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From Husemöller's 'Fiber Bundles' (slightly rephrased):

Proposition: Consider a bundle $\xi: E \to B$, and a mapping f: B' \to B. Then for any $s \in \Gamma(\xi)$ there is a \sigma: B' \to E \times_B B' defined by \sigma(b') = (b', sf(b')) that is a section of $f^*(\xi)$, and $f_\xi \sigma = s f$. If $f$ is an identification map and if $\sigma \in \Gamma(f^*(\xi))$ is such that $f_\xi \sigma$ is constant on $f^{-1}(b)$ for all $b \in B$, then there is a $s \in \Gamma(\xi)$ such that $sf = f_\xi \sigma$.

Proof: [...] For the second statement, we have a factorization of $f_\xi \sigma$ by $f$, giving a map $s: B \to E$ with $sf = f_\xi \sigma$. Moreover, $\xi sf = \xi f_\xi \sigma = f f^*(\xi) \sigma = f$ and $\xi s = 1_B$, since $f$ is surjective. Then $s$ is the desired cross section.


What does 'an identification mapping' mean? Does it merely mean a surjection? I'm sure it can't mean a homeomorphism or even a bijection because that would make the second part trivial.

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    It's also called a quotient map. It means the map is surjective and the codomain carries the quotient topology induced by the relation $x \sim y$ if and only if $f(x) = f(y)$.2011-09-04

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Given two topological spaces $(X,\tau_X)$ and $(Y,\tau_Y)$, a map $f:X\rightarrow Y$ is an identification map iff it is surjective and for every subset $U$ of $Y$, $U\in\tau_Y$ is equivalent to $f^{-1}(U)\in \tau_X$.

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    @Peter Yes, because, in particular, $f^{-1}(U)$ is open for all open subsets $U$ of $Y$.2018-05-23
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As Iasafro says, this means that $B$ has the quotient topology induced by the surjection f\colon B' \to B and the topology on B'. The implication of this that Husemöller is using is that any continuous map g\colon B' \to C that is constant on the fibres of $f$ will induce a unique continuous map $g_*\colon B \to C$ such that $g_* \circ f = g$.

[The existence and uniqueness of such a map of sets is trivial. To see that it's continuous, let $U$ be an open subset of $C$. Then $f^{-1}(g_*^{-1}(U)) = g^{-1}(U)$ is open as $g$ is continuous, and hence $g_*^{-1}(U)$ is open because $B$ has the quotient topology induced by $f$.]