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Erica Schultz Yakovetz
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The 5th Great G&S Sing-Out
This Labor Day weekend, while one option would admittedly have been to go to the 71st World Science Fiction Convention in San Antonio, TX… I instead chose to geek out with a different kind of fandom at the 5th Great Gilbert & Sullivan Sing-Out, organized by the Victorian Lyric Opera Company in Rockville, Maryland.
The premise of the Sing-Out is to sing through all 13 extant Gilbert & Sullivan operettas, more or less in their entirety (though without the dialogue), in one weekend. (I was told that the first Sing-Out was done all in one day, 8am to 2am. I am, on the whole, delighted that they've moved toward pacing it out over 3 days.)
Longtime readers may be aware that I first got into Gilbert & Sullivan in the fall of 1999, when I did my first show with the MIT Gilbert & Sullivan Players: Iolanthe. In that time, I've performed in fully staged productions of 10 of the 13 operettas (Patience, Mikado and Ruddigore being the remaining exceptions), and numerous additional concert performances. Somewhere along the way, I crossed over that line (as many a G&S fan knows) from "I've been in some of the shows" to "I know MOST of the shows well enough that it's fun to actually go sing a bunch of them in a row with fellow aficionados." So I've also attended marathons at Brown University in April 2011 and with the New England Gilbert & Sullivan Society in June 2011. But the Rockville Sing-Out only happens once every five years (making this the 20th anniversary… if you go by birthdays), and by now I'm hooked enough to actually travel out of state for this sort of thing.
The Sing-Out is truly more of a community singalong than a performance; all are welcome to come and sing chorus, with the option of sitting in the risers on the stage, or to simply watch and sing from the audience. However, the organizers do cast all the solos in advance (by application rather than by audition), and I had the honor of singing Dame Hannah in Ruddigore on Saturday afternoon, as well as one of the chorus leads (Olga) in The Grand Duke on Sunday morning. (Everyone thinks Mad Margaret is the plum women's role in Ruddigore, and maybe it is, but I'd rather sing Hannah any day. You get the awesome exposition song at the beginning, a feature in the gorgeous madrigal in the Act 1 finale, and — spoiler alert — your own out-of-nowhere love plot with adorable duet at the end.)
Hearing/singing all the shows in one big binge definitely lends some new perspective; for instance, I am assembling some new lists of favorite duets and other bits, which I will attempt to post in future. But mostly it feels oddly like coming home to places I haven't been since I first (or most recently) inhabited them: Ploverleigh, Pfennig-Halbpfennig, Tremorden Castle, Titipu, Utopia… and of course the House of Lords and the Tower of London. As I posted to Facebook after I got home last night: "Is it sick that I immediately wanted to do it all over again singing different parts?" :-)
And, needless to say, another big part of what made the weekend so awesome was catching up with old friends (from MITG&SP to the G&S Society of New York)… and meeting new ones. Many thanks to all the organizing committee, staff, and volunteers — particularly the accompanists! — who made this magical weekend possible.
September 2, 2013 by eschultz72	Categories: Gilbert & Sullivan, performance	| Tags: appearances, g&s, musical theater, ruddigore | 2 Comments
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2 thoughts on "The 5th Great G&S Sing-Out"
Sam says:	 September 2, 2013 at 11:40 pm
Erica's Dame Hannah was awesome!
eschultz72 says:	 September 3, 2013 at 12:04 am
<3 But not as awesome as Sam's Ko-Ko, Duke Rudolph, and even Arac!
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Home › Amandla › The Black Panther moment
The Black Panther moment
Posted on March 15, 2018 by AIDC — No Comments ↓
A critique of the critiques
Amandla! Ulutsha Newsletter Issue No.7 | Editorial | 15 March 2018
It's been a month since Black Panther hit cinemas and it has thus far grossed $1billion at the box office across the world. As the dust settles and the euphoria fizzles out, what do we have left? Many reviews have talked to the euphoria around the movie, but have they thoroughly expanded on it and highlighted its importance? Having read quite a few of them and having engaged in many conversations, one keeps hearing (of course in different variations) how the film could have been more and how the film perpetuates the ills of the capitalist society that we live in. But based on the movie is a Hollywood production – how could it even be "more"?
It is key to understand that the excitement about the movie stems from several things.
Before the movie was released, its marketing capitalised on the radicalisation of pop culture that has been challenging norms around black bodies in the world: It capitalised on the dissatisfaction of people around the representation of black folks in movies, it equally capitalised on #blacklivesmatter and the #woke mindset of people. The so-called excessive excitement after its release was based on how it was one of the first major blockbusters with a mostly black cast and with a black director, outside of the usual comedic genre that black Hollywood occupies.
The movie is one of the first major films in Hollywood that presents an African utopian society – a society that we superficially all have dreamt about seeing as opposed to the picture of war and devastation that mainstream media normally portrays of Africa. The film depicts black women in strong and powerful ways: as an inventor and scientist, undercover spy and an all-women army. The official language of Wakanda being isiXhosa, one of the official languages in South Africa, as well as the melting pot of African cultures and ethnic groups featuring in the movie – these are just some of the many themes that have also contributed to the excitement and created the "moment" that everyone was talking about.
The "moment" created expectations for the movie to not only depict a true story of the heterogeneous cultures and identities on the African continent but also represent various facets of differently abled people and queer folks and for it to lay the foundations for revolutionary transformed society – a society that is not plagued with the malaise of capitalism. But these expectations are somewhat unfounded considering that Black Panther is a Hollywood film – how could a Marvel superhero film ever tackle the issues that revolutionaries seek to deal with?
This is not a call not to challenge and critique the problematic themes that the movie reproduces. Themes like the "white-savior", the apparent necessity to have a white character in every mainstream movie, who in the end heroically saves the day. As if having a white male saviour in an all-black cast is not ironic enough – the climax of the absurdity lies in him being a CIA agent, since the CIA has historically wreaked havoc on the African continent. Another problematic theme is the depiction of the black radical man as a disruptive villain: Erik Killmonger, an American black ops soldier seeking to use the vibranium in Wakanda to liberate all oppressed black people around the world. His portrayal as a disruptive villain disregards his progressive and radical ideas, and the celebration of his death represents the demise of a radical black internationalism after which a sense of "order" can continue. In addition to that, we need to question the apparent xenophobia and crude nationalism in the film which becomes apparent when one of the characters states that they cannot allow refugees into Wakanda because they "come with their problems". It is also a matter of concern how the continued extraction of vibranium is not remotely challenged – at a time when the continent is in real life grappling with the social, economic and environmental destruction caused by mining. Regardless of the fact how sustainable or unsustainable vibranium could be extracted – which is not depicted in the film –, extractivism and the subsequent exploitation of the earth is toxic in itself – an issue which the movie does not sufficiently question.
That being said, the critique should not overweigh how phenomenal and breathtaking the movie is. For almost two hours, Black Panther takes us to a place that a young black child could have never imagined. In the cinema, one's chest swells up with joy that here is a feature film on the big screen which has black people that are not slaves, not struggling or supporting a white person. We are brought to a place where the politics around natural hair do not exist, and whilst we are busy criticizing the movie, we take for granted how visually fantastic it was to see black women being who they are on a daily basis: naturally fierce and powerful. Yes, one also has to question how they depict the accepted beauty standards of tall, skinny black able-bodied women. However, the movie is a step in the right direction in terms of the representation of black dark-skinned women in Hollywood.
It is extremely profound how the movie makes us crave a society that aesthetically looks like Wakanda. It was almost difficult to leave the cinema once the final credits started playing – one is still stuck in the mesmerizing vacuum of why can't we have something like this outside of the movie? The power of this feeling is extremely important: It lays the foundation of how we can change the socialization that leads us to believe that all superheroes are white, and therefore inspires us to look beyond just a superhero movie and to challenge the representation and position of black people in society.
The film should have dealt with many issues highlighted above. It could have sparked an inferno of radical change and introduced us to a just society, where the oppressed do not become the oppressor, where minorities are represented and are no longer "othered", where nationalism is questioned and where the norms of capitalism are challenged. But we cannot let these shortcomings be the only thing that we dwell on. We cannot let them cloud the importance of the film challenging the mainstream. Black can be beautiful, powerful and prosperous – and Black Panther is a celebration of that.
This Editorial is part of the youth lead Newsletter "Amandla!Ulutsha". Click to read the current and previous issues.
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E M Forster
E M Forster by Edward Gooch © Edward Gooch/Getty Images
The British novelist and literary critic E. M. Forster was born on New Year's Day 1879 in London. His architect father died young, leaving Forster and his mother enough money to be comfortable for the rest of their lives. In 1883, they moved to the house in Hertfordshire which would become the inspiration for Howards End (1910).
Forster was unhappy at Tonbridge School, but enjoyed himself far more at King's College Cambridge. Having no urgent need to make a living on graduation, he travelled with his mother in Italy, gathering the inspiration for A Room with a View (1908). His critically successful first novel, Where Angels Fear to Tread (1905) also has Italian themes; his favourite of his own novels was The Longest Journey (1907).
Between 1910 and 1913 he wrote Maurice, a novel which reflected his own – then illegal – homosexuality. But though he made a large donation to the Homosexual Law Reform Society in the 1960s and occasionally wrote articles advocating reform, Maurice was not published until the year after Forster's death (1971). Forster was at his happiest during a two-year relationship with a young policeman called Bob Buckingham who later married; despite Buckingham's marriage, the two men remained friends.
Forster also travelled twice to India in 1912 and 1921; the trips helped him begin and complete A Passage to India (1924), which has since been read as an important early document of post-colonialism.
Forster was a member of the literary 'Bloomsbury set', and a perceptive critic. Virginia Woolf wrote in her diary that 'he says the simple things that clever people don't say; I find him the best of critics for that reason'.
In the 1930s, he became an increasingly prominent public voice for what the academic Lord Annan described as 'liberal humanism'. On the eve of the Second World War he published one of his most famous essays, 'Two cheers for democracy', later called 'What I believe'.
He also collaborated with Eric Crozier on the libretto for Benjamin Britten's opera Billy Budd (1951); and though he refused permission during his life, enormously popular films were made of his books after his death.
Though Forster refused a knighthood in 1949, he accepted a Companion of Honour in 1953, eight honorary degrees, and the Order of Merit on his 90th birthday. He died in 1970, having spent the last 25 years of his life as an honorary fellow of King's College Cambridge.
Further information about the life of E M Forster can be found here via the Oxford Dictionary of National Biography.
Edward Morgan Forster
Novelist, Essayist
1 January 1879, Marylebone, London, England
7 June 1970, Coventry, West Midlands, England
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Exploring identity, Literature 1900–1950, Gender and sexuality
A year after E M Forster's death, his novel about a relationship between two men, Maurice, was published. Kate Symondson explores how Forster's sexuality shaped his writing and the long period during which he didn't publish anything at all.
The mystery and muddle of A Passage to India
Literature 1900–1950, Power and conflict
Kate Symondson explores the tensions and dualities at the heart of A Passage to India and the challenges E M Forster faced in writing the novel.
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The alienated modernist self is a product of the big city rather than the countryside or small town. Katherine Mullin describes how an interest in the sensibility associated with the city – often London, but for James Joyce, Dublin – developed from the mid-19th century to the modernist period.
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E M Forster's draft of Howards End (1908–10)
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E M Forster's A Room with a View: The question of view and viewpoint
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The Howards End of E M Forster's 1910 novel is a country house where the action begins and ends. The building ...
Maurice was published in 1971, a year after E M Forster's death. Forster began writing his 'gay ...
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News, info, and updates about SAID!
We are now accepting registrations for our Winter Session beginning in December. Check out our Class Schedule or Register today! We have classes for children of all ages and all skill levels, from beginners to champions. Join us!
Congratulations to our lovely dancers on a wonderful day at the Lenihan Feis!! From teams to Traditional Sets to Solos, you ALL made us extremely proud!! Too many medals, trophies, and smiles to keep track of. Check out our Facebook Page for pictures from the event!


