If $Hleq G$, then $BH to BG$ is a fiber bundle with fiber $G/H$

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Suppose $G$ is a topological group and $Hleq G$ is a closed subgroup. The inclusion $Hto G$ induces a map on classifying spaces $BHto BG$. I've seen in some sources that $BHto BG$ is actually a fiber bundle with fiber $G/H$, for example in the following notes PDF, Proposition 2.15 on the bottom of page 59.

I understand the reasoning to be as follows. It's a general fact that $EGto (EG)/H$ is a principal $H$-bundle, and since $EG$ is contractible, we know that $EGto (EG)/H$ is the universal $G$-bundle. Therefore $(EG)/H$ is homotopy equivalent to $BH$.

It's also a general fact that $(EG)/H to B$ is a fiber bundle with fiber $G/H$. However, I feel like we can't just replace $(EG)/H$ with $BH$ in this context, since they are only homotopy equivalent. And, in the PDF above, the author write $(EG)/H = BH$, which is perplexing.

Here's another confusion I have with this statement. Take $H$ to be the trivial group. Then $BHto BG$ should be a fiber bundle for all (reasonable) topological groups $G$. But this fishy to me. I find it hard to believe that, at the very least, $BH$ surjects onto $BG$ for any $G$.

How can I reconcile these issues?

general-topology algebraic-topology fiber-bundles principal-bundles classifying-spaces

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asked Jul 17 at 18:33

Ross

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up vote
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Suppose $G$ is a topological group and $Hleq G$ is a closed subgroup. The inclusion $Hto G$ induces a map on classifying spaces $BHto BG$. I've seen in some sources that $BHto BG$ is actually a fiber bundle with fiber $G/H$, for example in the following notes PDF, Proposition 2.15 on the bottom of page 59.

I understand the reasoning to be as follows. It's a general fact that $EGto (EG)/H$ is a principal $H$-bundle, and since $EG$ is contractible, we know that $EGto (EG)/H$ is the universal $G$-bundle. Therefore $(EG)/H$ is homotopy equivalent to $BH$.

It's also a general fact that $(EG)/H to B$ is a fiber bundle with fiber $G/H$. However, I feel like we can't just replace $(EG)/H$ with $BH$ in this context, since they are only homotopy equivalent. And, in the PDF above, the author write $(EG)/H = BH$, which is perplexing.

Here's another confusion I have with this statement. Take $H$ to be the trivial group. Then $BHto BG$ should be a fiber bundle for all (reasonable) topological groups $G$. But this fishy to me. I find it hard to believe that, at the very least, $BH$ surjects onto $BG$ for any $G$.

How can I reconcile these issues?

general-topology algebraic-topology fiber-bundles principal-bundles classifying-spaces

share|cite|improve this question

asked Jul 17 at 18:33

Ross

213

add a comment | 

up vote
2
down vote

favorite

up vote
2
down vote

favorite

Suppose $G$ is a topological group and $Hleq G$ is a closed subgroup. The inclusion $Hto G$ induces a map on classifying spaces $BHto BG$. I've seen in some sources that $BHto BG$ is actually a fiber bundle with fiber $G/H$, for example in the following notes PDF, Proposition 2.15 on the bottom of page 59.

I understand the reasoning to be as follows. It's a general fact that $EGto (EG)/H$ is a principal $H$-bundle, and since $EG$ is contractible, we know that $EGto (EG)/H$ is the universal $G$-bundle. Therefore $(EG)/H$ is homotopy equivalent to $BH$.

It's also a general fact that $(EG)/H to B$ is a fiber bundle with fiber $G/H$. However, I feel like we can't just replace $(EG)/H$ with $BH$ in this context, since they are only homotopy equivalent. And, in the PDF above, the author write $(EG)/H = BH$, which is perplexing.

Here's another confusion I have with this statement. Take $H$ to be the trivial group. Then $BHto BG$ should be a fiber bundle for all (reasonable) topological groups $G$. But this fishy to me. I find it hard to believe that, at the very least, $BH$ surjects onto $BG$ for any $G$.

How can I reconcile these issues?

general-topology algebraic-topology fiber-bundles principal-bundles classifying-spaces

share|cite|improve this question

asked Jul 17 at 18:33

Ross

213

Suppose $G$ is a topological group and $Hleq G$ is a closed subgroup. The inclusion $Hto G$ induces a map on classifying spaces $BHto BG$. I've seen in some sources that $BHto BG$ is actually a fiber bundle with fiber $G/H$, for example in the following notes PDF, Proposition 2.15 on the bottom of page 59.

I understand the reasoning to be as follows. It's a general fact that $EGto (EG)/H$ is a principal $H$-bundle, and since $EG$ is contractible, we know that $EGto (EG)/H$ is the universal $G$-bundle. Therefore $(EG)/H$ is homotopy equivalent to $BH$.

It's also a general fact that $(EG)/H to B$ is a fiber bundle with fiber $G/H$. However, I feel like we can't just replace $(EG)/H$ with $BH$ in this context, since they are only homotopy equivalent. And, in the PDF above, the author write $(EG)/H = BH$, which is perplexing.

Here's another confusion I have with this statement. Take $H$ to be the trivial group. Then $BHto BG$ should be a fiber bundle for all (reasonable) topological groups $G$. But this fishy to me. I find it hard to believe that, at the very least, $BH$ surjects onto $BG$ for any $G$.

