A local system of coefficients on a space $X$ is a functor $F\colon \Pi(X)\rightarrow Ab$ from the fundamental groupoid to the category of abelian groups. From this, one can define the homology groups of $X$ with local coefficients $H_*(X,F)$ as done e.g. in chapter VI of Whitehead's book "Elements of homotopy theory". Therein, it is also shown that this construction yields functors $H_i$ from the category of spaces with local coefficients (a morphism between (X,F) and (Y,G) is a continuous map $f\colon X\rightarrow Y$ with a natural transformation $\eta\colon F\rightarrow f^*G$) to the category of abelian groups.

Suppose now that $X$ is path-connected and has a base point $x\in X$. Then the inclusion $\pi_1(X,x)\rightarrow \Pi(X)$ is an equivalence of categories, so we can pick an inverse functor $i\colon \Pi(X)\rightarrow \pi_1(X,x)$. If we have an module $F$ over $\pi_1(X,x)$ (which is the same as a functor $\pi_1(X,x)\rightarrow Ab$), we get a functor $\Pi(X)\rightarrow Ab$ by $F\circ i$.

Hence, we can define the homology groups of $X$ with local coefficients in the $\pi_1(X,x)$-module $F$ as $H_*(X,F):=H_*(X,F\circ i)$.

First question: Why is this definition independent of the chosen inverse $i$?

It seems to be a common thing to identify local coefficient systems over a path-connected space with modules over the fundamental group at a chosen point, so this has to be true.

Second question: Does this definition extend to a functor $H_i$ from the category of pointed path-connected spaces $(X,x)$ with modules over $\pi_1(X,x)$, where morphisms between $(X,x,F)$ and $(Y,y,G)$ are continuous maps $f\colon X\rightarrow Y$ together with an $\pi_1(f)$-equivariant map $F\rightarrow G$.

As the identification mentioned above is abundant in books and papers within algebraic topology, this should be true as well, but I encounter difficulties in proving this. The main problem seems to be that it might not be possible to choose the inverses $i\colon\Pi(X)\rightarrow \pi_1(X,x)$, such that they form a natural transformation.


1 Answer 1

  1. Inverses are unique up to unique isomorphism: that is, if $F : C \to D$ is a functor and $G_1, G_2 : D \to C$ are two inverses to it, there's a unique natural isomorphism $G_1 \cong G_2$ compatible with the data of being an inverse. (This might require some subtle modifications to the naive notion of "the data of being an inverse": a definition that should be fine is a left adjoint where the unit and counit are isomorphisms.)

  2. Yes, everything is fine. Maybe I misunderstand your question, but I don't even see why you need to pick inverses for this statement: there are natural maps $\pi_1(X, x) \to \Pi_1(X)$ of groupoids and you just need to pull back local systems along these maps.

  • $\begingroup$ By an inverse $\Pi(X)\rightarrow \pi_1(X,x)$ I meant a functor, such that the composition with the natural functor $\pi_1(X,x)\rightarrow \Pi(X)$ are equivalent (not equal!) to the identity. The usual construction picks a path from each point to $x$ and conjugates with the value of this path, which involves a lot of choices. $\endgroup$
    – Bill
    Jan 25, 2016 at 17:46
  • $\begingroup$ math.stackexchange.com/questions/1613519/… seems to describe exactly this situation more abstractly. $\endgroup$
    – Bill
    Jan 25, 2016 at 17:50
  • 1
    $\begingroup$ @Bill: yes, of course. But for uniqueness you should say more. In the correct definition, the equivalences are part of the data. (Otherwise the best you can ask for is uniqueness up to isomorphism, and something stronger is true.) But in the even more correct definition you should ask for something slightly different. You can either ask for both a left and a right inverse (up to equivalence), which are not identified, or you can ask for an adjoint equivalence (which is what I said above). $\endgroup$ Jan 25, 2016 at 18:02
  • $\begingroup$ To say it another way: if $f : X \to Y$ is any map of groupoids whatsoever, it induces a pullback functor $f^{\ast} : \text{Loc}(Y) \to \text{Loc}(X)$ with both a left and a right adjoint (left and right Kan extension respectively). So far I have not chosen anything: all of this is entirely canonical. Now if $f$ is an equivalence, then so is $f^{\ast}$, and one way to describe what this means is that the unit and counit isomorphisms for either its left or its right adjoint are isomorphisms. This is a property and not a structure, and in particular I again don't need to make any choices. $\endgroup$ Jan 25, 2016 at 18:07
  • $\begingroup$ So if you want to canonically translate between $\pi_1(X, x)$-sets and local systems, instead of picking inverses you can take either left or right Kan extensions (although you should pick one and be consistent about it), and this will be natural in all of the senses you could want. $\endgroup$ Jan 25, 2016 at 18:08

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