Let's face it, the weather is rubbish, but that doesn't mean we can't brighten up the weekend with some fabulous high street fashion.
We're loving tropical prints on our street style site Look What I'm Wearing. Carla Benstead rocks them like a pro with her Red Herring dress, Forever 21 bag and pumps from high street fashion favourite Primark. Good work Carla!


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2018/09/01 14:40 kbilz [More Information] 2013/06/27 10:44 jvaughan [Thomas Mills] 2013/06/27 10:43 jvaughan [More Information] 2013/06/27 10:42 jvaughan [More Information] 2013/06/19 14:15 jvaughan [Thomas Mills] 2009/07/14 14:17 jgregory 2009/06/12 13:20 jgregory 2009/06/12 13:12 jgregory 2009/06/08 13:01 jgregory created
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thomas_mills [2013/06/27 10:42]
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jvaughan [Thomas Mills]
Thomas Mills was born on 29 September 1763, in Long Island, N.Y. He moved to South Carolina with his father and older brother and signed up with Lt. Col. Francis Marion during the [[Revolutionary War]]. Mills was only 12 or 13 years old at the time. He served with the 2nd South Carolina Regiment and Marion's group, the Swamp Fox Soldiers, from 1776 to 1779. He later traveled to Tennessee, Kentucky, and Ohio where he was a scout and Indian fighter. He met [[Daniel Boone]] and knew Simon Kenton well. In 1786, Mills was among several families who petitioned the Commonwealth of Virginia to establish the towns of Washington and Limestone, later known as Maysville. He served in the Corn Stalk Militia between 1791 and [[1799]] and fought in the [[War of 1812]].  Thomas Mills was born on 29 September 1763, in Long Island, N.Y. He moved to South Carolina with his father and older brother and signed up with Lt. Col. Francis Marion during the [[Revolutionary War]]. Mills was only 12 or 13 years old at the time. He served with the 2nd South Carolina Regiment and Marion's group, the Swamp Fox Soldiers, from 1776 to 1779. He later traveled to Tennessee, Kentucky, and Ohio where he was a scout and Indian fighter. He met [[Daniel Boone]] and knew Simon Kenton well. In 1786, Mills was among several families who petitioned the Commonwealth of Virginia to establish the towns of Washington and Limestone, later known as Maysville. He served in the Corn Stalk Militia between 1791 and [[1799]] and fought in the [[War of 1812]].
- Mills married Lydia Wilson in 1786; they had several children before she died in [[1805]]. The following year he married Lydia Cachel, 12 March [[1806]]. They moved from Clermont Co., Ohio to Cheviot and Mill Creek in Hamilton Co., Ohio. Mills had a total of 13 children. The family left Mill Creek in [[1835]] and eventually settled at [[Bullittsville]], in [[Boone County|Boone Co.]], Kentucky, where he died in [[1864]]. + Mills married Lydia Wilson in 1786; they had several children before she died in [[1805]]. The following year he married Lydia Cachel, 12 March [[1806]]. They moved from Clermont Co., Ohio to Cheviot and Mill Creek in Hamilton Co., Ohio. Mills had a total of 13 children. The family left Mill Creek in [[1835]] and eventually settled at [[Bullittsville]], in [[Boone County|Boone Co.]], Kentucky, where he died in 1864.
* [[http://bcplfusion.bcpl.org/Repository/DEDICATION_IN_HONOR_OF_THOMAS_AND_LYDIA_MILLScomb.pdf|Dedication in honor of Thomas and Lydia Mills, Oct. 20, 1990, Compiled by Fern H. Taylor.]] * [[http://bcplfusion.bcpl.org/Repository/DEDICATION_IN_HONOR_OF_THOMAS_AND_LYDIA_MILLScomb.pdf|Dedication in honor of Thomas and Lydia Mills, Oct. 20, 1990, Compiled by Fern H. Taylor.]]
* [[http://www.findagrave.com/cgi-bin/fg.cgi?page=pv&GRid=13404634&PIpi=46172378|Thomas Mills on Find A Grave.com]] * [[http://www.findagrave.com/cgi-bin/fg.cgi?page=pv&GRid=13404634&PIpi=46172378|Thomas Mills on Find A Grave.com]]
+ * [[http://bcplfusion.bcpl.org/Repository/SAR_application.pdf|SAR Application for Thomas Mills]]
===== Related Topics ===== ===== Related Topics =====
* [[Biographies]] * [[Biographies]]
thomas_mills.txt · Last modified: 2018/09/01 14:40 by kbilz


I thought this had been covered before, but I can't find the link, so maybe I can ask now/again?
I am a member (if that's the correct expression?) of the Andalucian health system and have been since becoming a UK pensioner and having had all the S1 forms and whatever else it took at the time - 5 years ago.
But I have a EHIC card which seems to have come from the UK and which expires in April 2018. I intend to travel next year in Europe and after the expiry of that EHIC card. Do I get another from the UK?...or do I have to get one from the Spanish system and if so what's it called please and from where do I source it?
Sorry for all the questions but the web pages (and there are many of them) are somewhat impenetrable when one is not a mainstream (ie one live abroad) enquirer.
An S1 pensioner has their EHIC card issued by the UK, not Spain, so renew it via the UK.
Thanks, Sid. My original post was on Sunday with limited ability to make meaningful phone calls to sort anything out. Now, it's Monday and a few phone calls have been made and I'm afraid, bureaucracy rules as you might expect.
1. You cannot apply for a renewal online!
2. Renewals can only be obtained within 6 months of the expiry of the original.
Actually, with all possible qualifying permutations spelled out in excruciating detail online, it becomes apparent how difficult it must be for Brexit negotiators to unravel years of complicated rules and regs......but I digress!
I seem to recall this was discussed some time ago. My personal experience of renewing an EHIC (original issued by the Spanish system) card was quite simple, telephone Newcastle and give details of current EHIC card. New one posted fairly quickly. Best to do this before current card expires.


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Our journey of the All Souls Trilogy locations in Oxford resumes after the hiatus due to the AllSoulsCon celebration. This time we explore some of the locations where Diana jogs or strolls with Matthew, as well as some landmarks briefly mentioned in A Discovery of Witches that are part of the Oxford charm and we think are worth a visit.
"I've been a member—once—of Merton, Magdalen, and University colleges. I've been a member of New College and Oriel twice each. And this is the first time All Souls has paid any attention to me." Matthew Clairmont, Chapter 14, ADOW.
Christ Church College
In A Discovery of Witches, several colleges are mentioned. While we don't cover them in this guide, they are truly beautiful and worth a visit time permitting. If you only have time for one college besides to the obligatory stops at All Souls and New College, we recommend Christ Church. You'll find some of the colleges mentioned in A Discovery of Witches on the "Other Colleges" photo album on Flickr.
"My familiar route took me past Oriel College and to the tall, black iron gates between Merton and Corpus Christi." Chapter 7, ADOW
Clairmont rolled along the drive to the gates of Christ Church, where he waited for an opening in the early-morning traffic dominated by delivery trucks, buses, and bicycles." Chapter 7, ADOW (Note: Christ Church is Phoebe Taylor's college).
Clairmont steered up the hill and took a right down the High Street. We passed the statue of George II's wife standing under the cupola at The Queen's College, then headed toward Oxford's botanical gardens. Chapter 7, ADOW.
Photos: Other Colleges photo album
Pitt Rivers Museum
Marcus had even checked Matthew's favorite Oxford haunt, the Pitt Rivers Museum, where the vampire could often be found dividing his attention between the skeleton of an iguanodon and a likeness of Darwin. Chapter 16, ADOW.
The Pitt Rivers Museum holds one of the world's finest collections of anthropology and archaeology. Note: The entrance to the Pitt Rivers Museum is through the Oxford University Museum Natural History (OUMNH) on Parks Road.
Address: 1 Banbury Rd, Oxford OX2 6NN
Website: http://www.prm.ox.ac.uk
Photos: Other Oxford sites photo album
Old Parsonage Hotel
Old Parsonage menu
"The chef from the Old Parsonage's restaurant made it-and provided precise instructions on how not to burn it to a crisp when I reheated it." Matthew, Chapter 14, ADOW.
It's the hotel where Matthew procures the memorable dinner he serves to Diana in his rooms at All Souls College (Ch. 14). At a very short walk from the Ashmolean Museum, this stylish hotel occupies a 17th-century parsonage, and it's ranked one of the best hotels in Oxford. It's no surprise, the restaurant is Matthew's pick.
Website: http://www.oldparsonage-hotel.co.uk
Sean, who'd shared many a drink with me in the pink-stuccoed pub across the street in our graduate-student days […] Chapter 1, ADOW.
The pink-stuccoed pub mentioned in the very first chapter of A Discovery of Witches is The King's Arms. We recommend it if you want to experience good, traditional English pub fare.
Address: 40 Holywell Street, Oxford, OX1 3SP
Website: http://kingsarmsoxford.co.uk
St. Helens Passage leading to the Turf Tavern
Turf Tavern
We navigated the uneven sidewalks of Holywell Street, past the entrance to the Turf pub, and through the New College gates. Ch. 14, ADOW.
Tucked in a short, narrow alley nestled against the old city wall, the Turf Tavern feels like finding a treasure. Look for the St. Helen's Passage under the Bridge of Sights to reach the tavern. And weather permitting, enjoy the outdoor beer garden.
Address: 4-5 Bath Place, Oxford, OX1 3SU
Website: http://www.turftavern-oxford.co.uk
Museum of the History of Science
Adjacent to the Bodleian and the Sheldonian Theatre, the Museum of the History of Science holds a collection of early scientific instruments mostly from Oxford illustrious "celebrities," including Einstein's chalkboard featuring his equations from a lecture in Oxford. This is the museum confirming the authenticity of the "re-discovered" telescope built by Thomas Harriot and Nicholas Vallin.
Astrolabes at the Museum of the History of Science
THE WORLD'S leading expert on early scientific instruments at Oxford University's Museum of the History of Science, Anthony Carter, confirmed today that a refracting telescope bearing the names of Elizabethan mathematician and astronomer Thomas Harriot and Nicholas Vallin, a Huguenot clockmaker who fled France for religious reasons, is indeed genuine. Chapter. 40, SON.
All Souls Trilogy fans will enjoy the collection of astrolabes, which are also detailed in the museum's online catalogue.
Address: Broad Street, Oxford OX1 3AZ
Website: http://www.mhs.ox.ac.uk
University Church of St. Mary the Virgin
Worth the 127 narrow, twisting stairs of the bell tower, the climb rewards visitors with a stunning view of All Souls College, the Radcliffe Camera, the Bodleian, and other Oxford landmarks.
Address: The High Street, Oxford OX1 4BJ
Website: http://www.university-church.ox.ac.uk
Sir Thomas Bodley portrait by Nicholas Hilliard
Weston Library (New Bodleian)
The New Bodleian, recently renamed Weston Library, is briefly mentioned in chapter 4 of A Discovery of Witches, when Mr. Johnson says that Matthew has some manuscripts sent up from the New Bodleian Reading Room.
Check the website of the newly reopened Weston Library to find out what exhibit is available. At the time of our visit, the awe-inspiring Marks of Genius exhibit showcased many All Souls Trilogy-related items, including a portrait of Thomas Bodley by Nicholas Hilliard, works of Shakespeare, Marlow, Vesalius, the Gutenberg Bible, Newton's Principia, and more.
Address: Broad Street, Oxford OX1 3BG
Website: http://www.bodleian.ox.ac.uk/our-work/estates-projects/weston
Posted in All Souls Locations	| Tags: A Discovery Of Witches, Deborah Harkness, Diana Bishop, Oxford, The Book of Life	|
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A Toast to The Serpent's Mirror »


