text:
Canadian Authors Association
Writers helping writers since 1921
About CAA
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Founded in 1921 by Stephen Leacock, Pelham Edgar, B.K. Sandwell, John Murray Gibbon (who became its first president), and other prominent writers of the time, to lobby for the protection of authors' rights, the Canadian Authors Association has continued to pursue this objective to this day. Instrumental in the 1924 copyright legislation, the Association began pressing for a new Copyright Act in the 1980s.
We also continue to participate in the national organizations we helped create to ensure that Canadian writers retain rights to their intellectual property and are paid for the use of their works. These organizations include Access Copyright, the Book and Periodical Council, the Canadian Copyright Institute, the International Authors Forum, and the Public Lending Right Commission.
Some 25,000 writers have been members including, in the early days, Stephen Leacock, Bliss Carman, Robert W. Service, Ralph Connor, E.J. Pratt, Mazo de la Roche and Sir Charles G. D. Roberts.
Timeline: Milestones and Achievements
Helped create The Canadian Writers' Foundation to provide financial assistance to indigent authors and their families. This charitable foundation continues to help eminent Canadian writers who have fallen on hard times.
Founded the Governor General's Literary Awards. (In 1959 the Awards were turned over to Canada Council and continues to provide recognition to outstanding Canadian authors.)
Published Canadian Author & Bookman, originally founded in 1919 as The Canadian Bookman (Canadian Poetry was incorporated in 1968, and the publication became simply Canadian Author with the Fall 1992 issue. Publication was suspended in 1998.)
Originated the first standard book contract to protect authors' rights in negotiations with publishers. Also began a campaign to bring a Public Lending Right program to Canadian authors. Other Canadian writing groups joined the campaign, and the PLR program was finally created in 1986.
Successfully lobbied to have Canada join the Universal Copyright Convention. This convention continues to protect writers whose countries are signatories.
Published the first edition of The Canadian Writer's Guide (13th edition published in 2002).
Established the Vicky Metcalf Awards for children's literature (suspended in 2002; currently administered by the Writers' Trust of Canada).
In keeping with the motto 'Writers helping writers', reaffirmed Canadian Authors' policy of welcoming non-published writers as members.
Established the Allan Sangster Award for meritorious service to the Canadian Authors Association (only members are eligible).
Established the CAA Literary Awards for Fiction, Nonfiction, Poetry and Drama (currently suspended).
Celebrated Canadian Authors Association's 60th anniversary with Syllables of Recorded Time: The Story of the Canadian Authors Association 1921–1981) by Lyn Harrington, published 1981.
Established the Lela Common Award for Canadian History and the Jubilee Award for Short Stories (both are currently suspended). Received tax exempt status as a registered national arts service organization (NASO).
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Welcome to my website, where you can view some of my photography as well as book a photo session! Feel free to look around and enjoy my content.
For senior photo's, I will charge $15 and do at least 45 minutes worth of photos and edit 10 of them. If you allow me to use them for my portfolio I will charge just $10.
Feb 2019 - Free photo's on Saturday's!
Book now, before February is over, go to the "Book" page to book a photo shoot today!


Ejabberd Community 13.10 supports YAML format for configuration file. Ejabberd newcomers were often frighten editing Erlang term file (ejabberd.cfg) and the YAML support is a way to help configuring ejabberd using a more familiar text format. It's possible to convert ejabberd configuration to YAML using convert_to_yaml command from ejabberdctl.
body: "Welcome to this Jabber server."
How to let ejabberd use the YAML configuration file?
If you installed ejabberd from binary installer, you must edit ejabberdctl script to change configuration file extension on the EJABBERD_CONFIG_PATH environment variable, at line 95.
Problem 'exit "Problem loading ejabberd config file /etc/ejabberd/ejabberdctl.cfg approximately in the line 2: syntax error before: '#'"' occurred executing the command.
So can you more specific about what problem you have? or is that resolved already?
you try to convert ejabberdctl.cfg, which is not ejabberd.cfg. that's the reason of your error cause ejabberdctl.cfg is not an erlang term based configuration file.
I noticed an issue showing up as "no function clause matching ejabberd_c2s" in my logs. It seems that some values are not converted from the old .cfg to the new .yml configuration. Check your yml configuration file for entries that show up as "val01,val02", break them out onto newlines, pre-pended with '-'.
As explained in the post, it is just a matter of changing environment variable EJABBERD_CONFIG_PATH.
I'm following following steps to configure ejabbed server on local system, but when open server URL on port number 5280, it's not loading. 'Page Not Found'.
Am I missing anything here, Please suggest.


Did Salih Khater Panic after being scared by the Ambulance?
After watching the Salih Khater video it looks very much as though he was driving normally, was cut up by an ambulance on an emergency call, swerved to the left, hit the cyclist on the pedestrian crossing and then just panicked.
Has this been talked about anywhere or is it definitely being treated as a terrorist incident?


We are pleased to announce that for the 11th year in a row, Altair Advisers has been named by Barron's magazine as one of the top 100 independent financial advisers. We are particularly proud of Altair's standing as the only Chicago firm, and one of only 12 nationwide, to be included in these prestigious rankings every year since the list's inception in 2007. Criteria for consideration include volume of assets overseen, revenue, and quality of practice.
Source: Barron's "Top 100 Independent Financial Advisors," September 18, 2017. Barron's is a registered trademark of Dow Jones & Company, L.P. All rights reserved. Neither Altair Advisers nor any of its employees pays a fee to Barron's in exchange for this rating.


As I sit here in my PJ's waiting to see if I stay awake to see midnight, with my children sleeping peacefully, my belly full of Chinese takeaway, I can reflect happily on a lovely year.
This year saw the arrival of our beautiful Termite, completing our family and best of all I achieved my dream of a home birth as she was born at home peacefully and happily.
Our peanut started reception and has blossomed and learned lots and lots including how to read which I'm puffed with pride to say she is really good at.
The Golden Child has started nursery and we've seen him grow in confidence developing friendships independent of his big sister, and turning into a real chatter box.
Personally I'm proud of my blog, it may not be huge yet but my followers are growing and I feel pleased to have been approached by some amazing companies to review their products.
I'm sure we must have been through some lows but all in all our 2013 has rocked and I hope we can continue through 2014 as happily as we have been. Happy New Year to you all. I hope all of your dreams come true.
Did you make midnight? I did just. .
Happy new year - what a fab 2013 you've had and Im sure 2014 is going to see everything continuing the same way. Heres to a fn filled 2014!


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Q: gdal installation for Python Visual Studio I want to develop GIS based GUI in python using Visual Studio 2013.
But when I am trying to using gdal it is showing error given below-
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
ImportError: No module named gdal

It would be helpful, if you let me know how to install gdal for python?
I am using Visual studio 2013 for python.

A: For windows, you may download from here.

*

*Add the installation directory bin folder to your system PATH, remember to put a semicolon in front of it before you add to the existing path.
C:\gdalwin32-1.6\bin

*Create a new user or system variable with the data folder from your installation.
Name : GDAL_DATA
Path : C:\gdalwin32-1.6\data

If you don't know how to do this, than suggested Microsoft KnowledgeBase documentation
For more details you may take look here for other platforms.

A: You should be able to install the components for various versions of MSVC and Python at this website:
http://www.gisinternals.com/
I've only used the above resource for running GDAL under Python. I've never used Visual Studio 2013 for Python, but hopefully the above resource provides the necessary components.

redpajama_set_name: RedPajamaStackExchange


For so many, this became a joke, but as you will see, it was no laughing matter and was a serious burn, which caused Michael Jackson continuous,debilitating pain with complications compounded by the auto immune disease Lupus that he already suffered from. It is due to these procedures and the enduring pain. that Michael became addicted to prescription painkillers, as he admitted himself and for which he sought help for. It is important to understand this and to SEE this. The following photos are graphic, and disturbing, and though I feel that the release of these is an invasion of Michael's privacy, I also realize that until many actually SEE the pain he was dealing with, the stigma will continue. Many thanks to TrueMichaelJackson.com for the following photos and information regarding Michael Jackson's scalp burn.
On January 27th 1984, Michal Jackson was at the Shrine Auditorium in Los Angeles, filming the Pepsi spot in front of fans. A pyrotechnic malfunction ignited the singer's hair, and handlers rushed to his aid. The incident left Michael with second-degree burns on his scalp.
"When they were trying to repair that burned spot, he had a balloon under his scalp that was inflated..He let me feel it. It was a huge mound. As the skin got stretched, they cut it out and stitched the scalp. He was in excruciating pain."
But (INAUDIBLE) would you — (INAUDIBLE) too much stretch back in the scar, you understand?
KING: Did you see him one other time?
KLEIN: Of course I did. But he would have a stretch back on the scar.
I mean the scar would get worse after they removed it.
And I had to put a stop to it. So I told Michael, we have to stop this. And that's when I fired this plastic surgeon altogether.
And I said I can't deal with this anymore. We're going to deal with me as your doctor or you're going to have to find another doctor if you want to work with him.
~ He was in pain and he did need painkillers. Everybody took it out of context.


