k+1})$ be the subcomplex of $S_H(M_{k+1})$ consisting of simplices with no special vertices. Then $S_H(M_{k+1})$ is obtained from $S'_H(M_{k+1})$ by attaching the stars of the special vertices to $S'_H(M_{k+1})$ along the links of the special vertices. These links can be identified with copies of $S_H(M_k)$ so they are contractible by induction on $k$.  The interiors of the stars are disjoint since there are no edges joining special vertices. Hence $S_H(M_{k+1})$ deformation retracts to $S'_H(M_{k+1})$ since it is obtained by attaching contractible complexes along contractible subcomplexes.

The proof will be completed by showing that $S'_H(M_{k+1})$ deformation retracts onto a subcomplex isomorphic to $S_H(M_k)$.  Let $\Sigma$ be a sphere in $M_{k+1}$ splitting off a three-punctured sphere $P$ having $\partial_0$ and $\partial_k$ as its other two boundary components.  After putting a system $S$ in $S'_H(M_{k+1})$ into normal form with respect to a maximal system containing $\Sigma$, $S$ will intersect $P$ in a set of parallel disks separating $\partial_0$ from $\partial_k$.  We can eliminate these disks from $S\cap P$ by pushing them across $\partial_k$ one by one and then outside $P$.  This can also be described as surgering $S$ along the circles of $S\cap \Sigma$ using the disks they bound in $\Sigma$ on the side of $\partial_k$.  Such a surgery on a sphere $s$ of $S$ produces a pair of sphere $s'$ and $s''$ with $s'$ the sphere we have pushed across $\partial_k$ and $s''$ a trivial sphere parallel to $\partial_k$ which is then discarded.

\begin{center}
  \begin{tikzpicture}[scale=.7]
   \tikzset{partial ellipse/.style args={#1:#2:#3}{insert path={+ (#1:#3) arc (#1:#2:#3)} }}
 \draw [thick] (0,0) ellipse (4cm and 2cm);
  \draw  [thick]   (4,0) ellipse (4cm and 2cm);
    \draw [thick] (0,0) [partial ellipse=-58:58:3.75cm and 1.75cm];
          \draw [thick] (4,0) [partial ellipse=122:238:3.75cm and 1.75cm];
        \draw [thick] (0,0) [partial ellipse=-48:48:4.25cm and 2.25cm];
  \draw [thick] (4,0) [partial ellipse=-108:108:3.75cm and 1.75cm];
 \draw[thick] (-2,0) circle(.5);\node (b0) at (-2,0) {$\partial_0$};
\fill (-2,-.5) circle(.075); \node (p) at (-2,-.9 ) {$p$};
\draw[thick] (2,0) circle(.5);\node (b1) at ( 2,0) {$\partial_k$};
\node (sigma) at (-3,-2) {$\Sigma$};
\node (sigma) at (2.5,-2.25) {$s$};
\node (sigma) at ( 3.9,-1.3 ) {$s'$};
\node (sigma) at ( 2,-1) {$s''$};
\node (P) at (-1,1) {$P$};
  \end{tikzpicture}
\end{center}

We claim that $s'$ is neither trivial nor special.
If it were either of these, it would separate $M_{k+1}$ into two components, hence $s$ would also separate. The effect of replacing $s$ by $s'$ is to move the puncture $\partial_k$ from one side of $s$ to the other side. It is then not hard to check that $s$ being neither trivial nor special implies the same is true for $s'$.

The surgery defines a path in $S(M_{k+1})$ from $S$ to $S\cup s'$ and then to $(S\cup s')-s$.  The fundamental group of the component of $M_{k+1}-S$ containing $p$ is unchanged during this process so the path lies in $S'_H(M_{k+1})$.  By eliminating all the disks of $S\cap P$ by surgeries in this way we see  that $S'_H(M_{k+1})$ deformation retracts onto its subcomplex of the systems disjoint from $\Sigma$. This subcomplex can be identified with $S_H(M_k)$ by identifying $M_{k+1}-P$ with $M_k$ and choosing a new basepoint $p$ in $\Sigma$.
\end{proof}

For the proof of the main theorem in the paper we will work with certain subcomplexes of $S(M_k)$ and $S_H(M_k)$,  the subcomplexes  $Y(M_k)\subset S(M_k)$ and $Y_H(M_k)\subset S_H(M_k)$ consisting of sphere systems $S$ with $M_k - S$ connected. Eventually only the case $k=1$ will be needed, but to prove the key property that $Y(M_1)$ and $Y_H(M_1)$ are highly connected we will need to consider larger values of $k$.  It will be convenient to extend the definition of $Y(M_k)$ and $Y_H(M_k)$ to allow $n=0$, with $M_k$ the sphere $S^3$ with $k$ punctures.  In this case $Y(M_k)$ and $Y_H(M_k)$ are empty since all spheres in $M_k$ are separating.

\begin{definition}  A simplicial complex $K$ is {\it $m$-spherical} if it is $m$-dimensional and $(m-1)$-connected.   A complex is {\it spherical\/} if it is $m$-spherical for some $m$.
\end{definition}

\begin{theorem}\label{YH} Let $H$ be a free factor of $F_n=\pi_1(M_k,p)$.  Then
$Y_H(M_k)$ is $(n-rk(H)-1)$-spherical, where $rk(H)$ is the rank of $H$.
\end{theorem}

In particular, when $H$ is trivial $Y(M_k)$ is $(n-1)$-spherical.  This special case is part of Proposition~3.1 of \cite{Hat} whose proof contained the same error that is corrected below.  A corrected proof of the special case already appeared in Proposition~3.2 of \cite{HW}.

\begin{proof}
$Y_H(M_k)$ has dimension $n-rk(H)-1$ since a maximal simplex of $Y_H(M_k)$ has $n-rk(H)$ spheres. This follows because all free factors of $F_n$ of the same rank are equivalent under automorphisms of $F_n$, and all sphere systems with connected complement and the same number of spheres are equivalent under orientation-preserving homeomorphisms of $M_k$. Thus it suffices to prove $Y_H(M_k)$ is $(n-rk(H)-2)$-connected.

Let $i\leq n-rk(H)-2$.  Any map  $g\colon S^i\to Y_H(M_k)$ can be extended to a map $\widehat g\colon D^{i+1}\to S_H(M_k)$ since $S_H(M_k)$ is contractible.  We can assume $\widehat g$ is a simplicial map with respect to some triangulation of $D^{i+1}$ compatible with its standard piecewise linear structure.  We will repeatedly redefine $\widehat g$ on the stars of certain simplices in the interior of $D^{i+1}$ until eventually the image of $\widehat g$ lies in $Y_H(M_k)$.

To each sphere system $S$ we associate a dual graph $\Gamma(S)$, with one vertex for each component of $M-S$ and one edge for each sphere in $S$.  The endpoints of the edge corresponding to $s\in S$ are the vertices corresponding to the component or components adjacent to $s$. We say a sphere system  $S$ is   {\it purely separating\/} if $\Gamma(S)$ has no edges which begin and end at the same vertex.  Each sphere system $S$ has a {\it purely separating core}, consisting of those spheres in $S$ which correspond to the core of $\Gamma(S)$, i.e., the subgraph  spanned by edges with distinct vertices.  The purely separating core of $S\in S_H(M_k)$ is empty if and only if $S$ is in $Y_H(M_k)$.

