# Showing map is vector bundle isomorphism

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.

• 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 $\pi_1$. Commented Dec 1, 2019 at 20:11
• @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 $E_2$. Commented Dec 2, 2019 at 1:35

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).

Edited:

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.

• 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). Commented Dec 6, 2019 at 4:29
• 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 $B$. Commented Dec 6, 2019 at 4:31
• I edited my answer. Note that we need both directions to see that $\psi$ is continuous. Commented Dec 6, 2019 at 6:27
• 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. Commented Dec 6, 2019 at 6:46