4
$\begingroup$

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?

$\endgroup$
0
$\begingroup$

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 $A$.

$\endgroup$
  • $\begingroup$ Yes, I think the same. Then, how should I proceed? $\endgroup$ – avati91 Aug 14 '15 at 12:39
  • $\begingroup$ 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. $\endgroup$ – Mr.P Aug 14 '15 at 12:51
  • $\begingroup$ I think I have to find the hypothesis under which it works... $\endgroup$ – avati91 Aug 14 '15 at 12:58
  • $\begingroup$ 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 $C$. $\endgroup$ – Mr.P Aug 14 '15 at 12:59
  • 1
    $\begingroup$ 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. $\endgroup$ – Mr.P Aug 14 '15 at 13:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.