Mixing
connected sum of surface
$begingroup$
When I read Massey's Algebraic Topology:An Introduction,page 9,he points out that the topological type of $S_1$#$S_2$(here $S_i$ is surface,# is connected sum,i.e.,cutting an open disc $D_i$ in each surface,and then gluing the boundary circle through a homeomorphism h) does not depend on the choice of discs $D_i$ or the choice of the homeomorphism h.
The independence of h is quite evident,but I don't know why the choice of discs is irrelevant.Is there any refference?
general-topology algebraic-topology
asked Feb 1 at 10:09
Daniel XuDaniel Xu
877
$endgroup$
add a comment |
$begingroup$
When I read Massey's Algebraic Topology:An Introduction,page 9,he points out that the topological type of $S_1$#$S_2$(here $S_i$ is surface,# is connected sum,i.e.,cutting an open disc $D_i$ in each surface,and then gluing the boundary circle through a homeomorphism h) does not depend on the choice of discs $D_i$ or the choice of the homeomorphism h.
The independence of h is quite evident,but I don't know why the choice of discs is irrelevant.Is there any refference?
general-topology algebraic-topology
asked Feb 1 at 10:09
Daniel XuDaniel Xu
877
$endgroup$
$begingroup$
(Obviously, the spaces $S_i$ should both be connected for this to be true)
$endgroup$
– Dan Rust
Feb 1 at 13:36
add a comment |
$begingroup$
When I read Massey's Algebraic Topology:An Introduction,page 9,he points out that the topological type of $S_1$#$S_2$(here $S_i$ is surface,# is connected sum,i.e.,cutting an open disc $D_i$ in each surface,and then gluing the boundary circle through a homeomorphism h) does not depend on the choice of discs $D_i$ or the choice of the homeomorphism h.
The independence of h is quite evident,but I don't know why the choice of discs is irrelevant.Is there any refference?
general-topology algebraic-topology
asked Feb 1 at 10:09
Daniel XuDaniel Xu
877
$endgroup$
When I read Massey's Algebraic Topology:An Introduction,page 9,he points out that the topological type of $S_1$#$S_2$(here $S_i$ is surface,# is connected sum,i.e.,cutting an open disc $D_i$ in each surface,and then gluing the boundary circle through a homeomorphism h) does not depend on the choice of discs $D_i$ or the choice of the homeomorphism h.
The independence of h is quite evident,but I don't know why the choice of discs is irrelevant.Is there any refference?
general-topology algebraic-topology
general-topology algebraic-topology
asked Feb 1 at 10:09
Daniel XuDaniel Xu
877
asked Feb 1 at 10:09
Daniel XuDaniel Xu
877
asked Feb 1 at 10:09
Daniel XuDaniel Xu
877
asked Feb 1 at 10:09
Daniel XuDaniel Xu
877
asked Feb 1 at 10:09
Daniel XuDaniel Xu
877
877
$begingroup$
(Obviously, the spaces $S_i$ should both be connected for this to be true)
$endgroup$
– Dan Rust
Feb 1 at 13:36
add a comment |
$begingroup$
(Obviously, the spaces $S_i$ should both be connected for this to be true)
$endgroup$
– Dan Rust
Feb 1 at 13:36
$begingroup$
(Obviously, the spaces $S_i$ should both be connected for this to be true)
$endgroup$
– Dan Rust
Feb 1 at 13:36
$begingroup$
(Obviously, the spaces $S_i$ should both be connected for this to be true)
$endgroup$
– Dan Rust
Feb 1 at 13:36
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Wikipedia has an extremely short "Disc Theorem" entry which references Palais' "Extending diffeomorphisms" paper. It says
Disc Theorem. For a smooth, connected, $n$-dimensional manifold $M$, if $f, f'colon D^n to M$ are two equi-oriented embeddings then they are ambiently isotopic.
This is one of the fundamental results in Differential Topology, and in particular it implies that connected sums are well-defined wrt choice of embeddings. There might be a simpler proof in $2$D, but this is the standard result which is typically cited for all dimensions.
