I'm currently trying to understand how the different incarnations of homology with local coefficients relate to one another. Let $X$ be a semi-locally simply connected space, and let $\pi_1 = \pi_1(X,x_0)$. Homology with local coefficients is usually built from one of the following three objects:

  1. A $\mathbb{Z}\pi_1$-Module
  2. A Bundle of discrete abelian groups $p:E\to X$

In other words, these are fiber bundles $G\hookrightarrow E\to X$ with fibers discrete abelian groups isomorphic to $G$, and whose structure group is some subgroup of $\operatorname{Aut}(G)$ so that the local trivializations $\phi_U:p^{-1}(U)\to U\times G$ restrict to homomorphisms on the fibers.

  1. A Functor $\mathcal{L}:\Pi_1(X)\to\textbf{Ab}$

where $\Pi_1(X)$ is the fundamental groupoid, and $\mathcal{L}(x)$ is always discrete abelian.

Given a $\mathbb{Z}\pi_1$-module $M$, one can construct a bundle $\widetilde{X}\times_{\pi_1}M\to X$ of discrete abelian groups using the Borel construction. Conversely, given a bundle of discrete abelian groups $p:E\to X$, this is really a covering space, and so there is an action of $\pi_1$ on the fiber $G$, giving it the structure of a $\mathbb{Z}\pi_1$-module.

This brings me to my questions:

A. How does a bundle $p:E\to X$ of discrete abelian groups give rise to a functor $\mathcal{L}:\Pi_1(X)\to\textbf{Ab}$?

Edit: So that the resulting homology groups $H_*(X;E)$ and $H_*(X;\mathcal{L})$ are isomorphic?

Here is my guess: for a bundle $p:E\to X$ we set $\mathcal{L}(x) = p^{-1}(x)$, and for a homotopy class of paths $[\omega:I\to X]$ (a morphism from $\omega(0)=x_0$ to $\omega(1)=x_1$) we set $\mathcal{L}[\omega]$ to be the map $p^{-1}(x_0)\to p^{-1}(x_1)$ built from using the homotopy lifting property on

$$h:p^{-1}(x_0)\times I\to X,\quad h(e,t) = \omega(t).$$

(lift this to $H:p^{-1}(x_0)\times I\to E$, and then $H(-,1):p^{-1}(x_0)\to p^{-1}(x_1)$ is the map I'm referring to.) However, I'm having a hard time showing that this is a homomorphism. This is probably not a good approach since there is no canonical identification of each fiber with $G$.

B. How does a local system $\mathcal{L}:\Pi_1(X)\to\textbf{Ab}$ give rise to a bundle $p:E\to X$ of discrete abelian groups? (or a $\mathbb{Z}\pi_1$-module?)

Edit: So that the resulting homology groups $H_*(X;E)$ and $H_*(X;\mathcal{L})$ are isomorphic?

I gather from this discussion that it is possible in this case but I'm not sure how that would work.


[1.] Hatcher, Algebraic Topology, pg. 330

[2.] Whitehead, Elements of Homotopy Theory, pg. 257 (note: he calls functors $\mathcal{L}:\Pi_1(X)\to\textbf{Ab}$ "bundles of groups")

