Mixing
Homotopy classes and CW approximation
$begingroup$
Suppose $X$ is a connected CW complex with $dim X = n$ and $Y$ is CW-complex which is $n$-connected. Is it true that the set of homotopy classes of maps $f colon X to Y$ is trivial, i.e. $[X,Y]={[text{constant map}]}$.
A second related question. Suppose $X$ is as above and $Y$ a CW complex. Can I say that
$[X,Y] cong [X,Y_n]$ (bijection of sets) where $Y_n$ is the $n$-skeleton of $Y$. If not, what goes wrong?
algebraic-topology homotopy-theory
asked Jan 23 at 13:37
Wilhelm L.Wilhelm L.
41117
$endgroup$
|
show 1 more comment
$begingroup$
Suppose $X$ is a connected CW complex with $dim X = n$ and $Y$ is CW-complex which is $n$-connected. Is it true that the set of homotopy classes of maps $f colon X to Y$ is trivial, i.e. $[X,Y]={[text{constant map}]}$.
A second related question. Suppose $X$ is as above and $Y$ a CW complex. Can I say that
$[X,Y] cong [X,Y_n]$ (bijection of sets) where $Y_n$ is the $n$-skeleton of $Y$. If not, what goes wrong?
algebraic-topology homotopy-theory
asked Jan 23 at 13:37
Wilhelm L.Wilhelm L.
41117
$endgroup$
3
$begingroup$
For the first question, yes by the cellular approximation theorem en.m.wikipedia.org/wiki/Cellular_approximation_theorem. For the second question, no; consider $S^1$ mapping to the disk $D^2$ obtained by gluing a 2-cell to the circle by the identity. It is true that $[X,Y] = [X,Y_{n+1}]$. Think of the $+1$ as accounting for homotopies (crossing with a 1-dimensional interval).
$endgroup$
– Aleksandar Milivojevic
Jan 23 at 14:23
1
$begingroup$
@AleksandarMilivojevic Why not an official answer?
$endgroup$
– Paul Frost
Jan 23 at 15:19
$begingroup$
I second @PaulFrost comment
$endgroup$
– Wilhelm L.
Jan 23 at 15:20
$begingroup$
@AleksandarMilivojevic : I don't understand what the first question has to do with cellular approximation... To me it's simply by induction on the skeleton of $X$ that you see that at each level the map is nullhomotopic by $n$-connectedness, and then it is globally nullhomotopic as $X$ has dimension $n$. In particular I don't see how you can use cellular approximation, care to explain ?
$endgroup$
– Max
Jan 23 at 15:52
$begingroup$
@Max Since $Y$ is $n$-connected, you can give it a CW decomposition with one 0-cell and no other cells in dimension $leq n$. Now any map from $X$ is, by the theorem, homotopic to a map to the $n$-skeleton of $Y$, which is now just a point. I agree that invoking the theorem was unnecessary, as you point out, but it ties the two questions together.
$endgroup$
– Aleksandar Milivojevic
Jan 23 at 15:56
|
show 1 more comment
$begingroup$
Suppose $X$ is a connected CW complex with $dim X = n$ and $Y$ is CW-complex which is $n$-connected. Is it true that the set of homotopy classes of maps $f colon X to Y$ is trivial, i.e. $[X,Y]={[text{constant map}]}$.
A second related question. Suppose $X$ is as above and $Y$ a CW complex. Can I say that
$[X,Y] cong [X,Y_n]$ (bijection of sets) where $Y_n$ is the $n$-skeleton of $Y$. If not, what goes wrong?
algebraic-topology homotopy-theory
asked Jan 23 at 13:37
Wilhelm L.Wilhelm L.
41117
$endgroup$
Suppose $X$ is a connected CW complex with $dim X = n$ and $Y$ is CW-complex which is $n$-connected. Is it true that the set of homotopy classes of maps $f colon X to Y$ is trivial, i.e. $[X,Y]={[text{constant map}]}$.
A second related question. Suppose $X$ is as above and $Y$ a CW complex. Can I say that
$[X,Y] cong [X,Y_n]$ (bijection of sets) where $Y_n$ is the $n$-skeleton of $Y$. If not, what goes wrong?
algebraic-topology homotopy-theory
algebraic-topology homotopy-theory
asked Jan 23 at 13:37
Wilhelm L.Wilhelm L.
41117
asked Jan 23 at 13:37
Wilhelm L.Wilhelm L.
