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European Outdoor Conservation Association Voting
CAST YOUR VOTE! Help decide which of these important projects will receive funding from the European Outdoor Conservation Association
The Great Outdoors magazine is again working with the European Outdoor Conservation Association (EOCA) to give the public the opportunity to choose where some of the Association's funding will go in 2016. EOCA is a membership body comprised of businesses throughout the outdoor industry, which raises money for conservation projects across the world. This year you can vote on four projects in the Alpine category, each of them nominated by a member of EOCA as a worthy recipient of vital funds. Read about each of the projects here, and then click at the bottom of the page to go to the voting site.
Caring for Coniston's Old Man, UK
Nominated by: George Fisher ltd
The route from Goat's Water to the summit of the Old Man of Coniston snakes around this serene lake and under the spectacular, brooding cliffs of Dow Crag. It is one of the most popular routes in the Lake District National Park bringing thousands of walkers every year, many en route to the summit of Coniston Old Man, one of the park's highest peaks. But the impact of millions of footsteps on the fragile terrain is causing massive erosion, scarring to the landscape and impacting on vulnerable vegetation. Sediment from erosion is causing problems in water bodies downstream and rare vegetation and rock features are being damaged when walkers stray from the path. The funding will enable repairs to the path using traditional techniques during the summer of 2016 and reseeding and ongoing maintenance work will keep this famous path in good repair for future generations to enjoy.
Amount Requested: €28,885
www.nurturelakeland.org
Conserving Central Asia's Ancient Fruit and Nut Forests
Nominated by: Trek & Mountain Magazine
Central Asia's endangered fruit and nut forests are vital to the communities living in the landscape as well
as being of international biodiversity importance. Home to the wild ancestor of the domestic apple, they
offer global adaptability to climate change, and provide vital ecosystem services such as water regulation and erosion control. Some of the poorest communities in Tajikistan and Kyrgyzstan depend heavily upon the forests for food, medicinal plants, firewood, hay and livestock grazing. This current use is unsustainable, severely degrading the forests. This project will promote more sustainable income-generation initiatives like honey selling and community-based tourism: training a network of hosts and guides, and developing visitor activities and leaflets. Forest protection will be encouraged by planting 4,000 fast-growing saplings for firewood and construction needs, fencing threatened tree species to protect from grazing and delivering creative outdoor educational activities in schools.
www.fauna-flora.org
Rewilding the Highlands, Scotland
Nominated by: Messe Friedrichshafen
Trees for Life's Rewilding the Highlands project aims to establish one of the UK's most inspiring examples of rewilding – featuring habitat creation for endangered or rare wildlife such as golden eagle, Scottish wildcat, red squirrel and pine marten, the planting of 50,000 trees, a boost for wildlife tourism, and 10,000 rare montane tree species being grown each year to re-establish higher altitude woodlands. Centrepiece of the project is Dundreggan Conservation Estate – a 'lost world' biodiversity hotspot where more than 3,000 species have been discovered, including 10 found nowhere else in the UK and others that are extremely rare. The 4,000-hectare estate welcomes over 300 volunteers annually, part of an exciting 250-year vision to save Scotland's ancient Caledonian Forest. Engaging with local communities is an important part of the initiative, including support to enhance biodiversity at nearby community project Glengarry Community Woodland.
www.treesforlife.org.uk
Saving Mountain Gorillas, Rwanda
Nominated by: Craghoppers
This world-reknowned long-term project strives to protect the world's remaining mountain gorillas. The population monitored by The Dian Fossey Gorilla Fund lives in the high altitude forests of the Rwandan section of the Virunga massif, featuring 8 volcanoes, dense forests and a high concentration and diversity of endangered species and plants. However the area is bordered by one of Africa's densest urban populations and extreme poverty means communities often use resources from within the national park including water, wood and bushmeat. The goal of this project is to increase daily monitoring and protection of mountain gorilla groups, remove snares, expand anti-poaching patrols, and help local communities benefit from conservation.
www.gorillafund.org
Head to the Cairngorms this April
Gallery: Sheffield Adventure Film Festival


Back in 1980, Willem bought a remote potato farm and transformed it into a butterfly paradise. He has worked the land with shovel and scythe for almost 40 years in order to preserve the butterflies. Willem's paradise has become one of the most important habitats in the region for these fluttering beauties, but there is a problem. Due to an overload of nitrogen in the air, a monoculture of grass slowly drives away this flower-rich butterfly paradise. Butterflies that used to be common just a few decades ago are rare these days, or have completely disappeared.


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'Gran Turismo Sport' is a blast even if you're terrible at it
I've never been into Gran Turismo (or racing games in general), but when my colleague Timothy J. Seppala asked if I wanted to play the franchise's next title, GT Sport, I didn't think twice about saying yes. Especially after seeing how gorgeous the latest trailer looks. And I'm glad I checked it out, even if I did end up in second to last place in the two races I played. The graphics in Gran Turismo Sport are as gorgeous as they look on screenshots; it's seriously hard to differentiate cars in the game from real ones. Put simply, it's a perfect way to showcase gameplay at 4K 60fps (don't forget about HDR, either), and even more fun if you're experiencing it the way I did: in a vibrating "sled" with a racing seat, steering wheels and pedals.
It didn't take long for me to figure out how to properly handle the car I was driving (aka not crashing into walls), but that's probably because I had the brake and steer assist on. Still, even with that helpful setting on, I only managed to grab the 7th spot out of the eight-player race I played against other members of the media, including IGN, Game Informer and Polygon, to name a few. Thankfully for me, my co-worker was even worse than I was, so I managed to salvage my brief Gran Turismo career. Unfortunately we both let Engadget down and I won't ever forgive myself for that -- I'll stick to FIFA from now on.
Gran Turismo Sport is set to arrive this fall for PlayStation 4, and you can bet I plan to make a fool of myself again when that happens.
Follow all the latest news from E3 2017 here!
Article 'Gran Turismo Sport' is a blast even if you're terrible at it compiled by Original article here
WD's next-gen SSDs add even more speed and capacity
Western Digital's new drives are the first to use 64-layer 3D NAND tech.
Sony's Chromecast soundbar and sub blast the bass under your sofa
Sony has five new soundbars for CES, including one that puts a subwoofer under your sofa.
Dell's Laser Projector Promises 100-inch 4K HDR Display
No bulb to replace, 8.3 million pixels, and as it's a ultra-short throw projector it can be positioned just 4-inches away from a wall.
The Morning After: Wednesday, June 14th 2017
Welcome to the middle of the week. It was Nintendo's turn to dominate E3, the world's biggest gaming show. Super Mario Odyssey is coming to the Switch this Octo...
TSA begins testing fingerprint check-ins at two US airports
The TSA will test the accuracy of its experimental biometric tech at two airports for four weeks.
Intel's wireless HTC Vive add-on is where VR is headed
This is how Intel wants to tackle wireless desktop VR.
Revealed: The Man Behind the Most Viral GIF in Politics
He gave us what is arguably the most famous dumpster fire in the world. But why?
Uber board member resigns after making a sexist remark
David Bonderman has left the company after making a sexist remark during a meeting meant to address the company's toxic culture.
Sony's fantastical Toio cubes combine robotics with papercraft
Sony Toio cubes are what happen when you give a Lego robot brains
Atari 2600 fans get the mod console they deserve
Get ready to blow into those Atari 2600 carts one more time.
The Mystery of What Millennials Really Want Is Solved at Last
A single conversation over a couple of beers reveals the prosaic truth: decent pay and benefits, a lot like their parents had.
The Next Big Opportunity for Tech Entrepreneurs? 'Smart' Homes
"Smart" homes these days means safe homes, in terms of security and health. The IoT opportunities are out there. Are you game?

redpajama_set_name: RedPajamaCommonCrawl


In many fields of science including population dynamics, the vast state spaces inhabited by all but the very simplest of systems can preclude a deterministic analysis. Here, a class of approximate deterministic models is introduced into the field of epidemiology that reduces this state space to one that is numerically feasible. However, these reduced state space master equations do not in general form a closed set. To resolve this, the equations are approximated using closure approximations. This process results in a method for constructing deterministic differential equation models with a potentially large scope of application including dynamic directed contact networks and heterogeneous systems using time dependent parameters. The method is exemplified in the case of an SIR (susceptible-infectious-removed) epidemiological model and is numerically evaluated on a range of networks from spatially local to random. In the context of epidemics propagated on contact networks, this work assists in clarifying the link between stochastic simulation and traditional population level deterministic models.


Three cards, each with a walnut shell printed on its back and only one with a pea on its front, are explained, then put into envelopes and mixed by a spectator. No matter how well the envelopes are mixed, the performer can always locate the pea! Everything is examinable and immediately repeatable! No reset! Easy to do!
*Includes a bonus routine where all the cards are shown and there is no pea, then a pea is shown under every shell! Finally, the cards are honestly shown, then placed in envelopes (so they can't be manipulated), and finished as above.
Note: The envelopes are identical and are not marked in any way. The performer cannot see into the envelopes and does not get any clue from any of the envelopes. Each card is identical in size, shape and color. The cards are cut from ordinary poster board.
The method is so subtle and fool-proof, that it is the perfect principle for many routines, from standard magic effects, to mentalism. The props can be thoroughly examined. Effects can be repeated. Even if a spectator tries to mess things up, you always know the correct answer!


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GOP Slams Pelosi for Jan 6 Committee Drama
(LibertySociety.com) – In House Minority Leader Kevin McCarthy's (R-CA) quest to create a bipartisan January 6th commission, he appointed five trusted Republicans to find the truth.
But when House Speaker Nancy Pelosi (D-CA) rejected Reps. Jim Jordan (R-Ohio) and Jim Banks (R-IN), it was apparent to McCarthy and others she was more interested in blocking Trump supporters than making an actual bipartisan commission.
In her feeble attempt to feign cooperation across party lines, Pelosi invited Reps. Liz Cheney (R-WY) and Adam Kinzinger (R-IL) to sit on the committee. But these particular Republicans previously voted to impeach former President Trump after the events at the Capitol on January 6, 2021. It's very likely their feelings about Trump may lead to a bias that could skew the truth.
House Minority Leader Kevin McCarthy announces Republicans will run their own investigation into the Capitol insurrection.
Senate Republicans blocked the creation of a bipartisan, independent commission to investigate January 6th back in May. pic.twitter.com/ATjdcG2vOj
— The Recount (@therecount) July 21, 2021
In response to the Speaker's bold move to stack the commission in her favor, McCarthy announced Republicans would complete their "own investigation." He plans to run a "reasonable and fair" analysis to compare his group's findings to those of Pelosi's heavily Democratic crew.
Going this route might be the only way to combat a partisan committee and ensure the American people know the real story of what happened that day at the Capitol.


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Sludka () is a rural locality (a village) in Ust-Alexeyevskoye Rural Settlement,  Velikoustyugsky District, Vologda Oblast, Russia. The population was 4 as of 2002.

Geography
Sludka is located 67 km southeast of Veliky Ustyug (the district's administrative centre) by road. Gavrino is the nearest rural locality.

