If the restriction of a vector bundle $\xi$ on $A \times [a,c]$ and $A \times [c,b]$ is trivial then $\xi$ is trivial I'm currently studying vector and fibre bundles from Husemaller. I don't understand in the following proposition the necessity of $(B_1 \cap B_2) \times F^k$ being closed. Is this a topology necessity or is it a vector bundle one?
Any help would be appreciated.

 A: This might be overkill, but I believe its helpfull.
If $B$ is a topological space with two subsets $B_1$, $B_2$ you can ask when
$$
\require{AMScd}
\begin{CD}
B_1 \cap B_2 @>>> B_1\\
@VVV @VV{\iota_2}V \\
B_2 @>{\iota_1}>> B
\end{CD}
$$
is a pushout diagramm. That is asking, when do two morphisms $f_i:B_i \to T$ to some other topological space which agree on the intersection $B_1 \cap B_2$ induce a unique morphism $f: B \to T$ with $f \circ \iota_i = f_i$ for $i=1,2$. On the level of maps of sets its quite immediate that for these two maps there always exists a unique map $f:B \to T$ with $f \circ \iota_i = f_i$. So the only question is, if this map is continuous.
As an example take $B = [0,2], B_1 = [0,1], B_2 = (1,2]$ (then $B_1 \cap B_2 = \emptyset$ is closed). Then the pushout property is not true.
If $B_1$ and $B_2$ are both open or both closed  in $B$, the statement will be true. (Check that preimages of open/closed sets are open/closed.)
So the point is that $B_1$ and $B_2$ are both closed (then their intersection is also closed), not that $B_1 \cap B_2$ is closed (see my counterexample).
