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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.

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

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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).

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