I have troubles solving the following exercise in the book on algebraic topology I'm reading.

Suppose $\xi = (E, B, F, p)$ is a fiber bundle, $f: B' \rightarrow B$ is a continuous map, $f^*\xi$ is the induced bundle. Suppose now that there is another fiber bundle $\eta = (E', B', F, p')$ and there is a continuous map $\phi: E' \rightarrow E$ such that it maps every layer over $b'$ homeomorphically onto the layer over $f(b')$ (that is $\phi|_{p'^{-1}(b')}$ is a homeomorphism between $p'^{-1}(b')$ and $p^{-1}(f(b'))$). Then $\eta$ is equivalent to $f^*\xi$.

It is easy to take $\phi \times p'$ as a map from $E'$ onto the total space of $f^*\xi$ (which is $\{(e, b') \in E \times B' | p(e) = f(b')\}$) and prove that it is bijective and (obviously) continuous. Moreover it obviously makes the diagram with two fiber projections commutative so all that's left to do to prove the equivalency is to show that this map is open and this is exactly what I have troubles doing.

There are no conditions on any of the spaces mentioned in this exercise whatsoever (I was hoping to somehow use the $(A^B)^C \simeq A^{B \times C}$ equivalency (or adapt its proof) but neither of the spaces are said to be locally compact). I was trying to prove this straightforward in terms of suitable neighborhoods but no luck. So any advice? How do I turn a "per-layer" property into a global one?

  • $\begingroup$ What you call "layer" is usually denoted with "fiber" $\endgroup$ – Riccardo Jan 29 '18 at 22:53

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