text:
Raúl Zibechi: New US Strategy Threatens Latin America
New US Strategy Threatens Latin America
By: Raúl Zibechi
In April, the United States government launched a potent counter-offensive to recuperate lost ground in a region that continues being vital for its global domination. No one in his right mind could imagine that the empire would let its influence in Latin America dissolve without playing all its cards. In the new world scenario, ruled by economic and financial crisis, and when the Pentagon needs to turn towards the Pacific, its presence on this continent cannot assume only a military profile.
On May 1, General Martin Dempsey, Chair of the Joint Chiefs of Staff, debated the new Defense Strategy at the Carnegie Endowment for International Peace, pointing out that it not only consists in "rebalancing" the armed forces towards the Asia-Pacific region, as Barack Obama pointed out in January. He defined the necessity of "constructing a network of alliances around the globe" for what will be necessary "to resolve the pending challenges, such as questions related with technology transfer, intelligence exchange and military sales to foreign countries." (Carnegieendowment.org)
In April the Secretary of Defense, León Panetta, made a South American tour that took him to Colombia, his principal military ally, later to Brazil and finally to Chile, where the Concón military base was just inaugurated. "The proposition for this trip is to participate in consultations with several of our associates in this part of the world and to attempt to foment innovative security alliances in the region" (http://spanish.chile.usembassy.gov).
The Concón base, in Valparaíso province, forms part of that "innovation policy." It was constructed within 60 days by the Southern Command and Chile's Navy as an urban warfare training camp, called Military Operations in Urban Territories (MOUT) contemplated in "humanitarian" and preventive missions. In September 2011, the Chilean Minister of Defense, Andrés Allamand, had signed a cooperation agreement that permits "the deployment of US troops on Chilean soil, faced with the eventuality that the national army is surpassed by some emergency situation." (El Ciudadano, 3/5/12.)
But the climax of Panetta's mini tour happened in Brazil, the day following the interview with the Minister of Defense, Celso Amorim, in which he offered ample technology transference if he opts for the purchase of F-18 Boeing Super Hornet pursuit planes, instead of the Rafale from the French Dassault (Aviation). On April 25, Panetta offered a conference at the War College, in Río de Janeiro, in which he detailed his proposal of broad strategic cooperation between the US and Brazil.
He talked to Brazil's military elites, business owners and politicians, not to the public at large. He began saying that both countries "find themselves at a critical point of common history" (Defesanet, 25/4/12). "It is the moment for strengthening ourselves in the birth of a new agreement, simultaneously strong and innovative, based on the mutual interests of two countries, as Western powers." He insinuated that Brazil could come to occupy its longed-for permanent seat on the UN Security Council, but it was not clear.
He called to establish a new dialogue for "transforming the Brazil-US relationship in the area of defense," involving the emerging nation in inter-national military questions and he assured that bilateral relations are at their best moment since 1945.
In one crucial paragraph it brought up the most thorny aspect of the bilateral relationship: "Brazil is an economic power and cooperation in high technology, which needs to flow in both directions, seems limited by the controls to currently existing exports. Responding to that, we made the decision to free 4, 000 export licenses for Brazil, a level similar to what we have with our best global allies."
Panetta added that the purchase of the 36 F-18 pursuit planes can "radically transform the relationship between both defense industries" and he concluded by assuring that: "Amorim is expected in Washington shortly to continue the dialogue."
How should this speech be interpreted? Without a doubt, it is produced at a key and delicate moment. The victory of François Hollande is analyzed in Brazil as an opportunity for strengthening the alliance with France, while China's presence in the region does not stop growing. Amorim assured months ago that the decision about the purchase of the pursuit planes would be made before the middle of the year, but logically after the French elections. This is the time. Nevertheless, the empire does not usually offer a broad technology transfer for the purchase of three-dozen airplanes. The objective appears more ambitious: the Pentagon makes its "generous" technological and diplomatic offer (the seat on the Security Council) in exchange for a military and strategic submission. In my way of seeing it, it's blackmail.
The cables revealed by Wikileaks point out that in 2009 the US tried to sabotage the transfer of space and nuclear technology from the Ukraine to Brazil (Defesanet, 13/5/12), two decisive aspects for the emerging country's strategic autonomy. But Brazil is already developing spatial technology with China and has its own advanced nuclear program. The message is clear: if Brasilia is not subordinate, the military circle will be tighter every day, as the new military base in Chile demonstrates.
It is not simple to anticipate the path that the Brazilian elites will take. For much less, Getulio Vargas was cornered until driving him to suicide. The coming weeks will reveal a good part of the enigma: the delayed decision on the purchase of the pursuit planes will show the reigning state of mind in the country that is proposing to unite the region to speak with the same voice in the world.
En español: http://www.jornada.unam.mx/2012/05/18/opinion/025a2pol
Category: Raúl Zibechi Tags: politics, US Defense Strategy & Trade
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text:
De TV Show was een langlopend praatprogramma van de TROS en vanaf 1 januari 2014 van AVROTROS. De eerste uitzending van de TV Show was op vrijdagavond  6 november 1981 bij de TROS op Nederland 2 met als presentator Ivo Niehe. In 1983 verhuisde het programma naar Nederland 1. Vanaf 1988 was het programma vast op Nederland 2 te zien. Later verhuisde het weer naar Nederland 1, later NPO 1 genaamd.
Op 1 januari 2022 werd de laatste aflevering uitgezonden, met als gast André van Duin.

Gasten
Sinds het begin van het programma zijn regelmatig wereldsterren te gast. In de loop der jaren zijn honderden gasten bij Niehe langsgekomen. Voor buitenlandse gasten werd een groot beeldscherm geïnstalleerd en deed Niehe alsof hij rechtstreeks contact had via de satelliet. Hij werd erg bekwaam in die techniek.

TV Show op Reis
In 1986 was de TV Show op Reis een nieuwe uitdaging. De gasten komen niet naar de studio, maar Niehe gaat naar de gasten toe. Doordat hij zoveel talen beheerst, is het mogelijk veel buitenlandse gasten te bezoeken.

Overgang naar Talpa
De bedoeling was dat het programma vanaf begin 2006 overging naar Talpa, maar hiervoor gaf de TROS geen toestemming. De TV Show ging door met presentator en producent Reinout Oerlemans en Niehe ging over naar Talpa met het programma Ivo. Na één seizoen maakte hij echter in oktober 2007 zijn rentree bij de TROS met de TV Show, omdat zijn nieuwe programma bij Talpa veel minder kijkers trok.

Externe links
 De TV Show (AVROTROS)
 De TV Show (Beeld en Geluid wiki)