References

Rural localities in Velikoustyugsky District

redpajama_set_name: RedPajamaWikipedia


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Basilea - Still capturing value from oncology
The decision by Basilea to focus on its well-established anti-infectives franchise at the expense of its nascent oncology pipeline has led to an improved financial expectation of sustainable profitability from 2023.
Various strategic actions have been taken to achieve this, including returning the rights of former lead programme derazantinib (an FGFR inhibitor) to originator Merck & Co (formerly Arqule). More recently, Basilea reported the sale of its preclinical DNA Damage Repair programme to Nodus Oncology. This programme was divested for a modest up-front and potentially substantial further payments of up to CHF241m in milestones and 5% of sales, should the programme ultimately prove successful.
The announcement today that Basilea has identified and secured a partner for its Phase 1 ready product candidate BAL0891 represents another successful component of Basilea's re-focus. BAL0891 was in-licensed from NTRC (a Dutch precision medicine company) in 2018 and represents a potentially novel mitotic checkpoint inhibitor targeting both threonine tyrosine kinase (TTK) and polo-like kinase 1 (PLK1), which together are responsible for activating the mitotic spindle assembly checkpoint (SAC). Disruption of this process by BAL0891 is expected to result in premature cell division and cancer cell death. Basilea had previously demonstrated the activity of BAL0891 in various cancer cell lines in vitro, and single agent efficacy in in-vivo human cancer models. The FDA approved an IND for BAL0891 in December 2021.
Under the terms of the asset purchase (and sub-license) agreement with oncology specialist SillaJen, Basilea will receive upfront and near-term milestone payments of $14m plus further potential payments of $320m in development, regulatory and sales milestones. Additionally, Basilea will receive up to double-digit % tiered sales royalties if BAL0891 is successfully commercialised. Basilea remains responsible for milestone and royalty payments to NRTC. As a specialist clinical-stage oncology company, we believe that Sillajen represents another safe pair of hands to entrust the future prospects for this important programme.
As a result of the agreement, Basilea has upgraded its financial guidance for 2022. The company has guided towards higher total revenues of CHF116m to CHF122m (from CHF106m to CHF112m), an improved operating result of CHF-10m to CHF-15m (from CHF -20m to CHF-25m) and improved cash flow from operating activities of CHF 0 to CHF-5m (from CHF-10m to CHF-15m).
For Basilea, this agreement represents another important step in exiting oncology and re-focusing on its long-standing and successful anti-infectives franchise. Further sales growth of the anti-fungal Cresemba and a successful US market introduction of the 5th generation cephalosporin antibiotic ceftobiprole will be key to delivering on the long-term potential. Additionally, we also look forward to Basilea progressing its recently acquired novel anti-fungal programme.
We calculate a discounted cash flow fair value of CHF120 per share for Basilea.
For all our research on Basilea, please use the link here.
Calvine Partners LLP is authorised and regulated by the Financial Conduct Authority.
Basilea Pharmaceutica is a client of Calvine Partners, and as such, this publication is not independent and should be considered a marketing communication under FCA Rules. None of the information contained in this publication should be considered as any form of advice.


If you listen to the heroes having their conversation, then when you start the mission you'll already be under attack by a Nazi gunner and a Nazi grenadier. So listen to the conversation once, and then restart the mission and press the escape key to skip it. That way you'll be able to clobber the Nazis as soon as they enter the room. Bullet is a good choice for this if you've purchased "fast healing" for him, because then whatever damage he sustains will heal itself.
After you've defeated the two Nazis, you'll learn about your main objective for this mission: to defeat two Nazi generals, collect their keys, and then use the keys to escape from the prison. The generals can be found at #4 and the exit is at #7. Nazi generals are an interesting enemy in that they don't attack anything themselves; they just improve the attack of their allies. However, this bonus isn't consequential enough to worry about, so attack damage-dealing Nazis first and save the generals for last.
At #3 you'll find a friendly kill-a-rilla in a cage. If you free it, then it will follow you for the rest of the mission, and it will help you defeat enemies. Better yet, after you've destroyed its cage, a Nazi rifleman and a Nazi gunner will charge at you and give you some extra prestige points.
At #6 you'll encounter a Nazi gunner, a Nazi grenadier, a Nazi rifleman, and two kill-a-rilla guns. These new and improved kill-a-rillas are nasty, but they're easier to kill than the original version. That's because they shoot guns, and Sky King deflects the piercing attacks from guns. (It also helps that the kill-a-rilla guns start out at half health.) So send Sky King in alone to face the enemies, and then when he gets close to them, have him use his "blast off" power. That should kill most of the enemies in the room, and whatever remains should be easy enough for him to mop up.
Once you've collected both keys from the generals, bringing Bullet to the door at #7 will end the mission. So make sure you've completed all of the secondary objectives first.
Clicking on the plans will complete one of the secondary objectives.


SONYA KAM MAKE-UP AND HAIR (UEN 53207210C) is a business entity registered with Accounting and Corporate Regulatory Authority (ACRA), Singapore. The UEN is issued in January 22, 2012. The entity status is issued in Live.
Please comment or provide details below to improve the information on SONYA KAM MAKE-UP AND HAIR.


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Q: How to avoid git sync of prisma .env files When I set up prisma in my project, by following this article, I can set up .env and set database url and password.
After that,DB connection was established and some tables were migrated. But I am wondering that Are there any problem to add .env to .gitignore ?
password is hardcoded and this credentials are not pushed.
If someone has experienced this problems or know other options, will you please let me know.

A: It is strongly recommended that the env files are added to .gitignore.
Executing npx prisma init should create a .gitignore file and a .env file along with creating prisma folder. The default behaviour is that the env file gets added to the .gitignore file so that the credentials are not accidentally pushed to source control.
Here is a reference for using multiple env variables or using prisma without any env variables.
There should be no issues on adding .env to .gitignore file.

redpajama_set_name: RedPajamaStackExchange


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Saudi Arabia arms: Opposition leaders call on next PM to launch inquiry
Image copyright EPA
Image caption Theresa May met Crown Prince Mohammed bin Salman at the G20 last week
UK opposition leaders are calling on the next PM to launch an inquiry into the sale of arms to Saudi Arabia.
In a letter to the two Tory contenders - Boris Johnson and Jeremy Hunt - the leaders accused the pair of a "lack of moral leadership" on the issue.
It is signed by Labour's Jeremy Corbyn, the SNP's Ian Blackford, Lib Dem Sir Vince Cable, Plaid Cymru's Liz Saville Roberts and Green MP Caroline Lucas.
Saudi Arabia is currently engaged in the war in Yemen.
The letter comes after the UK Court of Appeal ruled the sale of arms to the country was unlawful.
Judges said there was a clear risk the arms might be used in a serious violation of international humanitarian law and licences should be reviewed - though not immediately suspended.
A spokeswoman for Prime Minister Theresa May said the government was "disappointed" with the ruling and would be seeking permission to appeal against it.
Yemen's conflict in 400 words
Corbyn quizzes May on UK arms sales to Saudi Arabia
UK arms sales to Saudi Arabia unlawful, court rules
The leaders criticised Mr Johnson and Mr Hunt for their tenures as foreign secretary, saying the £4.6bn-worth of UK arms trade with Saudi Arabia since 2015 had supplied "a regime that is directly responsible for indiscriminate attacks on civilians".
According to UN statistics, there have been 17,640 civilians casualties in Yemen so far.
Their letter says: "The UK government, whoever leads it, cannot hide behind platitudes and must start dealing in reality."
The politicians want either a full parliamentary or public inquiry after the court ruling to "find out how [it] was allowed to happen, and which ministers were responsible".
"The UK needs a prime minister that acts and speaks with vision, wisdom and compassion, and that does so while reflecting the values at the heart of our society," they wrote.
"Your respective track records as foreign secretary do not inspire confidence that either of you are willing or capable of rising to that challenge".
'Clear risk'
Equipment sold to Saudi Arabia since the bombing of Yemen began in March 2015 includes Typhoon and Tornado fighter jets, as well as precision-guided bombs.
Under UK export policy, military equipment licences should not be granted if there is a "clear risk" that weapons might be used in a "serious violation of international humanitarian law".
Theresa May has insisted controls on UK arms exports are among the "most robust in the world" and the UK takes its responsibilities very seriously.
The government has defended the UK's wider trade and intelligence relationship with the Gulf state, saying it has helped save many lives in the UK.