Let $\sigma$ be a simplex of $D^{i+1}$ of maximal dimension among the simplices $\tau$ with $\widehat g(\tau)$ purely separating. Note that all such simplices $\tau$ lie in the interior of $D^{i+1}$ since the boundary of $D^{i+1}$ maps to $Y_H$.
Let $S = \widehat g(\sigma)$, and let $N_0,\cdots,N_r$ ($r\geq 1$) be the connected components of $M-S$, with $p\in N_0$.  A simplex $\tau$ in the link $lk(\sigma)$  maps to a system  $T$ in the link of $S$, so that each $T_j=T\cap N_j$ is a sphere system in $N_j$ and $H\leq \pi_1(N_0-T_0,p)$.  Furthermore $N_j-T_j$ must be connected for all $j$ since otherwise the core of $\Gamma(S\cup T)$ would have more edges than $\Gamma(S)$, contradicting the maximality of $\sigma$.  Thus $\widehat g$ maps  $lk(\sigma)$ into a subcomplex of $S_H(M_k)$ which can be identified with $Y_H(N_0)*Y(N_1)*\cdots *Y(N_r)$.  Some of the factors $Y_H(N_0)$ and $Y(N_j)$ can be empty if $rk(H)=rk(\pi_1(N_0))$ or $rk(\pi_1(N_j))=0$. Such factors are $(-1)$-spherical and contribute nothing to the join.

Since $\sigma$ is a simplex in the interior of $D^{i+1}$,   $lk(\sigma)$ is a sphere of dimension $i-dim(\sigma)$.  Each $N_j$ has fundamental group of rank $n_j\leq n$ with equality only if $N_j$ has fewer than $k$ boundary components, so by
induction on the lexicographically ordered pair $(n,k)$, $Y_H(N_0)$ is $(n_0 - rk(H) - 1)$-spherical and, for $j\geq 1$, $Y(N_j)$ is $(n_j-1)$-spherical.  The induction can start with the cases $(n,k)=(0,k)$ when the theorem is obvious.
For the join $Y_H(N_0)*Y(N_1)*\cdots *Y(N_r)$ it then follows that this is spherical of dimension $(\sum_{i=0}^r n_j) -rk(H)-1$.

Now  $  n=(\sum_j n_j) + rk(\pi_1(\Gamma(S))) = (\sum_j n_j) + m - r$ where $m$ is the number of spheres in $S$, i.e., edges in $\Gamma(S)$.  Since a simplicial map cannot increase dimension, we have $\dim(\sigma) \ge m-1$. Therefore

\vskip-22pt

\begin{align*}
i-\dim(\sigma)&\leq n-rk(H)-2-\dim(\sigma) \\
        &\le n-rk(H)-m-1 \\
         &=\Bigl(\sum_j n_j\Bigr)-rk(H)-1-r \\
         & <\Bigl(\sum_j n_j\Bigr)-rk(H)-1.
 \end{align*}

 \vskip-8pt

Hence the map $\widehat g\colon lk(\sigma)\to Y_H(N_0)*Y(N_1)*\cdots *Y(N_r)$ can be extended to a map of a disk $D^k$  into $Y_H(N_0)*Y(N_1)*\cdots *Y(N_r)$, where $k=i+1-dim(\sigma)$. The system $S$ is compatible with every system in the image of  $D^k$, so this map can be extended to a map $\sigma*D^k\to S_H$.   We replace the star of $\sigma$ in $D^{i+1}$ by the disk $\partial(\sigma)*D^k$, and define $\widehat g$ on $\partial(\sigma)*D^k$ using this map.

What have we improved?  The new simplices in the disk $\partial(\sigma)*D^k$ are of the form $\sigma^\prime*\tau$, where $\sigma^\prime$ is a face of $\sigma$ and $\widehat g(\tau)\subset Y_H(N_0)*Y(N_1)*\cdots *Y(N_r)$.  The image of such a simplex $\sigma^\prime*\tau$ is a system $S^\prime\cup T$ such that in $\Gamma(S^\prime\cup T)$ the edges corresponding to $T$ are all loops. Therefore any simplex in the disk $\partial(\sigma)*D^k$ with purely separating image must lie in the boundary of this disk, where we have not modified $\widehat g$.

We continue this process, eliminating purely separating simplices until there are none in the image of $\widehat g$.  Since every system in $S_H(M_k)-Y_H(M_k)$ has a non-trivial purely separating core, in fact the whole disk maps into $Y_H(M_k)$, and we are done.
\end{proof}

\section{Free factors}

We now turn to the poset  $FC_n$  of proper
free factors of the free group $F_n$, partially ordered by inclusion.   A $k$-simplex in the geometric realization $|FC_n|$ is a flag  $H_0<H_1<\cdots< H_k$ of proper free factors of $F_n$, each properly included in the next.  Each $H_i$ is also a free factor of $H_{i+1}$ (see ~\cite{MKS}, p. 117]), so that a maximal simplex of $|FC_n|$ has dimension
$n-2$.

We want to model free factors of $F_n$ by sphere systems in $Y=Y(M)$, by taking the fundamental group of the (connected) complement. Here $M$ is the manifold obtained from the connected sum of $n$ copies of $S^1\times S^2$ by removing the interior of a closed ball. A sphere system with $n$ spheres and connected complement, corresponding to an $(n-1)$-dimensional simplex of $Y$, in fact has simply-connected
complement.  But we only want to consider proper free factors, so instead we consider the $(n-2)$-skeleton $Y^{(n-2)}$.  Since $Y$ is $(n-2)$-connected by Theorem~\ref{YH}, $Y^{(n-2)}$ is $(n-2)$-spherical.

In order to relate $Y^{(n-2)}$ to $FC_n$,  we take the barycentric subdivision $B_n$ of $Y^{(n-2)}$.  Then $B_n$ is the geometric realization of a poset of isotopy classes of sphere systems, partially ordered by inclusion. If $S\subseteq S^\prime$ are sphere systems, we have
$\pi_1(M-S,p)\geq \pi_1(M-S^\prime,p)$, reversing the partial ordering.  Taking fundamental group of the complement thus gives a poset map $f\colon B_n\to (FC_n)^{op}$, where $(FC_n)^{op}$ denotes $FC_n$ with the opposite partial ordering.

\begin{proposition}\label{lift}   $f\colon B_n\to (FC_n)^{op}$ is
surjective.
\end{proposition}

\begin{proof}  Every simplex of $FC_n$ is contained in a simplex of dimension $n-2$ so it suffices to show $f$ maps onto all $(n-2)$-simplices. The group $Aut(F_n)$ acts transitively on $(n-2)$-simplices of $FC_n$, and all elements of $Aut(F_n)$ are realized by homeomorphisms of $M$, so $f$ will be surjective if its image contains a single $(n-2)$-simplex, which it obviously does.
\end{proof}

\begin{corollary}\label{connected} $FC_n$ is connected if $n\geq 3$.
\end{corollary}

\begin{proof}   Theorem~\ref{YH} implies that $B_n$ is connected for $n\geq 3$.  So, given any two vertices of $FC_n$, lift them to  vertices of $B_n$ by Proposition~\ref{lift},  connect the lifted vertices by a path, then project the path back down to $FC_n$.
\end{proof}

For any proper free factor $H$, let  $B_{\geq H}$ denote the fiber $f/H$, consisting of isotopy classes of sphere systems $S$ in $B_n$ with $\pi_1(M-S,p)\geq H$.

\begin{proposition}\label{fiber}
Let $H$ be a proper free factor of $F_n$.  Then $B_{\geq H}$ is $(n-rk(H)-1)$-spherical.  If $rk(H)=n-1$ then $B_{\geq H}$ is a single point.
\end{proposition}

\begin{proof} The fiber
$B_{\geq H}$ is the barycentric subdivision of $Y_H$, so is $(n-rk(H)-1)$-spherical by Theorem~\ref{YH}.