Televisieprogramma van de TROS
Nederlands praatprogramma
Programma van AVROTROS

redpajama_set_name: RedPajamaWikipedia


text:
The Journey of Xuanzang: An Introduction
Posted on May 5, 2011 December 27, 2013 by Dharma Documentaries
The Journey of Xuanzang is a lavishly illustrated 12-part biography of the great Chinese traveler and translator Xuanzang, based on Records of the Western World and A Biography of the Tripitaka Master.
Tomorrow I will start posting The Journey of Xuanzang which is a lavishly illustrated biography of the great Chinese traveler and translator Xuanzang, based on his own Records of the Western World, and on a book written by two of his disciples entitled A Biography of the Tripitaka Master.
The series was made by Fo Guang Shan temple's Beautiful Life TV (BLTV) in Taiwan and broadcast in 12 parts of a little under 30 minutes each. I will post them over the next few weeks in playlists each around an hour long, therefore containing two of the episodes.
The story in this biography is told in Chinese by a narrator (with subtitles), with the occasional reading of dialogue recorded in the books, which is accompanied by scene enactments. The documentary goes to many of the places Xuanzang visited, and sometimes reconstructs them.
A Page from A Biography of the Tripitaka Master
Finding that the Buddhist scriptures in China were incomplete and in something of a disarray in the 7th century AD, Xuanzang determined to travel overland to India in the hope of studying there and bringing back the Teaching.
Xuanzang's Handwriting
He left China in 629 AD, and after an incredibly difficult journey along the Silk Road and across the Hindu Kush, he eventually arrived three years later in Kashmir in 631, and spent the next two years studying there.
From there he traveled down into the Ganges Basin, taking in the pilgrimage sites, and eventually onto the great medieval Buddhist University at Nalanda, where he was to spend most of the next decade studying under its famous Abbot, Ven. Silabhadra, who taught the Yogacara school of Buddhism.
When he returned to China in 645, 17 years after leaving the country he was greeted as a hero by the Emperor Taizong of the newly founded Tang Dynasty.
He brought with him a treasure trove of scriptures, including the following:
Mahayana Sutra: 224 items
Mahayana Sastra: 192
Sthaviravada Sutra, Sastra and Vinaya: 14
Mahasa?gika Sutra, Sastra and Vinaya: 15
Mahisasaka Sutra, Sastra and Vinaya: 22
Sa?mitiya Sutra, Sastra and Vinaya: 15
Kasyapiya Sutra, Sastra and Vinaya: 17
Dharmaguptaka Sutra, Sastra and Vinaya: 42
Sarvastivadi Sutra, Sastra and Vinaya: 67
Yin-lun (Treatises on the Science of Inference): 36
Sheng-lun (Etymological Treatises): 13
and dedicated the rest of his life to supervising the translation of the texts into Chinese.
Xuanzang is one of the great heroes of Chinese Buddhism and anyone interested in the history of Buddhism in the Medieval period, especially as it was understood and practiced in India and in China, should try and see this series.
The difficulties Xuanzang had to overcome on his journey, the determination with which he fulfilled his self-appointed task, and his dedication to the cause of religious studies also make this a story of great human adventure and endeavour.
Click here to find the various posts and related material
Some Stills from the Documentary
Crossing the Desert at Night
Xuanzang in Nalanda University
Nalanda's Abbot, Ven. Silabhadra
Xuanzang's Skull Relic
Culture, Dharma, China, India, Mahayana, Monastics, Silk Road
3 thoughts on "The Journey of Xuanzang: An Introduction"
Milan Medic says:
Did Ven. Silabhadra had a beard?! Would Mahayana monk had a beard? Is that historically accurate?!
Anandajoti says:
I think myself this is probably a misrepresentation, maybe long beards in Chinese tradition are associated with venerability, but from a Buddhist monastic point-of-view, the rules stands that head-hair and beard must be shaven and then kept short.
Lazy buddhist says:
Nalanda University then was a thriving institution. it provided for all sects of Buddhist traditions not only those sects that we assume must be clean shaven or wearing robes of the same color or patterns. I think Teachers with beards may be a sign of great achievement that the courts during that era mandated all abbots to wear.
https://dharma-documentaries.net/b/2k8
Asian Corridor 6, Guge Mystery of the Lost Kingdom
Angkor, Land of the Gods: 01 Empire Rising
Devour the Earth
Shangri-la by Michael Wood
Borobudur, Pathway to Enlightenment
Caring for the Beyond, Two Lao Buddhist Festivals for the Deceased
The Secrets of the Mongolian Steppe
A Force More Powerful 05, Poland
Stone Pastures of Ladakh
The Journey of Xuanzang 7 & 8
Xuan Zang, Eminent Monk
Xuan Zang (2016 Film)
The Great Pilgrim Xuanzang
A Bright Moon, the Life of Hong-yi
Light of the Moon, Legacy of Xuan Zang


The Hualalai Golf Course, the first Jack Nicklaus Signature Course on the Big Island is home of the PGA Tour Champions Tour Mitsubishi Electric Championship. This carefully groomed course was designed with a sense of place. Special care was taken to preserve the historic King's Trail located on the course, and other significant cultural sites at the resort. Residents and residential guests of Hualalai along with guests of the Four Seasons Resort Hualalai may enjoy this great facility, which includes a nine-acre driving range with 27,000 square foot short game practice area.


text:
\section{Introduction}
A manifold is closed if it is compact and has no
boundary. In an orientable 3-manifold every 2-dimensional
embedded submanifold has two sides. Embedding $g$ of a submanifold $N$
to a manifold $M$ means that there is a homeomorphism between $N$ and
$g(N)\subset M$. If $g$ mapping a submanifold $N$ to $M$ is not an
embedding then $g(N)$ has self-intersections.

An embedded circle $S^1$ is called a loop. If it has self-intersections
it is only a closed curve. If $l$ and $l'$ are closed curves,
a homomorphism is a continuous mapping $h:I\times I\to M$ such that
$l=\{h(0,x)|x\in I\}$ and $l'=\{h(1,x)|x\in I\}$. We say that $l$ is
homotopic to $l'$ and write it $l\simeq l'$. Homotopic images of $S^1$
belong
to the same homotopy class and the set of homotopy classes is a group,
the fundamental group $\pi_1(M)$ of the manifold $M$. As a special case, if
$l'$ is a point, $l$ is said to be contractible.
If every loop in the manifold is contractible, $\pi_1(M)=1$ and $M$ is
said to be simply-connected.

The Poincar\'e Conjecture states that every
simply-connected closed 3-manifold is homeomorphic to the
3-dimensional sphere $S^3$.

A Morse function is a function $f:M\to \R$ such that in all
but isolated points there is a diffeomorphism $g$ from a neighborhood $V$
of a point $p\in M$ to $\R^3$ so, that if $g(p)=(x,y,z)$, $f(p)=z+c$
where $c$ is a constant.

In a closed 3-manifold there are finitely many points where there is no such
homeomorphism. The points are called critical points of $f$ and their
structure is well known: they are classified by the index $i(p)$. Critical
points of index 0 are points where there is a homeomorphism $g:V\to \R^3$
from a local neighborhood of a critical point $p$ such that if
$g(p)=(r,\theta,\phi)$, then $f(p)=r+c$ where $c$ is a constant. A critical
point of index 3 for $f$ is a critical point of index 0 for $-f$, so the
Morse levels $f^{-1}(x)$ are spheres and the level grows towards the center.
A critical point of index 1 is a point $p\in M$ where $f$ has a saddlepoint.
There is a local homeomorphism $g:V\to \R^3$ from a neighborhood $V$ of $p$
which maps the Morse levels $f^{-1}(x)$ to surfaces shown in Figure 1.
The level $x$ grows to the direction of the arrows in Figure 1. A
critical point of index 2 for $f$ is a critical point of index 1
for $-f$.

\includegraphics {pcfig1.ps}
\hskip 1in	Figure 1.

We will assume that $M$ is a closed and differentiable 3-manifold.

The following notations will be used: $I$ = a closed interval; $S^n$ = a closed n-sphere; $D^2$ = a closed disc;
$T^2$ = a torus; $B^3$ = a closed 3-ball. $D^2_r=\{(x,y)|x^2+y^2\leq r\}$
= a closed disc of radius $r$.
$U^3=\{(x,y,z)|x^2+y^2\leq 1, z\in [0,1], (x,y,1)=(x,y,0)\}$ = a filled torus.
A loop is an image of an embedding $g:S^1\to M$. A path is an image of an
embedding $g:I\to M$. A disc is an image of an embedding $g:D^2\to M$.
A collar is an image  of an embedding $g:D^2_1\backslash D^2_{1\over 2}\to M$.

A closed tubular neighborhood $V$ of a loop $l$ is an image of an embedding
$h:U^3\to M$ such that $l$ is the image of the points $x=y=0$ in $U^3$.
A closed tubular neighborhood $V$ of a path $l$ is an image of a homeomorphism
$h:D^2\times I\to M$ such that $l$ is the image of the points $x=y=0$ in $D^2$.
A closed cylinder neighborhood of a disc $D$ is an embedding of $D^2\times I$ to $M$
such that $D$ is the image of the points $(r,\theta,1/2)$.
When necessary, we assume that these embeddings are smooth.

Let $X\approx Y$ mean that the manifold $X$ (possibly with a
boundary) is homeomorphic to $Y$. Usually we will assume, that
the homeomorphism is a diffeomorphism. Let $\gamma \simeq \gamma'$ mean that
the loop $\gamma$ is homotopic to $\gamma'$.

We will assume, that
the Morse functions are smooth at all points except for a finite set of
critical points. Smooth means here that $f$ is sufficiently many times (say, 3
to be sure) continuously differentiable in local coordinates. Let us write
$M^a_f=\{ x\in M | f(x)\le a \}$, $\partial M^a_f=\{ x\in M | f(x)=a \}$
and if no confusion can arise we write $M^a=M^a_f$. $\partial M^a$ is here
called a Morse level, but when there is no confusion we also call $a$ a level.
We will write $M_a=M\backslash int(M^a)$. Let $a_{max}$ and $a_{min}$
denote the highest and lowest levels $a$ respectively such that $f^{-1}(a)$
is not empty.

\section{The idea of the proof}

It suffices to show that every differentiable simply-connected closed
3-manifold
is homeomorphic to the 3-dimensional sphere $S^3$ as proving the claim for
differentiable 3-manifolds proves it for all. As $M$ is simply-connected
it is orientable, so we start by proving some lemmas for closed, orientable
and differentiable 3-manifolds.

We do not change the manifold $M$ in the proof but construct different Morse
functions on it. First we recall from a theorem of Smale that it is
possible to find a Morse
function which has only one critical point of index 0 and of index 3.
In Lemma 2 we show, that we can change the Morse function in a closed tubular
neighborhood
of a path so, that in the new Morse function critical point of index 2
are on a higher level. We say, that a 2-handle is moved up by changing the
Morse function. In a similar way we can construct a new Morse function
where a selected critical point of index 1 is on a lower level. We say,
that a 1-handle is moved down by changing the Morse function.
The modification to the Morse function for moving a 2-handle up or a 1-handle
down requires changes only in a closed tubular neighborhood of a path.
The modification changes Morse levels in the way that is illustrated
in Figure 2 by showing a 2-dimensional section. The modification, given
precisely in Lemma 2, is radially symmetric with respect to the y-axis in Figure 2.

\includegraphics {pcfig2.ps}

\hskip 1in	Figure 2.

By moving all 2-handles up and all 1-handles down we get to the situation where
every critical point of index one is on a lower level than any critical point
of index two.
Let the lowest level of a critical point $p$ of index 2 be $c$ and
the highest level of a critical point of index 1 be $b$.

Let $a$ satisfy $b<a<c$. Then $a$ is a regular level, that is, not a level
where there are critical points.
We show that $M^a=\{q\in M|f(q)\le a\}$ and
$M_a=\{q\in M|f(q)\ge a\}$ are both handlebodies. That is, 3-manifolds with
a boundary obtained by attaching handles to a 3-ball. This is not automatically
true, we must show that the manifolds do not contain embedded
exotic homotopy balls, i.e., 3-manifolds, which
have the boundary $S^2$ and which are simply-connected but not homeomorphic
to 3-balls.

In Lemma 9 we show that inside $M^a$ and $M_a$ there are no exotic homotopy
balls. This is so
because if there is an exotic homotopy ball after the addition of
a 1-handle, there already was an exotic homotopy ball before the addition
of the 1-handle.
Figure 3 illustrates this: If the blackened part is a part
of an exotic ball, then there already existed an exotic ball before adding the
1-handle because we can connect the parts also by a tube that does not go
through the 1-handle.

\includegraphics [width=90mm] {pcfig3.ps}

\hskip 1in	Figure 3.

Let us notice that this is only true for the level $a$ such
that all 2-handles are above the level $a$ and all 1-handles are below the
level $a$. On other levels we cannot conclude that there are no
embedded exotic balls. We cannot even conclude that both sides are
handlebodies with possible embedded exotic homotopy balls, since they may have
nontrivial fundamental groups.
We can show these things only for the level $a$.
In order to show that $M^a$ does not contain exotic balls we use induction
by adding 1-handles one by one and conclude that no exotic ball exists.
We can show the same for $M_a$ by making a similar induction with the Morse
function $-f$.