The Eastern York softball team earned a 3-2 win over Dallastown on Wednesday.
The Golden Knights got a strong start from Maelynn Leber.
Morgan Dobbeck also enjoyed a strong day at the plate for Eastern.
The Golden Knights are hoping to return to the state playoffs this season.
WRIGHTSVILLE — Eastern York softball coach Brad Rohrbaugh has been around the sport a long time.
Twenty-nine years, to be exact.
Over the course of that time, the Golden Knights skipper never really gave much thought about leading the program to the state playoffs.
When Rohrbaugh and his team finally found their way to the PIAA postseason back in the spring of 2017, however, the veteran coach couldn't believe how much fun it was.
Now, just two years removed from that experience, Rohrbaugh is hopeful that his 2019 squad has what it takes to give the Eastern boss a chance to have a little more fun.
With more than half of that team still around for their senior seasons in 2019, Rohrbaugh knows that his squad is capable of getting to states. His players only need one thing — belief in themselves.
After dropping a sloppy regular-season opener Tuesday against reigning York-Adams League Division I champion Central York, the Knights rebounded with an inspired effort against another Y-A Division I contender in Dallastown on Wednesday. Behind a strong effort from starter Maelynn Leber and big night at the plate by Morgan Dobbeck, the Knights earned a 3-2 triumph.
Good first step: While it is much too early to make any predictions about where the Knights will be two months from now, the victory over Dallastown was certainly a good first step. Leber held the Wildcats to two runs over six innings, while Dobbeck collected two hits and scored two runs.
Just as important, however, was the work of Morgan Estes. Just a freshman, Estes relieved Leber to pitch the final inning and retired the top of the Wildcats lineup in order.
While Leber, Dobbeck and Estes all played key roles in Wednesday's victory, what really brought a smile to the Eastern coach's face is what he heard from his players after the game.
The triumph is also something that Rohrbaugh hopes will propel his team to greater success as the season moves on.
Rohrbaugh's sentiments were nearly identical to the ones that Dallastown coach Leanne Potts-Jones conveyed about her squad, which returns a solid battery in pitcher Kelsie Merriman and catcher Allison Hoffman.
Merriman struck out eight over six innings while scattering six hits, three of which came during a two-run first by the Knights.
Kennard-Dale 10, York Tech 0 (5 innings): At Fawn Grove, Lexie Kopko led the Rams, going 2 for 2 at the plate, including a triple, to go with three RBIs, four runs scored and two walks. Also for K-D, Crystal Mullins went 2 for 4 with a double and one RBI, Brooke Ashenfelter doubled and Alexis Hurley threw a five-inning shutout, striking out seven and walking one.
Central York 6, New Oxford 3: At New Oxford, Sarah White pitched a complete-game five-hitter to get the win, striking out three, while allowing two earned runs. Anna Saylor (three hits, three RBIs, two runs), Emma Keller (three hits, RBI), Natalie Craig (two hits, two runs) and Briana Smith (two hits, RBI, run) led the Central attack.
West York 10, York Catholic 5: At West York, Bre Harvey led the Bulldogs by going 3 for 4 with one run scored. Also for West York, Alexis Colon went 2 for 4 with a triple, two runs scored and one RBI, while Kourtney Hartzel went 2 for 4 with two runs scored. Hartzel also threw a complete game, striking out seven, walking one and allowing three earned runs. For the Irish, Rebecca Baum went 2 for 3, including a double, with one run scored and one RBI, while Kelleigh Pollock went 2 for 4 with one RBI and one run scored.
Spring Grove 16, Susquehannock 2 (5 innings): At Glen Rock, Hannah Gartrell powered the Rockets by going 3 for 4 at the plate, including a double, with three runs scored and one RBI. Also for the Rockets, River Schneeman homered with three RBIs and two runs scored, Krysten Moore had two doubles and two RBIs, Bailey Rapson had a double, two RBIs and two runs scored and Sarah Jones tripled, with one run scored and two RBIs. For the Warriors, Kendra Gemmill homered with two RBIs and one run scored.
South Western 15, York Suburban 1 (5 innings): At Suburban, Kendall Snyder led the Mustangs by going 4 for 4 with three RBIs and three runs scored. Teammate Ryleigh Sprague went 2 for 3 with three runs scored.
Dallastown 15, Gettysburg 3 (6 innings): At Gettysburg, the visitors rolled behind Julian Bailey (three hits, double, two RBIs), Cam Urey (two hits, three runs, RBI), Cole Bankert (two RBIs, hit, run), Chase Hoecke (double, two runs, RBI) and Peter Capobianco (three RBIs, run, hit). Morgan Smith got the complete-game win, allowing two earned runs, while striking out six and walking one.
Red Lion 13, Dover 1: At Dover, C.J. Czerwinski pitched a complete-game three-hitter to get the win, striking out 10 and walking one. Cole Daugherty (three hits, homer, three runs, two RBIs), Czerwinski (two hits, homer, four RBIs, run), Ben Price (two hits, three runs, RBI) and Nate Hodgkinson (three hits, double, two RBIs) led the Lions' attack. Andrew Chronister doubled and drove in Dover's run.
Central York 3, New Oxford 2: At New Oxford, Grant Smeltzer started the game on the mound, throwing 5 2/3 innings while allowing three hits and two runs, striking out two and walking two, to pick up the win. Smeltzer also helped himself at the plate by going 1 for 2 with one RBI and one walk. Teammate Landon Ness threw the final 1 1/3 innings of shutout relief to record the save.
West York 6, York Catholic 3: At West York, Justin Wetzel allowed one earned run over six innings to get the win, striking out six, while allowing two hits and four walks. Corey Wise (hit, two RBIs, run), Jeff Minot (hit, two RBIs), Trent Ketterman (two hits) and Christian Fletcher (hit, two runs) led the Bulldogs' offense. Luke Kordaz had two hits and an RBI for the Irish.
Biglerville 2, Eastern York 1: At Wrightsville, Evan Rishell pitched well for Eastern, allowing just two hits and no earned runs over six innings, striking out 10 and walking four. Emmit Silar had three hits (including a homer) with an RBI and run scored for the Golden Knights. Ben Bretzman pitched a complete game for Biglerville, striking out eight and walking one while allowing seven hits.
Delone Catholic 6, York High 2: At York, Kyle Denisch pitched four innings and allowed two runs to get the win. Jake Sherdel pitched three innings of shutout relief to get the save. Denisch struck out seven and Sherdel struck out four. Avery Kuntz (two hits, two runs, RBI), Josh Sherdel (double, two RBIs, run) and Jake Sherdel (double, run) led Delone's offense.
South Western 4, York Suburban 2 (9 innings): At Suburban, the Mustangs plated two runs in the top of the ninth. Ty Cromer led the Mustangs, going 1 for 2 with two RBIs and one run scored. Teammate Zach Reed threw three innings of relief, striking out seven, walking one and allowing no hits or runs to pick up the win. For the Trojans, pitchers Spencer Butz and Dawson Wood combined for three strikeouts and four walks, while allowing one earned run. Teammate Max Reinecker went 2 for 4.
Kennard-Dale 12, York Tech 0 (5 innings): At Fawn Grove, Jason Williams led the Rams by going 3 for 3, including a homer and a double, to go with two RBIs and two runs scored. Also for K-D, Garrett Lowe went 2 for 3, with a double, two runs scored and one RBI; Collin Griffin went 2 for 3 with one RBI and one run scored; and Cole Sinnott threw a five-inning no-hitter, striking out nine and walking none.
Susquehannock 6, Spring Grove 3: At Glen Rock, Sam Gardner led the Warriors by throwing five innings, striking out seven and walking two. He allowed four hits and no-earned runs to pick up the win. Also for the Warriors, Josh Schmidt went 2 for 2 with one RBI and one stolen base, Ben Moronic went 2 for 3 with one run scored and two stolen bases and Joe Gusherowski doubled. For the Rockets, Kevin Cruz and Dylan Shaffer each doubled.
Central York 5, Dover 0: At Dover, the Panthers received singles victories from Jacob Stewart, Alex Arrow and Michael Gilliland. In doubles, Ian Jones and Jadon Nardo combined to win the No. 1 match, while Thomas Elder and Grant Sommer combined to win the No. 2 match.
New Oxford 5, Spring Grove 0: At Spring Grove, the Colonials received singles victories from Justin Gruver, Brian Klunk and Michael Pagnanelli.
Gettysburg 4, Susquehannock 1: At Glen Rock, the visitors received singles victories from Drew Heinzelmann, Kwaku Berry, and Noe Oberholtzer-Hess. Josh Amara and Lucas Reck combined to win the No. 2 doubles match for the home team.
Biglerville 4, York Catholic 1: At Springettsbury Township, the Canners received straight-set singles victories from Noah Mattson and Aaron Orndorff. For the Irish, Joe Zelis won a singles match.
York Suburban 25, West Shore Christian 1: At Suburban, Kamille Wasilewski led the Trojans with 11 goals and one assist. Also for Suburban, Lulu Mooney had five goals and one assist, Regan Good had three goals, Julia Stell had two goals and two assists and Sofia Corto had one goal and two assists.


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Cláudio André Mergen Taffarel, brazilski nogometaš in trener, * 8. maj 1966, Santa Rosa, Brazilija.

Med letoma 1988 in 1998 je za brazilsko nogometno reprezentanco odigral 101 tekmo, od česar je 18 nastopov vpisal kot prvi vratar na Svetovnih prvenstvih v nogometu 1990, 1994 in 1998. Z Brazilijo je leta 1994 osvojil naslov svetovnega prvaka. Sodeloval je tudi na nogometnemu delu poletnih olimpijskih iger leta 1988.

Sklici

Glej tudi
 seznam brazilskih nogometašev

Zunanje povezave

Živeči ljudje
Brazilski nogometaši
Brazilski nogometni reprezentanti
Olimpijski tekmovalci za Brazilijo
Nogometaši Poletnih olimpijskih iger 1988
Prejemniki srebrnih olimpijskih medalj za Brazilijo
Nogometaši Galatasarayja
Nogometaši Parme F.C.
Nogometaši SC Internacionala
Nogometaši Clube Atlética Mineiro
Nogometaši A.C. Reggiane 1919
FIFA klub 100
Brazilski nogometni trenerji
Trenerji Galatasarayja


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Zircon, cadmium yellow
Regulatory process names 1 Other identifiers 1
Mol. formula:
According to the notifications provided by companies to ECHA in REACH registrations no hazards have been classified.
This substance is manufactured and/or imported in the European Economic Area in 10 - 100 tonnes per year.
This substance is used in articles, by professional workers (widespread uses), in formulation or re-packing, at industrial sites and in manufacturing.
Release to the environment of this substance can occur from industrial use: industrial abrasion processing with low release rate (e.g. cutting of textile, cutting, machining or grinding of metal) and of articles where the substances are not intended to be released and where the conditions of use do not promote release.
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It's every dog owner's worst nightmare. They come home or call their dog in from the yard, but their pooch is nowhere to be seen. They're gone. They're lost.
It's hard not to take it personally, even if you have one of the dog breeds that run away. But dogs are curious creatures and they love to explore. Your pooch doesn't understand the anxiety they put you through if they disappear. They also don't realize they may not be able to find their way back home or could be picked up by someone else, or worse.
Luckily, our tech-filled world gives dog owners more ways to find a lost dog, sometimes just minutes after they go missing. Here are four modern-day solutions that can reunite a dog owner with their missing pooch.
Any dog collar is better than none, but today collars go well beyond a tag with printed contact information. There are a number of tracking collars that leverage technology to find dogs fast.
Which tracking collar will work best depends on your dog and the situation where they're most likely to get lost. Basic GPS tracking collars that sync with your smartphone work in most scenarios. However, specialized devices like the Garmin Astro dog tracking combo may be more appropriate if you regularly take your dog hunting or camping.
They'll be more expensive than a traditional collar, but if your furry friend goes missing a tracking collar will be worth every penny. Do your research and read reviews before deciding which tracking collar will work best. Also, keep in mind the tracking collar will help you, but it won't help someone else who finds your dog. Your pooch still needs to have an ID tag attached to the tracking collar.
Hopefully you got your dog microchipped already. Microchipping your dog may sound inhumane, but it's actually one of the nicest things you could ever do for your dog. A microchip significantly increases the chances that you'll be reunited with your lost dog.
Many lost dogs are picked up by good samaritans or local animal control and brought to an animal shelter. There the workers try to figure out who the owner is. One quick and easy way to do this is check the dog's microchip.
A microchip is a tiny glass tube with a radio transmitter that's the size of a rice grain. The microchip also has an electronic device with the dog's ID code. When the microchip is implanted the ID is paired with the owner's name and contact information. All the animal shelter has to do is scan the dog for the microchip to pull up the information and give you a call.
Nextdoor is the modern day version of posting paper flyers around the neighborhood. It's a social site organized into neighborhood groups. You can send messages to people in the group (i.e. your neighborhood), and finding lost dogs is one of the best uses for the tool.
However, the one downside is that not everyone in the neighborhood will be signed up on Nextdoor. You can expand your search party by also posting information on social media. If your neighborhood or HOA has a Facebook page add a message with a picture asking for everyone to be on the lookout. Do the same on Instagram and Twitter. Utilize every platform you can to increase the odds that someone will get the notice and provide information.
This old-school tactic is still very useful today. A dog that's roaming around the neighborhood could be seen by someone just passing through and a flyer could be the only way they know the dog is lost.
You can go high-tech with your flyers by including a QR code with additional pictures and information. You may also want to set up an email account or social media account where people can contact you in addition to adding your phone number.
Finding a lost dog has evolved over the years. However, owners have to take the precautions above before they run off to make use of modern day dog finding technology.