41117
asked Jan 23 at 13:37
Wilhelm L.Wilhelm L.
41117
asked Jan 23 at 13:37
Wilhelm L.Wilhelm L.
41117
asked Jan 23 at 13:37
Wilhelm L.Wilhelm L.
41117
41117
3
$begingroup$
For the first question, yes by the cellular approximation theorem en.m.wikipedia.org/wiki/Cellular_approximation_theorem. For the second question, no; consider $S^1$ mapping to the disk $D^2$ obtained by gluing a 2-cell to the circle by the identity. It is true that $[X,Y] = [X,Y_{n+1}]$. Think of the $+1$ as accounting for homotopies (crossing with a 1-dimensional interval).
$endgroup$
– Aleksandar Milivojevic
Jan 23 at 14:23
1
$begingroup$
@AleksandarMilivojevic Why not an official answer?
$endgroup$
– Paul Frost
Jan 23 at 15:19
$begingroup$
I second @PaulFrost comment
$endgroup$
– Wilhelm L.
Jan 23 at 15:20
$begingroup$
@AleksandarMilivojevic : I don't understand what the first question has to do with cellular approximation... To me it's simply by induction on the skeleton of $X$ that you see that at each level the map is nullhomotopic by $n$-connectedness, and then it is globally nullhomotopic as $X$ has dimension $n$. In particular I don't see how you can use cellular approximation, care to explain ?
$endgroup$
– Max
Jan 23 at 15:52
$begingroup$
@Max Since $Y$ is $n$-connected, you can give it a CW decomposition with one 0-cell and no other cells in dimension $leq n$. Now any map from $X$ is, by the theorem, homotopic to a map to the $n$-skeleton of $Y$, which is now just a point. I agree that invoking the theorem was unnecessary, as you point out, but it ties the two questions together.
$endgroup$
– Aleksandar Milivojevic
Jan 23 at 15:56
|
show 1 more comment
3
$begingroup$
For the first question, yes by the cellular approximation theorem en.m.wikipedia.org/wiki/Cellular_approximation_theorem. For the second question, no; consider $S^1$ mapping to the disk $D^2$ obtained by gluing a 2-cell to the circle by the identity. It is true that $[X,Y] = [X,Y_{n+1}]$. Think of the $+1$ as accounting for homotopies (crossing with a 1-dimensional interval).
$endgroup$
– Aleksandar Milivojevic
Jan 23 at 14:23
1
$begingroup$
@AleksandarMilivojevic Why not an official answer?
$endgroup$
– Paul Frost
Jan 23 at 15:19
$begingroup$
I second @PaulFrost comment
$endgroup$
– Wilhelm L.
Jan 23 at 15:20
$begingroup$
@AleksandarMilivojevic : I don't understand what the first question has to do with cellular approximation... To me it's simply by induction on the skeleton of $X$ that you see that at each level the map is nullhomotopic by $n$-connectedness, and then it is globally nullhomotopic as $X$ has dimension $n$. In particular I don't see how you can use cellular approximation, care to explain ?
$endgroup$
– Max
Jan 23 at 15:52
$begingroup$
@Max Since $Y$ is $n$-connected, you can give it a CW decomposition with one 0-cell and no other cells in dimension $leq n$. Now any map from $X$ is, by the theorem, homotopic to a map to the $n$-skeleton of $Y$, which is now just a point. I agree that invoking the theorem was unnecessary, as you point out, but it ties the two questions together.
$endgroup$
– Aleksandar Milivojevic
Jan 23 at 15:56
3
3
$begingroup$
For the first question, yes by the cellular approximation theorem en.m.wikipedia.org/wiki/Cellular_approximation_theorem. For the second question, no; consider $S^1$ mapping to the disk $D^2$ obtained by gluing a 2-cell to the circle by the identity. It is true that $[X,Y] = [X,Y_{n+1}]$. Think of the $+1$ as accounting for homotopies (crossing with a 1-dimensional interval).
$endgroup$
– Aleksandar Milivojevic
Jan 23 at 14:23
$begingroup$
For the first question, yes by the cellular approximation theorem en.m.wikipedia.org/wiki/Cellular_approximation_theorem. For the second question, no; consider $S^1$ mapping to the disk $D^2$ obtained by gluing a 2-cell to the circle by the identity. It is true that $[X,Y] = [X,Y_{n+1}]$. Think of the $+1$ as accounting for homotopies (crossing with a 1-dimensional interval).