A Heegaard split is a general way of expressing 3-manifolds by glueing
two handlebodies at their boundaries. The genus $g$ of the split is the
number of handles in the handlebodies.
In Lemma 10 we show that we have a special kind of Heegaard
split $M^a$, $M_a$.
There are homeomorphisms
$g_1$ and $g_2$ such that $g_1(M^a)$ and $g_2(M_a)$ are handlebodies
embedded in $\R^3$, and a boundary homeomorphism $\psi: g_1(M^a) \to g_2(M_a)$.
We can take $g_1$ and $g_2$ differentiable if needed.
Standard noncontractible generators $x_{a,1},...,x_{a,g}$
and contractible generators $y_{a,1},...,y_{a,g}$ can be defined for
$\partial g_1(M^a)=\partial H_{a,g}$.
Similarily, the loops $x_{b,1},...,x_{b,g}$, $y_{b,1},...,y_{b,g}$ are
standard generators for $\partial g_2(M_a)=\partial H_{b,g}$. Figure 4 shows
these generator loops.

\includegraphics {pcfig4.ps}

\hskip 1in	Figure 4.

The boundary homeomorphism $\psi: \partial g_1(M^a) \to \partial g_2(M_a)$
induced by a Morse function has the property, that $\psi$ and $\psi^{-1}$ map
loops to loops.

The following example for the genus $g=1$
shows another important thing. Not all combinations
of standard generators can be represented as loops but only as closed curves
that have self-intersections. In particular, if a contractible
generator wraps several times around, there is needed a noncontractible
generator to provide a shift, or the curve is not a loop but has
self-intersections.
In Figure 5 a) the surface of a torus $\partial H_{b,1}$
is divided into four squares with side length 4.
There are identications
in the horizontal boundaries $((x,4)=(x,-4)$ and in the vertical boundaries
$(y,-4)=(y,4)$ making it a 2-dimensional torus.

\includegraphics[width=100mm]{pcfig5.ps}

\hskip 1in	Figure 5.

Let us stretch the part above $y=0$ into a strip of length 16 and width 2,
stretch the part below $y=0$ into a strip of length 4 and width 2 and wrap
the strip 5 times around a torus as in Figure 5 b). A loop drawn in Figure 5
a) maps to a loop drawn in Figure 5 b). Notice, that we need the shift in
Figure 5 a).

We can define the standard generators on the surface of the handlebody
$H_{a,1}$ as follows:
$y_{a,1}$ is the loop from $(1,0)$ to $(1,2)$, and $x_{a,1}$ is the loop from
$(1,1)$ to $(21,1)$ in Figure 5 b), and we can define the standard
generators on the surface of the other handlebody $H_{b,1}$ as follows:
$y_{b,1}$ is the loop from $(2,-4)$ to $(2,4)$, and $x_{b,1}$ is the loop from
$(0,2)$ to $(8,2)$ in Figure 5 a).
The loop drawn to Figure 5 a) is
$$l_b= (x_{b,1} {y_{b,1}}^{-\alpha} )^4 {y_{b,1}}^{-1+4\alpha}$$
where $\alpha =0.5$ is the offset the loop advances in each round, the
necessary shift needed to make a loop.

Instead of fundamental groups of $H_{a,1}$ and $H_{b,1}$ we will consider
the first homology groups $H_1(\partial H_{a,1})$ and $H_1(\partial H_{b,1})$
with integer coefficients.
The boundary map $\psi:\partial H_{a,1}\to \partial H_{b,1}$ induces a linear
map between the homology groups.
The first homology group is abelian, so the loop can be expressed
more simply as
$$l_b=(x_{b,1})^4{y_{b,1}}^{-1}.$$
As the group is abelian, we can also use the additive notation:
$$l_b=4x_{b,1}-y_{b,1}.$$

Clearly, we can use linear algebra for solving the generators $x_{a,1}$
and $x_{b,1}$, and if the manifold is simply-connected, we will manage
to solve $x_{a,1}$ and $x_{b,1}$ as linear combinations of $y_{a,1}$ and
$y_{b,1}$. However, this is not only linear algebra with rational coefficients
but there are the
properties that loops map to loops and not every closed curve can be modified
to be a loop. With these two properties we manage to show in
Lemma 13 that a loop
$g_1^{-1}(x_{a,1})$ can be associated with a loop $l$ which
is homotopic
to it in $g_1(M^a)$ and $l$ bounds an embedded disc in $g_2(M_a)$.
There is a critical point of index 2 in this disc, which we will move down.

What we want to do with this result is to reduce the genus of the Heegaard
split by canceling one 2-handle and 1-handle pair. Then induction on the genus
yields the result. The tricky thing is that in order to cancel a pair
we should move a 2-handle down. The modification that is needed for the Morse
function in order to move a 2-handle down is the inverse modification to the
one in Figure 2. The difficult part is that in order to move a 2-handle down
(or 1-handles up, which is the same thing for $-f$)
we need a closed cylinder neighborhood of a disc.
In order to move a 2-handle
up (or 1-handles down, which is the same thing for $-f$)
we only needed a closed tubular neighborhood of a path. Topologically these are
the same and that is why the modification is the same, but while it is easy
to find a 1-dimensional path which does not meet any critical points, it is not
easy to find a large disc which is avoiding critical points. It is where we
need the simply-connectedness of $M$.

We can of course find an embedded disc from $M_a$ that bounds any contractible
loop on $\partial M_a$, but this loop has to be moved down to a low Morse
level. We have to find an embedded collar that has the same boundary as
this embedded disc.
If we get such a collar then we can construct a Morse function where
the critical point of index 2 is on a level just above the level of the lowest
critical point of index 1.

We can try to push down a loop that is contractible in $M_a$
and see what happens.
Let us take the loop $l_a=\psi(g_2^{-1}(y_{b,1}))$ on $\partial M^a$
and try to push it down by a family of loops $l_d$
parametrized by $d$ where $d=a$ at the beginning,

We can move the loop $l_d$ towards the gradient of
the Morse function $-f$ so that the Morse level of $f$ decreases.
Thus, at each level $d$ holds $l_d\subset \partial M^d$.
In the beginning the loop $l_d$ creates an embedded collar when $d$ decreases
but if $l_d$ meets a critical point of index 1 there may happen several
things - the loop may get knotted, get self-intersections,
split, merge and so on.
This creates a complicated situation, and we try to avoid hitting any
critical points if possible.

In order to avoid critical points, we try to find a loop
that goes around one 1-handle only in $M^a$,
so that the loop can be pushed down
creating a collar that does not go through any other critical points. Thus,
we need a loop on $H_{a,1}$ that has only one noncontractible generator
$x_{a,1}$. If the genus $g=1$ this is of course true, but if $g>1$ we
need to find a suitable loop in Lemma 13. By using linear algebra and
the two properties that loops map to loops and not all closed curves can
be represented as loops we find a suitable loop that has only one
noncontractible generator $x_{a,1}$. Then we can push the collar down
all the way without hitting critical points because meeting these
critical points means going around a 1-handle that is created in a critical
point.
Our collar goes around the 1-handle that is created on the first
critical point of index 1. We will not continue the collar to this level
but stop just before the level on some regular level. Thus, our collar avoids
all critical points and stays as an embedded collar.

The embedded collar formed by moving the loop $l_d$ down can be connected
the the disc in order to form a larger disc. We take a closed cylinder
neighborhood of the disc. Inside the neighborhood
we can the change of the Morse function, using the inverse of the change
in Figure 2, and move the 2-handle to a lower level.
The change does not affect the Morse function outside the neighborhood
and will not change the levels of other critical points.
In this way we can move the 2-handle to a so
low level that the two first critical points after the
0-handle are a 1-handle and a 2-handle and the 2-handle is a disc contracting
the 1-handle. Together they make a 3-ball. We can replace the Morse
function by such a Morse function where these two critical points do
not appear at all. In Lemma 14 we show that this results to a reduction
of the genus of the Heegaard split, which proves by induction
that there are no exotic homotopy 3-spheres.

\section{The proof}

A surface $N\subset M$ is a connected, closed, embedded 2-manifold. For any
regular value $a$ the set $\partial M^a_f$ is a finite set of disjoint
surfaces. Let $N_a\subset \partial M^a$ be a surface.

\th Lemma 1. Let $M$ be a closed, orientable and differentiable 3-mani\-fold.
Each surface $N_a\subset \partial M^a$ for a regular
value $a$ is 2-sided and orientable, and separates $M$ into two orientable
3-manifolds with boundary $M_1$ and $M_2$, $\partial M_1=\partial M_2=M_1\cap M_2
=N_a$.

\proof As $M$ is orientable the surface $N_a$ separates $M$.$\eop $

\th Theorem D (Smale). Let $M$ be a closed simply-connected differentiable
3-manifold.
There exists a Morse function $f:M\to \R$ such that
$f$ has one critical point of index 0 only and one critical point of index 3.
The number of critical points is finite.

\proof
See the Theorem D of Smale in [2]. $\eop$

\th Lemma 2. Let $M$ be a closed orientable and differentiable 3-manifold.
Let $f:M\to \R$ be a Morse function such that $f$ has the critical points
$p_0,...,p_n$.

{\sl Let $k$ be such that $i(p_k)=2$. Let ${a'}_1>a_1=f(p_k)$,
${a'}_1<a_{max}$, be a noncritical
level such that there exists a path $l:I\to M$ satisfying }

{\sl $l(0.5)=p_k$, }

{\sl $f(l(1))>{a'}_1$, }

{\sl for each $x,y\in (0,1)$ holds that if $x<y$ then $f(l(x))<f(l(y))$ and}

{\sl $l(I)\cap \{ p_m | 0\le m\le n\}=\{p_k\}$.}

{\sl Then there is a Morse function $f':M\to \R$ such that:}

{\sl $f'$ and $f$ have the same critical points and they have the same indices,}

{\sl $f'(p_k)={a'}_1$ and $f'(p_m)=f(p_m)$ for all critical points $m \not =k$,}

\proof
Select a small closed tubular neighborhood
$V$ of $l(I)$ such that $V\cap \{p_m | 0 \le m \le n\}=\{p_k \}$.

Define
$A=\{ (r,\theta,z)\in \R^3 | 0 \le r \le 1, 0\le \theta < 2\pi , 0 \le z \le 1\}$.
Select $s\in (0,1/2)$. Let ${w'}_s:A\to A$ be defined as

${w'}_s(r,\theta,z) = (r,\theta,z+(1-r)(0.5-s){z \over s})$
if $0\le z\le s$, $0\le r\le 1$,

${w'}_s(r,\theta,z) = (r,\theta,z+(1-r)(0.5-s))$
if $s< z\le 0.5$, $1-{{0.5-z}\over {0.5-s}} \le r \le 1$.