(Here "ambiently isotopic" means there is an isotopy $H: Mtimes I to M$ which begins at the identity map and induces an isotopy between $f$ and $f'$; "equi-oriented" means that $f$ preserves orientation iff $f'$ does. In fact the proof of Palais' theorem shows a bit more: the ambient isotopy can be chosen to be fixed outside of a compact, contractible subspace containing the images of $f$ and $f'$.)
It's been a while since I looked at it, but a proof sketch sort of goes like this: First choose a small open tube around the unit interval $Isubset Usubset mathbb{R}^2$ and an embedding $gammacolon bar U to M$ where $gamma(0) = f(0)$, $gamma(1) = f'(0)$. Now pick a small disc $Dsubset U$ centred at $0$, and construct an ambient isotopy in the tube $Fcolon Utimes I to U$ which transports $D$ to a disc centred at $1$. Then you have to construct ambient isotopies $H_1, H_2$ on $M$ which shrink $f(D^n)$ and $f'(D^n)$ down to $gamma(D)$ and $gamma(F_1(D))$ respectively (I guess this step uses a linearization trick). Then these three isotopies are pieced together to give the result.
answered Feb 1 at 15:45
WilliamWilliam
3,2011228
$endgroup$
add a comment |
$begingroup$
If you consider two different locations/ sizes of the disc on one surface, you can define a homotopy on that surface that transforms one disc to the other. This works because all disks are homotopically trivial. Hence the topological type of the connected sum does not depend on the particular choice of discs.
answered Feb 1 at 10:26
quaraguequarague
692312
$endgroup$
$begingroup$
In order for the two connected sums to be diffeomorphic (or even homeomorphic) the embeddings need to be isotopic, and not just homotopic. I think the discs being contractible is not quite enough.
$endgroup$
– William
Feb 1 at 15:51
$begingroup$
@William I read 'topological type' to mean homotopic. IIRC for smooth closed surfaces being homotopic implies being diffeomorphic but proving diffeomorphic directly is probably better.
$endgroup$
– quarague
Feb 1 at 16:28
$begingroup$
You're right, in dimension 2 "homotopy equivalent" <=> "homeomorphic" <=> "diffeomorphic". However I think this follows from the classification theorem, which needs the fact that connected sum is well-defined.
$endgroup$
– William
Feb 1 at 16:45
1
$begingroup$
The connected sum operation is not a homotopical operation. A homotopy from one disc to another gives you nothing.
$endgroup$
– user98602
Feb 1 at 17:16
$begingroup$
To illustrate this point: $Bbb{CP}^2 # overline{Bbb{CP}}^2$ and $Bbb{CP}^2 # Bbb{CP}^2$ are not homotopy equivalent. You use the orientation of the disc in an essential way to determine the connected sum operation, and this is only sensible as you homotope through embeddings.
$endgroup$
– user98602
Feb 1 at 17:33
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%2f3096053%2fconnected-sum-of-surface%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Wikipedia has an extremely short "Disc Theorem" entry which references Palais' "Extending diffeomorphisms" paper. It says
Disc Theorem. For a smooth, connected, $n$-dimensional manifold $M$, if $f, f'colon D^n to M$ are two equi-oriented embeddings then they are ambiently isotopic.
This is one of the fundamental results in Differential Topology, and in particular it implies that connected sums are well-defined wrt choice of embeddings. There might be a simpler proof in $2$D, but this is the standard result which is typically cited for all dimensions.
(Here "ambiently isotopic" means there is an isotopy $H: Mtimes I to M$ which begins at the identity map and induces an isotopy between $f$ and $f'$; "equi-oriented" means that $f$ preserves orientation iff $f'$ does. In fact the proof of Palais' theorem shows a bit more: the ambient isotopy can be chosen to be fixed outside of a compact, contractible subspace containing the images of $f$ and $f'$.)