  • 1
    $\begingroup$ From functors $\mathcal{L}:\Pi_1(X)\to\mathbf{Ab}$ to $\mathbb{Z}\pi_1(X,x)$-module is easy. Just take $\mathcal{L}(x)$. Since $\operatorname{End}_{\Pi_1(X)}(x)=\pi_1(X,x)$, by functoriality, $\mathcal{L}(x)$ is endowed with a $\pi_1(X,x)$-module structure. $\endgroup$ – Roland Jun 2 '19 at 16:32
  • $\begingroup$ @Roland In that case I think I'm missing something. Just to be concrete, I'm wondering if I can take $\mathcal{L}$ and produce a bundle $p:E\to X$ so that $H_*(X;\mathcal{L})$ and $H_*(X;E)$ are isomorphic. From the linked discussion it seems like for general spaces, functors $\mathcal{L}:\Pi_1(X)\to\textbf{Ab}$ don't always give rise to bundles of groups. I could take the module $\mathcal{L}(x)$ and create a bundle $E$. The resulting homology groups $H_*(X;\mathcal{L})$ and $H_*(X;E)$ wouldn't be isomorphic then would they? $\endgroup$ – multi porpoise Jun 2 '19 at 16:48
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    $\begingroup$ The restriction is that $X$ should be locally path connected and semi-locally simply connected, this is required so that $\pi_1$ and $\Pi_1$ are meaningful. I mean, for weird topological spaces, the notion of path is not good (there may be way too many of them or way too few of them), whereas the notion of bundle still makes sense. See, for your direction 1. to 2. you used the universal cover of $X$, but for weird spaces this might not exist. $\endgroup$ – Roland Jun 2 '19 at 17:13
  • $\begingroup$ @Roland Ok that makes sense, so in the case that $X$ is locally path connected, I can take $\mathcal{L}(x)$ and form a bundle of groups $E$, and $H_*(X;E)\cong H_*(X;\mathcal{L})$? $\endgroup$ – multi porpoise Jun 2 '19 at 17:22
  • 1
    $\begingroup$ Yes I don't know exactly what is your definition of $H_*(X,E), H_*(X,\mathcal{L})$, but they should be isomorphic more or less by definition. $\endgroup$ – Roland Jun 2 '19 at 17:28

You have not yet made use of the fact that the local trivializations restrict to homomorphisms on the fibers. Also, perhaps you meant to write isomorphisms on the fibers. I'll assume so, and here's how to make use of that.

Consider your chosen path $\omega : I \to X$. The idea is to pull back open covers on $X$ via the continuous map $\omega$ to obtain open covers on $I$, to which the Lebesgue Number Lemma can be applied. (Perhaps you have encountered these kinds of applications of the Lebesgue Number Lemma before, but if not then good examples of such applications can be found in the fundamental group portions of Munkres' book "Topology".)

On $X$ you have an open cover consisting of the open sets $U$ over which you have local trivializations. The collection $\{\omega^{-1}(U)\}$ is an open cover of $I=[0,1]$. Applying the Lebesgue number lemma, there exists $\lambda > 0$ such that any subset of $I$ of diameter $<\lambda$ is contained in some element of that collection. Choose an integer $n$ so that $\frac{1}{n} < \lambda$ and consider the decomposition $$t_0=0, \quad t_1 = \frac{1}{n}, \quad t_2 = \frac{2}{n}, \quad ... \, , \quad t_n=\frac{n}{n}=1 $$ For each $i=1,..,n$ there is an open set $U_i$ over which you have a local trivialization such that $[t_{i-1},t_i] \subset \omega^{-1}(U_i)$ and so $\omega[t_{i-1},t_i] \subset U_i$. Using the local trivializations over $U_1,U_2,...,U_n$ in turn you obtain a sequence of isomorphisms:

  1. The first goes from the $x_0=\omega(t_0)$ fiber $p^{-1}(\omega(t_0))$ to the $\omega(t_1)$ fiber $p^{-1}(\omega(t_1))$;

  2. The second goes from $p^{-1}(\omega(t_1))$ to $p^{-1}(\omega(t_2))$;


$n$. The $n^{\text{th}}$ goes from $p^{-1}(\omega(t_{n-1}))$ to $p^{-1}(\omega(t_n)) = p^{-1}(x_1)$.

By composition you get your desired isomorphism from the $x_0$ fiber $p^{-1}(x_0)$ to the $x_1$ fiber $p^{-1}(x_1)$.

Now there is a further issue, namely why this isomorphism depends only on $[\omega]$ and not on $\omega$ itself. For that, you need another argument applying Lebesgue number lemma, but this time the argument is applied to a path homotopy $H : I \times I \to X$ between a path homotopic pair of paths $\omega_1,\omega_2$. That's more complicated, but perhaps this is enough to answer your question.

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  • $\begingroup$ Thanks Professor Mosher! Firstly, yes that should definitely be isomorphisms on the fibers, and secondly, I think I can show that this depends only on homotopy type from here. This only leaves question (B) unresolved, so I'll leave it open a bit longer. $\endgroup$ – multi porpoise Jun 2 '19 at 15:09

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