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NKOSI SIKELEL' iAFRIKA - GOD BLESS AFRICA. SOUTH AFRICAN ANTHEM - MANDELA R.I.P.
It is impossible to not think back over the life of Nelson Mandela on hearing of his death. My own involvement with South Africa goes back 45 years or more to the time when I was a teenager and a fledgling activist looking for good causes to support. South Africa was a place we knew very little about because it was an embarrassing secret to so called western democracies that were up to their elbows in it through investments. Slave Labour is one of the purest forms of capitalism there is and that was what provided the rich investment opportunities. It also provided a great standard of living with well paid jobs and all you had to do to get your share of this was to have a white skin. The whole stinking edifice was held together by that most inhuman brutal system called Apartheid, unparalleled in its evil and depravity.
That so many seemingly nice decent people, (relatives anyone?) Yes it's a horrible admission isn't it, were able to go there and make themselves a comfortable living is a horrible reminder of how frail our human condition really is. The Jewish Americam writer Hannah Arendt coined the phrase "The banality of evil" to describe Adolph Eichman when reporting his trial for war crimes and that sums up South Africa under white supremacist rule as they enjoyed the wages and lovely houses provided by the black population who were treated like beasts of burden. Now of course we see the disgusting spectacle of those who wanted Mandela not only to lose but to be executed declaring their admiration for him. Head of that list is the revolting and amoral Margaret Thatcher who declared him a terrorist and vehemently opposed sanctions against apartheid South Africa while conveniently forgetting to mention that her equally wretched husband Dennis had considerable sums of money invested in that corrupt and evil system, as did churches, The city and pillars of our superior society. Cameron was also an enthusiastic opponent of Sanctions as was Tebbit and the rest of them, a more thoroughly dishonest and revolting bunch would be hard to find in history's pages.
To take on these circumstances and triumph was one of history's greatest achievements and it is true in my opinion to say that without Mandela it would not have been done. His courage and dignity were quite amazing and he rightly became recognised as one of the greatest leaders in history. He was the mighty indestructible presence in the background as we marched and demonstrated and sometimes blundered our way through the Anti Apartheid struggle, I wonder how many times the South African White Supremacists must have bemoaned the fact that they did not just kill him when they had the chance. He became too big to kill and too well known to slip on the soap and fall down stairs in prison nor would the world believe a tale of suicide in prison, he became more powerful than those who jailed him. He was the inspiration for us and millions around the world. Glasgow was the first place to grant him freedom of the city, something to be proud of. I took my daughter Clare to see him in George Square as he thanked Glasgow for its unstinting support over the years, we were both young enough for me to lift her on to my shoulder to see him speak and then delight the crowd with an impromptu display of singing and dancing a precious memory for us both, as a delirious crowd embraced him while mother Glasgow poured down its relentless gentle rain, a memory she still talks about. He is quite simply irreplaceable but for South Africa there is no going back, the racist beast has been slain and so it must stay. There will be plenty of struggles ahead but they will win in the end and, as the legendary murdered American union man Joe Hill said "don't mourn me, organize" Madiba would agree with that sentiment.
Posted by Cllr Terry Kelly at Saturday, December 07, 2013
Before you go off on one about how evil the Tories were in opposing sanctions against South Africa's apartheid regime, let's look at what Maggie Thatcher really was involved in at that point in history....Via This genuine document
Sunday, December 08, 2013 10:42:00 pm
Byeck said...
'S Africa was a place we knew little about.'
Kelly, you knew nowt about the Falklands either..it occurs to me, you're knocking the Empire while knowing bugger all about it.
So, to add to your limited knowledge, as early as mid 19th Century,under British rule, everyone born in Cape Colony,, irrespective of colour,or race, was entitled to vote.
Monday, December 09, 2013 8:21:00 pm
Dazed and confused, tell us about Thatcher and sanctions then. Tell us your version.
Tuesday, December 10, 2013 1:17:00 pm
Apartheid wasn't all bad was it Byeck?. People knew their place didn't they and they didn't give cheek to their masters it were the good old days.
Kelly @ 19.00
Kelly, you rascal, I don't think I said anything good about apartheid, so nice try, but no coconut.
Now, Lady Thatcher and apartheid. Mandela disagrees with your view, the ANC disagree with you, recently released documents disagree with you and, for heavens sake, even I disagree with you.
D'you think, Sage of Renfrew, you might have got it wrong...?
Anonymous Byeck said...Tuesday, December 10, 2013 4:43:00 pm.
Mrs. Thatcher was opposed to sanctions, she supported the South African Government in it's opposition to the ANC and she regarded Mandela as a terrorist.
I am not aware of anything that I have written that Mandela and the ANC would disagree with. I'm not much concerned whether you agree with me or not.
Kelly @ 4.43
Kelly, you're in denial..She (Lady Thatcher) is an enemy of apartheid. We have much to thank her for,' Nelson Mandela, July 1990
'I believe releasing (Mandela) would have more impact than anything else.' Secret letter from Lady Thatcher to Botha, 1985, released by ANC.
I usually forgive your cavalier attitude to facts, laddie, but this is a fable too far. By the by, to shed a little more light on your darkness,, Botha was President of S Africa.
The Tories were against sanctions, David Cameron was involved with printing hang mandela posters, yet you with your mighty socialist mind remain in bed with them when it comes to Scottish Independence, please explain and discuss the meaning of principles you charlatan.
Wednesday, December 11, 2013 10:51:00 am
Byeck said...Tuesday, December 10, 2013 7:26:00 pm.
May 4 1979 On the steps of Number 10 Downing Street.
"Where there is discord, may we bring harmony. Where there is error, may we bring truth. Where there is doubt, may we bring faith. And where there is despair, may we bring hope." Thatcher quoting from the prayer of St. Francis of Assisi. That went well didn't it?.
"It was you who brought democracy to Chile" Thatcher praising General Pinochet.
"To the Leader of the Great First of September Revolution" 1981. Thatcher's Christmas greetings to Colonel Muammar Gaddafi.
Saddam Hussein supported in power by financial and military help by Thatcher's Government.
Thatcher, evil beyond belief.
Anonymous said...Wednesday, December 11, 2013 10:51:00 am.
According to deviants like you we were in bed with the Tories during World War Two as well when the snp were siding with Hitler and the nazis.
World war two was not political for the people of Britain and the independence issue is not political either it is constitutional and when it is over we will go back to opposing each other just as the snp will with their present independence allies, if they still have any by September 2014.
You are a cretin and you would be better employed polishing up your jackboots and ironing your tartan frock.
What exactly has this reply to do with Mandela and the ANC praising Baroness Thatcher?
Honestly, Terry lad, I despair, I really do. But I'm still fond of you...you put me in mind of my uncle, Mad Albert.....a fine man, but a carrot short of a casserole.
Wednesday, December 11, 2013 2:28:00 pm
Anonymous Byeck said..Wednesday, December 11, 2013 2:28:00 pm.
I shouldn't have to explain this and if you genuinely don't understand it then you have clearly inherited your mad uncle Albert's genes. You might as well say Thatcher was fond of praising Arthur Scargill and Tony Benn as well so patently dishonest is the suggestion.
As to why it was said I have no idea only those who said it can explain but I am convinced it was not genuine, there must have been mileage in it for them. Just as Thatcher thought there might be mileage in quoting St. Francis of Assisi. It's called politics and your pretence at not grasping it is wearing thin.
HA HA the snp were siding with Hitler and the Nazi's what like Oswald Mosely the infamous labour politician you moron, one from the lunatic fringe doesn't constitute party policy cretin, whereas currently five councils in Scotland are run by Labour/conservative partnerships, you couldn't squeeze a rizla between the policy's of Red Ed and Call me dave, and you, a tired old 1970's socialist just keep toeing the party line anyway, you are a disgrace to socialism you half wit, cast off the yoke of the london based labour party and think for yourself (it's not as difficult as dinosaurs like you imagine)
Thank you, Councillor. And now, my final word on this subject, a poem (after the style of W T McGonagall)
Maggie dear,
You showed no fear,
You stood against Apartheid,
As for the Chuff,
I called his bluff,
And showed him for a S**teheid!
Kelly @ 3.07 Tuesday, December 10, 2013 4:43:00 pm
Thatcher the Devil
now with the evil.
burning below for the seeds she did sew.
Torture and hatred is her legacy.
Millions grieve that she left in misery.
Revolting tyrant mad and bad.
without her the world is a little less sad.
May the hellish fire bring eternal pain.
And may we never see her like again.
Anonymous said...Wednesday, December 11, 2013 3:08:00 pm.
Moseley left the Labour Party before becoming a fascist the snp were in favour of Hitler and the Nazis, can you spot the difference. The snp are still in favour of the Nazis as we seen when they welcomed the Flemish fascists to march with them in Edinburgh a couple of months ago.
"we Scots did not mind Mrs. Thatcher's economic policies so much" (alex salmond).
Thursday, December 12, 2013 11:27:00 am
Just to educate you, the SNP was only founded in 1934, the leader of the SNP from 1942-1945 advised scottish people to refuse conscription and was consequently jailed for his protest, however Edward Mosely a labour member of parliament and a member of cabinet led fascist black shirts marching through the streets of london, and he fell out with the labour party over unemployment policy not because he was a fascist.
there you go free education SNP style, and by the by Tel the flemish nationalists weren't invited they just turned up and were given short shrift as one would expect, if you ever get around to organising a NO rally let me know how you will keep UKIP BNP and the orange order away you half wit.
Thursday, December 12, 2013 12:57:00 pm
Anonymous said...Thursday, December 12, 2013 12:57:00 pm.
The Mosley I'm referring to was Oswald not Edward? you obviously know a lot about this don't you?
He fell out with the Labour Party because he could not get his way mainly over unemployment; he was not a fascist while in the Labour Party he formed the British Union of Fascists later.
The snp tried to get people not to oppose Hitler and they were led by Arthur Donaldson at the time who was known as a Hitler appeaser and was rightly jailed, that's the snp. They were also marching side by side with the Flemish Fascists in Edinburgh a fact that was highlighted in newspaper pictures, you are either a liar or a fool or perhaps both.
The snp also has a squalid history of anti English and anti Irish racism as well as anti Catholic sectarianism. You are apparently proud of them; you are welcome to each other. You are Scotland's shame, anti English attacks rising right now fuelled by the same snp racist rhetoric used throughout history by Nazis and nationalists like Hitler, Franco and Mussolini, what a lovely bunch you people are.
Thursday, December 12, 2013 3:16:00 pm
"Arthur Donaldson at the time who was known as a Hitler appeaser and was rightly jailed, that's the snp."
Utter lies the ONLY MP interned during the war was Captain Archibald Henry Maule Ramsay a Scottish Unionist, the forerunners to the current Tory Party. That would be the SAME Tory party that Labour are now in bed with in the NO campaign. Let's also not forget that Oswald Mosley was a LABOUR MP at one point.
In contrast Arthur Donaldson was held for six days and then released without charge.
"The snp also has a squalid history of anti English and anti Irish racism as well as anti Catholic sectarianism."
Again sheer lies, I don't suppose you would care to cite any evidence would you? The Labour party have a long history of sectarianism in Scotland indeed just last year Gordon Matheson the Labour head of Glasgow Council gave a speech in front of a cheering crowd of Orange Order members days prior to the local elections.
According to eye witnesses, Mr Maheson claimed the 18 month old marches policy aimed at restricting the number and duration of marches was wrong and promised to review the policy should Labour win the upcoming local elections.
The Labour party also have a long history of being anti-Scottish and deliberately lying to the Scottish people. Indeed your OWN former UK Chancellor Denis Healey admitted that Labour lied over the value of North Sea Oil when he stated in an interview with Holyrood Magazine that the UK Government had "underplayed the value of the oil to the country" over fears that Scotland would decide to go independent.
He also added that Scotland would prosper under independence thanks to North Sea oil and Westminster politicians are "worried stiff" about losing revenues from offshore production.
"anti English attacks rising right now"
Again an outright LIE as analysis of the figures for racist attacks in 2012 revealed that attacks against 'white English' actually fell in the last year in all police force areas which were able to provide a breakdown of the statistics.
Anonymous said...Monday, December 16, 2013 9:40:00 pm
Donaldson was jailed and the snp were handing out pro Hitler propaganda and campaigning to persuade young Scotsmen not to oppose the Nazis, that is a fact. The snp welcomed the Flemish Nazis to their march in Edinburgh and that is also a fact. Moseley left the Labour Party before he became a fascist and set up the British Union of Fascists that is another fact.
The snp are guilty of anti English racism, every time they criticise "Westminster", or "down south" or "London" or "diktats from the home counties" they mean England and the English. They use weasel words in an attempt to hide their racism. They also have a history of anti catholic sectarianism. Their leader objected to the visit of Pope John Paul and I have personally seen propaganda from their members which was anti Catholic and they still are, that's another fact.
I recently helped to facilitate a council civic reception to commemorate the 150th. Anniversary of a local Orange Order branch in my ward, they are not illegal and they are also 'bona fide' members of the Church of Scotland, which was a respectable organisation the last time I looked. It is common knowledge that they have a problem with a minority of followers whom they try to discourage from behaving badly but the snp have an even greater problem with their members and followers and I expect that fact to be writ large after they take a humiliating defeat in the forthcoming laugharendum when the streets will not be safe for anyone with an English accent.
You are a liar and an apologist for evil, that evil is nationalism in all its ugly guises, as Einstein said you are "the measles of the planet Earth". Go off and iron your white hood and polish up your jackboots, you are a disgrace to the Scottish people.
A minority of followers? You want to try living with what we on the Gallowgate are subjected to during the marching season. The Famine Song played, spat at for being 'taigs'. It's like Northern Ireland when your respectable friends go a marching. The BNP is a legal organisation do you support that as well?
You should be ashamed supporting such bigotry on an otherwise excellent post on Mandela.
Saturday, December 21, 2013 11:02:00 pm
I think you are assuming too much here. I didn't say I supported the Orange order and for the sake of clarity I do not. There are lots of organisations I do not support but as long as they are legal I am prepared to tolerate them, the incidents you describe are police matters. I am a supporter of a united Ireland and I want to be able to express that, if I wasn't able to do that I would consider that an injustice. Freedom of assembly and freedom of expression comes at a price and part of that price is that it must be available to all within the law. People who know me would consider it a collectors item for someone to accuse me of "supporting" the Orange order.
SCOTTISH WORKERS BETRAYED BY SALMOND AND THE SNP. ...
NKOSI SIKELEL' iAFRIKA - GOD BLESS AFRICA. SOUTH A...
REMEMBERING ROSA PARKS, A TRUE AMERICAN HERO.
IS THE GLASGOW HERALD PLAYING THE GAME WITH THE IN...