text:
Lecture 33 - Chapter 2: Tying Up Loose Ends
I'm going to end my lectures on Chapter 2 now, since my plan is to spend three weeks on each chapter, and I'm eager to tell you about categories - the subject of the next chapter!
There are a few loose ends I'd like to tie up, though.
One is a trick for building \(\mathrm{Cost}\)-categories from pictures like this:
This is called a \(\mathbf{Cost}\)-weighted graph since it's a graph with edges labeled by costs. You can think of the vertices of the graph as cities, and the edges as ways to get from one city to another. Each of these ways has a cost.
Here's how we build a \(\mathrm{Cost}\)-category \(\mathcal{X}\) from this data. We take \(\mathrm{Ob}(\mathcal{X})\), the set of objects of \(\mathcal{X}\), to be the the set of vertices of our graph. For each pair of vertices \(x,y\) we then need to choose a cost \(\mathcal{X}(x,y)\). To do this, we find the cost of the cheapest path from \(x\) to \(y\). Here a "path" is a sequence of edges in our graph, each starting where the last one ended. The "cost" of a path is the sum of the costs of all the edges in this path.
For example, in the picture above there are exactly two paths from \(B\) to \(C\). One goes from \(B\) to \(A\) to \(C\) and its cost is \(2 + 3 = 5\). The other goes from \(B\) to \(D\) to \(C\) and its cost is \(5 + 6 = 11\). So,
$$ \mathcal{X}(B, C) = 5 . $$ All this should be familiar if you often travel, paying fares for planes, trains, buses and taxis. The nice thing is that this recipe really obeys the two rules of a \(\mathrm{Cost}\)-category! We have
$$ \mathcal{X}(x,x) = 0 $$ since to get from \(x\) to itself we can always use the path with no edges, which has cost \(0\). We also have the triangle inequality
$$ \mathcal{X}(x,z) \le \mathcal{X}(x,y) + \mathcal{X}(y,z) $$ since one way to get from \(x\) to \(z\) is to take the cheapest path from \(x\) to \(y\) and then the cheapest path from \(y\) to \(z\).
I'm being quick and sloppy here, but you can see more details in Section 2.3.3 of the book.
Puzzle 93. How should we define \(\mathcal{X}(x,y)\) when there is no path from \(x\) to \(y\)? How can we define \(\mathcal{X}(x,y)\), more rigorously than I've done, yet also very simply, so that this case is not an "exception"?
We can generalize this whole story, replacing \(\mathbf{Cost}\) by any unital commutative quantale. These are a very nice kind of symmetric monoidal poset - another example is our friend \(\mathbf{Bool}\). You can read about them in Section 2.5 of the book. When \(\mathcal{V}\) is one of these, you can build a \(\mathcal{V}\)-category from any \(\mathcal{V}\)-weighted graph, and the book gives a nice algorithm for doing it.
One other loose end involves maps between enriched categories. I explained last time what a map between two \(\mathcal{V}\)-categories is like - it's called a \(\mathcal{V}\)-functor. But I didn't tell you how to map a \(\mathcal{V}\)-category to a \(\mathcal{W}\)-category! You can only do this when we have a map from \(\mathcal{V}\) to \(\mathcal{W}\).
This involves a beautiful trick called change of base.
Puzzle 94. Suppose \(\mathcal{V}\) and \(\mathcal{W}\) are monoidal posets and \(f: \mathcal{V} \to \mathcal{W}\) is a monoidal monotone. Suppose \(\mathcal{X}\) is a \(\mathcal{V}\)-category. Show how to construct a \(\mathcal{W}\)-category \(\mathcal{X}_f\) with the same objects as \(\mathcal{X}\) and
$$ \mathcal{X}_f(x,x') = f(\mathcal{X}(x,x')) $$ for all objects \(x,x'\). Hint: follow your nose, don't do anything fancy.
Once you've done this, you can map your \(\mathcal{V}\)-category \(\mathcal{X}\) to a \(\mathcal{W}\)-category \(\mathcal{Y}\). First you use change of base to turn \(\mathcal{X}\) into a \(\mathcal{W}\)-category \(\mathcal{X}_f\), and then you pick a \(\mathcal{W}\)-functor
$$ F : \mathcal{X}_f \to \mathcal{Y} .$$ In short: first you move \(\mathcal{X}\) to the world where \(\mathcal{Y}\) lives, and then it can talk to \(\mathcal{Y}\)!
To understand the usefulness of this, you need to see examples. I'll get you started here:
Puzzle 95. Show that there is a monoidal monotone \(f : \mathbf{Cost} \to \mathbf{Bool} \) sending \(\infty\) to \(\texttt{false}\) everything else to \(\texttt{true}\). In economics, this means "if the cost of something is infinite, nobody can do it - otherwise it's possible, at least if you're rich enough".
Puzzle 96. Take this \(\mathbf{Cost}\)-graph
turn it into a \(\mathbf{Cost}\)-category \(\mathcal{X}\), and then use change of base to create the \(\mathbf{Bool}\)-category \(\mathcal{X}_f\). Describe \(\mathcal{X}_f\) in detail and also say what it means in terms of travel routes!
I also feel bad because I didn't get as deep into some applications as I'd like - for example, Tobias Fritz's theorems justifying the use of real numbers to measure the value of resources in some cases (but not in others). I'd be happy to talk about them in our discussions, though! You can get started by reading these blog articles:
Tobias Fritz, Resource convertibility (part 1), Azimuth, 7 April 2015.
Tobias Fritz, Resource convertibility (part 2), Azimuth, 10 April 2015.
If there is no path from \(x\) to \(y\), then we could view it as being the case that no matter how long we take, we will never get from from \(x\) to \(y\), i.e.\(\mathcal{X}(x,y) = \infty\).
Comment Source:>**Puzzle 93.** How should we define \\(\mathcal{X}(x,y)\\) when there is no path from \\(x\\) to \\(y\\)? How can we define \\(\mathcal{X}(x,y)\\), more rigorously than I've done, yet also very simply, so that this case is not an "exception"? If there is no path from \\(x\\) to \\(y\\), then we could view it as being the case that no matter how long we take, we will ***never*** get from from \\(x\\) to \\(y\\), i.e.\\(\mathcal{X}(x,y) = \infty\\).
That's the right answer! And that's one reason it's so nice to have \(\infty\) in our set of "costs" (or "times").
So, I invite others to answer the second part of the question! I was very sloppy about the definition of \(\mathcal{X}(x,y)\). If we do it carefully, the answer "\(\infty\)" should pop out in this case automatically, without any special extra rule.
Comment Source:That's the right answer! And that's one reason it's so nice to have \\(\infty\\) in our set of "costs" (or "times"). So, I invite others to answer the second part of the question! I was very sloppy about the definition of \\(\mathcal{X}(x,y)\\). If we do it carefully, the answer "\\(\infty\\)" should pop out in this case _automatically_, without any special extra rule.
The other goes from B to D to C and its cost is 5+6=1
Unless we're working in \(\mathbb{Z}/10\mathbb{Z}\), I think that should equal 11!
Comment Source:John: > The other goes from B to D to C and its cost is 5+6=1 Unless we're working in \\(\mathbb{Z}/10\mathbb{Z}\\), I think that should equal 11!
We should take \(\mathcal{X}(x, y)\) to be the meet -- the greatest lower bound -- over all paths in the graph from \(x\) to \(y\). When there are no such paths, every value is a lower bound of this set, so we just take the greatest -- in this case, that's \(\infty\).
More generally:
\[ \mathcal{X}(x, y) = \bigwedge \left\{ \bigotimes p \mid p \text{ is a path from } x \text{ to } y \right\} \]
Comment Source:> So, I invite others to answer the second part of the question! I was very sloppy about the definition of \\(\mathcal{X}(x,y)\\). If we do it carefully, the answer "\\(\infty\\)" should pop out in this case _automatically_, without any special extra rule. We should take \\(\mathcal{X}(x, y)\\) to be the meet -- the greatest lower bound -- over all paths in the graph from \\(x\\) to \\(y\\). When there are no such paths, every value is a lower bound of this set, so we just take the greatest -- in this case, that's \\(\infty\\). More generally: \\[ \mathcal{X}(x, y) = \bigwedge \left\\{ \bigotimes p \mid p \text{ is a path from } x \text{ to } y \right\\} \\]
Jonathan - right! Defining \(\mathcal{X}(x,y)\) in terms of a meet is also useful in the case when there are infinitely many paths from \(x\) to \(y\). As long as our symmetric monoidal preorder has all meets, we're fine. And a "quantale" has all meets.
Here's how I'd give the formula. In category theory, a graph is a set \(V\) of vertices and a set \(E\) of edges, together with maps \(s,t : E \to V\) giving the source and target (starting-point and ending-point) of each edge. Given a graph, a path from a vertex \(x\) to a vertex \(y\) is a finite sequence of edges \((e_1, \dots, e_n)\) with
$$ s(e_1) = x, \quad t(e_1) = s(e_2), \quad \dots, \quad t(e_{n-1}) = s(e_n), \quad t(e_n) = y. $$ For any symmetric monoidal poset \(\mathcal{V}\), a \(\mathcal{V}\)-weighted graph is a graph together with a map
$$ \ell: E \to \mathcal{V} $$ assigning each edge an element of \(\mathcal{V}\). We define the length of a path \(p = (e_1, \dots, e_n)\) in a \(\mathcal{V}\)-weighted graph to be
$$ L(p) = \ell(e_1) \otimes \cdots \otimes \ell(e_n) .$$ Then we define
$$ \mathcal{X}(x,y) = \bigwedge \{ L(p) : \; p \textrm{ is a path from } x \textrm{ to } y \} .$$ This will exist if \(\mathcal{V}\) has all meets.
Puzzle 97. Under what conditions on \(\mathcal{V}\) will this construction of \(\mathcal{X}(x,y) \) give a \(\mathcal{V}\)-category?
A unital commutative quantale will do the job, but it's more fun to check the two \(\mathcal{V}\)-category axioms and see what will make them work.
Comment Source:Jonathan - right! Defining \\(\mathcal{X}(x,y)\\) in terms of a _meet_ is also useful in the case when there are _infinitely many_ paths from \\(x\\) to \\(y\\). As long as our symmetric monoidal preorder has all meets, we're fine. And a "quantale" has all meets. Here's how I'd give the formula. In category theory, a **graph** is a set \\(V\\) of **vertices** and a set \\(E\\) of **edges**, together with maps \\(s,t : E \to V\\) giving the source and target (starting-point and ending-point) of each edge. Given a graph, a **path** from a vertex \\(x\\) to a vertex \\(y\\) is a finite sequence of edges \\((e_1, \dots, e_n)\\) with \[ s(e_1) = x, \quad t(e_1) = s(e_2), \quad \dots, \quad t(e_{n-1}) = s(e_n), \quad t(e_n) = y. \] For any symmetric monoidal poset \\(\mathcal{V}\\), a **\\(\mathcal{V}\\)-weighted graph** is a graph together with a map \[ \ell: E \to \mathcal{V} \] assigning each edge an element of \\(\mathcal{V}\\). We define the **length** of a path \\(p = (e_1, \dots, e_n)\\) in a \\(\mathcal{V}\\)-weighted graph to be \[ L(p) = \ell(e_1) \otimes \cdots \otimes \ell(e_n) .\] Then we define \[ \mathcal{X}(x,y) = \bigwedge \\{ L(p) : \; p \textrm{ is a path from } x \textrm{ to } y \\} .\] This will exist if \\(\mathcal{V}\\) has all meets. **Puzzle 97.** Under what conditions on \\(\mathcal{V}\\) will this construction of \\(\mathcal{X}(x,y) \\) give a \\(\mathcal{V}\\)-category? A unital commutative quantale will do the job, but it's more fun to check the two \\(\mathcal{V}\\)-category axioms and see what will make them work.
The greast lower bound of a \(\mathrm{Cost}\)-category has a name: \(\mathrm{min}\).
So to take Jonathan Castello's example above, I would modify it to give,
\[ \mathcal{X}(x, y) = \mathrm{min}\left\{\mathcal{d}(p) \mid p \text{ is a path from } x \text{ to } y \right\} \]
where \(\mathcal{d}(p) \) is a distance function on the path \(p\).
Comment Source:The greast lower bound of a \\(\mathrm{Cost}\\)-category has a name: \\(\mathrm{min}\\). So to take Jonathan Castello's example above, I would modify it to give, \\[ \mathcal{X}(x, y) = \mathrm{min}\left\\{\mathcal{d}(p) \mid p \text{ is a path from } x \text{ to } y \right\\} \\] where \\(\mathcal{d}(p) \\) is a distance function on the path \\(p\\).
Jonathan wrote:
Since \(\mathbb{Z}/10\mathbb{Z}\) is only a monoidal preorder if we use the discrete preordering, I'd better fix that!
Comment Source:Jonathan wrote: > Unless we're working in \\(\mathbb{Z}/10\mathbb{Z}\\), I think that should equal 11! Since \\(\mathbb{Z}/10\mathbb{Z}\\) is only a monoidal preorder if we use the discrete preordering, I'd better fix that! <img src = "http://math.ucr.edu/home/baez/emoticons/tongue2.gif">
Let \(f: \mathbf{Cost} \to \mathbf{Bool}\) be the function, \[ f(c) := [c > \infty], \]
then if \(c = \infty\), \([\infty > \infty] = \texttt{false} \) and \(\texttt{true}\) otherwise.
It is monoidal monotone because for all \( c \in \mathbf{Cost} \), \[ \begin{align} \texttt{true} = I_\mathbf{Bool} \\ \implies f(I_\mathbf{Cost}) \\ = f(d(c,c)) \\ = f(0) \\ = [0 > \infty], \end{align} \]
and for every \( c_1,c_2,c_3 \in \mathbf{Cost} \), \[ \begin{align} [d(c_1,c_2) > \infty] \wedge [d(c_2,c_3) > \infty] = f(d(c_1,c_2)) \wedge f(d(c_2,c_3)) \\ \implies f(d(c_1,c_2)+d(c_2,c_3)) \\ \implies f(d(c_1,c_3)) \\ = [d(c_1,c_3) > \infty] . \end{align} \]
