Local coefficient System and universal cover

We work with a topological space $B$ which is path-connected and locally path-connected.

I have troubles writing down a formal proof for the following proposition:

Prop: Any local coeeficient system $A\hookrightarrow E \to B$ is of the form $$ A \hookrightarrow \tilde{B}\times_{\pi_1B} A \to B $$ i.e. is associated
to the principal $\pi_1B$-bundle given by the universal cover
$\tilde{B}$ of $B$ where the action is given by a homomorphism $\pi_1B \to $Aut$(A)$.

I’m able to show that the natural monodromy action of the fundamental group on $A$ gives a group homomorphism $\pi_1B \to $Aut$(A)$. But I’m not able to show that this implies that $E \cong \tilde{B}\times_{\pi_1B} A $. Can you help me on that?

Note: The author of the book I’m reading says this should be easy to show using a standard covering space argument.

Here are some definitions:

Def: A local coefficient system is a fiber bundle $p:E\to B$ such that

  • The fiber is a discrete abelian group $A$
  • The structure group $G$ is a subset of Aut$(A)$

Def: Let $p:P\to B$ be a principal $G$-bundle, where $G$ acts on an abelian group $A$. The Borel contruction is the quotient
$$ P\times_G A = P\times A \;/ \sim $$
where the equivalence relation $\sim$ is defined by $(p,a) \sim (pg, g^{-1}a) $, for $g\in G$.

The map $q:P\times_G A \to B$ given by $q([p,a]) = p(p)$ gives a fiber bundle with fiber $A$.

Solutions Collecting From Web of "Local coefficient System and universal cover"

I think one can proceed as follows :

Let us call $p : E \to B$ the local coeffiscient system, and let $x$ be a point in $B$ Let us also call $\varphi : \pi_B \to \mbox{Aut}(A)$ the holonomy homomorphism. Every choice of a point $a\in p^{-1}(x)$ determines a sub-covering in $E$, so that $a$ and $b$ live in the same sub-covering if and only if they are related by an element of $\pi_1B$. So each of these sub-coverings has fibre $\mbox{Im} (\varphi) \cong \pi_1B/\ker \varphi$, with $\pi_1B$ acting on the global covering as one would expect, so they are all isomorphic (to $\tilde{B}/\ker(\varphi)$).
Furthermore there are obviously $A/\pi_1B$ ($=A/\mbox{Im}(\varphi)$) copies of this bundle in $E$, so that $E\cong \tilde{B}/\ker(\varphi) \times A/\mbox{Im}(\varphi)$.

\begin{align*}\tilde{B}\times_{\pi_1B}A = (\tilde{B}\times A)/\pi_1B &=((\tilde{B}\times A)/\ker(\varphi))/(\pi_1B/\ker(\varphi))\\ &=((\tilde{B}/\ker(\varphi))\times A)/(\pi_1B/\ker(\varphi))\end{align*} because the action of $\ker(\varphi)$ on $A$ is trivial (by definition). For the last step, choose a $y\in p^{-1}(x)$, and define the following map :
((\tilde{B}/\ker(\varphi))\times A)/(\pi_1B/\ker(\varphi)) \to \tilde{B}/\ker(\varphi) \times A/\mbox{Im}(\varphi) \\
[([z],a)] \mapsto ([\tilde{z}],[a])
where $[ – ]$ stands for `class of $-$’ and $\tilde{z}$ denotes any element of the $\pi_1G$-orbit of $z$ that is on the same sub-bundle as $y$. The $\tilde{z}$ are all related by an element of $\ker(\varphi)$, so this map is well defined, one can compute an inverse, so it is bijective, and one can check that it is a fibre bundle morphism.

I hope this helped, and wasn’t too messy.