redpajama_set_name: RedPajamaCommonCrawl


text:
Dubai bank Emirates NBD to make it easier for VAT-registered SMEs to get loans
The new programme will cover home loans, auto loans and business loan products
by Aarti Nagraj
Dubai's biggest bank Emirates NBD has announced the launch of a new VAT-based loan programme for small and medium-sized enterprises (SMEs) in the country.
SMEs can now apply for financing by providing copies of their VAT (Value Added Tax) returns filed with the UAE's Federal Tax Authority to serve as validation of business turnover and income, the bank said in a statement.
The new programme will cover home loans, auto loans and business loan products.
Suvo Sarkar, senior EVP and head of Retail Banking and Wealth Management, said the move supports the objective of Expo 2020 Dubai to foster growth of the SME sector.
The lender is the official banking partner of Expo 2020 Dubai.
The UAE implemented 5 per cent value added tax (VAT) on the supply of most goods and services in the country on January 1, 2018.
The country's Federal Tax Authority (FTA) has revealed that a total of 296,000 businesses have registered for VAT in its first year of implementation.
Read more: First year of VAT in the UAE: FTA reveals the key details of implementation
According to official data released last year, the UAE has 350,000 SMEs, which collectively provide 86 per cent of private sector jobs and account for 60 per cent of gross domestic product.
The Ministry of Economy estimates that in Dubai alone, SMEs will account for nearly 95 percent of companies, employ 42 per cent of the workforce and contribute 40 per cent of GDP by 2021.
Also read: UAE regulator to launch SME financing platform
Aarti Nagraj January 6, 2019


text:
coronavirus Food and agriculture Health Migrants30. June 2020
Fruit pickers sweat as virus flares in northeast Spain
Fraga, Spain – In the verdant orchards of northeastern Spain, Guinean national Said Doumbia is picking nectarines in a job that earns him just 5.25 euros (nearly $6) per hour.
After a long day in the sun, he's exhausted but when he finally clocks off, he must have his temperature taken before entering the sports hall where he's staying — the complex has been set up especially for seasonal workers following a flare-up of the coronavirus.
Thousands of immigrant workers come to this fertile valley in the Aragon region every year but a fresh outbreak of infections has worried the authorities.
To tackle the problem, the small town of Fraga on Thursday opened temporary accommodation for up to 50 workers who had been sleeping outside and to detect new cases of COVID-19.
When Doumbia arrives, a Red Cross worker scans his forehead with an infrared thermometer, then he must disinfect his shoes and his hands.
Spain's state of emergency had only just been lifted when a new outbreak was detected at a farm near Fraga. Photo: Pau Barrena / AFP
Spain's state of emergency had only just […]
After showering, he can have a hot meal then sleep in one of the halls' well-spaced beds, with lights out at 10:00 pm and breakfast served from 5:00 am.
"Some are working, others are looking for work," said Red Cross official Oscar Gracia.
"Some get very emotional, saying that at last they have a roof over their heads and a plate of hot food."
Many homeless fruit pickers travel around the country seeking work according to the season.
Mame Cisse, a lean 34-year-old in a grey t-shirt, arrived from Senegal two years ago and so far this season, he's earned earning six-euros-per-hour for picking onions in Albacete, and eight for picking olives in Huelva, both areas in the southeast.
Now he's in Fraga "looking for more work".
Fraga set up accommodation for workers in a sports hall in order to help detect new cases of COVID-19. Photo: Pau Barrena / AFP
Fraga set up accommodation for workers i […]
– 'Couldn't do it without them' –
Spain has been badly hit by the virus, suffering more than 28,300 deaths, and the state of emergency had only just been lifted on June 21 when a new outbreak was detected at a fruit farm near Fraga which has been forced to close.
On Thursday, the health ministry said 250 cases had been identified in the area, almost all among seasonal workers, as the World Health Organisation warned of a resurgence in Europe since restrictions were eased.
After 14 weeks of lockdown but only one death, Fraga has had to reimpose some restrictions, banning gatherings of more than 15 people and visits to elderly care homes.
But mayor Carmen Costa said it was important "not to stigmatise the seasonal workers".
There are around 1,000 agricultural holdings around Fraga, where 80% of produce is earmarked for export. Photo: Pau Barrena / AFP
There are around 1,000 agricultural hold […]
"We need them to come and work, we need them as much as they need us," she said.
"We can't have people sleeping in the street, especially not with the current health concerns."
There are around 1,000 agricultural holdings in the area, and 80 percent of produce is earmarked for export with a market value of 60 million euros, says Ignacio Gramunt, director of mercoFraga which helps market the produce.
"In Fraga alone, we employ 5,000 seasonal workers who pick fruit and some 12,000 in the area. We couldn't do it without them."
On his farm, he employs 10 Bulgarians, all wearing masks and carefully spaced as they pick nectarines destined for Germany.
Faye, a 43-year-old from Senegal, had been sleeping rough until authorities opened a sports hall in Fraga. Photo: Pau Barrena / AFP
Faye, a 43-year-old from Senegal, had be […]
Also staying at the sports hall is Faye, a 43-year-old Senegalese national who has been sleeping rough for a month.
All he wants is to be able to work legally, addressing his plea to Spanish Prime Minister Pedro Sanchez.
"Please, Pedro," he says. "Give us papers."
By Laurence Boutreux
agriculture, Bulgaria, coronavirus, farming, fruit, Guinea, Health, immigrant workers, migrant workers, migrants, seasonal workers, Senegal, Spain
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text:
SUFC Futsal round-up - 04
by James Moody
@jamesiron95
A look at the teams and the scores from last week's Futsal fixtures.
The 'A' and 'B' sides played their second games of the campaign last Wednesday in Sheffield.
It was the first time, though, that the 'A' side took to the field following a no-show from Sheffield United a fortnight ago.
Scunthorpe United 'A' (Jack Webster) 1-6 Grimsby Town 'A' (Alex Aldrich, Jan Meiwald (3), Harry Burke (2))
Jack Webster (back row, right) got the Iron's goal in defeat against Grimsby Town as Webster and Jack Grey (back row, left centre) were both cautioned.
CHAMPIONSHIP NORTH
Team P W D L F A + / - Pts
1 Grimsby Town 'A' 2 2 0 0 12 3 9 6
2 Grimsby Town 'B' 2 1 0 1 7 6 1 3
3 Bradford City 'A' 2 1 0 1 6 6 0 3
4 Scunthorpe United 'A' 2 1 0 1 6 6 0 3
5 Middlesbrough 'A' 2 1 0 1 6 8 -2 3
6 Sheffield United 'A' 2 0 0 2 2 10 -8 0
Scunthorpe United 'B' (Kory Brown (3), Harry Tindall, Ollie Steele (2), Jonty Jackson (3), Elso Junior (2)) 11-2 Sheffield United 'B' (Tyler Doman, Louie Barrow)
Kory Brown (back row, centre right) and Jonty Jackson (front row, second left) both hit hat-tricks, while Ollie Steele (back row, centre left) and Elso Junior (back row, right) hit braces in 11-2 demolition. Harry Tindall (front row, second right) also netted.
LEAGUE ONE NORTH
1 Grimsby Town 'C' 3 3 0 0 17 8 9 9
2 Huddersfield Town 'A' 2 1 0 1 20 7 13 3
3 Scunthorpe United 'B' 2 1 0 1 14 8 6 3
4 Sheffield United 'B' 1 0 0 1 2 11 -9 0
5 Middlesbrough 'B' 2 0 0 2 4 23 -19 0
ELEVEN-A-SIDE TEAM WIN 4-0
Not in action this week on a Futsal pitch, players from the C, D and E teams featured in an eleven-a-side match against North Lindsey College.
The side were always in control and emphatically won 4-0 thanks to goals from Ashley Simpson, Jordan Spencer, Ben Johnson and Daniel Newell.
SUFC Trust


text:
Q: How to add gyrator in a circuit using circuitikz? How do you add elements like resistor in a circuit that has a gyrator?

