# A fiber product is a fiber bundle

Let $F,B$ be topological spaces. A fiber bundle $E$ over the basis $B$ with fiber $F$ is a topological space $E$ endowed with a continuous surjection $\pi:E\to B$ such that there exists an open cover $\{U_\alpha\}_\alpha$ of $B$ and homeomorphisms $\phi_\alpha:\pi^{-1}(U_\alpha)\to U_\alpha\times F$ such that $\pi=\pi_1\circ\phi_\alpha$, where $\pi_1:U_\alpha\times F\to U_\alpha$ is the projection on the first factor.

Let $A,B,C$ general sets and $f:A\to C$, $g:B\to C$ be general maps. We define the fiber product of $f$ and $g$ as $$A\times_C B:=\{(a,b)\in A\times B:f(a)=g(b)\}$$ Then define the projections $g':A\times_CB\to A$ and $f':A\times_CB\to B$.

In my lecture notes it's written that if $B$ is a fiber bundle over $C$ with fiber $F$ and projection $g$, then also $A\times_C B$ is a fiber bundle over $A$ with fiber $F$ and projection $g'$.

I want to prove this last statement.

My idea is to suppose, first of all, $f,g$ continuous. Then, if I prove that $f\circ g'=g\circ f$ (it immediately follows by definition of fiber product), we have that there is a morphism of fiber bundles between $B$ and $A\times_C B$, i.e. there exist $f':A\times_C B\to B$, $f:A\to C$ continuous map such that $f\circ g'=g\circ f$.

Since there is a morphism of fiber bundles, then $f\circ g'=g\circ f$ must be a fiber bundle.

Right?

Ok so $A\times_C B$ by your definition can also be seen as an association to a point $a\in A$ of the whole fibre of $g$ at the point $f(a)\in C$.
We know that $B$ has a trivialization (those maps to $U\times F$) over C, and we need to show that $g' : A\times_C B \longrightarrow A$ has a trivialization.
To this end I believe $f$ really needs to be continuous so we can pull back (term) the trivialization over $C$ to $A$.
• Well since they ask you to prove the general case construct a counterexample. Pick a $A=\mathbb{R}$. Pick two distinct points in $c_1,c_2\in C$ such that the fibres are not mapped into eachother by $id$. Define $f_{|\mathbb{Q}}(x)=c_1$ and $f_{|\mathbb{R}\backslash \mathbb{Q}}(x)=c_2$. Then there is no homeomorphic trivialization. – Mr.P Aug 14 '15 at 12:51
• This would make sense if $A$, $B$ and $C$ had topology and $f$ and $g$ were continuous and $g$ was the submersion which makes $B$ into a fiber bundle over $C$. – Mr.P Aug 14 '15 at 12:59
• Let $\varphi_{U_\alpha}$ be the trivialization for a neighbourhood $U_\alpha\subset C$. Fix a cover $W_i\subset A$ such that $f: W_i \rightarrow C$ is injective. Now for this cover define the trivialization for $A\times_C B$ as the map which acts with $id$ on the first component and with the trivialization for $B$ on the second. Take care to define the homomorphisms correctly. – Mr.P Aug 14 '15 at 13:11