Example #1

In proving pullback bundle is homotopy invariant Ralph Cohen's notes on the topology of fiber bundles use the following proof based on the covering homotopy theorem (pp.47):

Let $p: E \to B$ be a fiber bundle and $f_0, f_1: X \to B$ be homotopic to each other with the homotopy given by $H: X \times I \to B$. By the covering homotopy theorem we have the following fiber bundle over $X \times I$. $\require{AMScd}$ \begin{CD} f_0^*(E) \times I @>\tilde{H}>> E \\ @VVV @VVpV \\ X \times I @>H>> B \end{CD}

Then he claims that by definition this defines a map of bundles over $X \times I$ $\require{AMScd}$ \begin{CD} f_0^*(E) \times I @>>> H^*(E) \\ @VVV @VVV \\ X \times I @= X \times I \end{CD} which is clearly a bundle isomorphism since it induces the identity map on both the base space and on the fibers.

I don't understand why $f_0^*(E) \times I \to H^*(E)$ is an isomorphism. I agree that it induces the identity on the base space but how can the map on fibers is also the identity? If both are identity wouldn't that imply the two total spaces are the same?

Example #2

On the same notes pp.53 during the proof of the bijection between $[X,B]$ and isomorphism classes of principal bundles over $X$ where $(h,f)$ defines a homeomorphism of principal bundle as below $\require{AMScd}$ \begin{CD} P @>h>> E \\ @VqVV @VVpV \\ X @>f>> B \end{CD} which induces an equivariant isomorphism on each fiber thus induces the following bundle isomorphism to the pullback $\require{AMScd}$ \begin{CD} P @>>> f^*(E) \\ @VqVV @VVpV \\ X @= X \end{CD} again I don't understand why $P \to f^*(E)$ is a bundle isomorphism.

Example #3

On the same notes pp.48 in showing K-theory is a homotopy invariant, the missing step is to show if $E \to Y$ and $E' \to Y$ are isomorphic bundles over $Y$ i.e. $\require{AMScd}$ \begin{CD} E @>\cong>> E' \\ @VVV @VVV \\ Y @>\cong>> Y \end{CD} then their pullback bundles along the same map $f: X \to Y$ is also isomorphic i.e. $\require{AMScd}$ \begin{CD} f^*(E) @>\cong>> f^*(E') \\ @VVV @VVV \\ X @>\cong>> X \end{CD} The global homeomorphism between $E$ and $E'$ induces homeomorphism on their local trivializations which defines a homeomorphism on local trivializations of $f^*(E)$ and $f^*(E')$ (as the local trivialization of the pullback is defined using the local trivialization of the bundle). Global homeomorphism between $f^*(E)$ and $f^*(E')$ then follows from the pasting lemma?


1 Answer 1


In the first case, the explanation given in the notes is indeed rather poor. The fact that the map $f_0^*(E) \times I \to H^*(E)$ is an isomorphism follows not from the statement of the covering homotopy theorem (which, as those notes state it, doesn't even guarantee that $\tilde{H}$ induces homeomorphisms on each fiber) but from its proof. When you examine the proof, you can see that it constructs a certain bundle map $f_0^*(E) \times I \to H^*(E)$ which is seen to be continuous by the pasting lemma (it is constructed by pasting together pieces which lie over a cover of $B\times I$ by certain closed sets $A_{ij}=\bar{W}_i'\times [t_j,t_{j+1}]$). But actually, over each of these pieces $A_{ij}$, there are trivializations of both $f_0^*(E)\times I$ and $H^*(E)$ given by $\phi_k$ (since each of them is pulled back from a bundle on $U_\alpha$ which is trivialized by $\phi_k$), and when you use these trivializations to identify both bundles with the trivial bundle $A_{ij}\times F$, the map is just the identity map. This is what the notes mean when they say the map is "the identity on fibers": when you choose these local trivializations induced by $\phi_k$ on each piece of the base, the map becomes the identity on fibers. It follows that the inverse map, similarly defined piecewise, is also continuous, so the map is a homeomorphism.

In the second case, the explanation is much simpler: any equivariant map between principal bundles which covers the identity map is automatically a homeomorphism. To see this, you can look locally on the base, so you can assume both your bundles are trivial. Then you have map $\alpha:X\times G\to X\times G$ of the form $\alpha(x,g)=(x,\beta(x)g)$ for some continuous map $\beta:X\to G$ (the map $\beta$ is continuous because it is the second coordinate of the map $\alpha(x,1)$). The inverse of $\alpha$ is then just $\alpha^{-1}(x,g)=(x,\beta(x)^{-1}g)$, which is continuous because the inverse map $G\to G$ is continuous.

  • $\begingroup$ Many thanks for so detailed explanation, Eric! Your explanation seems to reveal a common approach to showing bundle isomorphism. That is to pass to local trivialization neighborhood to examine whether the given bundle map defines a local homeomorphism. If yes then we can apply the pasting lemma to assert that the local homeomorphisms patch up to a global homeomorphism thus establish bundle isomorphism. Am I right? Actually I mimic your proof for Example #3 above (global $\to$ local $\to$ local $\to$ global) could you also please have a check for me? $\endgroup$ Commented May 22, 2016 at 19:44
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    $\begingroup$ Yes, that's correct. Actually, when you can look locally on an open subset of the base (as you do in Example #3), you don't even need to use the pasting lemma; you just use the fact that continuity is a local property. The only reason I mentioned the pasting lemma is that in the first example, you are cutting the base into closed sets, rather than open sets. $\endgroup$ Commented May 22, 2016 at 19:46
  • $\begingroup$ I see! That is very clear to me now! Thank you so much for teaching me how to prove one type of problems instead of one problem Eric! One last point to confirm with you. In your explanation to Example #2 you seem to want to emphasize that $\beta$ is continuous if and only if $\alpha$ is because $\beta: X \to G$ operates only on the second coordinate of $\alpha$. Am I guessing correctly? $\endgroup$ Commented May 22, 2016 at 19:52
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    $\begingroup$ You could say that: more precisely, if $\alpha:X\times G\to X\times G$ is an arbitrary map of the form $\alpha(x,g)=(x,\beta(x)g)$, then $\alpha$ is continuous iff $\beta:X\to G$ is continuous. This means that if $\alpha$ is continuous then its inverse is automatically continuous, since its inverse can be found by composing $\beta$ with the inverse map $G\to G$. $\endgroup$ Commented May 22, 2016 at 21:08
  • $\begingroup$ Thank you so much Eric! Your clarification provides a perfect close to this question! I really appreciate your teaching! $\endgroup$ Commented May 22, 2016 at 23:17

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