A: This can be achieved in two ways using either tikz with circuit library or circuitikz.
a. Using tikz with circuit library:
\documentclass[border=5pt]{standalone}

\usepackage{tikz}
\usetikzlibrary{circuits.ee.IEC}
\usepackage[european resistor,siunitx]{circuitikz}

\begin{document}
    \begin{tikzpicture}
        % input arrow and its current symbol
        \draw (0.3,2.1) node[inputarrow]{}
                    node[above]{i\textsubscript{s}};
        % leftmost circuit
        \draw (0,2.1) node[ocirc]{} % gyrator height is 2.1 unit
            to [R,l=R\textsubscript{E}] (2,2.1)
            to [L,l=L$_\text{E}$](4,2.1);
        \draw (0,0) node[ocirc]{} to (4,0);
        % gyrator node
        \draw (5,2.1) node[gyrator]{}
                     node[above]{T};
        % middle circuit elements
        \draw (6.1,2.1) node[inputarrow]{}
                    node[above]{v\textsubscript{c}};
        \draw (6,2.1)
            to [R,l=R\textsubscript{MS}, i=] (8,2.1)
            to [L,l=M$_\text{MS}$]  (10,2.1)
            to [C,l=C\textsubscript{MS}] (12,2.1);
        \draw (12,0) to (6,0);
        % rightmost circuit
        \draw (13,2.1) node[transformer]{}
            node[above]{S\textsubscript{S}:1};
        \draw (14,2.1) to (15,2.1) node[ocirc]{};
        \draw (14,0) to (15,0) node[ocirc]{};
        % output current
        \draw (14.5,2.1) node[inputarrow]{}
                    node[above]{Q\textsubscript{s}};
        % input output levels
        \draw[->] (0,1.4) to node[right]{u\textsubscript{s}} (0,0.6);
        \draw[->] (15,1.4) to node[left]{P\textsubscript{s}} (15,0.6);
    \end{tikzpicture}
\end{document}

b. Using circuitikz:
\documentclass[border=5pt]{standalone}

\usepackage[european resistor,siunitx]{circuitikz}

\begin{document}
    \begin{circuitikz}
        % input arrow and its current symbol
        \draw (0.3,2.1) node[inputarrow]{}
                    node[above]{i\textsubscript{s}};
        % leftmost circuit
        \draw (0,2.1) node[ocirc]{} % gyrator height is 2.1 unit
            to [R,l=R\textsubscript{E}] (2,2.1)
            to [L,l=L$_\text{E}$](4,2.1);
        \draw (0,0) node[ocirc]{} to (4,0);
        % gyrator node
        \draw (5,2.1) node[gyrator]{}
                     node[above]{T};
        % middle circuit elements
        \draw (6.1,2.1) node[inputarrow]{}
                    node[above]{v\textsubscript{c}};
        \draw (6,2.1)
            to [R,l=R\textsubscript{MS}, i=] (8,2.1)
            to [L,l=M$_\text{MS}$]  (10,2.1)
            to [C,l=C\textsubscript{MS}] (12,2.1);
        \draw (12,0) to (6,0);
        % rightmost circuit
        \draw (13,2.1) node[transformer]{}
            node[above]{S\textsubscript{S}:1};
        \draw (14,2.1) to (15,2.1) node[ocirc]{};
        \draw (14,0) to (15,0) node[ocirc]{};
        % output current
        \draw (14.5,2.1) node[inputarrow]{}
                    node[above]{Q\textsubscript{s}};
        % input output levels
        \draw[->] (0,1.4) to node[right]{u\textsubscript{s}} (0,0.6);
        \draw[->] (15,1.4) to node[left]{P\textsubscript{s}} (15,0.6);
    \end{circuitikz}
\end{document}

Both result same output:


Well, we are 2 days away from a brand-new baby entering our family. Our oldest daughter is set to deliver by cesarean on Wednesday and I can't help but be so excited and anxious about his arrival (that's right, it's a boy)! I'm not anxious in a nervous way, I am anxious in a "I can't wait for you to meet your family" kind of way! The first, and probably the single most important thing I need you to know, is that you have been blessed with two of the most incredible people as your parents. Your mom and dad have been patiently waiting for a long time to add you to the brood and now you are almost here! But before you arrive, I want to tell you a little about who you will be meeting in a couple of days.
First- your mommy and daddy – they are pretty cool peeps if I do say so myself. Your mom's capacity to express unconditional love has and continues to amaze me through challenging times and times when it would be easier to give up, she never does. She will never give up on you, that you can be sure of. She will be steady and guide you in ways that you won't even recognize until you are grown and ready to guide your own family, she's sneaky like that. There are times when she is stubborn and hard-headed but bear with her, show her love because she will learn as much from you as you will learn from her. What you learn through those times will help you as you grow older. Your mom has a weak stomach and she hates touching chicken, and you better grow to like your kale, because she puts that sh&^ in everything (come to Aiya and Poppas, we will never feed you Kale). Learn to like FRIENDS – because your mom has been obsessed with that show since she was in high school. She can recite every episode, but you probably already know that don't you? She likes things to be perfect seamless (she says she doesn't but she kinda does). She might drive you crazy trying to get you to places on time. Be patient with her. She's just trying to make your life easier, more organized, and…well, perfect seamless (I mean that in the most loving of ways). She can be hard – both on those she loves and others but mainly on herself. Make sure you tell her when she's being too hard and help her be kinder to herself, sometimes she may need a gentle reminder. She will love everyone that you love, even if she has reservations about them, if you love them, she will love them. She'll make your enemies her enemies and will tell you what she would do to get past it, how to forgive them and how you should not waste time and energy on having enemies in the first place.
Your daddy – he loves your mommy so much. He will teach you how to be a good man because he walks the walk. I hope you get his work ethic. He will teach you how to paint a straight line, how to measure twice and cut once, to always put your family first, and how to put up with your in-laws someday. He will teach you how to make things with your own 2 hands. He is talented, kind, funny and did I say that he loves your mommy? He's going to teach you how to slay the big halibut and provide for your family. He will show you how to treat a girl and how to show her respect, because that is how he treats your mom and sister. He is someone to look up to. Your dad will give you enough freedom to make mistakes of your own but will still be there to guide you. He's not perfect, so don't expect him to be, but he will be perfect for you. In a world where family has become something that has been pushed aside or changed altogether, having a dad who puts his family first is important and something that you shouldn't take for granted. Your dad will lead by example – remember, if he can't find the words, he will teach you what you need to know by showing you, so watch him. He will teach you the value of keeping your word and fulfilling promises you make. You will learn to be careful about who you give your word to, this will make you a man of integrity like your dad.
Your mom and dad will show you what it means to compromise, how to argue fairly, the value of an apology, how much can be said with a hug, to be a responsible pet owner, how to be a good friend, and to always take care of your sister.
And then we have your Auntie – I want you to know that by having her, you will always have someone to talk to. You'll always have someone to try to answer any questions you may have. You have someone you can trust and would never harm you. Even if you're not right next to her, she will always be looking out for you. She has had a lot of practice with little boys and knows how little boys act and think, she will always get you and what you are going through. There will be times that you think your mom doesn't understand, auntie will. She will teach you to not let your life be about circumstance but to be strong and fierce. She will share your triumphs, your joy and your pain and she will be beside you all the way.
You won't see your Uncle right away face to face, but he will be there sitting in class waiting patiently for us to let him know you are here! You can rely on him to let you know how powerful you are. He will remind you that you are capable of greatness and that you have full control over your own fate. He will lead by showing you that the journey is to be embraced. He will show you how to be empathetic and strong in your convictions. He will teach you how to speak your peace respectfully but powerfully but also to be sensitive and humble. And above all else, if you have the chance to go out of your comfort zone, do it because he will tell you to grab a hold of your courage, and don't ever lose it because sometimes life is too short.
Your big sister – Wow, I can not wait to see how you two bond over the years. She's only little now, but what I think you will see as you grow up together is that she will show you to not be afraid to take risks and to always be yourself, let your sparkle shine through. That's how you will find the greatest reward. She will remind you to be to kind to others and to reach out to people when they need help. She will show you how to be considerate. With your big sister, you will know that you are loved and that you are safe.
As well as these people, you have me (Aiya) and Poppa who are over the moon about you and you will learn that Poppa is the yes Poppa and Aiya is the one who tries to keep the Yes Poppa in line! You have another set of Nana and Gramps who can't wait to meet you and introduce you to life on a farm plus another Auntie who will be one more strong and steady presence in your life. You have cousins and a load of more aunties and uncles, your love tank will be filled regularly and last but not least, you have a Great Nana that has been the most amazing grandma to countless children and who can teach you a thing or two about families and loving those who cross your path. Through the years, you will see how much of an impact she has made on her own family, by watching how your mom, aunt and uncle and all your cousins stand for each other and lift each other up when the going gets tough. She loves you with all of her heart and is so blessed to watch as you, her 11th (?) great grandchild enters this world.
Little one, you have hit the family jackpot! Your family will teach you to follow your heart, to dream like there aren't any limits and to feel the unconditional love of the family around you. We promise to guide you through life in the best way we can. We will let you make your own way but will never be too far behind to catch you if you fall. We cannot wait to meet you.
Don't WaitFaithFamily HistoryFamily ValuesHappily Ever AfterHappy DaysJust BecauseLife As We Know ItLife Changing EventsLife is in the DetailsParentingSpiritWho Am I?
What wonderful news! I love babies and am pretty much baby crazy.
The photos of the feet are so so cute.
We are over the moon excited!
Of course you would be. Babies are just soooo cute. Congratulations!
It's so much fun! Congratulations!!
He can't wait to meet you!
Thanks so much for reading! He is now here and is beautiful in every way!