Suppose $rk(H)=n-1$, so that $\pi_1(M,p)=H*\langle x\rangle$ for some $x$. An element of $B_{\geq H}$ is represented by a sphere system containing exactly one sphere $s$, which is non-separating with $\pi_1(M-s,p)=H$.  Suppose $s$ and $s^\prime$ are two such spheres.  Since $s$ and $s^\prime$ are both non-separating,  there is a homeomorphism $h$ of $M$ taking $s$ to $s^\prime$. Since automorphisms of $H$ can be realized by homeomorphisms of $M$ fixing $s$, we may assume that the induced map on $\pi_1(M,p)$ is the identity on $H$.

\begin{claim}  The induced map $h_*\colon \pi_1(M,p)\to\pi_1(M,p)$ must send $x$ to an element of
the form
$Ux^{\pm 1} V$, with $U,V\in H$.
\end{claim}

\begin{proof}  Let $\{x_1,\cdots,x_{n-1}\}$ be a basis for $H$, and let $W$ be the reduced word representing $h_*(x)$ in the basis $\{x_1,\cdots,x_{n-1},x\}$ for
$\pi_1(M,p)$.  By looking at the map induced by $h$ on homology, we see that the exponent sum of $x$ in $W$ must be $\pm 1$.  Since $h_*$ fixes $H$, $\{x_1,\cdots,x_{n-1},W\}$ is a basis for $\pi_1(M,p)$.  If $W$ contained both $x$ and $x^{-1}$, we could apply Nielsen automorphisms to the set $\{x_1,\cdots,x_{n-1},W\}$ until $W$ was of the form $x^{\pm 1}W_0x^{\pm 1}$.    But $\{x_1,\cdots,x_{n-1},x^{\pm 1}W_0x^{\pm 1}\}$ is not a basis, since it is Nielsen reduced and not of the form $\{x_1^{\pm 1},\cdots,x_{n-1}^{\pm 1},x^{\pm 1}\}$ (see \cite{LySc}, Prop. 2.8)).
\end{proof}

The automorphism fixing $H$ and sending $x\mapsto Ux^{\pm 1} V$ can be realized by a homeomorphism $h^\prime$ of $M$ which takes $s$ to itself (see \cite{Lau}, Lemme 4.3.1).  The composition $h^\prime h^{-1}$ sends $s^\prime$ to $s$ and induces the identity on $\pi_1$, hence acts trivially on the sphere complex.  Thus $s$ and $s^\prime$ are isotopic.
\end{proof}

\begin{corollary}\label{oneconn}  $FC_n$ is simply connected for $n\geq 4$.
\end{corollary}

\begin{proof}  Let $e_0,e_1,\cdots,e_k$ be the edges of an edge-path loop in $FC_n$, and choose lifts $\tilde e_i$ of these edges to $B_n$. Let $e_{i-1}e_{i}$ or $e_ke_0$ be two adjacent edges of the path, meeting at the vertex $H$.  The lifts $\tilde e_{i-1}$ and $\tilde e_{i}$ may not be connected, i.e. $\tilde e_{i-1 }$ may terminate at a sphere system $S^\prime$ and $\tilde e_i$ may begin at a different sphere system $S$.  However, both $S$  and $S^\prime$ are in the fiber $B_{\geq H}$, which is connected by Proposition~\ref{fiber}, so we may  connect $S$ and $S^\prime$ by a path in $B_{\geq H}$.  Connecting the endpoints of each lifted edge in this way, we obtain a loop in $B_n$, which may be filled in by a disk if $n\geq 4$, by Proposition~\ref{fiber}.   The projection of this loop to $FC_n$ is homotopic to the original loop, since each extra edge-path segment we added projects to a loop in the star of some vertex $H$, which is contractible.  Therefore the projection of the disk kills our original loop in the fundamental
group.
\end{proof}

\begin{remark} It is possible to describe the complex $Y_H$ purely in terms of $F_n$.  Suppose first that $H$ is trivial. Define a simplicial complex $Z$ to have vertices the rank $n-1$ free factors of $F_n$, with a set of $k$ such factors spanning a simplex in $Z$ if there is an automorphism of $F_n$ taking them to the $k$ factors obtained by deleting
the standard basis elements $x_1, \cdots, x_k$ of $F_n$ one at a time. There is a simplicial map $f\colon Y \to Z$ sending a system of $k$ spheres to the set of $k$ fundamental groups of the complements of these spheres. These fundamental groups are equivalent to the standard set of $k$ rank $n-1$ factors under an automorphism of $F_n$ since the homeomorphism group of $M$ acts transitively on simplices of $Y$ of a given
dimension, and the standard $k$ factors are the fundamental groups of the complements of the spheres in a standard system in $Y$.  The last statement of Proposition~\ref{fiber} says that $f$ is a bijection on vertices, so $f$ embeds $Y$ as a subcomplex of $Z$. The maximal simplices in $Y$ and $Z$ have dimension $n-1$ and the groups $Homeo(M)$ and
$Aut(F_n)$ act transitively on these simplices, so $f$ must be surjective, hence an isomorphism. When $H$ is nontrivial, $f$ restricts to an isomorphism from $Y_H$ onto the
subcomplex $Z_H$ spanned by the vertices which are free factors containing $H$.
\end{remark}

We are now ready to apply Quillen's spectral sequence to compute the homology of $B_n$ and thus prove the main theorem.

\begin{theorem}\label{spherical}  $FC_n$ is $(n-2)$-spherical.
\end{theorem}

\begin{proof}  We prove the theorem by induction on $n$.  If $n\leq 4$, the theorem follows from Corollaries~\ref{connected} and \ref{oneconn}.

Quillen's spectral sequence (\cite{Qui}, 7.7)  applied to $f\colon B_n\to (FC_n)^{op}$ becomes
$$E^2_{p,q}=H_p(FC_n;H\mapsto H_q(B_{\geq H}))\Rightarrow H_{p+q}(B_n),$$
where the $E^2$-term is computed using homology with coefficients in the functor $H\mapsto H_q(B_{\geq H})$.

For $q=0$, Corollary~\ref{connected} gives $H_0(B_{\geq H})= {\mathbb{Z}}$ for all $H$, so $E^2_{p,0}=H_p(FC_n,{\mathbb{Z}})$.

For $q>0$, we have $E^2_{p,q}=H_p(FC_n;H\mapsto \widetilde H_q(B_{\geq H}))$, and we follow Quillen (\cite{Qui}, proof of Theorem 9.1) to compute this.

