Let $X$ be a space which is paracompact, Hausdorff, and sufficiently nice that it has a universal covering space (and map) $p:\tilde{X}\to X$. Also, let $\pi:=\pi_1(X)$ and $A$ some $\mathbb{Z}[\pi]$-module, or if preferred, some abelian group on which $\pi$ acts. Under these conditions, there is a notion of cohomology with local coefficients which can be computed using cochain groups $C^n_\pi(X,A)=Hom_{\mathbb{Z}[\pi]}(C_n(\tilde X),A)$ where $C_n(\tilde{X})$ is the singular chain groups on $\tilde X$ and has an action of $\pi$ on it "inherited" from the action on $\tilde X$ itself by post-composition, that is, for $g\in \pi$, $\sigma \in C_n(\tilde X)$, $g\cdot \sigma$ is just $\sigma$ followed by the action of $g$. Let's call this cohomology $H^n_\pi(X,A)$.

On the other hand, there is another notion of cohomology with local coefficients whereby one defines a locally constant sheaf $\mathcal A$ on $X$ using the constant sheaf $A_\tilde{X}$ on $\tilde X$ and the action of $\pi$ on $A$, and takes the sheaf cohomology of that. It appears to be well known that (1) these two theories are essentially the same, and (2) for paracompact Hausdorff spaces, the Čech cohomology of a sheaf is the same as the sheaf cohomology of that same sheaf. That means that the Čech cohomology $\check{H}^n(X,\mathcal A)$ should be isomorphic to $H^n_\pi(X,A)$.

With all of this background out of the way, I can begin to describe my actual question. First of all, I cannot comprehend Sheaf cohomology, no matter how hard I try. Every reference I have found that describes it just seems impenetrable to me. On the other hand, I can more or less make sense of Čech cohomology, at least a lot better than "direct" sheaf cohomology. So: I would like to define some explicit map, $\check H^*(X ,\mathcal A)\to H^*_\pi (X, A)$ (or going the other way), which can exhibit this isomorphism more directly.

What appears natural for a map like this is to perhaps restrict to some convenient cofinal collection of covers of $X$ to get maps $C^*(\mathscr U,\mathcal A)\to C^*_\pi(X,A)$. A good candidate comes from the property of a universal cover, whereby every element $x\in X$ has a neighborhood $U_x$ so that $p^{-1}(U_x)$ composed of a disjoint union of open sets, each of which $p$ makes homeomorphic to $U_x$. It's not hard to show covers of the form $\mathscr U=\lbrace U_x : x\in X \rbrace$ with each $U_x$ a neighborhood like this is a cofinal collection, and then $p^{-1}(\mathscr U)$ is an open cover of $\tilde X$. There is a possibly useful property of Singular cohomology, too, which for any open cover $\mathscr V$ of $\tilde X$ makes the inclusion of the subcomplex $C_n(\mathscr V)=\langle \sigma: \text{im} (\sigma) \subset V\text{ for some }V \in \mathscr V \rangle$ a chain equivalence. That carries over to a cochain equivalence between $Hom_{\mathbb{Z}[\pi]}(C_n(\tilde X),A)$ and $Hom_{\mathbb{Z}[\pi]}(C_n(\mathscr V),A)$, and I'll call the latter group $C^n_\pi (\mathscr V , A)$.

My most promising candidate after several attempts, using these ideas, is to consider covers of $X$ as above, and for each one make a map $\psi_\mathscr{U}: \check{C}^n(\mathscr U , \mathcal A) \to C^n(p^{-1}(\mathscr U),A)$ by first applying a map $\iota_\mathscr{U}:\check{C}^n(\mathscr U , \mathcal A)\to \check{C}^n(p^{-1}(\mathscr U ), A_\tilde{X})$ defined by $(\iota_\mathscr{U} f)(p^{-1}(U_{x_0}),...,p^{-1}(U_{x_n}))=f(U_{x_0},...,U_{x_n})$, then picking out for each $\sigma \in C_n(p^{-1}(\mathscr U))$ the open set $U_\sigma:=U_{p(\sigma(v_0))}\cap ... \cap U_{p(\sigma(v_n))}$ and setting $(\psi_\mathscr{U} f)(\sigma):=[(\iota_\mathscr{U} f)(p^{-1}(U_\sigma))](\sigma(v_0))$. I say this is promising because evaluating at some point is needed in order to get the $\pi$-equivariance expected of an element of $C^n_\pi(p^{-1}(\mathscr U),A)$, and because it seems likely to give "injectivity" and "surjectivity" at least at the cohomology level after taking the direct limit defining Čech cohomology.

The only problem is that I'm not entirely sure this actually gives a cochain map. Running through the calculations, I've found that:

$(\partial\psi_{\mathscr{U}}f)(\sigma)-(\psi_{\mathscr{U}}\partial f)(\sigma)=[(\iota_{\mathscr{U}}f)(p^{-1}(U_{p(\sigma(v_{1}))}),...,p^{-1}(U_{p(\sigma(v_{n+1}))}))](\sigma(v_{1}))-[(\iota_{\mathscr{U}}f)(p^{-1}(U_{p(\sigma(v_{1}))}),...,p^{-1}(U_{p(\sigma(v_{n+1}))}))](\sigma(v_{0}))$

This is more or less because the '0 boundary' of the simplex sends $v_0$ to the same place the original simplex sends $v_1$ to, while the rest of the boundaries send $v_0$ to $\sigma(v_0)$. Of course, this needs to be $0$ to get a cochain map, but I'm not sure these two terms are the same (or if they are, how to show it).