We extend ${w'}_s$ to $1\ge z>0.5$ by:

${w'}_s(r,\theta,z) = (r,\theta,1-w_s(r,\theta,1-z))$
if $0\le 1-z\le s$, $0\le r\le 1$,

${w'}_s(r,\theta,z) = (r,\theta,1-w_s(r,\theta,1-z))$
if $s< 1-z\le 0.5$, $1-{{0.5-(1-z)}\over {0.5-s}} \le r \le 1$.

Let $w_s:A\to A$ be a smooth fuction for $r<1$ approximating ${w'}_s$.
Let us find a diffeomorphisms $g$ of $A$ to the tubular neighborhood $V$ so that

the points with $z=0.5$ map to $l(I)$, $g(0,0,0.5)=p_k$,

$g(0,0,0)=l(0)$, $g(0,0,1)=l(1)$ and

$g^{-1}(\partial {M^a}_f\cap V)=\{ w_{a_1}(r,\theta ,a)|0\le r\le 1,0\le \theta<2\pi \}$.

Let us write $A_{a,a_1}=g^{-1}(\partial {M^a}_f\cap V)$.

Let us construct $f':M \to \R$ as follows:

$f'|_{M\backslash V}=f|_{M\backslash V}$,

$f':V\to \R$ is defined by the condition: if $p\in g^{-1}(A_{a,{a'}_1})$
then $f'(p)=a$.

Then $f'(p_k)={a'}_1$ since $g^{-1}(p_k)=(0,0,0.5)\in A_{{a'}_1,{a'}_1}$.

Since ${a'}_1>a_1$, $f'$ does not have other critical points than $p_k$ in
$V$. (If ${a'}_1<a_1$, $f'$ could have critical points in $\partial V$ since
there could be created a new saddlepoint.) Finally smoothen $f'$ on
$\partial V$. $\eop$

Figure 2 shows the construction of $f'$ in $V$. We simply wrote down an
explicit function for doing the modification in Figure 2.

\th Lemma 3. Let $M$ be a closed simply-connected differentiable
3-manifold. There exists a Morse function $f:M\to \R$ such that $f$ has
one critical point of index 0 only, one critical point of index 3 only and
on each critical level there is one critical point only.

\proof
By Theorem D of [2] there is a Morse function $f$ such
that $f$ has one critical point of index 0 only and one critical point of
index 3 only.

By applying the modification in Lemma 2 successively we can construct a
Morse function which has the same critical points (and of same indices)
as $f$ but such that all critical points are on distinct levels. $\eop$

Let us denote the minimum and maximum values of $f$ by
$f(p_0)=a_{min}$ and $p_n=a_{max}$.

\th Lemma 4. Let $M$ be a closed simply-connected differentiable
3-manifold. If a Morse function $f:M\to \R$ has
one critical point of index 0 only and one critical point of index 3 only,
then for each level $a$ $\partial M^a$ is connected.

\proof If $\partial M^a$ is not connected, then there exists a noncontractible loop
passing the critical point of index 0, the critical point of index 3 and two
points in two components of $\partial M^a$. $\eop$

\th Lemma 5. Let $M$ be a closed simply-connected and differentiable
3-manifold.
Let $f:M\to \R$ be a Morse function such that $f$ has the critical points
$p_0,...,p_n$.

{\sl Let $k$ be such that $i(p_k)=2$. Let ${a'}_1>a_1=f(p_k)$ be a noncritical
level such that ${a'}_1<f(p_n)=a_{max}$. }

{\sl Then there is a Morse function $f':M\to\R$ such that:}

{\sl $f'$ and $f$ have the same critical points and they have the same indices,}

{\sl $f'(p_k)={a'}_1$ and $f'(p_m)=f(p_m)$ for all critical points $m \not =k$,}

\proof

If there exists a path $l:I\to M$ as in Lemma 2, then the result follows
from Lemma 2. The only problem is to select $l:I\to M$ such that

$l(I)=\{ p_m | 0\le m\le n\}=\{p_k\}$

and that the path satisfies:
for each $x,y\in (0,1)$ holds that if $x<y$ then $f(l(x))<f(l(y))$.

The points $p_0$ and $p_n$ are on levels outside $l(I)$. If $i(p_m)=1$,
then $p_m$ can be avoided as a critical point of index 1 in the direction
of growing Morse levels does not force a path to go through the critical point,
see Figure 1. Let $l(I)$ be selected so that it does not
intersect with critical points of index 1.

It is not obvious that a critical point of index 2 can be avoided when moving
a 2-handle in the direction of growing Morse level. Let $p_m$ be
a critical point of index 2 with $f(p_m)>f(p_k)$, $p_m\in l(I)$ and
there is no critical point $p_j\in l(I)$ such that $f(p_k)<f(p_j)<f(p_m)$.

Assume that there is no path which omits $p_m$. Select a small
embedded circle in a neighborhood of $p_k$ on the level $f(p_k)+\epsilon$
such that a path from every point of the circle a path towards grad($-f$)
passes through $p_k$. Continuing the circle to the direction of $grad(f)$
defines a surface $N\subset M$ such that $N=g(I\times I)$ where

$g(0,x)=p_k$, $g(1,x)=p_m$ for $x\in [0,1]$,

if $y>z$, $y,z\in [0,1]$, then $f(g(y,x)) > f(g(z,x))$ for any $x\in [0,1]$,

$g(y,0)=g(y,1)$ for any $y\in [0,1]$.

If there is no such $N$, then $l(I)$ can be selected to omit $p_m$.

$N$ is embedded as in all points $x\in N$, $x\not\in \{p_m,p_k\}$
the gradient of $f$ is unquely defined.

The surface $N$ separates $M$ as $M$ is orientable. As $p_m$ is a saddlepoint,
there are two points $p$ and $q$ on the opposite sides of $N$ such that both
$p$ and $q$ belong to $\partial M^{f(p_m)+\epsilon}$ for some small
$\epsilon>0$. Let $a=f(p_m)+ \epsilon$. By Lemma 4 $\partial M^a$ is
connected, therefore there is a path from $p$ to $q$ in $\partial M^a$.
This path must intersect $N$ as $N$ separates $M$. However, all points
$x\in N$ have $f(x)\le f(p_m)<a$. This contradiction shows that there is a
path $l(I)$ omitting $p_m$. $\eop$

\th Lemma 6. Let $M$ be a closed simply-connected and differentiable
3-manifold.
Let $f:M\to \R$ be a Morse function such that $f$ has the critical points
$p_0,...,p_n$.

{\sl Let $k$ be such that $i(p_k)=1$. Let ${a'}_1<a_1=f(p_k)$ be a noncritical
level such that ${a'}_1>f(p_0)=a_{min}$. }

{\sl Then there is a Morse function $f':M\to\R$ such that:}

{\sl $f'$ and $f$ have the same critical points and they have the same indices,}

{\sl $f'(p_k)={a'}_1$ and $f'(p_m)=f(p_m)$ for all critical points $m \not =k$,}

\proof
A critical point of index 1 of $f$ is a critical point of index 2 of $-f$.
The result follows from Lemma 5. $\eop$

\th Lemma 7. Let $M$ be a closed
simply-connected and differentiable 3-manifold.
There exists a Morse function $f:M\to \R$ such that

{\sl $f$ has the critical points $p_0,...,p_n$,}

{\sl the indices of the critical points are $i(p_0)=0$, $i(p_n)=3$,
$i(p_m)=1$ for $0<m\le l$, $i(p_m)=2$ for $l<m<n$,}

{\sl for any $m<j$ holds $f(p_j)<f(p_m)$.}

\proof
By Lemma 5 there is a Morse function $f'$ such that one critical point $p_k$
of index 2 is on a regular level $a$ with $a_{max}>a>f(p_m)$ for $m<n$,
$m\not = k$.

By applying Lemma 5 sequentially to all critical points of index 2
we find after $n-l-1$ steps a Morse function where critical points of index 2
are on higher levels than all critical points of index 1 or 0. $\eop$

\th Lemma 8. Let $M$ be a closed simply-connected and differentiable
3-manifold.
Let $f:M\to \R$ be a Morse function such that

{\sl $f$ has the critical points $p_0,...,p_n$,}

{\sl the indices of the critical points are $i(p_0)=0$, $i(p_n)=3$,
$i(p_m)=1$ for $0<m\le l$, $i(p_m)=2$ for $l<m<n$,}

{\sl for any $m<j$ holds $f(p_j)<f(p_m)$.}

{\sl Let $a$ be a regular level. If $a<f(p_{i+1})$, then $\pi_1(M^a)$
is free. If $a>f(p_i)$, then $\pi_i(M_a)$ is free.}

\proof
By the Morse theory $\pi_i(M^a)=1$ for $a\in (a_{min}, f(p_1))$. In each
critical point $p_k$ of index 1 ($0<k\le i$) by the Morse theory
$\pi_1(M^{f(p_k)+\epsilon})=\Z \times \pi_1(M^{f(p_k)-\epsilon})$ for
a small $\epsilon>0$. By induction $\pi_1(M^a)$ is free for any
regular level $a<f(p_{i+1})$. As critical points of index 2 for $f$
are critical points of index 1 for $-f$, $\pi_1(M_a)$ is free
for any regular level $a>f(p_i)$. $\eop$

A 1-handle is a cylinder $H\approx D^2\times I$. A handlebody of genus $k$
is by definition obtained by adding $k$ 1-handles to $B^3$ so, that the
1-handle $H$ is fixed at the discs $D^2\times \{0,1\}$ to the boundary
of $B^3$. At a critical point of index 1 a 1-handle
is added to $M^a$. An exotic homotopy ball is an embedded
manifold $B^3_e$ which has the boundary $\partial B^3_e\approx S^2$
and $\pi_1(B^3_e)=1$.

\th Lemma 9. Assume that a 3-manifold $M_1$ (with boundary) is obtained
from a 3-manifold $M_2$ by addition of a 1-handle $H\approx D^2\times I$,
$M_1=M_2\cup H$. Assume, that $M_1$ and $M_2$ are orientable. If $M_1$
contains an exotic homotopy ball $B^3_e$. Then $M_2$ contains an embedded
exotic homotopy ball  ${B'}^3_e$.

\proof
The embedding of $B^3_e$
to $M_1$ is not wild as both manifolds $M_1$ and $B^3_e$ have the same
dimension 3. Wild embeddings happen only with submanifolds of lower dimension.
As $B^3_e$ is not wildly embedded, $\partial B^3_e\approx S^2$ is not
a wildly embedded sphere.

Let $F$ be a smoothly embedded disc cutting the 1-handle $H$. $F$ can be
assumed to be in a general position so, that $F\cap \partial B^3_e$ is
a set of disjoint circles $A_0=\{C_i | i\in I_0\}$, $C_i\approx S^1$, $i_0$ finite.

