Mixing
Submersion with contractible fibers
$begingroup$
Let $f: Mto N$ be a surjective submersion of manifolds with contractible fibers. Is it true that $M$ and $N$ are homotopy equivalent?
algebraic-topology manifolds homotopy-theory
asked Jan 23 at 12:53
No_wayNo_way
59118
$endgroup$
add a comment |
$begingroup$
Let $f: Mto N$ be a surjective submersion of manifolds with contractible fibers. Is it true that $M$ and $N$ are homotopy equivalent?
algebraic-topology manifolds homotopy-theory
asked Jan 23 at 12:53
No_wayNo_way
59118
$endgroup$
3
$begingroup$
If $f$ is proper then, as a surjective submersion, it is a fibration, and in this case it is true that it is a homotopy equivalence under the assumption on its fibres. On the other hand it is not true that every surjective submersion is proper, and there are surjective submersions which are not fibrations.
$endgroup$
– Tyrone
Jan 23 at 14:23
2
$begingroup$
@Tyrone: Surprisingly, in this setting $f$ is a homotopy equivalence. This is a rather nontrivial result due to Meigniez.
$endgroup$
– Moishe Kohan
Jan 25 at 17:09
add a comment |
$begingroup$
Let $f: Mto N$ be a surjective submersion of manifolds with contractible fibers. Is it true that $M$ and $N$ are homotopy equivalent?
algebraic-topology manifolds homotopy-theory
asked Jan 23 at 12:53
No_wayNo_way
59118
$endgroup$
Let $f: Mto N$ be a surjective submersion of manifolds with contractible fibers. Is it true that $M$ and $N$ are homotopy equivalent?
algebraic-topology manifolds homotopy-theory
algebraic-topology manifolds homotopy-theory
asked Jan 23 at 12:53
No_wayNo_way
59118
asked Jan 23 at 12:53
No_wayNo_way
59118
asked Jan 23 at 12:53
No_wayNo_way
59118
asked Jan 23 at 12:53
No_wayNo_way
59118
asked Jan 23 at 12:53
No_wayNo_way
59118
59118
3
$begingroup$
If $f$ is proper then, as a surjective submersion, it is a fibration, and in this case it is true that it is a homotopy equivalence under the assumption on its fibres. On the other hand it is not true that every surjective submersion is proper, and there are surjective submersions which are not fibrations.
$endgroup$
– Tyrone
Jan 23 at 14:23
2
$begingroup$
@Tyrone: Surprisingly, in this setting $f$ is a homotopy equivalence. This is a rather nontrivial result due to Meigniez.
$endgroup$
– Moishe Kohan
Jan 25 at 17:09
add a comment |
3
$begingroup$
If $f$ is proper then, as a surjective submersion, it is a fibration, and in this case it is true that it is a homotopy equivalence under the assumption on its fibres. On the other hand it is not true that every surjective submersion is proper, and there are surjective submersions which are not fibrations.
$endgroup$
– Tyrone
Jan 23 at 14:23
2
$begingroup$
@Tyrone: Surprisingly, in this setting $f$ is a homotopy equivalence. This is a rather nontrivial result due to Meigniez.
$endgroup$
– Moishe Kohan
Jan 25 at 17:09
3
3
$begingroup$
If $f$ is proper then, as a surjective submersion, it is a fibration, and in this case it is true that it is a homotopy equivalence under the assumption on its fibres. On the other hand it is not true that every surjective submersion is proper, and there are surjective submersions which are not fibrations.
$endgroup$
– Tyrone
Jan 23 at 14:23
$begingroup$
If $f$ is proper then, as a surjective submersion, it is a fibration, and in this case it is true that it is a homotopy equivalence under the assumption on its fibres. On the other hand it is not true that every surjective submersion is proper, and there are surjective submersions which are not fibrations.
$endgroup$
– Tyrone
Jan 23 at 14:23
2
2
$begingroup$
@Tyrone: Surprisingly, in this setting $f$ is a homotopy equivalence. This is a rather nontrivial result due to Meigniez.
$endgroup$
– Moishe Kohan
Jan 25 at 17:09
$begingroup$
@Tyrone: Surprisingly, in this setting $f$ is a homotopy equivalence. This is a rather nontrivial result due to Meigniez.
$endgroup$
– Moishe Kohan
Jan 25 at 17:09
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Surprisingly, your question has positive answer:
Theorem 1. Suppose that $f: Mto N$ is a surjective submersion with contractible fibers. Then $f$ is a fibration (in the sense of Serre). In particular, $f$ is a homotopy-equivalence.
See Corollary 13 in
G. Meigniez, Submersions, fibrations and bundles. Trans. Amer. Math. Soc. 354 (2002), no. 9, 3771–3787.
In fact, this result is proven under a weaker assumption, $f$ is only required to be a "homotopy submersion".
Remark. In the same paper it is proven (Corollary 31)
Theorem 2. Suppose that $f: Mto N$ is a surjective submersion with fibers diffeomorphic to $R^p$ for some $p$. Then $f$ is a locally trivial (in $C^infty$ category) fiber bundle.
