Suppose $X$ is a (paracompact, Hausdorff) topological space and we want to define its Cech cohomology with coefficients in $\mathbb Z$. Here is the way I have seen this constructed. For each open cover $\mathcal U$, we define the complex ${\check{\mathcal C}}^\bullet(\mathcal U,\mathbb Z)$ of Cech cochains relative the open cover with the usual Cech differential. Now, if $\mathcal V$ is another open cover of $X$ which is a refinement of $\mathcal U$, then we have a map ${\check{\mathcal C}}^\bullet(\mathcal U,\mathbb Z)\to {\check{\mathcal C}}^\bullet(\mathcal V,\mathbb Z)$ for each "presentation" of $\mathcal V$ as a refinement of $\mathcal U$. But the chain homotopy class of this map is well-defined and thus, we get a well-defined directed system of groups $\check{H}^\bullet(\mathcal U,\mathbb Z)$ indexed by the open covers $\mathcal U$ of $X$, and their direct limit is defined to be $\check{H}^\bullet(X,\mathbb Z)$. On the other hand I have heard people speak of the complex $\check{\mathcal C}^\bullet(X,\mathbb Z)$ of Cech cochains on a space $X$, which computes the Cech cohomology of $X$.

I am interested in knowing the definition of this complex which computes the Cech cohomology. It can't simply be the direct limit of the complexes ${\check{\mathcal C}}^\bullet(\mathcal U,\mathbb Z)$ since these do not form a directed system of complexes (as the maps are well-defined only up to homotopy).

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    $\begingroup$ If the space has an open cover for which all finite intersections are acyclic for the given sheaf of coefficients, then you can use just that one open cover and the cochain complex it gives rise to. $\endgroup$ – David Roberts Nov 3 '17 at 21:30
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    $\begingroup$ Yes, I am aware of that. I was wondering if there is a canonically defined complex computing the Cech cohomology in the general case (independent of the choice of any acylic cover). $\endgroup$ – Mohan Swaminathan Nov 4 '17 at 3:52
  • $\begingroup$ Not that I'm aware, sorry. These issues were part of why categories were invented, in that there wasn't just some canonical complex to compute Čech cohomology... $\endgroup$ – David Roberts Nov 4 '17 at 7:35
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    $\begingroup$ Can we somehow make sense of taking a homotopy colimit of these cochain complexes? The hocolim, if it exists, is probably the solution I'm after. $\endgroup$ – Mohan Swaminathan Nov 8 '17 at 12:08
  • $\begingroup$ that sounds about right, since you want to capture the mapping space in a certain model category whose objects are simplicial sheaves. $\endgroup$ – David Roberts Nov 8 '17 at 20:54

This is the kind of thing sieves are good for. For an open cover, let $S$ denote the sieve it generates, so $S$ is poset of open subsets $V$ such that $V$ is contained in some element of the cover. A quasi-isomorphic model for the Cech complex is given by the "homotopy limit over the sieve", so the $0$th term in the complex is the product of the $F(U)$ for all $U$ in $S$, the $1$st term is the product of the $F(U)$ for all $U \subset U'$ in $S$, etc. Basically, instead of modeling "a section on each element of the open cover, which agree on intersections", you model "a section on each element of the sieve, which agree on restriction". The point now is that a refinement of open covers gives an honest inclusion of sieves, and the sieve model for the Cech complex is just plain functorial in inclusions of sieves. So you can just pass to the filtered colimit.

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    $\begingroup$ This is quite clearly the best answer. $\endgroup$ – John Pardon Mar 3 '19 at 11:20

While I'm sure a hocolim solution also exists (and works in general), one can avoid it, at least for compact topological spaces, by using open covers which are indexed by the points of your space $X$ (this idea goes back at least to Godement). These are called "rigid open covers" by Carlsson–Pedersen, who make the following definition:

Definition: A rigid open cover of a compact Hausdorff space $X$ consists of open sets $\{U_x\subseteq X\}_{x\in X}$ such that $x\in U_x$, $\overline{\{x:U_x=U\}}\subseteq U$, and $\#\{U:U=U_x\text{ for some }x\}<\infty$.

Now we only consider refinements between rigid open covers which are the identity map on the index set (and hence there is at most one refinement between any pair of rigid open covers). Thus the Cech complexes form an honest direct system, and one can simply take the direct limit of complexes.

I don't know the maximum generality in which one can make such a construction, though probably the above generalizes at least to the locally compact Hausdorff setting.

  • $\begingroup$ Does there exist a canonical complex for Cech cohomology whose elements do not have very large dimension? For this one they do, apparently $\endgroup$ – user74900 Jun 24 '18 at 16:25

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