For a subposet $A$ of $FC_n$, let $L_A$ denote the functor sending $H$ to a fixed abelian group $L$ if $H\in A$ and to $0$ otherwise.  Set $U=FC_{\leq H}=\{H^\prime\in FC_n| H^\prime\leq H\}$ and $V=FC_{<H}=\{H^\prime\in FC_n| H^\prime< H\}$.  Then
\begin{align*}
 H_i(FC_n;L_V)&=H_i(V;L), \hbox{ and}\\
H_i(FC_n;L_U)&=H_i(U;L)=\begin{cases}L &\hbox{ if } $i=0$ \hbox{ and }\\ 0 &\hbox {otherwise},\end{cases}
\end{align*}
since $|U|$ is contractible.  The short exact sequence of functors
$$
1\to L_V\to L_U\to L_{\{H\}}\to 1
$$
gives a long exact homology sequence, from which we compute
$$
H_i(FC_n;L_{\{H\}})=\widetilde H_{i-1}(FC_{<H};L).
$$
Now, $H\mapsto \widetilde H_q(B_{\geq H})$ is equal to the functor
$$
\bigoplus_{rk(H)=n-q-1} \widetilde H_q(B_{\geq H})_{\{H\}}
$$
since $B_{\geq H}$ is $(n-rk(H)-1)$-spherical by Proposition~\ref{fiber}.  Thus
\begin{align*}
E^2_{p,q}&=H_p(FC_n;H\mapsto\widetilde H_q(B_{\geq H}))\\
                    &=\bigoplus_{rk(H)=n-q-1} H_p(FC_n;\widetilde H_q(B_{\geq H})_{\{H\}})\\
                    &=\bigoplus_{rk(H)=n-q-1} \widetilde H_{p-1}(FC_{<H};\widetilde H_q(B_{\geq H}))
\end{align*}

Free factors of $F_n$ contained in $H$ are also free factors of $H$.  Since $H$ has rank $<n$.  $FC_{<H}$ is $(rk(H)-2)$-spherical by induction.  Therefore $E^2_{p,q}=0$ unless  $p-1=(n-q-1)-2$, i.e. $p+q=n-2.$  Since all terms in the $E^2$-term of the spectral sequence are zero except the bottom row for $p\leq n-2$ and the diagonal $p+q=n-2$, all differentials are zero and we have $E^2=E^\infty$ as in the following diagram:

\begin{center}\begin{tikzpicture}
\matrix (m) [matrix of math nodes,
             nodes in empty cells,
             nodes={minimum width=2ex, minimum height=2ex,
                    text depth=1ex,
                    inner sep=0pt, outer sep=0pt,
                    anchor=base},
             column sep=1.5ex, row sep=1ex]%
{
n-2 & E^2_{0,n-2} &0 &0 & 0 & \cdots \\
n-3 &0& E^2_{1,n-3} & 0 & 0 & \cdots \\
n-4 &0& 0 & E^2_{2,n-4} & 0 & \cdots \\
\vdots    & \vdots       & \vdots       & \vdots & \ddots       &0\\
1    & 0 & 0& 0  &  \cdots &E^2_{n-3,1}& 0\\
0    & H_0(FC_n) & H_1(FC_n) & H_2(FC_n)  &  \cdots &H_{n-3}(FC_n)&H_{n-2}(FC_n)& 0\\
      &  0     &  1          &  2           & \cdots & n-3&n-2          & \strut \\
};
\draw[thick] (-6,-1.65) to (6 ,-1.65);
 \draw[thick] (-5,2.5) to (-5,-2.25);
\end{tikzpicture}
\end{center}

\vskip-6pt

Since $FC_n$ is connected and the spectral sequence converges to $H_*(B_n)$, which is $(n-3)$-connected, we
must have $\tilde H_i(FC_n)=0$ for $i\neq n-2$.  Since $FC_n$ is simply-connected by Corollary~\ref{oneconn}, this implies that $FC_n$ is $(n-3)$-connected by the Hurewicz theorem. The theorem follows since $FC_n$ is $(n-2)$-dimensional.\end{proof}

\section{The Cohen-Macaulay Property}

In a PL triangulation of an $n$-dimensional sphere, the link of every $k$-simplex is an $(n-k-1)$-sphere. A poset is said to be {\it Cohen-Macaulay} of dimension $n$ if its geometric realization is $n$-spherical and the link of every $k$-simplex is
$(n-k-1)$-spherical (see \cite{Qui}).   Spherical buildings are Cohen-Macaulay, and we remark that $FC_n$ also has this nice local property.

To see this, let $\sigma=\{H_0<H_1<\cdots<H_k\}$ be a $k$-simplex of $FC_n$.  The link of $\sigma$ is the join of subcomplexes  $FC_{H_i,H_{i+1}}$ of $FC_n $ spanned by free factors $H$ with $H_i<H<H_{i+1}$ ($-1\leq i\leq k$, with the conventions $H_{-1}=1$ and $H_{k+1}=F_n$).  Counting dimensions, we see that it suffices to show that for each $r$ and $s$ with $0\leq s <r$, the poset $FC_{r,s}$ of proper free factors $H$ of $F_r$ which properly contain $F_s$ is $(r-s-2)$-spherical.  The proof of this is identical to the proof that $FC_n$ is $(n-2)$-spherical, after setting $n=r$ and replacing the complex $Y$ by $Y_{F_s}$.

\section{The map to the building}

As mentioned in the introduction,
the abelianization map $F_n\to {\mathbb{Z}}^n$ induces a map from the free factor complex $FC_n$ to the building $X_n$, since summands of ${\mathbb{Z}}^n$ correspond to subspaces of ${\mathbb{Q}}^n$. Since the map $Aut(F_n)\to GL(n,{\mathbb{Z}})$ is surjective, every basis for ${\mathbb{Z}}^n$ lifts to a basis for $F_n$, and hence every flag of summands of ${\mathbb{Z}}^n$ lifts to a flag of free factors of $F_n$, i.e. the map $FC_n\to X_n$ is surjective.

Given a basis $\{v_1,\cdots,v_n\}$ for ${\mathbb{Q}}^n$,  consider the subcomplex of $X_n$ consisting of all flags of subspaces of the form $\langle v_{i_1}\rangle\subset \langle v_{i_1},v_{i_2}\rangle\subset\cdots\subset\langle v_{i_1},\cdots,v_{i_{n-1}}\rangle$.  This subcomplex can be identified with the barycentric subdivision of the boundary of an $(n-1)$-dimensional simplex, so forms an $(n-2)$-dimensional sphere in $X_n$, called an {\it apartment}.

The construction above applied to  a basis of $F_n$ instead of ${\mathbb{Q}}^n$ yields an $(n-2)$-dimensional sphere in $FC_n$.
In particular, $H_{n-2}(FC_n)$ is non-trivial.  This sphere maps to an apartment in $X_n$ showing that the induced map $H_{n-2}(FC_n)\to H_{n-2}(X_n)$ is also non-trivial.

