# 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.

• @NoelLundström This is not an answer to my question Commented Nov 4, 2021 at 19:02
• sorry I completely misread your question :) Commented Nov 5, 2021 at 1:35

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).