$endgroup$
– Aleksandar Milivojevic
Jan 23 at 14:23
1
1
$begingroup$
@AleksandarMilivojevic Why not an official answer?
$endgroup$
– Paul Frost
Jan 23 at 15:19
$begingroup$
@AleksandarMilivojevic Why not an official answer?
$endgroup$
– Paul Frost
Jan 23 at 15:19
$begingroup$
I second @PaulFrost comment
$endgroup$
– Wilhelm L.
Jan 23 at 15:20
$begingroup$
I second @PaulFrost comment
$endgroup$
– Wilhelm L.
Jan 23 at 15:20
$begingroup$
@AleksandarMilivojevic : I don't understand what the first question has to do with cellular approximation... To me it's simply by induction on the skeleton of $X$ that you see that at each level the map is nullhomotopic by $n$-connectedness, and then it is globally nullhomotopic as $X$ has dimension $n$. In particular I don't see how you can use cellular approximation, care to explain ?
$endgroup$
– Max
Jan 23 at 15:52
$begingroup$
@AleksandarMilivojevic : I don't understand what the first question has to do with cellular approximation... To me it's simply by induction on the skeleton of $X$ that you see that at each level the map is nullhomotopic by $n$-connectedness, and then it is globally nullhomotopic as $X$ has dimension $n$. In particular I don't see how you can use cellular approximation, care to explain ?
$endgroup$
– Max
Jan 23 at 15:52
$begingroup$
@Max Since $Y$ is $n$-connected, you can give it a CW decomposition with one 0-cell and no other cells in dimension $leq n$. Now any map from $X$ is, by the theorem, homotopic to a map to the $n$-skeleton of $Y$, which is now just a point. I agree that invoking the theorem was unnecessary, as you point out, but it ties the two questions together.
$endgroup$
– Aleksandar Milivojevic
Jan 23 at 15:56
$begingroup$
@Max Since $Y$ is $n$-connected, you can give it a CW decomposition with one 0-cell and no other cells in dimension $leq n$. Now any map from $X$ is, by the theorem, homotopic to a map to the $n$-skeleton of $Y$, which is now just a point. I agree that invoking the theorem was unnecessary, as you point out, but it ties the two questions together.
$endgroup$
– Aleksandar Milivojevic
Jan 23 at 15:56
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
(converting above comment to answer; will provide more details if needed)
For the first question, yes by the cellular approximation theorem. For the second question, no; consider $S^1$ mapping to the disk $D^2$ (obtained by gluing a 2-cell to the circle by the identity on the boundary). It is true that $[X,Y]=[X,Y^{n+1}]$, also by cellular approximation. Think of the $+1$ as accounting for homotopies (crossing with a 1-dimensional interval).
edited Jan 23 at 15:42
Javi
2,8812829
answered Jan 23 at 15:33
Aleksandar MilivojevicAleksandar Milivojevic
402312
$endgroup$
1
$begingroup$
See this answer for details on the proof of $[X, Y] = [X, Y^{(n+1)}]$.
$endgroup$
– Michael Albanese
Jan 29 at 4:06
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%2f3084480%2fhomotopy-classes-and-cw-approximation%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
$begingroup$
(converting above comment to answer; will provide more details if needed)
For the first question, yes by the cellular approximation theorem. For the second question, no; consider $S^1$ mapping to the disk $D^2$ (obtained by gluing a 2-cell to the circle by the identity on the boundary). It is true that $[X,Y]=[X,Y^{n+1}]$, also by cellular approximation. Think of the $+1$ as accounting for homotopies (crossing with a 1-dimensional interval).
edited Jan 23 at 15:42
Javi
2,8812829
answered Jan 23 at 15:33
Aleksandar MilivojevicAleksandar Milivojevic
402312
$endgroup$
1
$begingroup$
See this answer for details on the proof of $[X, Y] = [X, Y^{(n+1)}]$.
$endgroup$
– Michael Albanese
Jan 29 at 4:06
add a comment |
$begingroup$
(converting above comment to answer; will provide more details if needed)
For the first question, yes by the cellular approximation theorem. For the second question, no; consider $S^1$ mapping to the disk $D^2$ (obtained by gluing a 2-cell to the circle by the identity on the boundary). It is true that $[X,Y]=[X,Y^{n+1}]$, also by cellular approximation. Think of the $+1$ as accounting for homotopies (crossing with a 1-dimensional interval).
edited Jan 23 at 15:42
Javi
2,8812829
answered Jan 23 at 15:33
Aleksandar MilivojevicAleksandar Milivojevic
402312
$endgroup$
1
$begingroup$
See this answer for details on the proof of $[X, Y] = [X, Y^{(n+1)}]$.