The property of buildings which is missing in $FC_n$ is that given any two maximal simplices there is an apartment which contains both of them.  For example, for $n=3$ and $F_3$ free on $\{x,y,z\}$ there is no ``apartment" which contains both the one-cells corresponding to $\langle x \rangle\subset \langle x,y\rangle$ and
$\langle yxy^{-1}\rangle\subset \langle x,y \rangle$,
since $x$ and $yxy^{-1}$ do not form part of a basis of $F_3$.

redpajama_set_name: RedPajamaArXiv


text:
Rosie Alice Huntington-Whiteley (* 18. April 1987 in Plymouth, England) ist ein britisches Model und Schauspielerin.

Leben
Rosie Huntington-Whiteley ist die Tochter einer Fitnesstrainerin und eines Landvermessers. Sie hat einen jüngeren Bruder und eine jüngere Schwester. Ihr Ur-Ur-Großvater war der Politiker und Parlamentsabgeordnete Sir Herbert Huntington-Whiteley, 1st Baronet. Aufgrund ihrer vollen Lippen, für die sie heute berühmt ist, und ihres doppelten Nachnamens wurde sie in ihrer Kindheit oft gehänselt, jedoch auch von ihren Mitschülern im Alter von 13 Jahren zu der Person gewählt, die am ehesten ein Supermodel werden könnte.

Von August 2007 bis Oktober 2009 war sie mit Tyron Wood, dem jüngsten Sohn des "Rolling Stones"-Gitarristen Ron Wood, liiert. Anschließend hatte sie eine Beziehung mit dem französischen Schauspieler Olivier Martinez. Seit April 2010 ist sie mit dem britischen Schauspieler Jason Statham liiert. Das Paar verlobte sich 2016. Im Februar 2017 gab Rosie Huntington-Whiteley ihre Schwangerschaft bekannt. Am 24. Juni desselben Jahres kam ihr Sohn Jack Oscar Statham auf die Welt.
Am 2. Februar 2022 kam ihre Tochter Isabella James Statham zur Welt.

Karriere
2003 wurde sie von einem Modelscout entdeckt und stand anschließend mit erst 16 Jahren das erste Mal vor der Kamera, 2004 gab sie bereits ihr Laufstegdebüt in New York City und wurde zugleich Werbegesicht für Tommy Hilfiger. Seit 2006 modelt sie für das US-Unterwäschelabel Victoria's Secret.

2008 engagierte Christopher Bailey sie anstelle von Agyness Deyn für die Herbst-Winterkampagne von Burberry, worauf sie es auf das Cover der britischen Modezeitschrift Vogue schaffte. Im folgenden Jahr war sie das Gesicht der Frühjahrs-Sommerkampagne der Modekette Karen Millen sowie der Herbst-Winterkampagnen von Godiva und Miss Sixty. 2009 wurde sie mit dem Elle Style Award als "Model des Jahres" ausgezeichnet, im November wurde sie zudem offizieller Victoria's Secret Engel und damit fester Bestandteil der Victoria's Secret Fashion Shows in New York City.

2010 wurde sie vom Fotografen Terry Richardson für den weltbekannten Pirelli-Kalender abgelichtet und hatte unter anderem Laufsteg-Auftritte für Prada in Mailand und Giles Deacon in Paris. Weiterhin erschien sie auf den Titelblättern der russischen Ausgabe von Harper's Bazaar, der britischen Gentlemen's Quarterly und des LOVE Magazine. Sie präsentierte zudem die Frühling-Sommerkollektionen von Monsoon, Thomas Wylde, Full Circle und VS Online, sowie die Herbst-Winterkollektionen von Burberry, Loewe, Thomas Wylde und Leon Max. Der Porträt- und Modefotograf Rankin veröffentlichte anlässlich des fünften Jubiläums des britischen Modelabels Thomas Wylde den Fotobildband "Ten Times Rosie", bei dem sie zehn unterschiedliche Charaktere stellvertretend für die zehn bisherigen Saisons des Labels einnahm.

Im März 2011 zierte sie erneut das Titelblatt der britischen Vogue und im Juli die Titelblätter von Elle und Gentlemen's Quarterly. Im Juni 2011 posierte sie für den türkischen Bademodenhersteller Ay Yildiz.

Während der Dreharbeiten zu einem Kurzfilm des Dessous-Labels Agent Provocateur überzeugte sie schauspielerisch so sehr, dass der anwesende US-Regisseur Michael Bay sie zum Casting für die weibliche Hauptrolle in dem Film Transformers 3 einlud, der am 29. Juni 2011 in den deutschen Kinos anlief. Sie konnte das Casting gewinnen und ersetzte damit Megan Fox, die in den beiden Vorgängerteilen die weibliche Hauptrolle der Mikaela Banes gespielt hatte. Bei den 2011 CinemaCon Awards wurde sie dafür mit dem "Female Star of Tomorrow"-Award ausgezeichnet. Im Juli 2011 wurde sie das Werbegesicht für "Burberry-Body", dem ersten eigenen Duft der Modemarke Burberry. Ebenfalls 2011, wurde sie bei den Guys Choice Awards in der Kategorie "Our New Girlfriend" ausgezeichnet.

Von den internationalen Lesern der Männermagazine Maxim und FHM wurde sie jeweils zur Sexiest Woman in the World 2011 gewählt. Für eine Herbst-/Winter-Kollektion 2012/2013 des britischen Einzelhändlers Marks & Spencer, entwarf sie ihre erste eigene Unterwäsche-Kollektion, die sie noch durch Pyjama-Sets, Schlafanzüge, Seidennachthemden im Vintage-Look sowie Morgenmäntel ergänzte.

Mit der Beauty-Linie Rosie for Autograph und der Beauty-Website Rose Inc. ist sie inzwischen auch unternehmerisch tätig.

Filmografie
 2011: Transformers 3 (Transformers: Dark of the Moon)
 2015: Mad Max: Fury Road

Auszeichnungen
 2009: Elle Style Award als Model des Jahres
 2011: Female Star of Tomorrow Award als Nachwuchsschauspielerin des Jahres
 2011: Guys Choice Award in der Kategorie "Unsere Neue Freundin" (Our New Girlfriend)
 2011: FHM's Sexiest Women in the World 2011
 2012: Elle Style Award in der Kategorie Style Icon
 2015: Elle Style Award als Model des Jahres
 2016: Businesswoman of the Year bei den Harper's Bazaar Awards