The set $A_0$ contains a nonempty subset $B_0=\{S_j |j\in I_1\}\subset A_0\}$
where each $S_j\subset \partial B^3_e$
is a boundary of a disc $D_j \subset \partial B^3_e$ satisfying
$int (D_j) \cap A_0\}=\emptyset$. In order to see that $B_0$ is nonempty,
notice, that each $C_i$ separates $\partial B^3_e\approx S^2$. There must be
such circles on $\partial B^3_e$ that one of the separated sides does not
contain smaller circles. Such a circle $C_i$ is a circle $S_j$.

Take one $S_j$ and a circle ${S'}_j\subset F$ which is slightly bigger than
$S_j$. There is a disc ${D'}_j\subset F$ such that $\partial {D'}_j={S'}_j$.
There may be other circles $C_i$ than $S_j$ inside ${D'}_j$ but that does
not matter here. Replace the disc ${D'}_j$ by a disc ${D''}_j$ which is close
to $D_j$. Then $D_j$, ${D''}_j$ and the annulus $F_0\subset F$ between the
circles $S_j$ and ${S'}_j$ separates $M_1$ and one side is a 3-ball $B_0$, it is
the side which does not contain points of $int (B^3_e)$.
The disc ${D''}_j$ separates $M_2$ into two components.
Let $U$ be the component which contain points of $int (B_0)$.

Let $V$ be a closed neighborhood of $F$ in $H$ and $V_1$ a smaller closed
neighborhood of $F$ in $V$. Then $F$ separates $V$ and $V_1$ and $\partial V$
is the union of a collar, which is subset of $\partial H$, and two discs
$D_a$, $D_b$ which have boundary at $\partial H$.

By construction ${D''}_j$ and $D_j$ insersect with $F$ only at
${S'}_j$ and $S_j$. Therefore we can select $V$ so small, that
the component in $V\cap B^3_e$ which contains $D_j$ is a collar
$C=c(S^1\times I)$, for an embedding $c:S^1\times I \to H$ such that
$C\subset V$, $c(S^1\times \{0\})\subset D_a$, $c(S^1\times \{1\})\subset D_b$,
$C\cap F=S_j$, $C\subset \partial B^3_e$.

Similarily we can find a collar
$C'=c'(S^1\times I)$, for an embedding $c':S^1\times I \to H$ such that
$C'\subset V$, $c'(S^1\times \{0\})\subset D_a$, $c'(S^1\times \{1\})\subset D_b$,
$C'\cap F={S'}_j$, $C'\cap \partial B^3_e=\emptyset$ and
${D''}_j\cap V\subset C'$. ${D''}_j\cap V$ is on one side of $F$ in $V$,
let that be the side where $D_a$ is.

Let us define a homeomorphism $g':M_1\to M_1$ so, that $g'$ keeps the
disc ${D''}_j$ fixed, expands the 3-ball $B_0$ to ${B'}_0$so, that the
disc $D_j$ is pushed to the component $K_b$ of $V\backslash int (V_1)$ which contains $D_b$.
The mapping $g'$ pushes a part $A$ of the homotopy ball $B^3_e$
which was in $M_2$ to $K_b$.

We must move this part of $B^3_e$ back to $M_a$ but not through $F$.
Let $l$ be a path from a point in $V_1\cap K_b$ to a point in $V_1\cap K_a$
which does not go through $F$. We can select $l$ so that it does not
intersect with $B^3_e$ as $\partial B^3_e\approx S^2$. Let $V_2$
be a small closed tubular neighborhood of $l$ in $M_1$. We
connect the tube $V_2$ to $A$ and move $A$ through the tube $V_2$ to
the other component $K_a$ of $V\backslash (V_1)$
by expanding a 3-ball into $K_b$.
Finally we restore $A$ back to $M_2$ by decreasing the 3-ball in $U$.
Now the tube $V_2$ is filled with a 3-ball, $M_2$ is restored and the
handle $H$ is again a handle. Let us call this homeomorphism of $M_1$
to $M_1$ by $g''$. See Figure 3 where the mapping is shown.

Define $g$ as the combined homeomorphism $g:M_1\to M_1$, $g(p)=g''(g'(p))$.
Then $g(M_2)=M_2$.

Let us rename the exotic homotopy ball $g'(B^3_e)$ as $B^3_e$.

There are now less circles in the intersection set
$A_1=F\cup \partial B^3_e$. If $A_1$ is nonempty,
let us return to the step of selecting one $S_j$, repeat the procedure and
get a smaller set $A_2$.

We repeat the procedure as long as $F\cup \partial B^3_e \not = \emptyset$.
When the set is empty, we have obtained an exotic homotopy ball which
is contained in $M_2\cup K_1\cup K_2\approx M_2$. So, there is an exotic
homotopy ball ${B'}^3_e$ in $M_2$.
$\eop$

\th Lemma 10. Let $M$ be a closed simply-connected differentiable
3-manifold. Then there exists a Morse function $f:M\to \R$ such that

{\sl $f$ has the critical points $p_0,...,p_n$,}

{\sl the indices of the critical points are $i(p_0)=0$, $i(p_n)=3$,
$i(p_m)=1$ for $0<m\le l$, $i(p_m)=2$ for $l<m<n$,}

{\sl for any $m<j$ holds $f(p_j)<f(p_m)$.}

{\sl For any regular level $a$ holds: if $a<f(p_{l+1})$, then $M^a$
is a handlebody and if $a>f(p_l)$, then $M_a$ is a handlebody.}

\proof
By Lemma 7 there exists such a Morse function. By Lemma 8 the fundamental
groups of $M^a$ and $M_a$ are free. A compact orientable 3-manifold
with a nonempty connected boundary, which has a free fundamental group
is a handlebody with embedded homotopy balls.

If there is an embedded exotic homotopy ball in $M^a$ for a regular
$a<f(p_{l+1})$, then Lemma 9 used $l$ times shows
that there is an exotic homotopy ball in $M^a$ for $a_{min}<a<f(p_1)$.
This is a contradiction as for $a_{min}<a<f(p_1)$, $M^a$ is a 3-ball.
This shows that there are no embedded exotic homotopy balls in $M^a$.
Therefore $M^a$ is a handlebody if $a<f(p_{l+1})$.

The same argument for $-f$ shows that there are no embedded exotic homotopy
balls in $M_a$ for $a>f(p_l)$.
Therefore $M_a$ is a handlebody if $a>f(p_l)$. $\eop$

Let us now summarize the results obtained so far. These results do not require
the manifold $M$ to be simply-connected, and the results are very natural.
We have created a Heegaard split for the manifold, but this Heegaard split
has the advantage that it is induced by a Morse function and we learned how to
move handles by Lemma 2.

\th Definition 1.
Let $M$ be a closed, orientable and differentiable 3-manifold.
Let $f$ be a Morse function with critical points $p_0,...,p_n$.
Let $a$ be a regular level with $f(p_m)<a$ for all critical points $p_m$
with $i(p_m)\le 1$ and $a<f(p_m)$ for all critical points $p_m$ with $i(p_m)>1$.
The 3-manifold $M$ has a Heegaard split induced by the Morse function $f$
if there there are homeomorphisms $g_1$, $g_2$ of $M^a$ and $M_a$ to standard
handlebodies $g_1(M^a)$ and $g_2(M_a)$ in $\R^3$
and a boundary identification homeomorphism $\psi$ such that the following diagram
commutes

$$\vbox{\rm\halign {\hfil #\hfil && \quad \hfil #\hfil \cr
 $M^a$&  ${{g_1} \above 0pt {\to}}$ & $g_1(M^a)$ & $\hookrightarrow$ & $\R^3$\cr
 $\uparrow$  & & $\uparrow$\cr
 $\partial M^a$  & & $\partial g_1(M^a)$\cr
 $id$ $\downarrow$& &$\psi$ $\downarrow$\cr
 $\partial M_a$  & & $\partial g_2(M_a)$\cr
 $\downarrow$ &  & $\downarrow$\cr
 $M_a$&  ${{g_2} \above 0pt {\to}}$ & $g_2(M_a)$ & $\hookrightarrow$ & $\R^3$\cr
}}$$

\th Corollary 1.
Let $M$ be a closed differentiable and
simply-connected 3-manifold. Let $f$ be a Morse function as in Lemma 10.
Let $a$ be a regular level with $f(p_i)<a<f(p_{i+1}$, then
$M^a$, $M_a$ is a Heegaard split induced by the Morse function $f$.

\proof
The claim follows directly from Lemma 10. $\eop$

The homeomorphism $\psi$ of a Heegaard split is called a
gleuing mapping. It is sufficient within a homeomorphism to define
the glueing mapping by defining how the standard generators are mapped.
A Heegaard split induced by a Morse function is a Heegaard split but
our goal is to use the function $f$ and to move a 2-handle down.

The following property is useful.

\th Lemma 11. Let $\psi$ be the glueing mapping of a Heegaard split induced by
a Morse function $f$. If $l\in \partial M_a$ is a loop, then
$l'={g_1}^{-1}(\psi^{-1}(g_2(l)))$ is a loop in $\partial M^a$.
If $l\in \partial M^a$ is a loop, then
$l'={g_2}^{-1}(\psi(g_1(l)))$ is a loop in $\partial M_a$.

\proof
The Morse function $f$ has $a$ as a regular level. Therefore
the glueing mapping $\psi:\partial M^a\to \partial M_a$ is bijective.
It follows that if $l$ is a loop, $l'$ is a loop. $\eop$

A Heegaard split of genus 1 can be defined
by glueing two standard handlebodies $H_{a,g}$ and $H_{b,g}$ of genus g
together so, that the standard generators
$x_{a,i},y_{a,i}$ of $H_{a,g}$ are mapped to words in generators
$x_{b,1},y_{b,1}$ of $H_{b,g}$ by $\psi$.
A fundamental group of the boundary $\partial H_g$ of a handlebody $H_g$
is a group generated by
$$x_1,...,x_g,y_1,...,y_g$$
with one relation $x_1y_1x_1^{-1}y_1^{-1}\cdots x_gy_gx_g^{-1}y_g^{-1}$.

\th Lemma 12. Let $M$ be simply-connected and obtained as a Heegaard split of
genus $g$. The boundary map
$\psi:\partial H_{a,g}\to \partial H_{b,g}$
induces a linear map between the first homology groups
$H_1(\partial H_{a,g};\Z)$ and $H_1(\partial H_{b,g};\Z)$.
These groups are generated by $2g$ generator loops
$$x_{c,1},...,x_{c,g},y_{c,1},...,y_{c,g} \hskip 2em c\in \{a,b\}$$
where $x_{c,j}$ are noncontractible in the handlebody $H_{c,g}$
and $y_{c,j}$ are contractible in the handlebody $H_{c,g}$.
Every $x_{c,j}$ can be expressed as
$$x_{c,j}=A_j{Y_a}^T+B_j{Y_b}^T \eqno(2)$$
where $A_j$ and $B_j$ are row vectors with rational entries
and $Y_c=[y_{c,1},\dots, y_{c,g}]^T$.