However, you do not need this stronger conclusion.
answered Jan 25 at 17:08
Moishe KohanMoishe Kohan
47.9k344110
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3084432%2fsubmersion-with-contractible-fibers%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Surprisingly, your question has positive answer:
Theorem 1. Suppose that $f: Mto N$ is a surjective submersion with contractible fibers. Then $f$ is a fibration (in the sense of Serre). In particular, $f$ is a homotopy-equivalence.
See Corollary 13 in
G. Meigniez, Submersions, fibrations and bundles. Trans. Amer. Math. Soc. 354 (2002), no. 9, 3771–3787.
In fact, this result is proven under a weaker assumption, $f$ is only required to be a "homotopy submersion".
Remark. In the same paper it is proven (Corollary 31)
Theorem 2. Suppose that $f: Mto N$ is a surjective submersion with fibers diffeomorphic to $R^p$ for some $p$. Then $f$ is a locally trivial (in $C^infty$ category) fiber bundle.
However, you do not need this stronger conclusion.
answered Jan 25 at 17:08
Moishe KohanMoishe Kohan
47.9k344110
$endgroup$
add a comment |
$begingroup$
Surprisingly, your question has positive answer:
Theorem 1. Suppose that $f: Mto N$ is a surjective submersion with contractible fibers. Then $f$ is a fibration (in the sense of Serre). In particular, $f$ is a homotopy-equivalence.
See Corollary 13 in
G. Meigniez, Submersions, fibrations and bundles. Trans. Amer. Math. Soc. 354 (2002), no. 9, 3771–3787.
In fact, this result is proven under a weaker assumption, $f$ is only required to be a "homotopy submersion".
Remark. In the same paper it is proven (Corollary 31)
Theorem 2. Suppose that $f: Mto N$ is a surjective submersion with fibers diffeomorphic to $R^p$ for some $p$. Then $f$ is a locally trivial (in $C^infty$ category) fiber bundle.
However, you do not need this stronger conclusion.
answered Jan 25 at 17:08
Moishe KohanMoishe Kohan
47.9k344110
$endgroup$
add a comment |
$begingroup$
Surprisingly, your question has positive answer:
Theorem 1. Suppose that $f: Mto N$ is a surjective submersion with contractible fibers. Then $f$ is a fibration (in the sense of Serre). In particular, $f$ is a homotopy-equivalence.
See Corollary 13 in
G. Meigniez, Submersions, fibrations and bundles. Trans. Amer. Math. Soc. 354 (2002), no. 9, 3771–3787.
In fact, this result is proven under a weaker assumption, $f$ is only required to be a "homotopy submersion".
Remark. In the same paper it is proven (Corollary 31)
Theorem 2. Suppose that $f: Mto N$ is a surjective submersion with fibers diffeomorphic to $R^p$ for some $p$. Then $f$ is a locally trivial (in $C^infty$ category) fiber bundle.
However, you do not need this stronger conclusion.
answered Jan 25 at 17:08
Moishe KohanMoishe Kohan
47.9k344110
$endgroup$
Surprisingly, your question has positive answer:
Theorem 1. Suppose that $f: Mto N$ is a surjective submersion with contractible fibers. Then $f$ is a fibration (in the sense of Serre). In particular, $f$ is a homotopy-equivalence.
See Corollary 13 in
G. Meigniez, Submersions, fibrations and bundles. Trans. Amer. Math. Soc. 354 (2002), no. 9, 3771–3787.
In fact, this result is proven under a weaker assumption, $f$ is only required to be a "homotopy submersion".
Remark. In the same paper it is proven (Corollary 31)
Theorem 2. Suppose that $f: Mto N$ is a surjective submersion with fibers diffeomorphic to $R^p$ for some $p$. Then $f$ is a locally trivial (in $C^infty$ category) fiber bundle.
However, you do not need this stronger conclusion.
answered Jan 25 at 17:08
Moishe KohanMoishe Kohan
47.9k344110
edited Jan 25 at 17:17
answered Jan 25 at 17:08
Moishe KohanMoishe Kohan
47.9k344110
answered Jan 25 at 17:08
Moishe KohanMoishe Kohan
47.9k344110
answered Jan 25 at 17:08
Moishe KohanMoishe Kohan
47.9k344110
47.9k344110
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3084432%2fsubmersion-with-contractible-fibers%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
3
$begingroup$
If $f$ is proper then, as a surjective submersion, it is a fibration, and in this case it is true that it is a homotopy equivalence under the assumption on its fibres. On the other hand it is not true that every surjective submersion is proper, and there are surjective submersions which are not fibrations.
$endgroup$
– Tyrone
Jan 23 at 14:23
2
$begingroup$
@Tyrone: Surprisingly, in this setting $f$ is a homotopy equivalence. This is a rather nontrivial result due to Meigniez.
$endgroup$
– Moishe Kohan
Jan 25 at 17:09