Weblinks

 Website über Rosie Huntington-Whiteley
 Biographie bei Gala.de

Einzelnachweise

Model
Filmschauspieler
Engländer
Brite
Geboren 1987
Frau

redpajama_set_name: RedPajamaWikipedia


text:
Without the Hudson River Bulkhead, New York would never have become the primary port in America…
Manhattan Waterfront, 1879 Print, Library of Congress
From Archaeology:
The ship lay buried in mud for two centuries as New York City grew up around it. The last time it sailed along the East Coast and pulled into the harbor on the west side of the island, Manhattan's population was probably less than 20,000. Just five blocks from the ship's final resting place, George Washington had recently been inaugurated at Federal Hall on Wall Street as president of the newly formed United States, of which New York was the first capital. After its last voyage, its owners pulled the ship up onto the shore in order to perform routine maintenance. But it's believed that when they removed the outer sheathing and saw that the lower timbers were too far gone to repair, they saved what was still useable and left the rest behind.
At the end of the eighteenth century, Lower Manhattan—the area around the original 1624 Dutch settlement—was too cramped and the harbor too small for the volume of trade that was coming into New York. In the late 1790s, the city began deliberate efforts to extend the shoreline farther out into the Hudson River, thereby expanding the size and usefulness of the harbor. By the 1830s, the area had been completely filled and the new shoreline lay 200 yards west of its original location at modern Greenwich Street. Over the decades, the earth brought in for the shoreline extension—and the trash discarded there—completely covered up the ship. The thick, clay-rich soil sealed the vessel in the fill, creating an anaerobic (oxygen-free) environment that saved the ship from decay due to bacteria that grow in aerobic (oxygen-rich) environments.
"The Hidden History of New York's Harbor", Jarrett A. Lobell, Archaeology
Archaeology, Hudson River, New York, New York Harbour, Sailing
Susan Fox Rogers: Paddling the Hudson River
The story of the subway is the urtext of any book about New York…
When the ground was first broken in 1900 to build a subway under the crowded streets of New York, some 25,000 of the city's residents gathered to watch the ceremonies and cheer
Power and New York's Forgotten Waterfronts
Essays Literature
In Jennifer Egan's The Keep, a former New York club promoter boasts a sixth sense for authority: "Danny could walk in a room and know who had power the way some people know from the feel of the air that it's going to snow."
In New York, a city of heavy drinkers, Theodore Roosevelt's diligence didn't go down easily…


Strangely, despite watching a bunch of cartoons in Russian, I still don't know why they're the "pilot" brothers.
Gamers who haven't summered in Vladivostok may not recognize the comical pair in G5 Entertainment's upcoming iOS adventure, Pilot Brothers. Despite enjoying reasonable fame in Russia as a Sherlock-and-Watson-inspired comic duo, our heroes—Chief and Colleague—are fairly unknown in the western hemisphere. That may quickly change with their upcoming point-and-tap iOS excursion.
The Pilot Brothers have taken on a number of adventures in their time as cartoon, film, and PC game heroes, many of which are titled after relatively mundane events that are gone about in ridiculous fashion. In the cartoon short, "The Pilot Brothers Make Spaghetti for Breakfast," for instance, Colleague cuts Chief's hair for noodles and a cactus provides tequila for the pasta sauce. The slightly less insane "The Pilot Brothers Sometimes Go Fishing," shows the duo teamed up with their striped elephant pal, fishing in a tiny puddle that becomes a lake when the elephant stands in it.
This same elephant is the focus of the pair's adventure in Pilot Brothers, an iOS port of the PC title, Pilot Brothers: On the Track of the Striped Elephant. Taking on a more exciting challenge than making breakfast or going fishing, the brothers are tasked with tracking down the rare, striped elephant "Baldakhin," who has gone missing from the local zoo. You'll control the two detectives as they traverse the zoo, sewers, hotel, and more in search of clues to Baldakhin's whereabouts. In classic point-and-click fashion, you'll need to collect items hidden in each scene and use them to solve puzzles which will unlock the next area. Arcade-style mini-games will also test your precision and the brothers' special abilities.
Even if you're not familiar with the Pilot Brothers cartoons and recognizable characters, like their ever-present pet cat, the iOS game is a standalone experience that any player can enjoy. The colorful, sharply outlined graphics of the cartoons is replicated here, making delineating between objects and puzzles very easy. The brothers' trademark humor and general wackiness will likely also be present, which should create a memorable and laid-back adventure for any fan of the genre.
If this first sampling of the brothers does well, there are plenty of other games to bring to iOS: maybe the 3D Kitchen Garden Wrecker Case? Too soon to tell, but we'll have a better idea when Pilot Brothers is released later this month.


text:
https://apnews.com/article/arizona-executions-f999919f50df1158b8dc2f4c03915842
Arizona executes man for 1980 killings of 2 people
By BOB CHRISTIE and JACQUES BILLEAUDNovember 17, 2022 GMT
FILE - This undated file photo provided by the Arizona Department of Corrections, Rehabilitation and Reentry shows Murray Hooper. Hooper is scheduled to be executed by lethal injection on Nov. 16, 2022, for his convictions in the 1980 killings of Pat Redmond and his mother-in-law, Helen Phelps, in Phoenix. On Monday, Nov. 14 a federal judge denied Hooper's bid to postpone the execution. (Arizona Department of Corrections, Rehabilitation and Reentry via AP, File)
FLORENCE, Ariz. (AP) — An Arizona man convicted of murdering two people in 1980 was put to death Wednesday in the state's third execution since officials resumed carrying out the death penalty in May after a nearly eight-year hiatus.
Murray Hooper, 76, received a lethal injection at the state prison in Florence for the killings of William "Pat" Redmond and his mother-in-law, Helen Phelps, at Redmond's home in Phoenix. Redmond's wife, Marilyn, also was shot in the head in the attack but survived and testified against Hooper at trial.
Authorities say the killings were carried out at the behest of a man who wanted to take over Redmond's printing business.
In Texas late Wednesday, prison officials executed an inmate for killing his pregnant ex-girlfriend and her 7-year-old son more than 17 years ago. Stephen Barbee was given a lethal injection at the state penitentiary in Huntsville.
Hooper's death was announced by Frank Strada, a deputy director of the Arizona Department of Corrections, Rehabilitation and Reentry.
Hooper chuckled several times while interacting with the execution team. It took more than 20 minutes from the time the execution team members walked into the room until they inserted IV lines in his right leg and right forearm to administer the sedative pentobarbital.
Marshall: Execution review should happen quickly
Missouri man put to death for killing police officer in 2005
Alabama 'execution survivor' reaches settlement with state
Judge mulls arguments in Mississippi death penalty protocol
There also was one unsuccessful attempt to insert a line in his right arm. The IV in his leg was inserted through his femoral artery.
After the execution warrant was read aloud, Hooper said, "It's all been said. Let it be done."
He then told his attorneys, friends and family "don't cry for me — don't be sad."
Hooper turned to look out the windows at about 30 witnesses seated on bleacher-type benchers, including an Associated Press reporter and two other journalists. He made eye contact, smiled broadly and waved.
During the first few minutes after the warden read the death warrant, nothing happened, and Hooper said "what are we waiting on?"
Once the drug began flowing, Hooper's fingers quivered, and he yawned. After that, he made no movement. About 15 minutes passed between when a warden said the execution was beginning and when Hooper was pronounced dead.
At one point while trying to insert the IV lines, a medical professional present couldn't find a syringe with the anesthetic used to numb the area, so one was brought in and used on Hooper.
"It will hurt less," the medical professional said. Hooper said, "OK, all right." He later said, "I can't believe this" and shook his head.
Arizona did not carry out the death penalty for nearly eight years after criticism that a 2014 execution was botched and because of difficulties obtaining execution drugs. No other executions are currently scheduled in Arizona, where 110 people are on death row.
Hooper was executed within hours of the U.S. Supreme Court rejecting without comment a last appeal over his claim that authorities had until recently withheld that Marilyn Redmond had failed to identify him in a photo lineup.
Authorities said that claim was based on a mistake a prosecutor made in a letter to the state's clemency board and now insist no such lineup was shown to her. She later identified Hooper in an in-person lineup.
Courts also rebuffed attempts by Hooper's lawyers to order fingerprint and DNA testing on evidence from the killings.
Authorities say Hooper and two other men forced entry into the Redmond home on Dec. 31, 1980. The three victims were bound, gagged, robbed and shot in the head.
Two other men, William Bracy and Edward McCall, were convicted in the killings. Both died before their death sentences could be carried out.
Authorities say Robert Cruz, who was alleged to have had ties to organized crime, hired Hooper, Bracy and McCall to kill Pat Redmond, who co-owned a printing business. They said Cruz wanted to take over the business and was unhappy Redmond rejected his offers to enter several printing contracts with Las Vegas hotels, according to court records. Cruz was acquitted of murder charges in both deaths in 1995.
Hooper's lawyers say Marilyn Redmond's description of the assailants changed several times before she identified their client, who said he was not in Arizona at the time. They also raised questions about the benefits received by witnesses who testified against Hooper, including favorable treatment in other criminal cases.
Barbee was executed in Texas for the February 2005 suffocation deaths of Lisa Underwood, 34, and her son Jayden at their home in Fort Worth. Barbee was the fifth inmate put to death this year in Texas. His was the 15th execution in the U.S. this year, up from last year's three-decade low of 11.
The executions come despite waning support in recent years for the death penalty across all political parties. About 6 in 10 Americans favor the death penalty, according to the General Social Survey, a major trends survey conducted by NORC at the University of Chicago. While a majority continue to express support for the death penalty, the share has declined steadily since the 1990s, when nearly three-quarters were in favor.
Billeaud reported from Phoenix.
More of AP's coverage of executions can be found at https://apnews.com/hub/executions


