Mixing
Does maps between fundamental groups induces a continuous map between spaces?
$begingroup$
Main question is this: Suppos $M$ is a manifold and $G$ is a finite group. If there is a group homomorphism $phi:pi_1 Mto G$, is there a continuous map $f:Mto BG$, where $BG$ is a classifying space?
One may generalize this question to ask if the functor $pi_1$ is full, but I think this is too good to be true.
This question arises from the following context. When I was reading some article, I read that there is a bijection between the isomorphism class of principal $G$-bundles over $M$ and the homotopy class of maps from $M$ to $BG$. I wanted to verify this, so I found out that (1) if two maps $f,g:Mto BG$ are in the same homotopy class, $f_*, g_*:pi_1 Mto G$ are conjugate, (2) a map $Mto BG$ induces a principal $G$-bundle, which is a pull back of $EGto BG$, (3) if $f_*, g_*$ are in the same conjugacy class, the principal $G$-bundles induced by each of them are isomorphic to each other, where the isomorphism is $pmapsto hph^{-1}$ where $hf_*h^{-1}=g$, and (4) a principal $G$-bundle $Nto M$ induces a group homomorphism $pi_1(M)to G$, since every loops on $M$ induces a $G$-action on $M$. I can deduce that the homotopy class of maps from $M$ to $BG$ gives an isomorphism class of principal $G$-bundles over $M$, but I am stuck on the converse statement. I think it is enough to prove that $pi_1(M)to M$ gives a continuous map $Mto BG$, and I presume that there is something with classifying space.
Thanks in advance.
algebraic-topology covering-spaces principal-bundles
asked Feb 3 at 4:52
J1UJ1U
617310
$endgroup$
add a comment |
$begingroup$
Main question is this: Suppos $M$ is a manifold and $G$ is a finite group. If there is a group homomorphism $phi:pi_1 Mto G$, is there a continuous map $f:Mto BG$, where $BG$ is a classifying space?
One may generalize this question to ask if the functor $pi_1$ is full, but I think this is too good to be true.
This question arises from the following context. When I was reading some article, I read that there is a bijection between the isomorphism class of principal $G$-bundles over $M$ and the homotopy class of maps from $M$ to $BG$. I wanted to verify this, so I found out that (1) if two maps $f,g:Mto BG$ are in the same homotopy class, $f_*, g_*:pi_1 Mto G$ are conjugate, (2) a map $Mto BG$ induces a principal $G$-bundle, which is a pull back of $EGto BG$, (3) if $f_*, g_*$ are in the same conjugacy class, the principal $G$-bundles induced by each of them are isomorphic to each other, where the isomorphism is $pmapsto hph^{-1}$ where $hf_*h^{-1}=g$, and (4) a principal $G$-bundle $Nto M$ induces a group homomorphism $pi_1(M)to G$, since every loops on $M$ induces a $G$-action on $M$. I can deduce that the homotopy class of maps from $M$ to $BG$ gives an isomorphism class of principal $G$-bundles over $M$, but I am stuck on the converse statement. I think it is enough to prove that $pi_1(M)to M$ gives a continuous map $Mto BG$, and I presume that there is something with classifying space.
Thanks in advance.
algebraic-topology covering-spaces principal-bundles
asked Feb 3 at 4:52
J1UJ1U
617310
$endgroup$
add a comment |
$begingroup$
Main question is this: Suppos $M$ is a manifold and $G$ is a finite group. If there is a group homomorphism $phi:pi_1 Mto G$, is there a continuous map $f:Mto BG$, where $BG$ is a classifying space?
One may generalize this question to ask if the functor $pi_1$ is full, but I think this is too good to be true.
This question arises from the following context. When I was reading some article, I read that there is a bijection between the isomorphism class of principal $G$-bundles over $M$ and the homotopy class of maps from $M$ to $BG$. I wanted to verify this, so I found out that (1) if two maps $f,g:Mto BG$ are in the same homotopy class, $f_*, g_*:pi_1 Mto G$ are conjugate, (2) a map $Mto BG$ induces a principal $G$-bundle, which is a pull back of $EGto BG$, (3) if $f_*, g_*$ are in the same conjugacy class, the principal $G$-bundles induced by each of them are isomorphic to each other, where the isomorphism is $pmapsto hph^{-1}$ where $hf_*h^{-1}=g$, and (4) a principal $G$-bundle $Nto M$ induces a group homomorphism $pi_1(M)to G$, since every loops on $M$ induces a $G$-action on $M$. I can deduce that the homotopy class of maps from $M$ to $BG$ gives an isomorphism class of principal $G$-bundles over $M$, but I am stuck on the converse statement. I think it is enough to prove that $pi_1(M)to M$ gives a continuous map $Mto BG$, and I presume that there is something with classifying space.