How can I reconcile these issues?

general-topology algebraic-topology fiber-bundles principal-bundles classifying-spaces

share|cite|improve this question

asked Jul 17 at 18:33

Ross

213

share|cite|improve this question

share|cite|improve this question

asked Jul 17 at 18:33

Ross

213

asked Jul 17 at 18:33

Ross

213

asked Jul 17 at 18:33

Ross

213

213

add a comment | 

add a comment | 

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The statement "$BH to BG$ is a fiber bundle" is a type error; "fiber bundle" is a word you apply to a map between topological spaces, but $BH$ and $BG$ are homotopy types. Being a fiber bundle is not a homotopy-invariant notion, and in fact every map between, say, connected CW complexes is a fiber bundle up to homotopy equivalence.

The homotopy-invariant statement is that the homotopy fiber of the map $BH to BG$ is $G/H$; this statement does not depend on a choice of a point-set model for $BH$ or $BG$. An equivalent statement is that you can always pick a choice of model for $BH$ and $BG$ such that $BH to BG$ (which here now refer to spaces) is a fiber bundle with fiber $G/H$.

When $H$ is the trivial group, $BH$ is (homotopy equivalent to) a point, and so we conclude that the map $bullet to BG$ has homotopy fiber $G$. It's more generally true that if $X$ is path-connected then the homotopy fiber of $bullet to X$ is the loop space $Omega X$.

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answered Jul 17 at 18:38

Qiaochu Yuan

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The statement "$BH to BG$ is a fiber bundle" is a type error; "fiber bundle" is a word you apply to a map between topological spaces, but $BH$ and $BG$ are homotopy types. Being a fiber bundle is not a homotopy-invariant notion, and in fact every map between, say, connected CW complexes is a fiber bundle up to homotopy equivalence.

The homotopy-invariant statement is that the homotopy fiber of the map $BH to BG$ is $G/H$; this statement does not depend on a choice of a point-set model for $BH$ or $BG$. An equivalent statement is that you can always pick a choice of model for $BH$ and $BG$ such that $BH to BG$ (which here now refer to spaces) is a fiber bundle with fiber $G/H$.

When $H$ is the trivial group, $BH$ is (homotopy equivalent to) a point, and so we conclude that the map $bullet to BG$ has homotopy fiber $G$. It's more generally true that if $X$ is path-connected then the homotopy fiber of $bullet to X$ is the loop space $Omega X$.

share|cite|improve this answer

answered Jul 17 at 18:38

Qiaochu Yuan

269k32564900

add a comment | 

up vote
2
down vote

The statement "$BH to BG$ is a fiber bundle" is a type error; "fiber bundle" is a word you apply to a map between topological spaces, but $BH$ and $BG$ are homotopy types. Being a fiber bundle is not a homotopy-invariant notion, and in fact every map between, say, connected CW complexes is a fiber bundle up to homotopy equivalence.

The homotopy-invariant statement is that the homotopy fiber of the map $BH to BG$ is $G/H$; this statement does not depend on a choice of a point-set model for $BH$ or $BG$. An equivalent statement is that you can always pick a choice of model for $BH$ and $BG$ such that $BH to BG$ (which here now refer to spaces) is a fiber bundle with fiber $G/H$.

When $H$ is the trivial group, $BH$ is (homotopy equivalent to) a point, and so we conclude that the map $bullet to BG$ has homotopy fiber $G$. It's more generally true that if $X$ is path-connected then the homotopy fiber of $bullet to X$ is the loop space $Omega X$.

share|cite|improve this answer

answered Jul 17 at 18:38

Qiaochu Yuan

269k32564900

add a comment | 

up vote
2
down vote

up vote
2
down vote

The statement "$BH to BG$ is a fiber bundle" is a type error; "fiber bundle" is a word you apply to a map between topological spaces, but $BH$ and $BG$ are homotopy types. Being a fiber bundle is not a homotopy-invariant notion, and in fact every map between, say, connected CW complexes is a fiber bundle up to homotopy equivalence.

The homotopy-invariant statement is that the homotopy fiber of the map $BH to BG$ is $G/H$; this statement does not depend on a choice of a point-set model for $BH$ or $BG$. An equivalent statement is that you can always pick a choice of model for $BH$ and $BG$ such that $BH to BG$ (which here now refer to spaces) is a fiber bundle with fiber $G/H$.

When $H$ is the trivial group, $BH$ is (homotopy equivalent to) a point, and so we conclude that the map $bullet to BG$ has homotopy fiber $G$. It's more generally true that if $X$ is path-connected then the homotopy fiber of $bullet to X$ is the loop space $Omega X$.

share|cite|improve this answer

answered Jul 17 at 18:38

Qiaochu Yuan

269k32564900

The statement "$BH to BG$ is a fiber bundle" is a type error; "fiber bundle" is a word you apply to a map between topological spaces, but $BH$ and $BG$ are homotopy types. Being a fiber bundle is not a homotopy-invariant notion, and in fact every map between, say, connected CW complexes is a fiber bundle up to homotopy equivalence.

The homotopy-invariant statement is that the homotopy fiber of the map $BH to BG$ is $G/H$; this statement does not depend on a choice of a point-set model for $BH$ or $BG$. An equivalent statement is that you can always pick a choice of model for $BH$ and $BG$ such that $BH to BG$ (which here now refer to spaces) is a fiber bundle with fiber $G/H$.

When $H$ is the trivial group, $BH$ is (homotopy equivalent to) a point, and so we conclude that the map $bullet to BG$ has homotopy fiber $G$. It's more generally true that if $X$ is path-connected then the homotopy fiber of $bullet to X$ is the loop space $Omega X$.

share|cite|improve this answer

answered Jul 17 at 18:38

Qiaochu Yuan

269k32564900

share|cite|improve this answer

share|cite|improve this answer

answered Jul 17 at 18:38

Qiaochu Yuan

269k32564900

answered Jul 17 at 18:38

Qiaochu Yuan

269k32564900

answered Jul 17 at 18:38

Qiaochu Yuan

269k32564900

269k32564900

add a comment | 

add a comment | 

 

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