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I'm a bit confused about the following part in Sir Michael Atiyah's "K-Theory."

Let $E = X \times V$ and $F = X\times W$, and let $\phi: E\to F$ be a vector bundle homomorphism. Why is the induced map $\Phi: X \to Hom(V,W)$ continuous? $V$ and $W$ are finite dimensional vector spaces (over $\mathbb{C}$).

To be a bit more precise, for $(x,v) \in E$, $\phi(x,v) = (x,\Phi(x)v)$.

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    Okay, so I figured out a way to prove it by choosing a basis on $V$ and $W$, then rewriting the map $\Phi$ as a map from $X$ into the complex matrices. The continuity of $\phi$ can be used to show that the map from $X$ into the entries of the matrix $\Phi$ are continuous. But I don't like this argument because it uses a basis and it doesn't generalize nicely to the cases where $V$ and $W$ are infinite dimensions. Is there a nicer, a more topological/abstract argument?2011-01-07
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    Ah, you just beat me to that very argument.2011-01-07

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The map $\Phi':X\times V\to W$ such that $\phi(x,v)=(x,\Phi'(x,v))$ is continuous. By properties of the the compact-open topology of function spaces, it follows that the map $\Phi:X\to\mathrm{Map}(V,W)$ such that $\Phi(x)(v)=\Phi'(x,v)$ is also continuous. Now $\Phi$ actually takes values in $\hom(V,W)$, whose usual topology actualy coincides with the restriction of the compact-open topology of $\mathrm{Map}(V,W)$, so the restriction $\Phi:X\to\hom(V,W)$ is continuous.