Comment Source:>**Puzzle 95.** Show that there is a monoidal monotone \\(f : \mathbf{Cost} \to \mathbf{Bool} \\) sending \\(\infty\\) to \\(\texttt{false}\\) everything else to \\(\texttt{true}\\). In economics, this means "if the cost of something is infinite, nobody can do it - otherwise it's possible, at least if you're rich enough". Let \\(f: \mathbf{Cost} \to \mathbf{Bool}\\) be the function, \\[ f(c) := [c > \infty], \\] then if \\(c = \infty\\), \\([\infty > \infty] = \texttt{false} \\) and \\(\texttt{true}\\) otherwise. It is monoidal monotone because for all \\( c \in \mathbf{Cost} \\), \\[ \begin{align} \texttt{true} = I_\mathbf{Bool} \\\\ \implies f(I_\mathbf{Cost}) \\\\ = f(d(c,c)) \\\\ = f(0) \\\\ = [0 > \infty], \end{align} \\] and for every \\( c_1,c_2,c_3 \in \mathbf{Cost} \\), \\[ \begin{align} [d(c_1,c_2) > \infty] \wedge [d(c_2,c_3) > \infty] = f(d(c_1,c_2)) \wedge f(d(c_2,c_3)) \\\\ \implies f(d(c_1,c_2)+d(c_2,c_3)) \\\\ \implies f(d(c_1,c_3)) \\\\ = [d(c_1,c_3) > \infty] . \end{align} \\]
Keith - nice! Just one nitpick: I think you mean to use \(f(c) = [c \lt \infty]\), not \([c \gt \infty]\)... unless you are playing a sneaky trick and defining \(\lt\) to mean the usual \(\gt\). Of course, this is all very confusing, given the strange way we are making \(\mathbf{Cost}\) into a poset.
Comment Source:Keith - nice! Just one nitpick: I think you mean to use \\(f(c) = [c \lt \infty]\\), not \\([c \gt \infty]\\)... unless you are playing a sneaky trick and defining \\(\lt\\) to mean the usual \\(\gt\\). Of course, this is all very confusing, given the strange way we are making \\(\mathbf{Cost}\\) into a poset.
Oh wait, I missed that.
Yes, I want numbers less than infinity. Numbers greater than infinity don't exist.
Edit: Actually, I do mean to use \(f(c) = [c \lt \infty]\), since if \(c\) is finite, then \(f(c) = [c \lt \infty] = \texttt{true}\), and \(f(\infty) = [\infty \lt \infty] = \texttt{false}\).
Keep in mind, \([x < y]\) is shorthand for \(\neg[x \geq y]\).
Comment Source:Oh wait, I missed that. Yes, I want numbers **less than** infinity. Numbers greater than infinity don't exist. Edit: Actually, I *do* mean to use \\(f(c) = [c \lt \infty]\\), since if \\(c\\) is finite, then \\(f(c) = [c \lt \infty] = \texttt{true}\\), and \\(f(\infty) = [\infty \lt \infty] = \texttt{false}\\). Keep in mind, \\([x < y]\\) is shorthand for \\(\neg[x \geq y]\\).
Dealing with \(\mathbf{Bool}\)-categories is like learning left and right all over again, but for every \(\mathbf{Bool}\)-category.
Comment Source:Dealing with \\(\mathbf{Bool}\\)-categories is like learning left and right all over again, but for every \\(\mathbf{Bool}\\)-category.
Category theory involves a lot of tricky arrow reversal tricks! It can be amusing if you just accept that there's no getting around it. Otherwise it's annoying.
Comment Source:Category theory involves a lot of tricky arrow reversal tricks! It can be amusing if you just accept that there's no getting around it. Otherwise it's annoying.
A quantale, with commutativity, \[ \begin{align} x\star \left (\bigvee_{i \in I} y_i \right) = \bigvee_{i \in I}(x \star y_i) \\ = \bigvee_{i \in I} (y_i \star x) \\ = \left (\bigvee_{i \in I} y_i \right ) \star x, \end{align} \] is a lot like the higher-order function \(\mathrm{map}\),
\[ \begin{align} \mathrm{map}(f,\{x_1,x_2,x_3,\cdots,x_n\} ) \\ = \{f(x_1),f(x_2),f(x_3),\cdots,f(x_n)\} \end{align} \]
except it's nicer in that it's commutative.
Comment Source:A quantale, with commutativity, \\[ \begin{align} x\star \left (\bigvee_{i \in I} y_i \right) = \bigvee_{i \in I}(x \star y_i) \\\\ = \bigvee_{i \in I} (y_i \star x) \\\\ = \left (\bigvee_{i \in I} y_i \right ) \star x, \end{align} \\] is a lot like the higher-order function [\\(\mathrm{map}\\)](https://en.wikipedia.org/wiki/Map_(higher-order_function)), \\[ \begin{align} \mathrm{map}(f,\\{x_1,x_2,x_3,\cdots,x_n\\} ) \\\\ = \\{f(x_1),f(x_2),f(x_3),\cdots,f(x_n)\\} \end{align} \\] except it's *nicer* in that it's commutative.
Another take on the same theme of reachability and cheapest-paths on graphs, and hacker-friendly, can be found in Stephen Dolan's Fun With Semirings (slides, paper). He describes the adjacency (or edge-cost) matrix of a graph. If it has boolean-valued entries (i.e., in the boolean semiring), the sum of its powers solves reachability. If the values are in the tropical semiring (min, +), the same sum of powers gives the cost of cheapest paths! (and changing (min, +) for (max, +) the most expensive ones). The matrix multiplication in the paper corresponds the one in section 2.5 of the book, where semirings are generalized to quantales ("A unital quantale is an idempotent semiring under join and multiplication", and both the boolean and tropical semiring here are idempotent).
Would it be correct to say that a resuorce theory sprung from a weighted graph as in the top post would have the limitation of single-ingredient, single-product boxes that the meringue pie chart doesn't verify?
Comment Source:Another take on the same theme of reachability and cheapest-paths on graphs, and hacker-friendly, can be found in Stephen Dolan's Fun With Semirings ([slides](http://www.cl.cam.ac.uk/~sd601/papers/semirings-slides.pdf), [paper](https://www.cl.cam.ac.uk/~sd601/papers/semirings.pdf)). He describes the adjacency (or edge-cost) matrix of a graph. If it has boolean-valued entries (i.e., in the boolean semiring), the sum of its powers solves reachability. If the values are in the tropical semiring (min, +), the same sum of powers gives the cost of cheapest paths! (and changing (min, +) for (max, +) the most expensive ones). The matrix multiplication in the paper corresponds the one in section 2.5 of the book, where semirings are generalized to quantales ("A unital [quantale](https://en.wikipedia.org/wiki/Quantale) is an idempotent semiring under join and multiplication", and both the boolean and tropical semiring here are idempotent). Would it be correct to say that a resuorce theory sprung from a weighted graph as in the top post would have the limitation of single-ingredient, single-product boxes that the meringue pie chart doesn't verify?
Jesus - yes, this sum of powers algorithm is one thing I wish I'd had time to explain. Luckily it's nicely explained in Seven Sketches.
Yes, that's another thing I didn't have time to get into. To understand multiple-input, multiple-output processes where each process has a cost (or time) we should use PERT charts or - more or less equivalently I think - timed Petri nets. I believe these give rise not to mere \(\mathcal{V}\)-categories but "symmetric monoidal \(\mathcal{V}\)-categories". However, I don't know of anybody who has developed the theory here yet in a systematic way. I think I could do it pretty quickly, but since I'm only allowing myself 3 weeks on each chapter I decided this would eat up too much of our limited time!
But that's one good thing about this course: it's giving me idea for more research projects.
Comment Source:Jesus - yes, this sum of powers algorithm is one thing I wish I'd had time to explain. Luckily it's nicely explained in _[Seven Sketches](http://math.mit.edu/~dspivak/teaching/sp18/7Sketches.pdf)_. > Would it be correct to say that a resuorce theory sprung from a weighted graph as in the top post would have the limitation of single-ingredient, single-product boxes that the meringue pie chart doesn't verify? Yes, that's another thing I didn't have time to get into. To understand multiple-input, multiple-output processes where each process has a cost (or time) we should use PERT charts or - more or less equivalently I think - timed Petri nets. I believe these give rise not to mere \\(\mathcal{V}\\)-categories but "symmetric monoidal \\(\mathcal{V}\\)-categories". However, I don't know of anybody who has developed the theory here yet in a systematic way. I think I could do it pretty quickly, but since I'm only allowing myself 3 weeks on each chapter I decided this would eat up too much of our limited time! But that's one good thing about this course: it's giving me idea for more research projects.
For those who are into the sort of stuff described by Jesus, here's another puzzle.
Start with the adjacency matrix of a graph and take matrix powers in the usual sense. In other words, consider the entries to live in the semiring \((\mathbb{N},+,\cdot)\). Then it's well-known that the \(n\)-th power is the matrix whose \((i,j)\)-entry counts the number of paths from the \(i\)th vertex to the \(j\)th vertex of length \(n\). So is there a way to phrase this in terms of enriched category theory as well?
As a warm-up to this question, let me thus pose:
Puzzle TF4: We know that we can keep track of whether it's possible to get from one vertex to another using a \(\mathbf{Bool}\)-enriched category; we can also keep track of how many steps it takes using a \(\mathbf{Cost}\)-enriched category. Can we also keep track of how many paths there are using an enriched category? In which monoidal poset would we have to enrich? And can we do this in such a way that we count paths of length \(n\) separately for each \(n\)?
Comment Source:For those who are into the sort of stuff described by Jesus, here's another puzzle. Start with the adjacency matrix of a graph and take matrix powers in the usual sense. In other words, consider the entries to live in the semiring \\((\mathbb{N},+,\cdot)\\). Then it's well-known that the \\(n\\)-th power is the matrix whose \\((i,j)\\)-entry counts *the number of paths* from the \\(i\\)th vertex to the \\(j\\)th vertex of length \\(n\\). So is there a way to phrase this in terms of enriched category theory as well? As a warm-up to this question, let me thus pose: **Puzzle TF4:** We know that we can keep track of *whether it's possible* to get from one vertex to another using a \\(\mathbf{Bool}\\)-enriched category; we can also keep track of *how many steps it takes* using a \\(\mathbf{Cost}\\)-enriched category. Can we also keep track of *how many paths there are* using an enriched category? In which monoidal poset would we have to enrich? And can we do this in such a way that we count paths of length \\(n\\) separately for each \\(n\\)?
Keith, I had some trouble following your solution to puzzle 95, because I'm a tad confused about how you defined \(d\). Could you please expand on this? From the current exposition, it seems that \(d : \mathbf{Cost} \times \mathbf{Cost} \to \mathbf{Cost}\); is that right?
Comment Source:Keith, I had some trouble following your [solution to puzzle 95](https://forum.azimuthproject.org/discussion/comment/18634/#Comment_18634), because I'm a tad confused about how you defined \\(d\\). Could you please expand on this? From the current exposition, it seems that \\(d : \mathbf{Cost} \times \mathbf{Cost} \to \mathbf{Cost}\\); is that right?
Yes. \(d\) is the distance metric of whatever \(\mathbf{Cost}\)-category we happen to be working in (let's call it \(\mathcal{C}\)). It must satisfy the equations,
\[ 0 \geq d(x,x), \text{ for all } x\in Ob(\mathcal{C}), \\ d(x,y) + d(y,z) \geq d(x,z), \text{ for all } x,y,z \in Ob(\mathcal{C}), \\ \text{where } Ob(\mathcal{C}) = [0,\infty]. \]
Comment Source:Yes. \\(d\\) is the distance metric of whatever \\(\mathbf{Cost}\\)-category we happen to be working in (let's call it \\(\mathcal{C}\\)). It must satisfy the equations, \\[ 0 \geq d(x,x), \text{ for all } x\in Ob(\mathcal{C}), \\\\ d(x,y) + d(y,z) \geq d(x,z), \text{ for all } x,y,z \in Ob(\mathcal{C}), \\\\ \text{where } Ob(\mathcal{C}) = [0,\infty]. \\]
Question, how do we define a \(\mathbf{Cost}\)-category that uses units?
For instance, if you're traveling around the US, and you're picking various destinations, you don't want a number when considering various hotels, you want the price (usually in US dollars).
Comment Source:Question, how do we define a \\(\mathbf{Cost}\\)-category that uses units? For instance, if you're traveling around the US, and you're picking various destinations, you don't want a *number* when considering various hotels, you want the *price* (usually in US dollars).
Keith, thanks for the explanation, but I'm failing to see where does the \(\mathbf{Cost}\)-category come into play in puzzle 95. As far as I understood, we need to prove that \(f\) is a mapping between two monoidal preorders; why do we need enriched categories to prove that \(f\) is a monoidal monotone? Or did I misunderstand the problem statement?
Comment Source:Keith, thanks for [the explanation](https://forum.azimuthproject.org/discussion/comment/18657/#Comment_18657), but I'm failing to see where does the \\(\mathbf{Cost}\\)-category come into play in puzzle 95. As far as I understood, we need to prove that \\(f\\) is a mapping between two monoidal preorders; why do we need enriched categories to prove that \\(f\\) is a monoidal monotone? Or did I misunderstand the problem statement?
