If $E_1$ and $E_2$ are vector bundles over the same base space $B$, then in Characteristic Classes by Milnor and Stasheff, in chapter 2, they show:

If $F:E_1\rightarrow E_2$ is a continuous map such that $F_b:\pi_1^{-1}(b)\rightarrow\pi_2^{-1}(b)$ is a linear isomorphism for each $b\in B$, then $F$ is a homeomorphism.

Can someone tell me what is wrong with the following proof?

$F$ is assumed continuous, and since each $F_b$ is bijective, so is $F$. If $U\subset B$ is open, then $F(\pi_1^{-1}(U)) = F(\pi_2^{-1}(U))$. Thus $F$ is also an open map, and so $F$ is a homeomorphism.

Something must be wrong here, because I have not used that $F_b$ is linear at all.

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    $\begingroup$ To show, that $F$ is open, you have to show, that every open subset of $E_1$ is mapped to an open subset of $E_2$. You only show it for the open sets, that contain entire fibers of $\pi_1$. $\endgroup$ Commented Dec 1, 2019 at 20:11
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    $\begingroup$ @JulianQuast: oh yeah, of course! Silly mistake on my part. This actually helped me see what was wrong with my argument: an open set in $E_1$ could just contain a small piece of the fiber, and we need that to go to something open in $E_2$. $\endgroup$ Commented Dec 2, 2019 at 1:35

1 Answer 1


Let us first see what happens if both bundles are product bndles, i.e. $E_1 = E_2 = B \times \mathbb R^n$.

Let $\pi : B \times \mathbb R^n \to B$ and $p : B \times \mathbb R^n \to \mathbb R^n$ denote the projections and let $\phi : B \times \mathbb R^n \to B \times \mathbb R^n$ be any function whose restriction to each fiber $\pi^{-1}(b)$ is a linear isomorphism onto itself. Define $\phi' = p \circ \phi : B \times \mathbb R^n \to \mathbb R^n$. Then $\phi(b,x) = (b,\phi'(b,x))$. Clearly $\phi$ is continuous iff $\phi'$ is continuous.

It is well-known that a function $\phi' : B \times \mathbb R^n \to \mathbb R^n$ is continuous iff the function $\phi'' : B \to GL(n,\mathbb R), \phi''(b)(x) = \phi'(b,x)$ is continuous. Here $GL(n,\mathbb R)$ is the group of vector space automorphisms on $\mathbb R^n$ endowed with subspace topology topology inherited from the finite-dimensional normed linear space $End(\mathbb R^n)$ of linear endomorphisms on $\mathbb R^n$. Thus $\phi$ is continuous iff $\phi''$ is continuous.

It is also well-known the the inversion function $\iota : GL(n,\mathbb R) \to GL(n,\mathbb R), \iota (f) = f^{-1}$, is continuous.

This shows that if $\phi$ is continuous, then the fiberwise inverse $\psi : B \times \mathbb R^n \to B \times \mathbb R^n$ of $\phi$ is also continuous. In fact, we have $\psi'' = \iota \circ \phi''$.

This generalizes of course from product bundles to trivial bundles.

Therefore, if $U \subset B$ is open such that both $E_i \mid_U$ are trivial, then the restriction of $F : E_1 \mid_U \to E_2 \mid_U$ is a bundle isomorphism. It is now an easy exercise to show that $F$ is a bundle isomorphism (which you can do by showing that $F$ is an open map).


To see that $\phi' \mapsto \phi''$ is a bijection for any $B$, note that the exponential map $E : Z^{X \times Y} \to (Z^Y)^X, E(f)(x)(y) = f(x,y)$, is a bijection for all $X,Z$ provided $Y$ is locally compact, where the function space $Z^Y$ is endowed with the compact-open topology. Here we have $X = B, Y = Z = \mathbb R^n$, so this applies. It is obvious that $\phi' : B \times \mathbb R^n \to \mathbb R^n$ has the property that $\phi'(b,-) : \mathbb R^n \to \mathbb R^n$ is a linear automorphism for all $b \in B$ iff $\phi'' = E(\phi')$ maps $B$ into $GL(n,\mathbb R^n)$. But the subspace topology on $GL(n,\mathbb R^n)$ inherited from $(\mathbb R^n)^{\mathbb R^n}$ agrees with the the above topology inherited from $End(\mathbb R^n)$. See Compact open topology on $\operatorname{GL}(n, \mathbb{R})$ coincides with Euclidean topology.

  • $\begingroup$ This is essentially the proof I ended up working out. However, I'm not sure about the first sentence of your third paragraph. I think you might need some restrictions on $B$ (locally compact, paracompact, or something). $\endgroup$ Commented Dec 6, 2019 at 4:29
  • $\begingroup$ Sorry, I should clarify my previous comment. The direction we need for this argument is always true, but the other direction of your iff might need some assumption on $B$. $\endgroup$ Commented Dec 6, 2019 at 4:31
  • $\begingroup$ I edited my answer. Note that we need both directions to see that $\psi$ is continuous. $\endgroup$
    – Paul Frost
    Commented Dec 6, 2019 at 6:27
  • $\begingroup$ Yes, you're right, we need evaluation to be continuous as well. Thanks for the reference to the other answer as well, I was only thinking in terms of the compact-open topology. $\endgroup$ Commented Dec 6, 2019 at 6:46

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