Isomorphic modules of sections imply isomorphic bundles For reference this is about a part of question 3-F in Characteristic Classes by Milnor and Stasheff, which discusses the $C(B)$-module structure of the space of sections of a topological vector bundle $\xi$ over some base space $B$, which is assumed to be Tychonoff. If we write $S(\xi)$ to be the sections of the vector bundle $\xi$, then problem asks the reader, among other things, to show that $\xi\cong\eta$ as vector bundles iff $S(\xi)\cong S(\eta)$ as $C(B)$-modules.
The first direction isn't bad. For the other, we assume that there is an isomophism $$\phi:S(\xi)\rightarrow S(\eta)$$
My progress is as follows: Writing $E(\xi)$ for the total space of the vector bundle $\xi$, and similarly for $\eta$, I have thus far defined a map $$\psi:E(\xi)\rightarrow E(\eta)$$ by sending $(p,v)\in E(\xi)$ to $(\phi(s))(p)$ where $s\in S(\xi)$ is any old section such that $s(p)=(p,v)$. For this construction to work one has to prove that there is indeed a section $s$ such that $s(p)=(p,v)$, and I did this using local triviality and the fact that B is Tychonoff. Additionally, one has to verify that this is well-defined, which requires showing that if we have two sections $s_1$ and $s_2$ such that $s_1(p)=s_2(p)=(p,v)$, then $(\phi(s_1))(p)=(\phi(s_2))(p)$. This also requires using the Tychonoff-ness of B and the fact that $\phi$ is injective.
What I am struggling with is showing this map (which is the only one I can think of) is continuous. It seems that one needs to make a jump from the continuity of individual sections to the continuity of the map as a whole. If anyone can help, it would be appreciated.
 A: Here is how to show continuity of $\psi:E(\xi)\rightarrow E(\eta)$ once you have proved that it is a well-defined map, linear on the fibers:  
Call $p$ the projection $p:E(\xi)\to B$.
Fix an arbitrary point $e\in E(\xi)$ and call  $b=p(e)\in B$ its projection.
Choose a neighbourhood $U$ of $b$ so small that there exist global sections $s_1,...,s_r\in S(\xi)$ such that the $s_1(b'),...,s_r(b')$ form a basis of the fiber $E_{b'}=p^{-1}(b')$ over every $b'\in U$.
[This uses local triviality of $\xi$,  and Tychonoffness in order to ensure that the $s_i$'s can be modified to global sections]
Then any point $e'\in p^{-1}(U)$ with projection $b'=p(e')$ can be written uniquely as $$e'=\sum_{i=1}^r f_i(b')s_i(b')$$ and the $f_i$'s are continuous  functions  $f_i\in C(U)$ [They are not supposed to be extendable to $B$]  
Now, applying your recipe  to the section $s=\sum_{i=1}^r f_i(b')s_i\in S(\xi)$  which satisfies  $s(b')=e'$ , we obtain  $$\psi (e')=\psi (s(b'))=[\phi(s)](b')=[\sum_{i=1}^r f_i(b')\phi(s_i)](b')=\sum_{i=1}^r f_i(b')[\phi(s_i)](b')$$ which shows that $\psi$ is continuous on $p^{-1}(U)$.
Conclusion
The map  $\psi:E(\xi)\rightarrow E(\eta)$ is continuous since it is continuous at any $e\in E(\xi)$.
Edit
The recipe alluded to above is the nice formula $\psi (s(b))=[\phi(s)](b)$, valid for all $s\in S(\xi)$ and all $b\in B$.