That would answer Puzzle 94.
In fact, trying answering Puzzle 94.
(Hint: Check Page 56 of the book)
Comment Source:That would answer **Puzzle 94**. In fact, trying answering **Puzzle 94**. (Hint: Check Page 56 of the book)
Here's another loose end that I thought might be worth tying up – why are we insisting on symmetric monoidal preorders? The reason for this is in section 2.4.3 of Seven Sketches, and in particular exercise 2.52 – we need symmetry in order to define the product of two \(\mathcal{V}\)-enriched categories:
Suppose \(\mathcal{X}\) and \(\mathcal{Y}\) are \(\mathcal{V}\)-enriched categories.
Define \(\mathcal{X} \times \mathcal{Y}\) by
\[\textrm{Ob}(\mathcal{X} \times \mathcal{Y}) = Ob(\mathcal{X}) \times Ob(\mathcal{Y})\]
\[(\mathcal{X} \times \mathcal{Y})(\langle x_0, y_0\rangle, \langle x_1, y_1\rangle) = \mathcal{X}(x_0, x_1) \otimes \mathcal{Y}(y_0, y_1)\]
Checking the unit law holds:
\[(\mathcal{X} \times \mathcal{Y})(\langle x, y\rangle, \langle x, y\rangle) = \mathcal{X}(x, x) \otimes \mathcal{Y}(y, y) \geq I \otimes I = I\]
Checking the composition law holds:
\[(\mathcal{X} \times \mathcal{Y})(\langle x_0, y_0\rangle, \langle x_1, y_1\rangle) \otimes (\mathcal{X} \times \mathcal{Y})(\langle x_1, y_1\rangle, \langle x_2, y_2\rangle)\]
\[= \mathcal{X}(x_0, x_1) \otimes \mathcal{Y}(y_0, y_1) \otimes \mathcal{X}(x_1, x_2) \otimes \mathcal{Y}(y_1, y_2)\]
(we now use symmetry to swap round the middle two terms)
\[\leq \mathcal{X}(x_0, x_1) \otimes \mathcal{X}(x_1, x_2) \otimes \mathcal{Y}(y_0, y_1) \otimes \mathcal{Y}(y_1, y_2)\]
\[\leq \mathcal{X}(x_0, x_2) \otimes \mathcal{Y}(y_0, y_2)\]
\[= (\mathcal{X} \times \mathcal{Y})(\langle x_0, y_0\rangle, \langle x_2, y_2\rangle)\]
So \(\mathcal{X} \times \mathcal{Y}\) is itself a \(\mathcal{V}\)-enriched category.
Comment Source:Here's another loose end that I thought might be worth tying up – why are we insisting on *symmetric* monoidal preorders? The reason for this is in section 2.4.3 of Seven Sketches, and in particular exercise 2.52 – we need symmetry in order to define the *product* of two \\(\mathcal{V}\\)-enriched categories: Suppose \\(\mathcal{X}\\) and \\(\mathcal{Y}\\) are \\(\mathcal{V}\\)-enriched categories. Define \\(\mathcal{X} \times \mathcal{Y}\\) by \\[\textrm{Ob}(\mathcal{X} \times \mathcal{Y}) = Ob(\mathcal{X}) \times Ob(\mathcal{Y})\\] and \\[(\mathcal{X} \times \mathcal{Y})(\langle x_0, y_0\rangle, \langle x_1, y_1\rangle) = \mathcal{X}(x_0, x_1) \otimes \mathcal{Y}(y_0, y_1)\\] Checking the unit law holds: \\[(\mathcal{X} \times \mathcal{Y})(\langle x, y\rangle, \langle x, y\rangle) = \mathcal{X}(x, x) \otimes \mathcal{Y}(y, y) \geq I \otimes I = I\\] Checking the composition law holds: \\[(\mathcal{X} \times \mathcal{Y})(\langle x_0, y_0\rangle, \langle x_1, y_1\rangle) \otimes (\mathcal{X} \times \mathcal{Y})(\langle x_1, y_1\rangle, \langle x_2, y_2\rangle)\\] \\[= \mathcal{X}(x_0, x_1) \otimes \mathcal{Y}(y_0, y_1) \otimes \mathcal{X}(x_1, x_2) \otimes \mathcal{Y}(y_1, y_2)\\] (we now use symmetry to swap round the middle two terms) \\[\leq \mathcal{X}(x_0, x_1) \otimes \mathcal{X}(x_1, x_2) \otimes \mathcal{Y}(y_0, y_1) \otimes \mathcal{Y}(y_1, y_2)\\] \\[\leq \mathcal{X}(x_0, x_2) \otimes \mathcal{Y}(y_0, y_2)\\] \\[= (\mathcal{X} \times \mathcal{Y})(\langle x_0, y_0\rangle, \langle x_2, y_2\rangle)\\] So \\(\mathcal{X} \times \mathcal{Y}\\) is itself a \\(\mathcal{V}\\)-enriched category.
This reminds me of the fact that if \(M\) is an ordinary monoid, then the multiplication function \(M\times M\to M\) isn't a monoid homomorphism unless \(M\) is commutative. Is that just a coincidence, or is there something deeper connecting both of these ideas?
Comment Source:This reminds me of the fact that if \\(M\\) is an ordinary monoid, then the multiplication function \\(M\times M\to M\\) isn't a monoid homomorphism unless \\(M\\) is commutative. Is that just a coincidence, or is there something deeper connecting both of these ideas?
Can't we reduce \(\mathcal{V}\)-functors to changes-of-base, mainly a change-of-base between a \(\mathcal{V}\)-category and itself?
Comment Source:Can't we reduce \\(\mathcal{V}\\)-functors to changes-of-base, mainly a change-of-base between a \\(\mathcal{V}\\)-category and itself?
Ah, that makes the "Periodic Table" of n,k categories make more sense.
Comment Source:Ah, that makes the "Periodic Table" of n,k categories make more sense.
Category theorists have a slick way to express the fact you're mentioning, and a bit more: a monoid in the category of monoids is exactly the same as a commutative monoid.
The proof is called the 'Eckmann-Hilton argument', and it was first used to note that a group in the category of groups is exactly the same as an abelian group.
It admits lots of generalizations. It's related to the fact that Anindya mentioned: the category of categories enriched over a symmetric monoidal preorder is itself monoidal. But the relation is a bit elusive.
It becomes slightly clearer (?) if we push the Eckmann-Hilton argument up one level, and also replace preorders by categories. Then we have these nice facts:
A monoid in the category of monoidal categories is a braided monoidal category.
The category of \(\mathcal{V}\)-enriched categories is monoidal if \(\mathcal{V}\) is a braided monoidal category.
These statements are closely connected. But I can't quite put my finger on the connection.
I could go on, listing more facts that illustrate this connection, but this is probably enough for now!
Comment Source:<img width = "150" src = "http://math.ucr.edu/home/baez/mathematical/warning_sign.jpg"> Owen wrote: > This reminds me of the fact that if \\(M\\) is an ordinary monoid, then the multiplication function \\(M\times M\to M\\) isn't a monoid homomorphism unless \\(M\\) is commutative. Is that just a coincidence, or is there something deeper connecting both of these ideas? Category theorists have a slick way to express the fact you're mentioning, and a bit more: _a monoid in the category of monoids is exactly the same as a commutative monoid._ The proof is called the ['Eckmann-Hilton argument'](https://en.wikipedia.org/wiki/Eckmann%E2%80%93Hilton_argument#Remarks_2), and it was first used to note that _a group in the category of groups is exactly the same as an abelian group_. It admits lots of generalizations. It's _related_ to the fact that Anindya mentioned: _the category of categories enriched over a symmetric monoidal preorder is itself monoidal_. But the relation is a bit elusive. It becomes slightly clearer (?) if we push the Eckmann-Hilton argument up one level, and also replace preorders by categories. Then we have these nice facts: 1. A monoid in the category of monoidal categories is a braided monoidal category. 2. The category of \\(\mathcal{V}\\)-enriched categories is monoidal if \\(\mathcal{V}\\) is a braided monoidal category. These statements are closely connected. But I can't quite put my finger on the connection. I could go on, listing more facts that illustrate this connection, but this is probably enough for now!
No, you're making a classic level slip here. There's no such thing as change-of-base between a \(\mathcal{V}\)-category and itself! Change-of-base is not a kind of map between \(\mathcal{V}\)-categories.
It's great that you made this mistake, because everybody does it. Please think about this! It's easy to get confused... but it's very important to get unconfused.
It's like "a function from an element of a set to itself."
There's no such thing as a function from an element of a set to itself. A function \(f: X \to Y\) goes between sets \(X\) and \(Y\). If \(X = Y\) this function may map an element \(x \in X\) to itself: \(f(x) = x\). But it's still a function from \(X\) to \(X\), not a function from \(x\) to \(x\).
Similarly, there's no such thing as a change-of-base from a \(\mathcal{V}\)-category to itself. Change-of-base turns \(\mathcal{V}\)-categories into \(\mathcal{W}\)-categories, so it's a functor \(\Phi: \mathcal{V}\mathrm{Cat}\to \mathcal{W}\mathrm{Cat} \). If \(\mathcal{V} = \mathcal{W}\) this functor may map a \(\mathcal{V}\)-category \(\mathcal{C}\) to itself: \(\Phi(\mathcal{C}) = \mathcal{C}\). But it's still a functor from \(\mathcal{V}\mathrm{Cat}\) to \(\mathcal{V}\mathrm{Cat}\), not an enriched functor from \(\mathcal{C}\) to \(\mathcal{C}\).
This is why change-of-base and enriched functors are both crucial, indispensable concepts.
I've been through this confusion myself a couple of times! It only takes 8 hours of agony to figure out what's going on.
Let me review the whole story:
First I told you about enriched functors between \(\mathcal{V}\)-categories. They're called \(\mathcal{V}\)-functors.
Then I told you about enriched functors from a \(\mathcal{V}\)-category to a \(\mathcal{W}\)-category. But these only makes sense when you first have a trick for turning \(\mathcal{V}\)-categories into \(\mathcal{W}\)-categories. This trick is "change of base".
Suppose \(\mathcal{V}\) and \(\mathcal{W}\) are monoidal posets and \(f: \mathcal{V} \to \mathcal{W}\) is a monoidal monotone. Suppose \(\mathcal{X}\) is a \(\mathcal{V}\)-category. Then there's a \(\mathcal{W}\)-category \(\Phi(\mathcal{X})\) with the same objects as \(\mathcal{X}\) and
$$ \Phi(\mathcal{X})(x,x') = f(\mathcal{X}(x,x')). $$ This trick, \(\Phi\), is called change of base.
Once you've got change of base, you can get an enriched functor from a \(\mathcal{V}\)-category \(\mathcal{X}\) to a \(\mathcal{W}\)-category \(\mathcal{Y}\). First you use change of base to turn \(\mathcal{X}\) into a \(\mathcal{W}\)-category \(\Phi(\mathcal{X})\). Then you pick a \(\mathcal{W}\)-functor
$$ F : \Phi(\mathcal{X}) \to \mathcal{Y} .$$ In short: first you move \(\mathcal{X}\) to the world where \(\mathcal{Y}\) lives, and then it can talk to \(\mathcal{Y}\).
Comment Source:Keith wrote: > Can't we reduce \\(\mathcal{V}\\)-functors to changes-of-base, mainly a change-of-base between a \\(\mathcal{V}\\)-category and itself? No, you're making a classic level slip here. There's _no such thing_ as change-of-base between a \\(\mathcal{V}\\)-category and itself! Change-of-base is not a kind of map between \\(\mathcal{V}\\)-categories. It's great that you made this mistake, because everybody does it. Please think about this! It's easy to get confused... but it's very important to get unconfused. It's like "a function from an element of a set to itself." There's _no such thing_ as a function from an element of a set to itself. A function \\(f: X \to Y\\) goes between _sets_ \\(X\\) and \\(Y\\). If \\(X = Y\\) this function may _map_ an element \\(x \in X\\) to itself: \\(f(x) = x\\). But it's still a function from \\(X\\) to \\(X\\), not a function from \\(x\\) to \\(x\\). Similarly, there's _no such thing_ as a change-of-base from a \\(\mathcal{V}\\)-category to itself. Change-of-base turns \\(\mathcal{V}\\)-categories into \\(\mathcal{W}\\)-categories, so it's a functor \\(\Phi: \mathcal{V}\mathrm{Cat}\to \mathcal{W}\mathrm{Cat} \\). If \\(\mathcal{V} = \mathcal{W}\\) this functor may _map_ a \\(\mathcal{V}\\)-category \\(\mathcal{C}\\) to itself: \\(\Phi(\mathcal{C}) = \mathcal{C}\\). But it's still a functor from \\(\mathcal{V}\mathrm{Cat}\\) to \\(\mathcal{V}\mathrm{Cat}\\), not an enriched functor from \\(\mathcal{C}\\) to \\(\mathcal{C}\\). This is why change-of-base and enriched functors are both crucial, indispensable concepts. I've been through this confusion myself a couple of times! It only takes 8 hours of agony to figure out what's going on. <center><img src = "http://math.ucr.edu/home/baez/emoticons/spiral_eyes.gif"></center> Let me review the whole story: First I told you about enriched functors between \\(\mathcal{V}\\)-categories. They're called \\(\mathcal{V}\\)-functors. Then I told you about enriched functors from a \\(\mathcal{V}\\)-category to a \\(\mathcal{W}\\)-category. But these only makes sense when you first have a trick for turning \\(\mathcal{V}\\)-categories into \\(\mathcal{W}\\)-categories. This trick is "change of base". Suppose \\(\mathcal{V}\\) and \\(\mathcal{W}\\) are monoidal posets and \\(f: \mathcal{V} \to \mathcal{W}\\) is a monoidal monotone. Suppose \\(\mathcal{X}\\) is a \\(\mathcal{V}\\)-category. Then there's a \\(\mathcal{W}\\)-category \\(\Phi(\mathcal{X})\\) with the same objects as \\(\mathcal{X}\\) and \[ \Phi(\mathcal{X})(x,x') = f(\mathcal{X}(x,x')). \] This trick, \\(\Phi\\), is called **change of base**. Once you've got change of base, you can get an enriched functor from a \\(\mathcal{V}\\)-category \\(\mathcal{X}\\) to a \\(\mathcal{W}\\)-category \\(\mathcal{Y}\\). First you use change of base to turn \\(\mathcal{X}\\) into a \\(\mathcal{W}\\)-category \\(\Phi(\mathcal{X})\\). Then you pick a \\(\mathcal{W}\\)-functor \[ F : \Phi(\mathcal{X}) \to \mathcal{Y} .\] In short: _first you move \\(\mathcal{X}\\) to the world where \\(\mathcal{Y}\\) lives, and then it can talk to \\(\mathcal{Y}\\)_.