$endgroup$
– Michael Albanese
Jan 29 at 4:06
add a comment |
$begingroup$
(converting above comment to answer; will provide more details if needed)
For the first question, yes by the cellular approximation theorem. For the second question, no; consider $S^1$ mapping to the disk $D^2$ (obtained by gluing a 2-cell to the circle by the identity on the boundary). It is true that $[X,Y]=[X,Y^{n+1}]$, also by cellular approximation. Think of the $+1$ as accounting for homotopies (crossing with a 1-dimensional interval).
edited Jan 23 at 15:42
Javi
2,8812829
answered Jan 23 at 15:33
Aleksandar MilivojevicAleksandar Milivojevic
402312
$endgroup$
(converting above comment to answer; will provide more details if needed)
For the first question, yes by the cellular approximation theorem. For the second question, no; consider $S^1$ mapping to the disk $D^2$ (obtained by gluing a 2-cell to the circle by the identity on the boundary). It is true that $[X,Y]=[X,Y^{n+1}]$, also by cellular approximation. Think of the $+1$ as accounting for homotopies (crossing with a 1-dimensional interval).
edited Jan 23 at 15:42
Javi
2,8812829
answered Jan 23 at 15:33
Aleksandar MilivojevicAleksandar Milivojevic
402312
edited Jan 23 at 15:42
Javi
2,8812829
edited Jan 23 at 15:42
Javi
2,8812829
edited Jan 23 at 15:42
Javi
2,8812829
2,8812829
answered Jan 23 at 15:33
Aleksandar MilivojevicAleksandar Milivojevic
402312
answered Jan 23 at 15:33
Aleksandar MilivojevicAleksandar Milivojevic
402312
answered Jan 23 at 15:33
Aleksandar MilivojevicAleksandar Milivojevic
402312
402312
1
$begingroup$
See this answer for details on the proof of $[X, Y] = [X, Y^{(n+1)}]$.
$endgroup$
– Michael Albanese
Jan 29 at 4:06
add a comment |
1
$begingroup$
See this answer for details on the proof of $[X, Y] = [X, Y^{(n+1)}]$.
$endgroup$
– Michael Albanese
Jan 29 at 4:06
1
1
$begingroup$
See this answer for details on the proof of $[X, Y] = [X, Y^{(n+1)}]$.
$endgroup$
– Michael Albanese
Jan 29 at 4:06
$begingroup$
See this answer for details on the proof of $[X, Y] = [X, Y^{(n+1)}]$.
$endgroup$
– Michael Albanese
Jan 29 at 4:06
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%2f3084480%2fhomotopy-classes-and-cw-approximation%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
$begingroup$
For the first question, yes by the cellular approximation theorem en.m.wikipedia.org/wiki/Cellular_approximation_theorem. For the second question, no; consider $S^1$ mapping to the disk $D^2$ obtained by gluing a 2-cell to the circle by the identity. It is true that $[X,Y] = [X,Y_{n+1}]$. Think of the $+1$ as accounting for homotopies (crossing with a 1-dimensional interval).
$endgroup$
– Aleksandar Milivojevic
Jan 23 at 14:23
1
$begingroup$
@AleksandarMilivojevic Why not an official answer?
$endgroup$
– Paul Frost
Jan 23 at 15:19
$begingroup$
I second @PaulFrost comment
$endgroup$
– Wilhelm L.
Jan 23 at 15:20
$begingroup$
@AleksandarMilivojevic : I don't understand what the first question has to do with cellular approximation... To me it's simply by induction on the skeleton of $X$ that you see that at each level the map is nullhomotopic by $n$-connectedness, and then it is globally nullhomotopic as $X$ has dimension $n$. In particular I don't see how you can use cellular approximation, care to explain ?
$endgroup$
– Max
Jan 23 at 15:52
$begingroup$
@Max Since $Y$ is $n$-connected, you can give it a CW decomposition with one 0-cell and no other cells in dimension $leq n$. Now any map from $X$ is, by the theorem, homotopic to a map to the $n$-skeleton of $Y$, which is now just a point. I agree that invoking the theorem was unnecessary, as you point out, but it ties the two questions together.
$endgroup$
– Aleksandar Milivojevic
Jan 23 at 15:56