Thanks in advance.
algebraic-topology covering-spaces principal-bundles
asked Feb 3 at 4:52
J1UJ1U
617310
$endgroup$
Main question is this: Suppos $M$ is a manifold and $G$ is a finite group. If there is a group homomorphism $phi:pi_1 Mto G$, is there a continuous map $f:Mto BG$, where $BG$ is a classifying space?
One may generalize this question to ask if the functor $pi_1$ is full, but I think this is too good to be true.
This question arises from the following context. When I was reading some article, I read that there is a bijection between the isomorphism class of principal $G$-bundles over $M$ and the homotopy class of maps from $M$ to $BG$. I wanted to verify this, so I found out that (1) if two maps $f,g:Mto BG$ are in the same homotopy class, $f_*, g_*:pi_1 Mto G$ are conjugate, (2) a map $Mto BG$ induces a principal $G$-bundle, which is a pull back of $EGto BG$, (3) if $f_*, g_*$ are in the same conjugacy class, the principal $G$-bundles induced by each of them are isomorphic to each other, where the isomorphism is $pmapsto hph^{-1}$ where $hf_*h^{-1}=g$, and (4) a principal $G$-bundle $Nto M$ induces a group homomorphism $pi_1(M)to G$, since every loops on $M$ induces a $G$-action on $M$. I can deduce that the homotopy class of maps from $M$ to $BG$ gives an isomorphism class of principal $G$-bundles over $M$, but I am stuck on the converse statement. I think it is enough to prove that $pi_1(M)to M$ gives a continuous map $Mto BG$, and I presume that there is something with classifying space.
Thanks in advance.
algebraic-topology covering-spaces principal-bundles
algebraic-topology covering-spaces principal-bundles
asked Feb 3 at 4:52
J1UJ1U
617310
asked Feb 3 at 4:52
J1UJ1U
617310
edited Feb 3 at 4:58
J1U
asked Feb 3 at 4:52
J1UJ1U
617310
asked Feb 3 at 4:52
J1UJ1U
617310
asked Feb 3 at 4:52
J1UJ1U
617310
617310
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Yes, such a map $f$ does exist. In fact, it exists even when $M$ is a CW complex (and every smooth manifold has a CW complex structure) and when $G$ is an arbitrary group (not just a finite groups).
The proof is a straightforward obstruction theory argument. First pick a maximal tree in the 1-skeleton of $M$ and map that to the base point of $BG$. Next, each remanining 1-cell in $M$ represents an element of $pi_1(M)$, map that edge to a loop in $BG$ representing the $phi$ image of that element. Now extend over the 2-skeleton: the boundary of each 2-cell represents the trivial element of $pi_1(M)$, so the image loop represents the trivial element of $pi_1(BG)$, so you can extend the map continuously over that 2-cell. Now induct up the dimensions: assuming the map is defined on the $n$-skeleton for $n ge 2$, for each $n+1$ cell the restriction of the map to the boundary of that cell is homotopically trivial, because $pi_n(BG)$ is trivial, so the map can be extended continuously over the whole $n+1$ cell.
answered Feb 3 at 5:55
Lee MosherLee Mosher
52.4k33891
$endgroup$
1
$begingroup$
When you say "$G$ is an arbitrary group" it still has to be discrete, otherwise $pi_n(BG)$ isn't necessarily trivial for $n > 1$.
$endgroup$
– William
Feb 3 at 17:39
add a comment |
Your Answer
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%2f3098196%2fdoes-maps-between-fundamental-groups-induces-a-continuous-map-between-spaces%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
$begingroup$
Yes, such a map $f$ does exist. In fact, it exists even when $M$ is a CW complex (and every smooth manifold has a CW complex structure) and when $G$ is an arbitrary group (not just a finite groups).
The proof is a straightforward obstruction theory argument. First pick a maximal tree in the 1-skeleton of $M$ and map that to the base point of $BG$. Next, each remanining 1-cell in $M$ represents an element of $pi_1(M)$, map that edge to a loop in $BG$ representing the $phi$ image of that element. Now extend over the 2-skeleton: the boundary of each 2-cell represents the trivial element of $pi_1(M)$, so the image loop represents the trivial element of $pi_1(BG)$, so you can extend the map continuously over that 2-cell. Now induct up the dimensions: assuming the map is defined on the $n$-skeleton for $n ge 2$, for each $n+1$ cell the restriction of the map to the boundary of that cell is homotopically trivial, because $pi_n(BG)$ is trivial, so the map can be extended continuously over the whole $n+1$ cell.
answered Feb 3 at 5:55
Lee MosherLee Mosher
52.4k33891
$endgroup$
1
$begingroup$
When you say "$G$ is an arbitrary group" it still has to be discrete, otherwise $pi_n(BG)$ isn't necessarily trivial for $n > 1$.