@John – thanks for that, I remember proving that Eckmann-Hilton result about two binary operations on the same set in an exercise many years ago, but I had no idea about its significance or that it had a name!
I was thinking that the "swap round the middle two terms" line in that proof reminded me of how you prove that \(f + g\) is a homomorphism given homomorphisms \(f\) and \(g\) of Abelian groups (or commutative monoids more generally):
\[(f + g)(x + y) = f(x + y) + g(x + y) = fx + fy + gx +gy = fx +gx +fy + gy = (f + g)x + (f + g)y\]
It strikes me that the Eckmann-Hilton argument might be some kind of converse to this, but I can't quite see how.
Comment Source:@John – thanks for that, I remember proving that Eckmann-Hilton result about two binary operations on the same set in an exercise many years ago, but I had no idea about its significance or that it had a name! I was thinking that the "swap round the middle two terms" line in that proof reminded me of how you prove that \\(f + g\\) is a homomorphism given homomorphisms \\(f\\) and \\(g\\) of Abelian groups (or commutative monoids more generally): \\[(f + g)(x + y) = f(x + y) + g(x + y) = fx + fy + gx +gy = fx +gx +fy + gy = (f + g)x + (f + g)y\\] It strikes me that the Eckmann-Hilton argument might be some kind of converse to this, but I can't quite see how.
Anindya - yes, the Eckmann-Hilton argument is a converse. I strongly urge you to prove it yourself, because it's delightful. I suggest proving this:
Eckmann–Hilton Theorem. If \(M\) is a monoid with multiplication \(\cdot : M \times M \to M\), and \(\cdot : M \times M \to M\) is a homomorphism of monoids, then \(M\) is commutative.
Here I'm making \(M \times \) into a monoid in the obvious way, with multiplication
\[ (a,b) \cdot (c,d) = (a \cdot c , b \cdot d).\]
You can prove it for groups if you feel more comfortable with them, but I recommend proving it for monoids instead of groups because inverses are not used in the argument. As far as I know, the identity element of the monoid does play an essential role in the argument!
This is one of the most satisfying short proofs I know.
Comment Source:Anindya - yes, the Eckmann-Hilton argument is a converse. I strongly urge you to prove it yourself, because it's delightful. I suggest proving this: **Eckmann–Hilton Theorem.** If \\(M\\) is a monoid with multiplication \\(\cdot : M \times M \to M\\), and \\(\cdot : M \times M \to M\\) is a homomorphism of monoids, then \\(M\\) is commutative. Here I'm making \\(M \times \\) into a monoid in the obvious way, with multiplication \\[ (a,b) \cdot (c,d) = (a \cdot c , b \cdot d).\\] You can prove it for groups if you feel more comfortable with them, but I recommend proving it for monoids instead of groups because inverses are not used in the argument. As far as I know, the identity element of the monoid _does_ play an essential role in the argument! This is one of the most satisfying short proofs I know.
\(\textrm{dot} : M \times M \rightarrow M\) is a homomorphism
\(\iff \textrm{dot}((a, b)\cdot(c, d)) = \textrm{dot}(a, b)\cdot\textrm{dot}(c, d)\)
\(\iff \textrm{dot}(a\cdot c, b\cdot d) = \textrm{dot}(a, b)\cdot\textrm{dot}(c, d)\)
\(\iff (a\cdot c)\cdot (b\cdot d) = (a\cdot b)\cdot (c\cdot d)\)
Set \(a = d = 1\), then \(c\cdot d = (1\cdot c)\cdot (d\cdot 1) = (1\cdot d)\cdot (c\cdot 1) = d\cdot c\)
OK I'm now thinking – a monoid is a category with one object, and a monoid homomorphism is a functor between monoids-as-categories. So can we generalise this to say something about categories-with-perhaps-many-objects when composition is functorial in some sense?
Comment Source:\\(\textrm{dot} : M \times M \rightarrow M\\) is a homomorphism \\(\iff \textrm{dot}((a, b)\cdot(c, d)) = \textrm{dot}(a, b)\cdot\textrm{dot}(c, d)\\) \\(\iff \textrm{dot}(a\cdot c, b\cdot d) = \textrm{dot}(a, b)\cdot\textrm{dot}(c, d)\\) \\(\iff (a\cdot c)\cdot (b\cdot d) = (a\cdot b)\cdot (c\cdot d)\\) Set \\(a = d = 1\\), then \\(c\cdot d = (1\cdot c)\cdot (d\cdot 1) = (1\cdot d)\cdot (c\cdot 1) = d\cdot c\\) OK I'm now thinking – a monoid is a category with one object, and a monoid homomorphism is a functor between monoids-as-categories. So can we generalise this to say something about categories-with-perhaps-many-objects when composition is functorial in some sense?
Yes, that's the proof, Anindya! Isn't it cute? You can see why identity elements are important: it doesn't work for mere semigroups.
I don't think there's a generalization to categories in the direction you're suggesting. The really nice generalizations to categories work a bit differently. You can read about them in my paper Higher-dimensional algebra and topological quantum field theory.
Even if you don't want to read the paper, look at the picture at the top of page 25, and then the picture below that! This reveals the topological underpinnings of the Eckmann-Hilton argument.
One of the first spinoffs of this viewpoint is:
A monoidal category in the 2-category of monoidal categories is a braided monoidal category.
A monoidal category in the 2-category of braided monoidal categories is a symmetric monoidal category.
See, it goes on for one more step when we categorify it! This eventually leads to the "periodic table of \(n\)-categories", which is one of the main subjects of my paper.
Comment Source:Yes, that's the proof, Anindya! Isn't it cute? You can see why identity elements are important: it doesn't work for mere semigroups. <img width = "150" src = "http://math.ucr.edu/home/baez/mathematical/warning_sign.jpg"> I don't think there's a generalization to categories in the direction you're suggesting. The really nice generalizations to categories work a bit differently. You can read about them in my paper [Higher-dimensional algebra and topological quantum field theory](https://arxiv.org/pdf/q-alg/9503002.pdf). Even if you don't want to read the paper, look at the picture at the top of page 25, and then the picture below that! This reveals the topological underpinnings of the Eckmann-Hilton argument. One of the first spinoffs of this viewpoint is: * A monoidal category in the 2-category of monoidal categories is a braided monoidal category. and then * A monoidal category in the 2-category of braided monoidal categories is a symmetric monoidal category. See, it goes on for one more step when we categorify it! This eventually leads to the "periodic table of \\(n\\)-categories", which is one of the main subjects of my paper.
cheers – will take a look at that
Comment Source:cheers – will take a look at that
Pete Morcos
In my ongoing attempts to catch up with the class, I've just finished Chapter 2! As I did with Chapter 1, I've made a series of picture posts to help me get better intuition for the material.
The full index of picture posts is at my wiki page. If you spot any errors or conceptual misunderstandings, I hope you'll let me know.
For those who read my Chapter 1 posts in May, there have been a few changes.
Update:18 June The two forward images - For some reason I had a lot of trouble with these concepts. I've fixed some errors that were in my first version.
New: 18 June Computing adjoints - A visual interpretation of John's formulas from Lecture 6.
New: 18 June Picture proof that left adjoints preserve joins - The proof of this fact dances back and forth between sets, joins, and adjunctions, so it's a bit hard to follow. I put the building blocks in picture form, and show how to glue them together in a "picture proof." It turned out quite pretty!
Here are the new posts for Chapter 2. Go to the index page for links.
Picture proof that oplax left adjoint means lax right adjoint - Another proof built up from picture fragments, to help keep track of all the properties involved.
Posets are subsets of a total order - I wanted to verify that you can always put a poset in some linear order, so that I could do so in later pictures. This post has little bearing on Chapter 2 concepts.
Visualizing product orders - Chapter 2 makes heavy use of order products, so I wanted a visual reference to help my intuition.
Monoidal total orders - I start with the simplest case for my first pictures of how monoidal products work. I quickly discover that I have no idea how to represent associativity visually.
Monoidal partial orders - Poset products are harder to draw, but in a very simple case I can manage to make a picture and learn a few things. I remain stumped about the question of freedom in choosing the monoidal unit.
Adjunction plots - As I was working on this chapter, I realized that the behavior of an adjunction can be drawn in function plot form, not just in sets-and-arrows form. This gave me an interesting new perspective on the Chapter 1 material.
Many products: abstract - Most of chapter two involves binary relations. I decided to put them all in product form to make it easier to spot the similarities and differences. Topics are: monoids, monoidal preorders, enriched categories, and closed preorders.
Many products: Cost - Using Cost as a concrete example lets me draw more helpful pictures of the various products. At the end, I combine several of the ideas above into a big 3D "adjunction plot" that captures most of the behavior of the Cost closed preorder.
Comment Source:In my ongoing attempts to catch up with the class, I've just finished Chapter 2! As I did with Chapter 1, I've made a series of picture posts to help me get better intuition for the material. The [full index](http://www.azimuthproject.org/azimuth/show/Pete+Morcos) of picture posts is at my wiki page. If you spot any errors or conceptual misunderstandings, I hope you'll let me know. For those who read my Chapter 1 posts in May, there have been a few changes. 1. *Update:18 June* **The two forward images** - For some reason I had a lot of trouble with these concepts. I've fixed some errors that were in my first version. 1. *New: 18 June* **Computing adjoints** - A visual interpretation of John's formulas from Lecture 6. 1. *New: 18 June* **Picture proof that left adjoints preserve joins** - The proof of this fact dances back and forth between sets, joins, and adjunctions, so it's a bit hard to follow. I put the building blocks in picture form, and show how to glue them together in a "picture proof." It turned out quite pretty! Here are the new posts for Chapter 2. Go to the index page for links. 1. **Picture proof that oplax left adjoint means lax right adjoint** - Another proof built up from picture fragments, to help keep track of all the properties involved. 1. **Posets are subsets of a total order** - I wanted to verify that you can always put a poset in some linear order, so that I could do so in later pictures. This post has little bearing on Chapter 2 concepts. 1. **Visualizing product orders** - Chapter 2 makes heavy use of order products, so I wanted a visual reference to help my intuition. 1. **Monoidal total orders** - I start with the simplest case for my first pictures of how monoidal products work. I quickly discover that I have no idea how to represent associativity visually. 1. **Monoidal partial orders** - Poset products are harder to draw, but in a very simple case I can manage to make a picture and learn a few things. I remain stumped about the question of freedom in choosing the monoidal unit. 1. **Adjunction plots** - As I was working on this chapter, I realized that the behavior of an adjunction can be drawn in function plot form, not just in sets-and-arrows form. This gave me an interesting new perspective on the Chapter 1 material. 1. **Many products: abstract** - Most of chapter two involves binary relations. I decided to put them all in product form to make it easier to spot the similarities and differences. Topics are: monoids, monoidal preorders, enriched categories, and closed preorders. 1. **Many products: Cost** - Using **Cost** as a concrete example lets me draw more helpful pictures of the various products. At the end, I combine several of the ideas above into a big 3D "adjunction plot" that captures most of the behavior of the **Cost** closed preorder.
Thanks for all these pictures, Pete! I've been too busy struggling to keep up production of my own lectures to spend enough time talking about your contributions.
Comment Source:Thanks for all these pictures, Pete! I've been too busy struggling to keep up production of my own lectures to spend enough time talking about your contributions.
Yeah, thank you Pete. Maybe at the end of this Pete could gather them up and make them into a book.
Edit: Or at least give them a nice home somewhere online.
Comment Source:Yeah, thank you Pete. Maybe at the end of this Pete could gather them up and make them into a book. Edit: Or at least give them a nice home somewhere online.