So: Is this the wrong map again? If it is, what's the right map? Or, is there some really simple way to see that those two terms are the same and the difference really is $0$? Are there any references that have attempted something similar?

  • 3
    $\begingroup$ While the equivalence between sheaf and Cech cohomology uses paracompactness, the equivalence of the former with singular cohomology uses that your space is locally contractible: arxiv.org/abs/1602.06674 The Cech complex of singular cochains is a bicomplex, and you can compute the homology of its totalization by two spectral sequences. If you do the Cech differential first, you get singular cohomology groups since singular cochains have vanishing higher hypercohomology. In the other direction you get Cech cohomology of your locally constant sheaf if the space is locally contractible. $\endgroup$ Nov 25, 2020 at 10:21
  • 4
    $\begingroup$ In other words, one naturally gets a zig-zag $\check C^*(\mathcal U,\mathcal A)\to \check C^*(\mathcal U,\mathrm{Sing}^\bullet(A)) \leftarrow \mathrm{Hom}_{\mathbb Z[\pi]}(C_n(\widetilde X),A)$ where both maps are reasonably explicit quasiisomorphisms, but that does not guarantee the existence of an explicit chain homotopy inverse. For the left map, such an inverse should in principle be given by choosing a good cover (all intersections are empty or contractible) and using the contractions to give a chain homotopy retraction of $C^*(U_1\cap\dots\cap U_n,A)$ to locally constant functions. $\endgroup$ Nov 25, 2020 at 10:34
  • $\begingroup$ Because notation for sheaf theory stuff seems to be all over the place, I feel I should ask what exactly you mean by "$Sing^{\bullet}(A)$". Is that just $U \to C^*(U,A)$, or something more complicated? And if it's the former, I don't really see how it "captures" the "$\pi$-equivariance" that local coefficients seem to require? $\endgroup$
    – Xindaris
    Nov 25, 2020 at 17:33
  • 1
    $\begingroup$ It's $U\mapsto \operatorname{Hom}_{\mathbb Z\pi}(C_*(p^{-1}(U),A)$, where $p:\widetilde X\to X$ is the universal cover. The point is that this is an acyclic resolution of the local system determined by $A$ and its $\pi$-action if your space is locally contractible, and that the usual argument using barycentric subdivision shows that the "Cech-first" spectral sequence degenerates on the $E_2$-page with everything in Cech-degree $0$ (note that this is weaker than the usual requirement that the resolution is by acyclic sheaves, which is not the case here - singular cochains aren't a sheaf). $\endgroup$ Nov 26, 2020 at 10:33
  • $\begingroup$ I'm reasonably sure I have it together now. It seems like the local contractibility is, in some sense, what "corrects" the equivalent issue to the one in my question that arises for the map going to the right in your comment. I wonder if you could "upgrade" your comments to an answer so I could approve it (and it won't be forever buried in the comments of a question with "0 answers"?) $\endgroup$
    – Xindaris
    Nov 27, 2020 at 22:29

1 Answer 1


Like I explained in the comments, there is no reason to expect the existence of a particularly explicit direct map between the Cech and singular complexes, since we naturally obtain a zig-zag $$ \check C^*(\{U_i\}_{i\in I},\mathcal A)\to \check C^*(\{U_i\}_{i\in I},\operatorname{Sing}^\bullet(\mathcal A))\leftarrow \operatorname{Sing}^\bullet(\mathcal A)(X) = \operatorname{Hom}_{\mathbb Z\pi}(\operatorname{Sing}^\bullet(\widetilde X),A) $$ where $\operatorname{Sing}^\bullet(\mathcal A)(-)$ is the presheaf which assigns to $U\in X$ the $\pi$-equivariant maps from $p^{-1}(X)$ to $A$ and $p:\widetilde X\to X$ is the universal cover. Note that the individual terms $\operatorname{Sing}^\bullet(\mathcal A)(-)$ are not sheaves: if you know the values on small simplices, i.e. those inside an open set of a cover, you can't recover the values on all simplices. However, the inclusion of small simplices is a quasiisomorphism since every simplicial chain is homologous to its barycentric subdivision, which will eventually be small. This shows that the left-pointing map is a quasiisomorphism; the right-pointing map is a quasiisomorphism if all intersections are empty or contractible, i.e. $\{U_i\}_{i\in I}$ is a good cover. For nice spaces (e.g. CW complexes) the Cech cohomology groups are computed by the Cech complex of a good cover. In that case, it is possible to get an explicit chain map from the singular cochain complex to the Cech complex as follows:

The good cover $\{U_i\}_{i\in I}$ determines an abstract simplicial complex $S_U$, i.e. a collection of finite subsets of $I$ (namely, those for which the intersection is nonempty) which is closed under taking subsets. We may inductively choose for $J\in S_U$ a simplex $\sigma_J:\Delta^{|J|-1}\to X$ whose image lands in $\bigcup_{j\in J} U_j$ and whose restriction to a face is given by the previous choice in the obvious sense (send the barycentre to a point in $\bigcap_{j\in J} U_j$ and iteratively "cone off" the maps on the lower-dimensional simplices, which is possible by contractibility). These singular simplices form a subcomplex of the singular chains, and it's straightforward to show that restricting singular cochains to this subcomplex gives a cochain map from these to the Cech complex.

In the other direction, one would have to pick a chain homotopy inverse to the inclusion of small simplices, which is not possible to do explicitly as far as I know.


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