It's been a while since I looked at it, but a proof sketch sort of goes like this: First choose a small open tube around the unit interval $Isubset Usubset mathbb{R}^2$ and an embedding $gammacolon bar U to M$ where $gamma(0) = f(0)$, $gamma(1) = f'(0)$. Now pick a small disc $Dsubset U$ centred at $0$, and construct an ambient isotopy in the tube $Fcolon Utimes I to U$ which transports $D$ to a disc centred at $1$. Then you have to construct ambient isotopies $H_1, H_2$ on $M$ which shrink $f(D^n)$ and $f'(D^n)$ down to $gamma(D)$ and $gamma(F_1(D))$ respectively (I guess this step uses a linearization trick). Then these three isotopies are pieced together to give the result.
answered Feb 1 at 15:45
WilliamWilliam
3,2011228
$endgroup$
add a comment |
$begingroup$
Wikipedia has an extremely short "Disc Theorem" entry which references Palais' "Extending diffeomorphisms" paper. It says
Disc Theorem. For a smooth, connected, $n$-dimensional manifold $M$, if $f, f'colon D^n to M$ are two equi-oriented embeddings then they are ambiently isotopic.
This is one of the fundamental results in Differential Topology, and in particular it implies that connected sums are well-defined wrt choice of embeddings. There might be a simpler proof in $2$D, but this is the standard result which is typically cited for all dimensions.
(Here "ambiently isotopic" means there is an isotopy $H: Mtimes I to M$ which begins at the identity map and induces an isotopy between $f$ and $f'$; "equi-oriented" means that $f$ preserves orientation iff $f'$ does. In fact the proof of Palais' theorem shows a bit more: the ambient isotopy can be chosen to be fixed outside of a compact, contractible subspace containing the images of $f$ and $f'$.)
It's been a while since I looked at it, but a proof sketch sort of goes like this: First choose a small open tube around the unit interval $Isubset Usubset mathbb{R}^2$ and an embedding $gammacolon bar U to M$ where $gamma(0) = f(0)$, $gamma(1) = f'(0)$. Now pick a small disc $Dsubset U$ centred at $0$, and construct an ambient isotopy in the tube $Fcolon Utimes I to U$ which transports $D$ to a disc centred at $1$. Then you have to construct ambient isotopies $H_1, H_2$ on $M$ which shrink $f(D^n)$ and $f'(D^n)$ down to $gamma(D)$ and $gamma(F_1(D))$ respectively (I guess this step uses a linearization trick). Then these three isotopies are pieced together to give the result.
answered Feb 1 at 15:45
WilliamWilliam
3,2011228
$endgroup$
add a comment |
$begingroup$
Wikipedia has an extremely short "Disc Theorem" entry which references Palais' "Extending diffeomorphisms" paper. It says
Disc Theorem. For a smooth, connected, $n$-dimensional manifold $M$, if $f, f'colon D^n to M$ are two equi-oriented embeddings then they are ambiently isotopic.
This is one of the fundamental results in Differential Topology, and in particular it implies that connected sums are well-defined wrt choice of embeddings. There might be a simpler proof in $2$D, but this is the standard result which is typically cited for all dimensions.
(Here "ambiently isotopic" means there is an isotopy $H: Mtimes I to M$ which begins at the identity map and induces an isotopy between $f$ and $f'$; "equi-oriented" means that $f$ preserves orientation iff $f'$ does. In fact the proof of Palais' theorem shows a bit more: the ambient isotopy can be chosen to be fixed outside of a compact, contractible subspace containing the images of $f$ and $f'$.)