text:
Princes of the Palace - The Royal British Family
This film charts the lives of the British Royal Family's princes and their importance within the Royal House of Windsor. Profiling Princes Phillip, Charles, William, Harry and the new arrival Prince George. Made by A2B Media, a documentary specialist on the British Royal Family (Queen Elizabeth: The Diamond Celebration, Diana Princess of Wales - A Life on the Edge), this high-definition programme features exclusive interviews with royal biographers, correspondents and exclusive access to those who have worked with the royal family to give an inside view of how the royal Princes will shape the British monarchy and its transition from Queen Elizabeth, the longest reigning British monarch, to what will undoubtedly be a long succession of Princes becoming Kings.
The Inside Story of the Men Behind Queen Elizabeth
Geners Documentary
Production Countries
Budget: $100,000
King Charles III of the United Kingdom
Camilla Parker-Bowles
Queen Elizabeth II of the United Kingdom


text:
Cloud Predictions 2017 - Daily News Egypt
Business Cloud Predictions 2017
Cloud Predictions 2017
Cloud computing is moving at a dizzying pace. With cloud technologies evolving at such an irrepressible rate, it's difficult to keep track of where it's all headed. How will your relationship with the cloud change over the course of the coming year? A modern cloud must support and energise the cloud journey from any starting …
Ahmed Adly May 24, 2017 Be the first to comment
Cloud computing is moving at a dizzying pace. With cloud technologies evolving at such an irrepressible rate, it's difficult to keep track of where it's all headed. How will your relationship with the cloud change over the course of the coming year?
A modern cloud must support and energise the cloud journey from any starting point, adapt fluidly to changing needs, and ultimately realise the potential of genuine business transformation.
Here are our predictions for how the cloud will impact your business processes in 2017 and beyond.
Cloud-based mission-critical workloads will take off
Cloud has long promised the migration of all enterprise production workloads. But that migration has yet to happen.
The chief barrier to cloud migration remains a lack of commitment and recourse to support production service-level agreements. On one hand, cloud providers are limiting their accountability as they lack the talent to support custom portfolios. On the other, they're failing to provide sufficient control into the public data centre to self-manage service-level agreements.
The Infrastructure as a Service (laaS) provider best equipped to take more responsibility and deliver the control tenants' demand will be the one to drive cloud migration in 2017.
Corporate-owned data centre numbers will plummet
Just a few years ago, this statement would have seemed outrageous. But now, it seems all but inevitable.
As organisations focus their IT spending on cloud computing, they'll begin to migrate their workloads from corporate-owned data centres to purpose-built facilities, managed and run by enterprise cloud providers. Mark Hurd predicts that we'll see corporate-owned data centre numbers fall 80 percent by 2025, and that the same percentage of IT spending will be devoted to cloud services.
While corporate data centre numbers may not fall straight away, we do expect an immediate reduction in direct investment for computer capacity, storage, and networking services.
Enterprise cloud becomes the most secure place for IT processing
This year's threat landscape will be highly changeable. External threats—coupled with the need for better governance and privacy mandates—will make security a key priority for all lines of business.
In years past, security was a major barrier to cloud investment. Data sovereignty, data privacy, and control issues deterred many organisations from pursuing cloud adoption. But in the future, those very same concerns will be the things that draw new organisations to the cloud.
Established cloud vendors with solid security track records have the expertise and resources to deploy layers of defence that many companies simply cannot duplicate in-house.
Digital transformation becomes the norm
Our world is becoming increasingly digitally connected, and it's transforming the way we live, work, and play.
These same technological advancements provide unprecedented opportunities for businesses to expand, innovate, and create new value. Sectors including healthcare, manufacturing, and even urban planning have been reimagined and redefined by the cloud.
To realise these opportunities, today's enterprises must not only develop new cloud-ready tools, but also put digital at the centre of their businesses. Hidden within today's digital connections are the solutions to our most urgent business challenges.
This year, we'll see more companies successfully embrace new integrated cloud technologies.
The rise of intelligent applications
Artificial Intelligence (Al) might sound like science fiction, but many of us use it every day. The software behind many online shopping sites and on-demand music services, for example, is a highly successful and highly pervasive form of Al.
These systems depend on technology infrastructure capable of importing, analysing, and interpreting huge volumes of data before acting on it—all without human intervention. And the next step for such technologies? To become an established part of customer service and other business operations.
Soon, we'll see intelligent applications capable of automatically recommending individualised actions and streamlining business tasks.
Nothing artificial about it: Al gets real
Al and robotics have carved out a niche in the manufacturing sector, and now these technologies are poised to bring their exciting benefits to a host of new industries.
The Al space is white-hot, and it's being fueled by the data explosion. Machine learning algorithms find patterns in enormous volumes of digital information and use that data to train, learn, and become even smarter.
ClOs ignore the Al wave at their peril. According to Toby Redshaw, consultant and former American Express CIO, the company that ignores Al-powered technology will be "the guy at the gunfight with a knife."
Developers do more with less coding
This year, a new tool looks set to join cloud app developers' toolkits. "Visual" or "low" coding will be everywhere in 2017.
For many organizations, the real-time enterprise has meant a rethinking of application development. IT teams are often stuck with a backlog of work, preventing them from delivering applications quickly enough to capitalise on new opportunities. Visual coding enables quick, straightforward development and extension of enterprise applications.
"More than ever before, application development and delivery professionals must obsess over their UI designs," say Forrester analysts John Rymer and Clay Richardson. Low-code vendors employ familiar drag-and-drop, WYSIWYG techniques to speed user interface creation!
The cloud empowers small business innovation
Cloud has become a catalyst for small business growth, allowing them to innovate freely, carve out new markets, and disrupt the status quo.
The digital economy demands that companies of all sizes compete based on technology-enabled value. While some seek to evolve existing business practices, others are striving to launch new services that exploit extensive, low-cost computational power. Traditionally, access to such high-performance resources has been too expensive for smaller businesses. But what once cost $100 million up front is now available for $10 per hour.
The cloud is allowing small businesses to innovate, experiment, and sustain ongoing profitability.
60% of IT organizations move systems management to the cloud
More than 90% of companies have multiple systems management tools, but just 6% trust their incomplete data. Consequently, IT operations professionals struggle to create effective management approaches.
The pace of business is increasing. As more organizations adopt DevOps practices and focus on digital experience, they'll need to eliminate management data silos and embrace machine learning just to keep up.
Some have already embraced systems management in the cloud, unifying management data across multiple clouds and on premises. Others are benefiting from data science applied to the operational management problem. Only Oracle Management Cloud provides an intelligent, unified, cloud-based approach that applies machine learning to the complete operational data set. And while many cloud tools are built exclusively for cloud systems, ours does both.
We expect that by 2020, 60% will have moved their most critical systems management use cases to the cloud.
50% of DevTest will move to the cloud
Last year, we predicted that DevTest workloads would have all but completely migrated to the public cloud by 2025.
At Oracle OpenWorld 2016, Mark Hurd revealed that "we are nearly halfway there" already.
With on-premise hardware and software, IT teams have to buy, license, and configure everything to create development environments that hopefully match production environments. Hurd estimates that the industry could save $150 billion by migrating DevTest to the cloud.
This year, we expect widespread adoption of cloud DevTest. We'll see new application development capabilities, including low-code approaches, tools for microservices, and support for all platforms and languages.
Ahmed Adly, ECEMEA Enterprise Cloud Computing Leader and Senior Director, Oracle
Topics: Cloud Predictions Computer digital transformation IT processing
Ahmed Adly
More in Ahmed Adly
https://dailyfeed.dailynewsegypt.com/2017/05/24/cloud-predictions-2017/
Robotics and business: A dim future?
Nexford University creating technological learning experience to serve Egyptian students
EGP stabilisation aids in raw material prices': OMSI CEO
BE ONE Software Solutions launch services in the Egyptian market
CBE-auto industry committee submits final report to CBE Governor next week
US first-quarter growth much stronger than expected
Catalyst targets business volume of $16m in Middle East in 2017
Leaders of Oracle consider Egypt one of the most promising investment markets
Educated guess vs analysed data: The importance of analytics for strategic CHRO
Al-Qadi discusses cooperation with US companies' officials
May 24, 2017 Breaking News


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text:
\section{preliminaries}

All manifolds in this paper are closed, orientable, connected, topological manifolds of dimension $\ge 6$. Since ordinary homology does not behave well for non-ANRs, we will use the Steenrod extension of a generalized homology theory $H_*$,\cite{Milnor1},\cite{F1}, \cite{CP}, \cite{EH}, which satisfies the usual Eilenberg-Steenrod axioms
for (generalized) homology theories, together with the union axiom. We denote by $\widetilde H_*$ its reduced version and we define it for pairs by setting $\widetilde H_*(X,Y)=\widetilde H_*(X/Y)=H_*(X,Y)$. Recall the axioms of a reduced Steenrod homology theory:
\\
\\(a) Exactness: For a compact metrizable pair $(X,Y)$ there's a long exact sequence
$$\cdot\cdot\cdot\to \widetilde H_i(Y) \to \widetilde H_i(X) \to \widetilde H_i(X/Y) \to \widetilde H_{i-1}(Y)\to\cdot\cdot\cdot$$
\\ (b) Milnor's additivity axiom:  Given a countable collection $X_i$ of pointed compact metric spaces and letting
$\bigvee X_i \subset \prod X_i$ be the null wedge, we have an isomorphism
$$\widetilde H_*(\bigvee X_i) \cong \prod \widetilde H_*(X_i).$$
\\
\\We need the following results \cite{AH},\cite{BM},\cite{Y}. Denote by $M(p)$ the $\mathbb Z_p$ Moore spectrum.
\begin{thm} If $p>1$ is an integer and $n \ge 3$, $\widetilde{H_*}(K(\pi,n);KO\wedge\Z_p)=0$
for any group $\pi$. If $\pi$ is torsion,  $\widetilde{H_*}(K(\pi,n);KO)=0$, for $n= 2$.
\end{thm}
\noindent For odd $p$, we have a chain of homotopy equivalences of spectra:
$$
\mathrm{KO}\wedge M(p)\sim \widetilde{KO}[\frac{1}{2}]\wedge M(p)\sim\mathbb L[\frac{1}{2}]\wedge M(p)\sim
\mathbb L\wedge M(p).
$$
\\ \noindent We recall the Universal Coefficient Formula
for coefficients  in $\Z_p$ for an extraordinary homology theory given by a spectrum $\mathbb E$
of CW complexes.
$$0\to H_n(K;\mathbb E)\otimes\Z_p\to H_n(K;\mathbb E\wedge M(p))\to
\mathrm{Tor}(H_{n-1}(K;\mathbb E),\,\Z_p)\to 0$$
where $\mathrm{Tor}(H,\Z_p)=\{c\in H|pc=0\}$.

\begin{defn}
Given a CW-complex $K$, denote by $P_2(K)$ the
\\$second\  stage\  of\  the\  Postnikov\  tower\  of\  K$. It is the CW-complex obtained by attaching to $K$ cells in dimensions 4 and higher to kill the homotopy groups of $K$ in dimensions 3 and above. Thus, $P_2(K)-K$ consists of cells of dimension $\geq4$ and $\pi_n(P_2(K))=0$ for $n\geq3$.
\end{defn}

\begin{defn}
A compact subset $X$ of an ANR $M$ is $\emph{cell-like}$ if for each neighborhood $U$ of $X$ the inclusion $i:X\hookrightarrow U$ is nullhomotopic.
\end{defn}
Cell-likeness is an intrinsic property of the space $X$. It can be easily shown that if $X$ is a compact ANR, then being cell-like is equivalent to being contractible.
A classical example of a space that is cell-like but not contractible is the topologist's sine curve. For expositions on cell-like maps, see \cite{Edwards}, \cite{La1}, \cite{La2}.

\begin{defn}
A map $f:X\to Y$ between compact metric spaces is cell-like provided that each point inverse is cell-like.
\end{defn}

It follows from the definition of cell-like spaces that a cell-like map is surjective.

\begin{defn}
A homotopy equivalence $f:N\to M$ between closed manifolds $\emph{factors through cell-like maps}$ if there is a space $X$ and cell-like maps $c_1:N\to X$ and $c_2:M\to X$ such that the following diagram
 \[
    \xymatrix{ N \ar[rr]^{f}\ar[dr]_{c_1}& &M\ar[dl]^{c_2}\\
    & X &
    }
    \]

\noindent homotopy commutes. We say that $N$ and $M$ are $CE-related$.

