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 $A$.

  • $\begingroup$ May I ask what book this was from? $\endgroup$
    – ziggurism
    Oct 7, 2014 at 4:11
  • $\begingroup$ Davis and Kirk, lemma 4.7 $\endgroup$
    – ziggurism
    Oct 7, 2014 at 5:53

1 Answer 1


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 $E\cong \tilde{B}/\ker(\varphi) \times A/\mbox{Im}(\varphi)$.

Now \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.

  • $\begingroup$ Thank you very much! I think it is a bit messy, but it really helped a lot! $\endgroup$
    – Abramo
    May 29, 2013 at 11:03
  • $\begingroup$ Sorry for late commenting, but may I ask you how this sub-covering is build? I mean, we choose an element in each fibre and to the construction? why do we obtain at the end a covering? $\endgroup$
    – Luigi M
    Aug 1, 2016 at 8:28

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