Quality means working with customers to define the real requirements for a material and then meeting these requirements reliably, timely and economically. More than just meeting the technical requirements for materials, quality is a goal in personal relationships, accurate and efficient documentation, on-time performance, and effective follow-up. We pursue a total quality management program that includes meeting ISO 9001 requirements at each and every operating unit of the company.
* We meet and exceed our customers' expectations in terms of specifications and delivery respectively.
* We are open and honest about our commitment and products.
improve our level of quality assurance.


text:
APPEARS IN News Extras Handguns
Suppressed Ruger Mark III: The "Star Wars" Gun
by Dave Campbell - Friday, August 26, 2016
Those who recognize this byline know that I am very much a traditionalist when it comes to guns—and a lot of other things as well. I prefer walnut stocks, bolt-action rifles—well, lever actions, too—N-frame Smith & Wesson revolvers, single-action revolvers and the 1911. While I recognize certain engineering and manufacturing improvements spearheaded by the gun industry, you would almost have to waterboard me before I'd admit that I own an AR-15. Actually, I have three but that's a moot point right now. So when I showed up at a prairie dog shoot with my Ruger Mark III Hunter pistol topped with an Aimpoint and a KDM suppressor, my longtime buddies asked me if I was feeling well or on some new meds. To understand this anomaly it's best to start at the beginning.
Starting along about 2002 a small remuda of ne'er-do-wells calling themselves outdoor or gunwriters would gather annually at the Tejon Ranch outside of Los Angeles for a few days of pig hunting a nd ground squirrel shooting. The details of these confabs should probably remain as tough to unearth as Obama's college records, but if I recall correctly, in 2004 Dick Williams—who was my handgun editor when I was Editor in Chief of Shooting Illustrated—brought a then-new Ruger Mark III Hunter pistol to the Tejon. Atop it he had mounted an Aimpoint Comp C3 red-dot sight.
After a little cajoling, I set aside my S&W K-22 and gave this modern upstart a whirl. My first four shots resulted in four kills of the little dirt rodents at ranges out to 60 or 70 yards. Hmm…I thought, I'm gonna have to get me one of these. I did, and just as Dick did, I mounted an Aimpoint Comp C3 sight on it, after removing the factory iron sights. It didn't replace my K-22, but I found myself including it on more and more rodent shoots. The Mark III Hunter is already a very accurate .22 LR pistol with its 6 7/8" bull barrel; the Aimpoint helps these old eyes a great deal in extracting the most from that accuracy. Happily, this particular pistol has a very crisp trigger pull of 3 1/2 lbs. which also contributes to its ability to hit small targets way out there.
I am an unrepentant gun geek, so as a geek I am always thinking of ways to improve my guns. Prairie doggin' is a summertime activity, and I hate clamping ear pro to my noggin and sweating underneath it. Enter John Killebrew of Rutledge, Ala.
John and I met via Facebook and some mutual friends. He spent 27 years as an electrical engineer in the aerospace industry. It was during that time that he cultivated a mechanical and machining background while troubleshooting various projects. When his employer shut its doors in 2012 John decided to combine his two loves: guns and engineering. He began designing suppressors in his head, and in 2013 he secured a manufacturing license from BATFE. John's specialty is making small improvements on basic designs to suit an individual customer—sort of a custom suppressor gunsmith—within reason.
Last fall I sent him the Ruger for barrel threading and placed an order for one of his cans, as suppressors are affectionately called. The threading job was superb; you really have to look hard to find the joint between the thread protector and the barrel. After the Feds got paid off and were satisfied I wasn't a terrorist, I took possession of the can in December.
Everyone—including me—who has shot one of John's cans says it is incredibly quiet. John wanted to shorten the barrel on my Hunter to ensure subsonic performance and the utmost in noise reduction, but I resisted because this is primarily a hunting gun and I want all the velocity I can get out of it. To be sure, there is still the supersonic crack of the bullet, but it is no more an irritant than the report of a pellet gun.
Some lessons learned: Rimfires are dirty calibers, and I mean really dirty. The carbon buildup on the baffles necessitates cleaning every 200 to 300 rounds. After I returned from a prairie dog safari in early June I took the can apart to clean it and needed a wooden dowel to beat the aluminum baffles out of it. In order to get the baffles clean, I resorted to soaking them in Kroil for a couple of hours followed by a couple of more hours cooking in my ultrasonic cleaner. John told me this is typical of rimfires. Center-fires operate at much higher temperatures and pressures and, consequently, are more-or-less self-cleaning.
Performance on the prairie dog safari was fully up to expectations, though my buddies and I did more target shooting than prairie doggin' with the Ruger this time because of incessant 30 to 40 mph winds that plagued us for three of the four days we shot. Still, I did the dirty deed on a dozen or so of the little rodents. It's a pleasure to be able to shoot sans ear pro.
Recently, another pal had his kids and grandkids visit and asked me to join them for a day of shooting. The first gun they all wanted to shoot was what they called my "Star Wars" gun. They probably put 200 rounds through it and fell in love with it. Brad, my buddy and their grandpa, has a similar Ruger in his collection and said he just may go the Aimpoint-and-suppressor route with it. Yeah, it ain't a classic, but this Ruger is definitely a keeper gun.
John Killebrew is fine tuning his .30-cal. can to bring it to market. Like the rimfire can, everyone who has shot it is impressed with its quietness. I'm already on the list for one. The rimfire can is priced at $250 with aluminum baffles; $275 for stainless steel baffles. If you decide to go with one of his cans, be prepared to be put on a waiting list. Killebrew Design and Manufacturing is a one-man show, and every part, including the baffles is hand turned on a lathe by John. You can see more about these superb suppressors at kdmcans.com.
Ruger Mark III Hunter Ruger Mark III Hunter Aimpoint KDM suppressor pistol Dave Campbell
A Guide to Precision-Rifle Shooting on a Budget
SHOT Show 2020: Walther Arms Q4 Steel Frame