\end{defn}

\begin{rem} Note that if $c_1:N\to X$ and $c_2:M\to X$ are cell-like maps, then there exists a homotopy equivalence $f:N\to M$ so that $c_2\circ f\sim c_1$. This follows from \cite{La2} Lemma 2.3.
\end{rem}

We recall the definition of the simple topological structure set of a manifold $M$ and the algebraic surgery exact sequence.
\begin{defn}

A $\emph{simple topological structure}$ $(N,f)$ on an $n$-dimensional manifold $M$ is an $n$-dimensional manifold $N$ together with a simple homotopy equivalence $f:N\to M$.

\end{defn}

\begin{defn}
The $\emph{simple topological structure set}$, for which we will omit decorations and denote here by $S(M)$, of an $n$-dimensional manifold $M$ is the set of equivalence classes of simple manifold structures on $M$. That is, $(N,f)\sim (N',f')$ if there is a homeomorphism $h$ so that the diagram
 \[
    \xymatrix{ N \ar[rr]^{h}\ar[dr]_{f}& &N'\ar[dl]^{f'}\\
    & M &
    }
    \]
homotopy commutes.
\end{defn}

Denote by $S^{CE}(M)$ the set of topological structures that factor through cell-like maps. All homotopy equivalences in $S^{CE}(M)$ are simple \cite{Dr1}, Proposition 2.3.
\noindent The structure set of a closed topological $n$-manifold, $n\ge5$ fits into the Sullivan-Wall surgery exact sequence \cite{Wall}

$$\cdots  L_{n+1}(\mathbb{Z}\pi_{1}(M)) \to S(M) \to [M,G/TOP] \stackrel{\theta}\to
L_{n}(\mathbb{Z}\pi_{1}(M)).$$

In this paper, weconsider the algebraic surgery exact sequence \cite{Dr1}:
$$\xymatrix@C.15in{
\cdots \ar[r] & L_{n+1}(\mathbb{Z}\pi_{1}(M))\ar[r] & S_n(M) \ar[r]^{\eta'}& H_n(M;\mathbb{L})\ar[r]^{A'} \ar[r] & L_{n}(\mathbb{Z}\pi_{1}(M))\ar[r] &\cdots}$$

\noindent where $H_n(M;\mathbb{L})=H^0(M;\mathbb{L})=[M,G/TOP\times\mathbb{Z}]$, $S(M)\subset S_n(M)$.
\\
\\ \noindent For $M$ closed and connected, there's a split monomorphism \cite{Dr1}
$$\xymatrix {0\ar[r] & S(M)\ar[r]^{i}& S_n(M)\ar[r]& \mathbb{Z}}.$$

\noindent So, $S_n(M)$ differs from $S(M)$ by at most a $\mathbb{Z}$.

\begin{prop}\cite{Wa}Let $E$ be a CW complex with trivial homotopy groups $\pi_i(E)=0$,
$i\ge k$ for some $k$, and $q:X\to Y$ a cell-like map
between compacta. Then $q$ induces a bijection of the homotopy
classes $q^*:[Y,E]\to[X,E]$.
\end{prop}
Let $q:M\to X$ be a cell-like map and $j:M\to P_2(M)$ be the inclusion map. Then by the above proposition, there is a map $g:X\to P_2(M)$ such that $g\circ q\sim j$. Denote by $i$ the induced map on their mapping cylinders, $i:M_q\to M_j$ where $i|_M=\mathrm{id}|_M$ and $i|_X=g$, and by $i_*:H_*(M_q,M;\mathbb L)\to H_*(P_2(M),M;\mathbb L)$ the induced homomorphism
on the Steenrod \( \mathbb L \)-homology groups \cite{F1}\cite{KS}.
\\
\\
\noindent We state the main results of \cite{Dr1}.
\\
\\ \noindent Let $M$ be a closed $n$-manifold, there's a map
$$\delta :H_{n+1}(P_2(M),M;\mathbb{L}) \cong S_{n+1}( P_2(M),M)\stackrel{\partial}\to
S_n(M)\stackrel{p}\to S(M).
$$
where $p$ is any splitting of $i$. For details on the maps above, refer to \cite{Dr1}, section 2.

\begin{thm} (\cite{Dr1}, Theorems 2.4 and 2.7) Let $M^n$ be a closed topological manifold, $n\geq6$. Let $T_{odd}(H_{n+1}(P_2(M),M;\mathbb{L}))$ be the odd torsion subgroup of $H_{n+1}(P_2(M),M;\mathbb{L})$. \\ Then $S^{CE}(M)=\delta (T_{odd}(H_{n+1}(P_2(M),M;\mathbb{L})))$. In particular, $S^{CE}(M)$ is a subgroup of the odd torsion subgroup of $S(M)$.
\\ Moreover, if $M$ is simply connected with finite $\pi_2(M)$, then $S^{CE}(M)$ is the odd torsion subgroup of $S(M)$.
\end{thm}

\section {Main Results}

In \cite{Dr1} Corollary 2.15 the authors constructed a closed simply connected 6-dimensional manifold $M$ with $H_6(M;\mathbb{L})=\Z_2\oplus\Z_p\oplus\Z$ and showed, using the Browder-Novikov-Wall surgery exact sequence and a bundle theoretic argument, that for any element $[N,f]\in S^{CE}(M)$, $N$ is not homeomorphic to $M$ (We note that in~\cite{Dr1}, in the computation of $H_6(M;\mathbb{L})$, the
$\mathbb Z$ and $\Z_2$ summands were mistakenly omitted. However, that does not change the final result since $S^{CE}(M)=\Z_p$ regardless). The manifold $M$ was constructed by PL-embedding the Moore complex $P=S^1\cup_pB^2$ in $\mathbb{R}^6$, then suspending to obtain $P'=S^2\cup_p B^3$ and taking $M=\partial W$, where $W$ is the regular neighborhood of $P'=S^2\cup_p B^3$ in $\mathbb{R}^7$.
  We state this result for completeness.
\begin{thm}[\cite{Dr1} Corollary 2.15] There are closed non-homeomorphic 6-dimensional manifolds $M$ and $N$ which are CE-related.
\end{thm}

In the next corollary, we generalize the authors' construction to any $n$-dimensional, $n\geq6$, manifolds $M$ and $N$. First, we need the following fact.
\begin{prop}(\cite{Ra}, Proposition 20.3)
For $n\geq4$ the structure set $S(M)$ of a simply connected $n$-dimensional topological manifold $M$ is such that $S(M)=S_{n+1}(M)=\mathrm{ker}(A:H_n(M;\mathbb{L}_\cdot)\to L_n(\mathbb{Z}))$ if $n\equiv0\ \mathrm{mod}\ 2$, and $S(M)=S_{n+1}(M)=H_n(M;\mathbb{L}_\cdot)$ if $n\equiv1\ \mathrm{mod}\ 2$.
\end{prop}

\begin{rem}Note that in this paper we use $H_n(M;\mathbb{L})$ and $S_n(M)$ which differ from $H_n(M;\mathbb{L}_\cdot)$ and $S_{n+1}(M)$ by at most a $\mathbb{Z}$ (See preliminaries and \cite{Dr1}, Proposition 4.5). Also, recall that $L_n(\mathbb{Z})=\mathbb{Z}$ if $n=4k$, $L_n(\mathbb{Z})=\mathbb{Z}_2$ if $n=4k+2$, and $L_n(\mathbb{Z})=0$ for $n$ odd.
\end{rem}