It's been a while since I looked at it, but a proof sketch sort of goes like this: First choose a small open tube around the unit interval $Isubset Usubset mathbb{R}^2$ and an embedding $gammacolon bar U to M$ where $gamma(0) = f(0)$, $gamma(1) = f'(0)$. Now pick a small disc $Dsubset U$ centred at $0$, and construct an ambient isotopy in the tube $Fcolon Utimes I to U$ which transports $D$ to a disc centred at $1$. Then you have to construct ambient isotopies $H_1, H_2$ on $M$ which shrink $f(D^n)$ and $f'(D^n)$ down to $gamma(D)$ and $gamma(F_1(D))$ respectively (I guess this step uses a linearization trick). Then these three isotopies are pieced together to give the result.
answered Feb 1 at 15:45
WilliamWilliam
3,2011228
$endgroup$
Wikipedia has an extremely short "Disc Theorem" entry which references Palais' "Extending diffeomorphisms" paper. It says
Disc Theorem. For a smooth, connected, $n$-dimensional manifold $M$, if $f, f'colon D^n to M$ are two equi-oriented embeddings then they are ambiently isotopic.
This is one of the fundamental results in Differential Topology, and in particular it implies that connected sums are well-defined wrt choice of embeddings. There might be a simpler proof in $2$D, but this is the standard result which is typically cited for all dimensions.
(Here "ambiently isotopic" means there is an isotopy $H: Mtimes I to M$ which begins at the identity map and induces an isotopy between $f$ and $f'$; "equi-oriented" means that $f$ preserves orientation iff $f'$ does. In fact the proof of Palais' theorem shows a bit more: the ambient isotopy can be chosen to be fixed outside of a compact, contractible subspace containing the images of $f$ and $f'$.)
It's been a while since I looked at it, but a proof sketch sort of goes like this: First choose a small open tube around the unit interval $Isubset Usubset mathbb{R}^2$ and an embedding $gammacolon bar U to M$ where $gamma(0) = f(0)$, $gamma(1) = f'(0)$. Now pick a small disc $Dsubset U$ centred at $0$, and construct an ambient isotopy in the tube $Fcolon Utimes I to U$ which transports $D$ to a disc centred at $1$. Then you have to construct ambient isotopies $H_1, H_2$ on $M$ which shrink $f(D^n)$ and $f'(D^n)$ down to $gamma(D)$ and $gamma(F_1(D))$ respectively (I guess this step uses a linearization trick). Then these three isotopies are pieced together to give the result.
answered Feb 1 at 15:45
WilliamWilliam
3,2011228
edited Feb 1 at 18:30
answered Feb 1 at 15:45
WilliamWilliam
3,2011228
answered Feb 1 at 15:45
WilliamWilliam
3,2011228
answered Feb 1 at 15:45
WilliamWilliam
3,2011228
3,2011228
add a comment |
add a comment |
$begingroup$
If you consider two different locations/ sizes of the disc on one surface, you can define a homotopy on that surface that transforms one disc to the other. This works because all disks are homotopically trivial. Hence the topological type of the connected sum does not depend on the particular choice of discs.
answered Feb 1 at 10:26
quaraguequarague
692312
$endgroup$
$begingroup$
In order for the two connected sums to be diffeomorphic (or even homeomorphic) the embeddings need to be isotopic, and not just homotopic. I think the discs being contractible is not quite enough.
$endgroup$
– William
Feb 1 at 15:51
$begingroup$
@William I read 'topological type' to mean homotopic. IIRC for smooth closed surfaces being homotopic implies being diffeomorphic but proving diffeomorphic directly is probably better.
$endgroup$
– quarague
Feb 1 at 16:28
$begingroup$
You're right, in dimension 2 "homotopy equivalent" <=> "homeomorphic" <=> "diffeomorphic". However I think this follows from the classification theorem, which needs the fact that connected sum is well-defined.
$endgroup$
– William
Feb 1 at 16:45
1
$begingroup$
The connected sum operation is not a homotopical operation. A homotopy from one disc to another gives you nothing.
$endgroup$
– user98602
Feb 1 at 17:16
$begingroup$
To illustrate this point: $Bbb{CP}^2 # overline{Bbb{CP}}^2$ and $Bbb{CP}^2 # Bbb{CP}^2$ are not homotopy equivalent. You use the orientation of the disc in an essential way to determine the connected sum operation, and this is only sensible as you homotope through embeddings.