$endgroup$
– William
Feb 3 at 17:39
add a comment |
$begingroup$
Yes, such a map $f$ does exist. In fact, it exists even when $M$ is a CW complex (and every smooth manifold has a CW complex structure) and when $G$ is an arbitrary group (not just a finite groups).
The proof is a straightforward obstruction theory argument. First pick a maximal tree in the 1-skeleton of $M$ and map that to the base point of $BG$. Next, each remanining 1-cell in $M$ represents an element of $pi_1(M)$, map that edge to a loop in $BG$ representing the $phi$ image of that element. Now extend over the 2-skeleton: the boundary of each 2-cell represents the trivial element of $pi_1(M)$, so the image loop represents the trivial element of $pi_1(BG)$, so you can extend the map continuously over that 2-cell. Now induct up the dimensions: assuming the map is defined on the $n$-skeleton for $n ge 2$, for each $n+1$ cell the restriction of the map to the boundary of that cell is homotopically trivial, because $pi_n(BG)$ is trivial, so the map can be extended continuously over the whole $n+1$ cell.
answered Feb 3 at 5:55
Lee MosherLee Mosher
52.4k33891
$endgroup$
1
$begingroup$
When you say "$G$ is an arbitrary group" it still has to be discrete, otherwise $pi_n(BG)$ isn't necessarily trivial for $n > 1$.
$endgroup$
– William
Feb 3 at 17:39
add a comment |
$begingroup$
Yes, such a map $f$ does exist. In fact, it exists even when $M$ is a CW complex (and every smooth manifold has a CW complex structure) and when $G$ is an arbitrary group (not just a finite groups).
The proof is a straightforward obstruction theory argument. First pick a maximal tree in the 1-skeleton of $M$ and map that to the base point of $BG$. Next, each remanining 1-cell in $M$ represents an element of $pi_1(M)$, map that edge to a loop in $BG$ representing the $phi$ image of that element. Now extend over the 2-skeleton: the boundary of each 2-cell represents the trivial element of $pi_1(M)$, so the image loop represents the trivial element of $pi_1(BG)$, so you can extend the map continuously over that 2-cell. Now induct up the dimensions: assuming the map is defined on the $n$-skeleton for $n ge 2$, for each $n+1$ cell the restriction of the map to the boundary of that cell is homotopically trivial, because $pi_n(BG)$ is trivial, so the map can be extended continuously over the whole $n+1$ cell.
answered Feb 3 at 5:55
Lee MosherLee Mosher
52.4k33891
$endgroup$
Yes, such a map $f$ does exist. In fact, it exists even when $M$ is a CW complex (and every smooth manifold has a CW complex structure) and when $G$ is an arbitrary group (not just a finite groups).
The proof is a straightforward obstruction theory argument. First pick a maximal tree in the 1-skeleton of $M$ and map that to the base point of $BG$. Next, each remanining 1-cell in $M$ represents an element of $pi_1(M)$, map that edge to a loop in $BG$ representing the $phi$ image of that element. Now extend over the 2-skeleton: the boundary of each 2-cell represents the trivial element of $pi_1(M)$, so the image loop represents the trivial element of $pi_1(BG)$, so you can extend the map continuously over that 2-cell. Now induct up the dimensions: assuming the map is defined on the $n$-skeleton for $n ge 2$, for each $n+1$ cell the restriction of the map to the boundary of that cell is homotopically trivial, because $pi_n(BG)$ is trivial, so the map can be extended continuously over the whole $n+1$ cell.
answered Feb 3 at 5:55
Lee MosherLee Mosher
52.4k33891
answered Feb 3 at 5:55
Lee MosherLee Mosher
52.4k33891
answered Feb 3 at 5:55
Lee MosherLee Mosher
52.4k33891
answered Feb 3 at 5:55
Lee MosherLee Mosher
52.4k33891
52.4k33891
1
$begingroup$
When you say "$G$ is an arbitrary group" it still has to be discrete, otherwise $pi_n(BG)$ isn't necessarily trivial for $n > 1$.
$endgroup$
– William
Feb 3 at 17:39
add a comment |
1
$begingroup$
When you say "$G$ is an arbitrary group" it still has to be discrete, otherwise $pi_n(BG)$ isn't necessarily trivial for $n > 1$.
$endgroup$
– William
Feb 3 at 17:39
1
1
$begingroup$
When you say "$G$ is an arbitrary group" it still has to be discrete, otherwise $pi_n(BG)$ isn't necessarily trivial for $n > 1$.
$endgroup$
– William
Feb 3 at 17:39
$begingroup$
When you say "$G$ is an arbitrary group" it still has to be discrete, otherwise $pi_n(BG)$ isn't necessarily trivial for $n > 1$.
$endgroup$
– William
Feb 3 at 17:39
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%2f3098196%2fdoes-maps-between-fundamental-groups-induces-a-continuous-map-between-spaces%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
