Extend a bundle "trivially"

Question

Suppose I have a fibre bundle $E\to B$ with compact fibre. Furthermore, $B$ is open in a larger, compact space, e. g. $B\subseteq B'$. I want to get a map $E'\to B'$ (not a bundle any more!) with

1. $E'|_B = E$
2. For $x\in B'\setminus B$ each fibre is only one point.
3. $E'$ is a compact space.
4. Each section $\omega:B\to E$ can be extended to $\omega':B'\to E'$.

The idea is to »collapse« the whole fibre when reaching the boundary of $B$.

For example, if $E:=[0,1]\times (0,1)$, $B:=(0,1)\subseteq [0,1]:=B'$ and the bundle $E\to B$ is the projection, this should be extended on $B':=[0,1]$ to $E'=\Sigma [0,1]$.

Is there a precise way to give this topological construction explicitely?

Answer 1

You should be able to construct this 'directly' in a similar way to the construction of the one-point compactification. Explicitly, you put $E' = E \cup (B'\setminus B)$, and take the open sets as follows: for each open set $U$ of $B'$, and each open set $V$ of $E$, take $V \cup \pi^{-1}(U \cap B) \cup (U \setminus B)$ as an open set.

The proof that this defines a valid topology and that $E'$ is compact should go similarly to the analogous proofs for the one-point compactification.

EDIT: I just realized that the first thing I wrote for the open sets was wrong and just corrected it. I believe I've confirmed that what I have now is correct, I can write out a proof if you like.