$endgroup$
– user98602
Feb 1 at 17:33
add a comment |
$begingroup$
If you consider two different locations/ sizes of the disc on one surface, you can define a homotopy on that surface that transforms one disc to the other. This works because all disks are homotopically trivial. Hence the topological type of the connected sum does not depend on the particular choice of discs.
answered Feb 1 at 10:26
quaraguequarague
692312
$endgroup$
$begingroup$
In order for the two connected sums to be diffeomorphic (or even homeomorphic) the embeddings need to be isotopic, and not just homotopic. I think the discs being contractible is not quite enough.
$endgroup$
– William
Feb 1 at 15:51
$begingroup$
@William I read 'topological type' to mean homotopic. IIRC for smooth closed surfaces being homotopic implies being diffeomorphic but proving diffeomorphic directly is probably better.
$endgroup$
– quarague
Feb 1 at 16:28
$begingroup$
You're right, in dimension 2 "homotopy equivalent" <=> "homeomorphic" <=> "diffeomorphic". However I think this follows from the classification theorem, which needs the fact that connected sum is well-defined.
$endgroup$
– William
Feb 1 at 16:45
1
$begingroup$
The connected sum operation is not a homotopical operation. A homotopy from one disc to another gives you nothing.
$endgroup$
– user98602
Feb 1 at 17:16
$begingroup$
To illustrate this point: $Bbb{CP}^2 # overline{Bbb{CP}}^2$ and $Bbb{CP}^2 # Bbb{CP}^2$ are not homotopy equivalent. You use the orientation of the disc in an essential way to determine the connected sum operation, and this is only sensible as you homotope through embeddings.
$endgroup$
– user98602
Feb 1 at 17:33
add a comment |
$begingroup$
If you consider two different locations/ sizes of the disc on one surface, you can define a homotopy on that surface that transforms one disc to the other. This works because all disks are homotopically trivial. Hence the topological type of the connected sum does not depend on the particular choice of discs.
answered Feb 1 at 10:26
quaraguequarague
692312
$endgroup$
If you consider two different locations/ sizes of the disc on one surface, you can define a homotopy on that surface that transforms one disc to the other. This works because all disks are homotopically trivial. Hence the topological type of the connected sum does not depend on the particular choice of discs.
answered Feb 1 at 10:26
quaraguequarague
692312
answered Feb 1 at 10:26
quaraguequarague
692312
answered Feb 1 at 10:26
quaraguequarague
692312
answered Feb 1 at 10:26
quaraguequarague
692312
692312
$begingroup$
In order for the two connected sums to be diffeomorphic (or even homeomorphic) the embeddings need to be isotopic, and not just homotopic. I think the discs being contractible is not quite enough.
$endgroup$
– William
Feb 1 at 15:51
$begingroup$
@William I read 'topological type' to mean homotopic. IIRC for smooth closed surfaces being homotopic implies being diffeomorphic but proving diffeomorphic directly is probably better.
$endgroup$
– quarague
Feb 1 at 16:28
$begingroup$
You're right, in dimension 2 "homotopy equivalent" <=> "homeomorphic" <=> "diffeomorphic". However I think this follows from the classification theorem, which needs the fact that connected sum is well-defined.
$endgroup$
– William
Feb 1 at 16:45
1
$begingroup$
The connected sum operation is not a homotopical operation. A homotopy from one disc to another gives you nothing.
$endgroup$
– user98602
Feb 1 at 17:16
$begingroup$
To illustrate this point: $Bbb{CP}^2 # overline{Bbb{CP}}^2$ and $Bbb{CP}^2 # Bbb{CP}^2$ are not homotopy equivalent. You use the orientation of the disc in an essential way to determine the connected sum operation, and this is only sensible as you homotope through embeddings.
$endgroup$
– user98602
Feb 1 at 17:33
add a comment |
$begingroup$
In order for the two connected sums to be diffeomorphic (or even homeomorphic) the embeddings need to be isotopic, and not just homotopic. I think the discs being contractible is not quite enough.