text:
The EU said the lack of progress in the implementation of the Minsk agreements
According to the head of the European Council Donald Tusk, the EU, like all countries in the G7, continues to believe that the crisis in the South-East of Ukraine can only be resolved "in full accordance with international law", in particular, with "respect for the sovereignty, territorial integrity and independence of Ukraine".
"I want to be clear that our position in relation to Russia, including economic sanctions, will remain unchanged as long as the Minsk agreement are not met in full", — said Tusk, speaking on the eve of the G7 summit in Japan.
At the same time, the European Council noted that the implementation of these agreements is moving much slower than anticipated.
"Unfortunately, [we see] a lot less progress in the implementation of the Minsk agreements than we had hoped a year ago in Ellmau [at the G7 summit in 2015]," said Tusk.
Earlier, on 20 may, Bloomberg, citing six unnamed officials from the European Union reported that the EU agreed to extend imposed sanctions against Russia for another six months. The news Agency stressed that representatives of all 28 member States of the EU, agreed that the lifting of the sanctions should follow the full implementation of the Minsk agreements. While Bloomberg emphasized that among the interlocutors including representatives of the countries that are "less likely to support sanctions".
The Agency noted that the extension of the sanctions, scheduled for June, will postpone the decision about a possible weakening them until December, when the United States changes presidential administration.
PreviousPutin said about the lack of "intractable problems" between Russia and the EU
NextThe labour Ministry has proposed to extend the "survival" for funded pensions
Putin: the format of "big eight" was not a useless tool
Media: LDPR proposes to deprive of mandates of deputies for absenteeism
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In this lecture ´´Electrocardiography´´ is explained. Section one is about ''Overview''. In this section, educator focuses on what is ECG? In addition to this, normal conduction pathway is discussed. Likewise, heart excitation related to ECG comes under consideration along with the significance of ECG. Furthermore, indications for performing ECG are pursued.
Section two is about ''Procedure''. Initially, educator sheds light on patient position for the conduction of procedure. After that duration of procedure is explained. Then ECG leads are discussed thoroughly. This is followed by highlighting standard leads, unipolar leads, and chest leads.
Section three is about ''Normal Electrocardiogram''. Educator begins by explaining wave forms which include the elaboration of P wave, PR interval, QRS complex, and ST segment. Moreover, cardiac axis is the next subject of discussion in this section.
Section four is about ''Interpretations''. In the beginning, information about interpretations is given. After that comprehensive discussion continues on rate. Then rhythm is explained. Alongside, educator also focuses on sinus tachycardia, atrial flutter, atrial fibrillation, and ventricular fibrillation.
Section five is about ''Abnormalities on Electrocardiogram''. Educator starts this section by clarifying the concept of abnormal wave forms. After that abnormalities in PR interval are discussed along with the elucidation of first degree heart block, second degree heart block, and third degree heart block. Subsequently, abnormalities in QRS complex and abnormalities in ST segment comes under consideration. Moving onward, electrolyte imbalances are pursued.


Two Chinese teachers were kidnapped from the Jinnah Town area of the provincial capital of Balochistan on Wednesday afternoon while another Chinese national reportedly escaped the abduction.
Some unidentified abductors had forced the two foreigners into a vehicle at gunpoint and driven away, according to the police. A passerby was injured after one of the abductors opened fire during the abduction.
"We collected evidence from the spot and efforts are underway to recover the Chinese nationals unhurt," said Aitzaz Goraya, the acting deputy inspector general of police. The Chinese nationals were on their way to the Jinnah Town, where many offices of the non-government organisations (NGO) are based.
Balochistan government spokesperson Anwarul Haq Kakar said that the men pretending to be 'police officers' kidnapped the Chinese teachers and wounded a passerby who tried to stop them. "A Chinese couple has been kidnapped," he confirmed.
He said that officials had earlier mistaken the wounded passerby for a security guard. "The passerby inquired why they were doing this and they said they were from a law enforcement agency, but when he asked for their identification cards, they shot him," he said.
"I was walking on the road when I saw three men forcing a Chinese woman in a white car and she was refusing and crying, I stopped to observe the situation but they had forced the woman in the car by then and were pushing the man," said Muhammad Zahir, the passer-by who got injured.
China's state media also confirmed the kidnappings, while deputy chief of mission in Islamabad Lijian Zhao said that the embassy was working towards their release. Quetta police chief Razzaq Cheema said another Chinese woman narrowly evaded the kidnappers outside a language centre in the Jinnah town.
He said that the armed men took the couple into custody at gunpoint when they were coming out from the centre. Balochistan Chief Minister Sanaullah Zehri strongly condemned the kidnapping of the foreign nationals, and ordered the Inspector General Police to take all measures for recovery of abductees as soon as possible.
He directed the authorities that a standard operating procedure (SOP) be designed for ensuring the safety and security for all the foreigners working in the province. In Balochistan, the Chinese interest has spiked in recent years after the announcement of the China-Pakistan Economic Corridor (CPEC) to connect Pakistan's ports to mainland China.


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Mom of 4, one with cerebral palsy wins part of $564 million Powerball
Posted 11:08 am, February 13, 2015, by Michelle Pekarsky, Updated at 11:10AM, February 13, 2015
SHALLOTTE, N.C. — One of the three winning tickets in Wednesday's Powerball drawing belongs to Marie Holmes, a 26-year-old mother with four kids, one of whom has cerebral palsy, WECT reported.
Holmes worked at McDonald's and Walmart but recently had to quit to care for her children, one boy and three girls.
The North Carolina mom bought the ticket in Shallotte, N.C. and will take home roughly $188 million, according to the North Carolina Education Lottery.
She shares a piece of the $564 million jackpot with two others who matched the Powerball numbers.
"First I'm going to pay my tithes because I wouldn't have none of it if it wasn't for God," she said.
Holmes also plans to set up college funds for her children and buy a new house.
"I don't think it's really hit me," she said. "I guess when it hits my account is when it's really going to hit me, but I'm thankful that I can bless my kids with something that I didn't have."
Holmes said she's ready for all the changes the money will bring.
"I'm ready for it. I'm ready to embrace the change. I'm very grateful for what's about to happen for my family."
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In 2013, GFM acquired accounting firm P. Gruchy and Associates. This acquisition has been critical for GFM as it allows the offering of not only comprehensive financial advice but also high-quality accounting solutions.
It had been a long term ambition of GFM to have the ability to offer clients full in-house financial services encompassing both Financial Planning and Accounting. This desire was very much born out of how complementary proper Financial Planning and Accounting advice is.
P. Gruchy and Associates, which has now been rebranded GFM Gruchy Accounting, is a specialist accounting firm whose business has been in operation for more than 40 years. The business has a broad client base but specialises in income tax returns for individuals, companies, trusts and SMSFs.
As is the case with GFM Wealth, GFM Gruchy has many clients who have been with them for more than 20 years. This is an indication of the quality of their advice and service as well as the stability of the business.