\proof
The first homology group is the abelianization of the fundamental group and
thus it is abelian and torsion free. We can use the same generators for
the first homology group, or select another set of generators.
As the group is abelian, we can use an additive notation and write a
word in the generators as

$w=\sum_i (a_ix_i+b_iy_i)$.

The pullback of the boundary map $\psi:\partial H_{a,g}\to \partial H_{b,g}$
is a linear map between the homology groups.
Thus, the map between the first homology groups
$H_1(\partial H_{a,g};\Z)$ and $H_1(\partial H_{b,g};\Z)$.
induced by the glueing mapping
can be described in a matrix form as

$Z_a=SZ_b$ , where

$Z_c=[X_c,Y_c]^T=[x_{c,1},...,x_{c,g},y_{c,1},...,y_{c,g}]^T$,
$c\in \{a,b\}$.

The entries of the matrix $S$ are rational numbers.
There are 2$g$ equations to solve $x_{a,j}$ and $x_{b,j}$.
If the matric $S$ is invertible,
every $x_{c,j}$ can be expressed as

$$x_{c,j}=A_j{Y_a}^T+B_j{Y_b}^T$$

where $A_j$ and $B_j$ are row vectors with rational entries.
Since $y_{c,j}$ is contractible in $H_{c,g}$ ($c\in \{a,b\}$),

$y_{a,j}\simeq 1$ for all $j=1,...,g$ in $H_{a,g}$,

$y_{b,j}\simeq 1$ for all $j=1,...,g$ in $H_{b,g}$.

It follows that if the matrix is invertible, then

$x_{a,j}\simeq 1$ for all $j=1,...,g$ in $M$,

$x_{b,j}\simeq 1$ for all $j=1,...,g$ in $M$.

This is as we expect since the manifold $M$ is simply-connected.

However, if the matrix $S$ is not invertible, then we cannot solve every
$x_{c,j}$ as a linear combination of $y_{a,k}$ and $y_{b,k}$.
In this case there are too few linearly independent equations and at least
one $x_{a,j}$ can only be expressed as a funtion of
$y_{a,k}$, $y_{b,k}$, $x_{b,k}$ and $x_{b,m}$ for some indices $k$ and
$m\not=j$. In this case $x_{a,j}$ cannot be contractible in $M$. Then
$M$ is not simply-connected. We conclude that the matrix $S$ must be
invertible.

In order to illustrate this conclusion by an example, let us take $g=1$ and
let the mapping be $x_{a,1}=x_{b,1}$, $y_{a,1}=y_{b,1}$. Here we cannot
solve $x_{a,1}$ as a linear combination of $y_{a,1}$ and $y_{b,1}$ and
the manifold is also clearly not simply-connected.
$\eop$

\th Lemma 13. If $M$ obtained as a Heegaard split of genus $g$
induced by a Morse function is simply-connected and $x_{a,j}$ is
a standard noncontractible generator loop of $H_{a,g}=g_1(M^a)$
then there exists
a loop $l$ homotopic to $x_{a,g}$ in $H_{a,g}$ which bounds a disc
in $H_{b,g}$.

\proof
We will select the generators of the first homology groups of
$\partial H_{c,g}$, $c\in \{a,b\}$ slightly differently as $y_j$ in Figure 4.
Let us first take a large filled torus
and let its noncontractible standard generator
be called $x_{c,1}$ and contractible standard generator be called $y_{c,1}$.
Then we glue $g-1$ small handles to the outer sphere of this filled torus.
This gives a handlebody of genus $g$. We select $x_{c,j}$ as noncontractible
generators as in Figure 4, but the contractible generators $y_{c,j}$ we
select such that they are on the surface of the big torus. Thus, they go
through the holes created by the small handles. In Figure 4 we have marked
these loops by $z_j$.
We can use these generators
in Lemma 12 and
as $M$ is simply-connected, there is a solution

$x_{a,j}=A_j{Y_a}^T+B_j{Y_b}^T$

where $A_j$ and $B_j$ are row vectors with rational entries.

In the example of Figure 5 a) we needed a shift $\alpha$
in order to create a curve without making self-intersections, that is, to
make a loop.

Moving the generators of $H_{a,g}$ to the left-hand side and
the generators of $H_{b,g}$ to the right-hand side gives

$x_{a,1}-A_1{Y_a}^T=B_1{Y_b}^T$

Let the vector $B_1$ have the components
$[b_1,\dots, b_g]$. Six observations can be made.

1) If the coefficients of $A_1$ are integers, the left-hand side can be
presented as a loop. That is, we can use $x_{a,1}$ to shift multiple
windings of $y_{a,k}$ in such a way that there are no intersections,
similarly as is done in Figure 5 a).

2) Assume that the coefficients of $A_1$ are rational numbers but not integers.
In that case the left-hand side is a set of disjoint paths that have no
self-intersections. We can find an integer $k$ such that when
the equation is
multiplied by $k$, the left-hand side can be presented as a loop.

3) It is generally true that if there is no shift, we do not get a loop
if a curve winds up multiple times. This can be seen in Figure 5 a).
Thus, if $b_j$ is an integer, the curve $b_jy_j$ on the surface
of a handlebody is not a loop unless every nonzero $b_j=\pm 1$.
It follows that the
right-hand side can be presented as a loop only if every nonzero $b_j=\pm 1$.

4) In the case 2) we had to multiply the equation by an integer $k$ in order
that the left-hand side could be presented as a loop. In order to see that
we get a loop in this case proceed as follows. Consider Figure 4 and notice
that we could add any number of handles to the torus in the center of Figure 4.
Thus, we have the generator $x_2$ and a number of generators $z_j$.
Take a loop where $x_2$ appears one time only and $z_j$
appears $c_j$ times where $c_j\in \Z$.
Cut the central torus in Fig. 4 along a loop parallel to $x_2$
and deform it into a strip $S^1\times [0,1]$.
Glue two copies of the strip together by identifying $S^1\times \{0\}$
with $S^1\times \{1\}$ of the second copy. Repeat this $k$
times and finally identify the sides $S^1\times \{0\}$
and $S^1\times \{1\}$. This yields a loop where $x_2$
appears $k$ times and $z_j$ appears $c_j$ times. This loop is mapped to a loop
and because of 3) every nonzero $b_j$ after multiplying by $k$ is $\pm 1$.
It follows that the integer $k$ is $\pm 1$.

5) As the left-hand side is not contractible in $H_{a,g}$,
the right-hand side has at least one coefficient $b_j$ which is not zero.

6) Loops map to loops in $\psi$, therefore the right-hand side is a loop
in $g_2(M_a)$. As it
is loop, every $y_{b,i}$ must have coefficient $\pm 1$ as there is
no term $x_{b,i}$. So

$l'=B_1{Y_b}^T=\sum_{1\le i\le g}b_iy_{b,i}$

where $b_i=\pm 1$ or zero.
This loop is contractible and bounds an embedded
disc in $H_{b,g}$.

We conclude that the preimage $g_1^{-1}(l)$ is
a loop in $\partial M^a$, homotopic to $g_1^{-1}(x_{a,1})$ in $M_a$
and bounds an embedded disc in $M_a$. $\eop$

\th Lemma 14. Let $M$ and $f$ be as in Lemma 10 and let $M$ have a Heegaard
split of genus $g$ induced by $f$. There is a Morse function $f'$ such that
$f'$ satisfies the conditions of Lemma 10 and the Heegaard split induced by
$f'$ has genus $g-1$.

\proof
Let $x_{a,1}$ be as in Lemma 13. By Lemma 13 there is a Morse fuction
$f$ satisfying Lemma 10 such that
$x_{a,1}$ bounds a disc in the handlebody $H_{b,g}$. We can assume that
the loop corresponding to the generator $x_{a,1}$ is created by the
addition of the first 1-handle in a critical point $p_1$.
On a regular Morse level $c$ where $c$ is only slightly higher than
$f(p_1)$ the manifold is divided into two parts $M^c$ and $M_c$. The
boundaries of $M^c$ and $M_c$ are toruses. The lower manifold $M^c$ is
a filled torus but we cannot know if the upper manifold $M_c$ is a filled
torus.
Fortunately, we do not need to
know that. We are only interested in the gleuing mapping between
these manifolds.
If $M_c$ is not a filled torus, we replace it by a filled torus.
Now we have two filled toruses glued together by a boundary map. Thus,
we have obtained a Heegaard split of genus 1, though not for the original
manifold. Every simply connected closed 3-manifold
which has a Heegaard split of genus 1 is homeomorphic to $S^3$. This means
that the modified $M_c$ and the original $M^c$ make $S^3$. That means that
the boundary map must be trivial: a noncontractible generator is mapped to a
contractible generator.
Let us now restore $M_a$. It was only temporarily replaced by a filled torus
in order to investigate the boundary mapping. The boundary mapping is of course
still the same. We know now that the boundary mapping is trivial.

We have the disc in $H_{b,g}$ that has $l$ as a boundary,
and a collar from $l$ to the level $c$. This collar does not meet any critical
points. We can make a similar move of a handle as is done in Figure 2
but this time we move a 2-handle down.

The collar can be added to the disc to make a bigger disc.
The disc contains a critical point $p_m$ of index 2.
We take a closed cylinder neighborhood of the disc.
Inside this neighborhood we make the inverse of the
modification given in Lemma 2.
This move lowers the level of the 2-handle. This lowering can be continued
until the 2-handle is moved to the level $c$. Thus, moving 2-handles down
is possible if there is an embedded collar that does not meet any
critical points.

We have on the level $c$ the 2-handle and on the level $f(p_1)$ the 1-handle.
The boundary mapping is trivial. These two handles cancel each other, i.e.,
the 2-handle fills the hole created by the 1-handle.
We can construct a new Morse function where these handles do not appear.
The new Morse function $f'$ satisfies Lemma 10
and the genus of $\partial {M^a}$ for $f'$ is $g-1$. $\eop$

The first generally accepted proof of the Poincar\'e conjecture was given
by Grigory Perelman in the year 2002. The proof applies advanced techniques.
We have obtained another proof by using elementary methods only:

\th Theorem 1. Every simply connected closed 3-manifold
is homeomorphic to the 3-sphere.

\proof
We may assume, that $M$ is differentiable since if the claim
holds for differentiable 3-manifolds it holds for all, see e.g. [1].
Assume, that $M$ is an exotic
homotopy sphere. Let $f:M\to \R$ be a Morse function as in Lemma 10.
Assume, that there is a minimum genus $g$ for the Heegaard split induced
by a Morse function satisfying Lemma 10. There is a lower bound to $g$
as there is no nontrivial Heegaard split of genus 1.
By Lemma 14 $M$ admits a Heegaard split as in Lemma 10 which has genus $g-1$.
This contradiction shows, that $M$ is $S^3$. $\eop$

redpajama_set_name: RedPajamaArXiv