\begin{cor}
For any $n\geq6$, for any $m$, there are $m$ closed non-homeomorphic simply connected $n$-manifolds $M$ and $N$ which are CE-related.
\end{cor}
\begin{proof}
  Let $p$ be a prime number such that $p\geq m$ and $p$ does not divide $\mathrm{order}(x_i)$ for every $x_i\in \pi_i(BG), i\leq n+1$. This is possible since the groups $\pi_*(BG)$ are finite. Let $n\neq 7, 8$. By general position, the Moore complex
$P=S^1\cup_pB^2$ can be PL-embedded in $\mathbb{R}^6$. Suspending $n-5$ times
embeds $P'=S^{n-4}\cup_{p}B^{n-3}$ in $\mathbb{R}^{n+1}$. Let $W$ be a regular neighborhood
of $P'$ in $\mathbb{R}^{n+1}$ and let
$\partial W = M$. The manifold $M$ is stably parallelizable because it is a closed codimension one submanifold of euclidean space $\mathbb{R}^{n+1}$ and it is simply connected.
By Lefschetz duality, $H_{n-4}(W,M)=H^5(W)=H^5(P')=0$ and
$H_{n-3}(W,M)=H^4(W)=H^4(P')=0$. Hence, the exact sequence of the pair
$(W,M)$ implies that $H_{n-4}(M)=\mathbb{Z}_p$. Similarly, we can find that $H_3(M)=\Z_p$ and $H_k(M)=0$ for $k\neq 0, 3, n-4, n$.
By the Atiyah-Hirzebruch spectral sequence $H_{n-4}(M;\mathbb{L})\cong L_n(\Z)\oplus H_3(M,L_{n-7}(\Z))\oplus \mathbb{Z}_p\oplus\mathbb Z $. Therefore, $H_{n}(M;\mathbb{L})\cong L_n(\Z)\oplus H_3(M,L_{n-7}(\Z))\oplus \mathbb{Z}_p\oplus\mathbb Z$. For $n=7$, from the exact sequence of pairs $(W,M)$, we get $H_{3}(M)=\mathbb{Z}_p\oplus\Z_p$ and $H_k(M)=0$ for $n\neq 0,3,7$. By the Atiyah-Hirzebruch spectral sequence $H_{3}(M;\mathbb{L})\cong \mathbb{Z}_p\oplus\Z_p\oplus\mathbb Z $. Hence, $H_{7}(M;\mathbb{L})\cong\Z_p\oplus\Z_p\oplus \mathbb{Z}$. For $n=8$, consider the PL-embedding of the Moore complex $P=S^1\cup_pB^2$ in $\mathbb{R}^7$ and suspend twice to embed $P'=S^{3}\cup_{p}B^{4}$ in $\mathbb{R}^{9}$. As before, let $M=\partial W$, where $W$ is the regular neighborhood of $P'$ in $\mathbb{R}^{9}$. By a similar argument, we get $H_3(M)=H_4(M)=\Z_p$, $H_k(M)=0$ for $k\neq 0,3,4,8$ and $H_{4}(M;\mathbb{L})\cong H_{8}(M;\mathbb{L}) \cong\mathbb Z\oplus \mathbb{Z}_p\oplus\mathbb Z $.
Hence, by Proposition 3.2, Remark 3.3, and Theorem 2.11, $S^{CE}(M)=\Z_p\oplus\Z_p$ for $n\equiv 3\ \mathrm{mod}\ 4$ and $S^{CE}(M)=\mathbb{Z}_p$ otherwise.

We get the following commutative diagram from the  Sullivan-Wall and the Quinn-Ranicki exact sequences.
$$
\xymatrix{
L_{n+1}(\mathbb{Z})\ar[r]\ar[d]^{=} &S(M) \ar[r]^(.4){\eta}\ar[d]^{\subset}& [M;\,G/TOP]\ar[d]\ar[r] &L_n(\mathbb{Z})\ar[d]^{=}\\
L_{n+1}(\mathbb{Z})\ar[r]& S_n(M) \ar[r]^(.4){\eta'} & H_n(M,\mathbb{L}) \ar[r] &L_n(\mathbb{Z})
}
$$

Choose a nontrivial $p$-torsion element $\beta=[N,f]\in S^{CE}(M)$.
We next show that $N$ is not homeomorphic to $M$ by showing that $N$ has a nontrivial topological stable normal bundle.
Let $[\gamma]=\eta(\beta)$.
The class $[\gamma]$ represents the difference between topological stable normal bundles on $M$ and $N$ which are defined by two lifts
$\nu_M:M\to BTOP$ and $\sigma:M\to BTOP$  of the Spivak map $M\to BG$ with respect to the fibration $j:BTOP\to BG$.
Here $\nu_M$ denotes a classifying map for the topological stable normal bundle on $M$.
Note that $\nu_N=\sigma\circ f$. Thus, the lifts $\nu_M$ and $\sigma$ are not fiberwise homotopic.
We need to show that $\nu_M$ and $\sigma$ are not homotopic in $BTOP$.

Since the stable normal bundle of $M$ is trivial, the map $\nu_M: M\to BTOP$ is nullhomotopic.
Note that the map $\sigma$ is homotopic to $i\circ\gamma$ where
$i:G/TOP\to BTOP$ is the inclusion of the fiber into the total space of the fibration $j$. The homotopy exact sequence of the fibration $j$ implies that after inverting by finitely many primes $p_1, p_2, ..., p_k$, where $p_i\neq p$, we get that the inclusion $i$  is an $(n+1)$-equivalence. In particular, $i_*: [M,G/TOP]\to [M,BTOP]$ is a bijection. Therefore, the map $i\circ\gamma$ is not nullhomotopic.

 Thus, $\nu_N$ is not nullhomotopic, the topological stable normal bundle of $N$ is nontrivial and, hence, $N$ is not homeomorphic to $M$.
\end{proof}

\begin{defn}
We define the reduced group $\widetilde S^{CE}(M)$ of $S^{CE}(M)$ to be the quotient group $S^{CE}(M)/\langle[N,f] -[N,\psi\circ f]\rangle$ where $f:N\to M$ and  $\psi : M\to M$ are orientation preserving homotopy equivalences that factor through cell-like maps.
\end{defn}
\noindent The above definition makes sense since $h_1\circ h_2$ factors through cell-like maps whenever $h_1$ and $h_2$ factor through cell-like maps \cite{Dr}, Corollary 2.11. Note that two different classes in $\widetilde S^{CE}(M)$ contain nonhomemorphic manifolds, for let $[N,f]$ and $[N,g]$ be elements of $S^{CE}(M)$ then $g=\psi\circ f$, where $\psi=gf^{-1}, \psi:M\to M$.

\begin{cor} Let $M$ be as in Corollary 3.4. Any two classes [N,f] and [P,g] in $S^{CE}(M)$ are such that $N$ and $P$ are non-homeomorphic.
\end{cor}
\begin{proof}
From Corollary 3.4, we have $S^{CE}(M)=\Z_p\oplus\Z_p$ for $n\equiv 3\ \mathrm{mod}\ 4$ and $S^{CE}(M)=\mathbb{Z}_p$ otherwise and nontrivial classes of  $S^{CE}(M)$ are classes of manifolds not homeomorphic to $M$. Hence, for any self homotopy equivalence $\psi$ of $M$, $[M,\psi]=[M,id]$ and therefore $\psi$ is homotopic to a homeomorphism. Therefore, $[N,f]=[N,\psi\circ f]$ and it follows that $S^{CE}(M) = \widetilde S^{CE}(M)$.
\end{proof}

Next, we generalize a result due to Dranishnikov (\cite{Dr2}, Theorem 7.2) which we need for the proof of Proposition 4.1. Dranishnikov showed that, under some assumptions, for a compact polyhedral pair $(P,L)$ and a non-zero element $\alpha\in\widetilde{H}_*(P,L)$, there is a compact metric space $Y\supset L$ and a map $f:(Y,L)\to (P,L)$ such that $\alpha\in \mathrm{Im}(f_*)$.

\begin{defn}
The cohomological dimension of a compactum $X$ with coefficients in the group $G$, denoted $\mathrm{dim}_GX$, is the largest integer $n$ such that $H^n(X,A;G)\neq0$ for some closed subset $A$ of $X$.
\end{defn}
\smallskip

\begin{defn}
A map $f:(X,L)\to (Y,L)$ is called strict if $f(X-L)=Y-L$ and $f|_L=\mathrm{id}_L$.
\end{defn}
\smallskip

\begin{thm}\cite{Dr2}(Theorem 7.2)Let $H_*$ be a generalized homology theory. Suppose that $\widetilde H_*(K(G,n))=0$. Then for any finite polyhedron pair $(K,L)$ and any element $\alpha\in H_*(K,L)$ there is a compactum $Y\supset L$ and a strict map $f:(Y,L)\to (K,L)$ such that
\\(a) $\mathrm{dim}_G(Y-L)\leq n$
\\(b) $\alpha\in \mathrm{Im}(f_*)$
\end{thm}
\smallskip
\begin{lem}Let $H_*$ a generalized homology theory. Suppose that $\widetilde H_*(K(G,n))=0$. Then for any finite polyhedron pair $(K,L)$ and $\alpha_1,...,\alpha_k\in H_*(K,L)$ there is a compactum $Y\supset L$ and a strict map $f:(Y,L)\to (K,L)$ such that
\\(a) $\mathrm{dim}_G(Y-L)\leq n$
\\(b) $\alpha_1,...,\alpha_k\in \mathrm{Im}(f_*)$
\end{lem}
\begin{proof}
Let $\alpha_1,...,\alpha_k\in H_*(K,L)$. By the above theorem, there are compacta $Y_i\supset L$ and strict maps $f_i:(Y_i,L)\to (K,L)$ such that $\mathrm{dim}_G(Y_i-L)\leq n$ and $\alpha_i\in \mathrm{Im}(f_{i_*})$, $i\in\{1,...,k\}$.
\\Let $Y=\coprod_L Y_i$, the spaces $Y_i$ attached along $L$ via the identitiy maps $\mathrm{id}:L\subset Y_i\to L\subset Y_j$.
Consider $f:(Y,L)\to (K,L)$, where $f|_{Y_i}=f_i$. Clearly, $f$ is a strict map.
Since $Y-L=\coprod (Y_i-L)$ we obtain
$\mathrm{dim}_G(Y-L)\leq n$.
\\For part (b), we have $H_*(Y,L)=\widetilde H_*(Y/L)=\widetilde H_*(\vee (Y_i/L))=\oplus\widetilde H_*(Y_i/L)=\oplus H_*(Y_i,L)$ and the induced map on homology \\$f_*:H_*(Y,L)=\oplus H_*(Y_i,L)\to  H_*(K,L)$ is such that $f_*|_{H_*(Y_i,L)}={f_i}_*$.