$endgroup$
– William
Feb 1 at 15:51
$begingroup$
@William I read 'topological type' to mean homotopic. IIRC for smooth closed surfaces being homotopic implies being diffeomorphic but proving diffeomorphic directly is probably better.
$endgroup$
– quarague
Feb 1 at 16:28
$begingroup$
You're right, in dimension 2 "homotopy equivalent" <=> "homeomorphic" <=> "diffeomorphic". However I think this follows from the classification theorem, which needs the fact that connected sum is well-defined.
$endgroup$
– William
Feb 1 at 16:45
1
$begingroup$
The connected sum operation is not a homotopical operation. A homotopy from one disc to another gives you nothing.
$endgroup$
– user98602
Feb 1 at 17:16
$begingroup$
To illustrate this point: $Bbb{CP}^2 # overline{Bbb{CP}}^2$ and $Bbb{CP}^2 # Bbb{CP}^2$ are not homotopy equivalent. You use the orientation of the disc in an essential way to determine the connected sum operation, and this is only sensible as you homotope through embeddings.
$endgroup$
– user98602
Feb 1 at 17:33
$begingroup$
In order for the two connected sums to be diffeomorphic (or even homeomorphic) the embeddings need to be isotopic, and not just homotopic. I think the discs being contractible is not quite enough.
$endgroup$
– William
Feb 1 at 15:51
$begingroup$
In order for the two connected sums to be diffeomorphic (or even homeomorphic) the embeddings need to be isotopic, and not just homotopic. I think the discs being contractible is not quite enough.
$endgroup$
– William
Feb 1 at 15:51
$begingroup$
@William I read 'topological type' to mean homotopic. IIRC for smooth closed surfaces being homotopic implies being diffeomorphic but proving diffeomorphic directly is probably better.
$endgroup$
– quarague
Feb 1 at 16:28
$begingroup$
@William I read 'topological type' to mean homotopic. IIRC for smooth closed surfaces being homotopic implies being diffeomorphic but proving diffeomorphic directly is probably better.
$endgroup$
– quarague
Feb 1 at 16:28
$begingroup$
You're right, in dimension 2 "homotopy equivalent" <=> "homeomorphic" <=> "diffeomorphic". However I think this follows from the classification theorem, which needs the fact that connected sum is well-defined.
$endgroup$
– William
Feb 1 at 16:45
$begingroup$
You're right, in dimension 2 "homotopy equivalent" <=> "homeomorphic" <=> "diffeomorphic". However I think this follows from the classification theorem, which needs the fact that connected sum is well-defined.
$endgroup$
– William
Feb 1 at 16:45
1
1
$begingroup$
The connected sum operation is not a homotopical operation. A homotopy from one disc to another gives you nothing.
$endgroup$
– user98602
Feb 1 at 17:16
$begingroup$
The connected sum operation is not a homotopical operation. A homotopy from one disc to another gives you nothing.
$endgroup$
– user98602
Feb 1 at 17:16
$begingroup$
To illustrate this point: $Bbb{CP}^2 # overline{Bbb{CP}}^2$ and $Bbb{CP}^2 # Bbb{CP}^2$ are not homotopy equivalent. You use the orientation of the disc in an essential way to determine the connected sum operation, and this is only sensible as you homotope through embeddings.
$endgroup$
– user98602
Feb 1 at 17:33
$begingroup$
To illustrate this point: $Bbb{CP}^2 # overline{Bbb{CP}}^2$ and $Bbb{CP}^2 # Bbb{CP}^2$ are not homotopy equivalent. You use the orientation of the disc in an essential way to determine the connected sum operation, and this is only sensible as you homotope through embeddings.
$endgroup$
– user98602
Feb 1 at 17:33
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%2f3096053%2fconnected-sum-of-surface%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
$begingroup$
(Obviously, the spaces $S_i$ should both be connected for this to be true)
$endgroup$
– Dan Rust
Feb 1 at 13:36