While Burrito Bison was quietly shopping at the local supermarket, a terrible boom suddenly resounded in the small Mexican town! All the town's bags of gummy bears exploded at the same time, releasing an army of living teddies ready to invade the world. Help our bison, along with his faithful friends El Pollo and Pineapple Spank, get rid of these vermin by flying as high and as far as possible to crush them all.


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Eddie Murphy >
Eddie Murphy To Be A Father For The Ninth...
Eddie Murphy To Be A Father For The Ninth Time
By Ed Biggs in Lifestyle / Showbiz on 05 November 2015
Follow Eddie Murphy
Murphy and his girlfriend Paige Butcher will welcome their first baby into the world in May next year.
Comedy legend Eddie Murphy is to become a father for the ninth time, a new report from the MailOnline confirms.
The stand-up comedian and Oscar-nominated actor, 54, and his current girlfriend Paige Butcher, 36, will welcome a baby in May 2016, a source close to Murphy confirmed on Thursday (November 5th).
"He couldn't be happier. Eddie really is a family man and loves all of his children, so he's so excited about having another one on the way," the unnamed source said.
Eddie Murphy is to become a father for the 9th time
Murphy had five children with his ex-wife Nicole Murphy, whom he divorced in 2006 after 13 years of marriage. Ranging in age from 13 to 26, their names are Bria, Myles, Shayne, Zola and Bella. Outside of that marriage, he had a son, Eric, with Paulette McNeely in 1989, and then a son Christian, with former girlfriend Tamara Hood in 1990.
After his divorce, he had a daughter with former Spice Girl Mel B in 2007, called Angel, after dating for a short time. Murphy was famously confirmed to be the father after a DNA test.
More: Eddie Murphy makes his first onstage joke in 28 years, and it's about Bill Cosby
Despite these circumstances, the entire family remains close as a result of Murphy's efforts to keep in regular contact with all of them. "Even though he has eight children by four different women, they all come together on holidays and for sporting events and hang out at his mansion as one big blended family," the source continued.
"All of the children love each other dearly and the women, having all moved on with their own relationships, co-exist very well together."
Murphy, who has always kept his private life away from the spotlight, met Australian model Butcher three years ago, and it's this relationship that is believed to have inspired him to return to the spotlight of stand-up comedy after nearly three decades.
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