\end{proof}

In what follows we show that for $S^{CE}(M)$ finite, there exists a space $X$ and a cell-like map $q:M\to X$ such that for any element $\gamma\in S^{CE}(M)$, $\gamma$ can be traced back to $H_{n+1}(M_q,M,\mathbb L)$ by the following sequence:
$$H_{n+1}(M_q,M;\mathbb L)\stackrel{i_*}\to H_{n+1}(P_2(M),M;\mathbb{L})\stackrel{\delta}\to S(M)
$$

\begin{prop}
Let $M^n$ be a closed connected topological manifold, $n\geq6$, $p$ odd and $\alpha_1, ...,\alpha_k\in H_*(P_2(M),M;\mathbb{L}\wedge M(p))$, where $\mathbb{L}\wedge M(p)$ is $\mathbb L$-theory with coefficients in $\mathbb Z_p$. Then there is a cell-like map $q:M\to X$ and elements $\hat\alpha_1,...\hat\alpha_k\in H_*(M_q,M;\mathbb{L}\wedge M(p))$ such that $i_*(\hat\alpha_i)=\alpha_i$.
\end{prop}
\begin{proof}  Let $n\geq7$ and $\alpha_1, ...,\alpha_k\in H_*(P_2(M),M;\mathbb{L}\wedge M(p))$. Then there is a finite complex $K$, $M\subset K\subset P_2(M)$ and elements $\gamma_1,...\gamma_k\in H_*(K,M;\mathbb{L}\wedge M(p))$ such that $\gamma_i$ is taken to $\alpha_i$ by the inclusion homomorphism. By Theorem 1.1 $\widetilde H_*(K(\mathbb Z,3);\mathbb{L}\wedge M(p))=0$, hence it follows from Lemma 3.8 that there is a compactum $Y\supset M$ and a strict map $f:(Y,M)\longrightarrow (K,M)$ such that $\mathrm{dim}_\Z (Y-M)\leq 3$ and $\alpha_1,...,\alpha_k\in \mathrm{Im}(f_*).$ The rest of the proof and the case $n=6$ is exactly like the proof of Proposition 3.6 in \cite{Dr1}.
\end{proof}

\begin{thm} Let $M^n$ be a closed connected topological manifold, $n\geq6$. If $\beta_1,...,\beta_k\in H_*(P_2(M), M;\mathbb{L})$ are odd torsion elements of order$(\beta_i)=p_i$, then there exist a cell-like map $q:M\to X$
and elements $\hat\beta_1,...,\hat\beta_k\in H_*(M_q,M,;\mathbb{L})$ such that $i_*(\hat\beta_i)=\beta_i$.
\end{thm}
\begin{proof} Let $l=p_1p_2\cdot\cdot\cdot p_k$. Consider the following commutative diagram of universal coefficient formulas
$$
{\xymatrix@C.0in{H_{*+1}(M_q,M;\mathbb{L}\wedge M(l))\ar[r]^{\phi'}\ar[d]^{i_*}&  \mathrm{Tor}(H_*(M_q,M;\mathbb{L}),\,\Z_l)
\ar[r]^{\ \ \subset}\ar[d] & H_*(M_q,M;\mathbb{L})\ar[d]^{i_*}\\
H_{*+1}(P_2(M),M;\mathbb{L}\wedge M(l))\ar[r]^{\phi} &  \mathrm{Tor}(H_*(P_2(M),M;\mathbb{L}),\,\Z_l)\ar[r]^{\ \ \ \ \ \subset} &
H_*(P_2(M),M;\mathbb{L})\\
}}
$$
where $\phi$ and $\phi'$ are epimorphisms. Let $\beta_1,...,\beta_k\in H_*(P_2(M), M;\mathbb{L})$, thus  $\beta_1,...,\beta_k\in\mathrm{Tor}(H_k(P_2(M),M;\mathbb{L}),\,\Z_l)$. Pick $\alpha_i\in\phi^{-1}(\beta_i)$ for each $i$. By Proposition 3.9, there is a cell-like map $q:M\to X$ and $\hat\alpha_1,...\hat\alpha_k\in H_{*+1}(M_q,M;\mathbb{L}\wedge M(l))$ such that $i_*(\hat\alpha_i)=\alpha_i$. It follows from the commutativity of the diagram that $i_*(\hat\beta_i)=\beta_i$ where $\hat\beta_i=\phi'(\hat\alpha_i)$.
\end{proof}

\begin{defn} Let $f:(Y,L)\to(X,L)$ be a strict map. A homotopy $f_t:Y\to X$ which is strict at each level is called $\emph{strict}$ if the homotopy $f_t:(Y,L)\to(X,L)$ is continuous.
\end{defn}

\begin{defn}\cite{Dr1}
    Let $Y$ be an open manifold and let $\bar Y$ be a compactification of an end of $Y$ by $X$, $Y=\bar Y-X$ . A $\emph{strict homotopy equivalence}$ \\${near}\ X$ is
        a strict map $\bar f:(\bar W,X)\to (\bar Y,X)$, where $\bar W$ is a compactification of an end of $W$ by $X$, such that there are neighborhoods $\bar U \supset \bar V$ of $X$ in $\bar W$ and $\bar U' \supset \bar V'$ of $X$ in $\bar Y$ such that $\bar f(\bar U) \subset \bar U'$ and there is a strict map $\bar g:(\bar U',X) \to (\bar U,X)$ such that
        \begin{enumerate}
        \item [(a)] $\bar g \circ \bar f|_{\bar V}$ is strict homotopic in $\bar U$ to $id_{\bar V}.$
        \item [(b)] $\bar f \circ \bar g |_{V'}$ is strict homotopic in $\bar U'$ to $id_{\bar V'}$.
        \end{enumerate}
        \end{defn}

\begin{defn}\cite{Dr1} The set of {\it germs of continuously controlled structures} on $Y$ at $X$, denoted $S^{cc}(\bar Y,X)_{\infty}$, is the set of equivalence classes of strict homotopy equivalences of manifolds near $X$. Two strict homotopy
        equivalences near $X$, $\bar f:(\bar W,X)\to (\bar Y,X)$
        and $\bar f':(\bar W',X)\to (\bar Y,X)$ are
        {\it equivalent} if there exist a neighborhood $\bar V$ of $X$ in
        $\bar W$
        and a strict map $\bar h:(\bar V,X) \to (\bar W',X)$ which is
        an open  imbedding
        and  $\bar f'\circ \bar h:\bar V\to \bar Y$ is  strict homotopic to
        $\bar f|_{\bar V}$.
\end{defn}

\noindent Let $q:M\to X$ be a cell-like map of a closed connected manifold $M$ and $M_q$ its mapping cylinder. Let $\stackrel\circ M_q=M_q-M\times\{0\}$.

 \begin{rem}(refer to \cite{Dr1}, Corollary 4.6 and section 5): Let $[\bar g]$ be an element of $S^{cc}(\stackrel\circ M_q,X)_{\infty}$. Then $\bar g:(\bar W,X)\to (\stackrel\circ M_q,X)$ is a strict homotopy equivalence near $X$. We can assume that $W=N\times(0,1)$ and $\bar W=\stackrel\circ M_p$, where $p:N\to X$ is cell-like. The forgetful map $\phi:S^{cc}(\stackrel\circ M_q,X)_{\infty}\to S(M)$ sends $[\bar g]$ to $[f]$, where $f:N\to M$ is a homotopy equivalence that factors through the cell-like maps $p$ and $q$.
\end{rem}

\begin{prop}(\cite{Dr1}, Proposition 5.4) Let $M$ be a closed connected $n$-manifold and $q:M\to X$ a cell-like map. Then the forgetful map $\phi:S^{cc}(\stackrel\circ M_q,X)_{\infty}\to S(M)$ factors as
$$
S^{cc}(\stackrel\circ M_q,X)_{\infty}\stackrel{j}\to
H_{n+1}(M_q,M;\mathbb{L})\stackrel{i_*}\to
H_{n+1}( P_2(M),M;\mathbb{L})\stackrel{\delta}\to S(M)
$$
where $j$ is a monomorphism with cokernel $\mathbb{Z}$ or $0$.
\end{prop}

\begin{thm}
Let $M$ be an $n$-dimensional closed connected topological manifold, $n\ge6$ and $G$ be a finitely generated subgroup of  $S^{CE}(M)$. Then there exists a space $X$ such that for any $[N,f]\in G$, $f:N\to M$ factors through cell-like maps to $X$.
\end{thm}

\begin{proof} It follows from Theorem 2.10 that $G$ is finite. Let $\gamma_1,...,\gamma_k\in G$. Then there are odd torsion elements $\gamma'_1,...\gamma'_k$ in $H_{n+1}(P_2(M),M;\mathbb{L})$ such that $\partial(\gamma_i)=\gamma'_i$. By Theorem 3.12, there exist a cell-like map $q:M\to X$ and elements $\beta_1,...,\beta_k\in H_*(M_q,M,;\mathbb{L})$ such that $i_*(\beta_i)=\gamma'_i$. The result follows from Proposition 3.17.
\end{proof}

\begin{thm}
For $n\geq 6$ and any $m$, there are $m$ closed non-homeomorphic $n$-manifolds that can be mapped by cell-like maps onto the same space $X$.
\end{thm}
\begin{proof}
Let $p$ and $M$ be as in Corollary 3.4. Then, by Corollary 3.6, any two classes in $S^{CE}(M)$ contain non-homeomorphic manifolds. The result follows from Theorem 3.18.
\end{proof}

redpajama_set_name: RedPajamaArXiv


Nursing Assistants also provide invaluable aid to nurses and doctors on the floor by completing important skilled tasks such as checking vital signs and much more.
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Provides instruction to patients and their families under the direction of the RN. May apply, monitor, and remove patient restraints as per physician or hospital protocol.
Cares for patients and self by supporting safety in the workplace. Ensures equipment malfunctions are reported to the appropriate department.
Identifies and communicates learning needs to the unit director, and takes action to improve knowledge, skills, and performance. Requests assistance in planning and prioritizing activities as needed.
Constructively receives feedback and direction, and participates in self-review as requested by the unit director.
Demonstrates awareness of behavior on the efficient functioning of the department. Supports department-based projects and quality initiatives.
Performs assigned work in a timely and productive manner. Completes department clerical functions as needed.
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Suggests and supports changes within the department. Assumes responsibility for assignments and views problems as challenges.
Demonstrates understanding of cultural diversity, horizontal violence, and impairment in the health professions.


text:
BOFURI: I Don't Want to Get Hurt, so I'll Max Out My Defense.
Episode 1 - Defense and First Battle Episode 2 - Defense and Friends Episode 3 - Defense and Reaching Second Level Episode 4 - Defense and Second Event Episode 5 - Defense and Spoils of War Episode 6 - Defense and Reinforcements Episode 7 - Defense and Upgrades Episode 8 - Defense and Third Event Episode 9 - Defense and Fourth Event Episode 10 - Defense and Attack Episode 11 - Defense and Mighty Foe Episode 12 - Defense and Bonds
Episode 1 - TBA
Honjou Kaede is invited by her friend Shiramine Risa to play a virtual reality MMO game with her. While Kaede doesn't dislike games, what she really, truly dislikes is being in pain. She creates a character named Maple, and decides to put all her points in VIT to minimize pain. As a result, she moves slowly, can't use magic, and even a rabbit can get the best of her. But as it turns out, she acquires a skill known as "Absolute Defense" as a result of her pumping points into VIT, as well as a "Counter Skill" that works against special moves. Now, with her ability to nullify all damage, she goes on adventures.
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Studio: SILVER LINK.
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Trailer: BOFURI: I Don't Want to Get Hurt, so I'll Max Out My Defense.