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Let $X$ be a topological space, with open covers $\mathfrak{U} = (U_i)_{i\in I}$ and $\mathfrak{V} = (V_j)_{j\in J}$ such that $\mathfrak{V}$ is a refinement of $\mathfrak{U}$ for totally ordered sets $I$ and $J$, i.e. there exists some map of sets $\lambda: J\to I$ such that for all $j\in J$, $V_j\subseteq U_{\lambda(j)}$. Recall that we define $U_{i_0,\ldots,i_p} = U_{i_0}\cap\cdots\cap U_{i_p}$ for $i_0,\ldots,i_p\in I$ and we define, for an abelian sheaf $\mathscr{F}$ on $X$, $$\check{C\,}^p(\mathfrak{U},\mathscr{F}) = \prod_{i_0<\cdots <i_p}\mathscr{F}(U_{i_0,\ldots,i_p})$$ with the boundary maps as usual, so that taking homology gives us $\check{H\,}^i(\mathfrak{U},\mathscr{F})$, and likewise for $\mathfrak{V}$. Now, I wish to define natural maps $\lambda^i:\check{H\,}^i(\mathfrak{U},\mathscr{F})\to \check{H\,}^i(\mathfrak{V},\mathscr{F})$, and I suspect that the way to do this is by defining a map of complexes $\lambda:\check{C\,}^\bullet(\mathfrak{U},\mathscr{F})\to \check{C\,}^\bullet(\mathfrak{V},\mathscr{F})$. However, I'm having some difficulty defining the map $\lambda^p: \check{C\,}^p(\mathfrak{U},\mathscr{F})\to \check{C\,}^p(\mathfrak{V},\mathscr{F})$. If I fix some element $$\alpha = (\alpha_{i_0,\ldots,i_p})_{i_0<\cdots<i_p}\in\check{C\,}^p(\mathfrak{U},\mathscr{F})$$ then I wish to define something along the lines of $$\lambda^p(\alpha)_{j_0,\ldots,j_p} = \alpha_{\lambda(j_0),\ldots,\lambda(j_p)}$$ but this may not be defined. The fix is most likely to extend the definition of $\check{C\,}^\bullet(\mathfrak{U},\mathscr{F})$ to include nonincreasing tuples with the usual sign convention, but will this give the same homology?

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When I learned about Cech cohomology, I was told that $\check{C^p}(\mathfrak U, \mathcal F)$ consists of skew-symmetric elements of $\prod_{i_0, \dots, i_p}\mathcal F(U_{i_0 \dots i_p}),$ i.e. elements satisfying $\sigma_{i_0, \dots, i_k, i_{k+1}, \dots i_p} = - \sigma_{i_0, \dots, i_{k+1}, i_k, \dots, i_p}$. So $\sigma_{i_0, \dots, i_p}$ does exist for $\{i_0, \dots, i_p\}$ non-increasing. There is even no need to define a total ordering on the indexing set in the first place. Of course, including $\sigma_{i_0 \dots i_p}$ with non-increasing $\{i_0, \dots, i_p\}$ makes no difference to the cohomology groups.

Your map, $\lambda^p(\alpha)_{j_0, \dots, j_p} = \alpha_{\lambda(j_0), \dots, \lambda(j_p) }|_{V_{j_0 \dots j_p}}$ is a perfectly good definition for the restriction map. Since this restriction map commutes with the boundary operator, it descends to a well-defined map on cohomology groups.

One thing we should check is that the restriction map $\check H^p(\mathfrak U, \mathcal F) \to \check H^p(\mathfrak B, \mathcal F)$ is independent of the choice of $\lambda : J \to I$. Indeed, for a given $V_j$, there may well be more than one $U_i$ such that $V_j \subset U_i$. Fortunately, the restriction map on cohomology is independent of the choice of $\lambda$, and this can be proved using a chain homotopy argument. I've never bothered to go through the argument, but according to the lecture notes from my masters course, a proof is given in Kodaira's "Complex Manifolds and Deformations of Complex Structures" as lemma 3.2 on page 117.

KReiser
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Kenny Wong
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  • The map $j\to \lambda(j)$ may not be injective, therefore, in $\alpha_{\lambda(j_0), ...,\lambda(j_p)}$ there could be repeated indices. In this case, should $\alpha_{\lambda(j_0), ...,\lambda(j_p)}$ be understood as zero? – Yuval Mar 16 '24 at 07:16
  • @Yuval You can and probably should define Cech cohomology with all indices, you care not if there are repetitions or non-alternating simplices. These would all disappear in cohomology anyway, as any cocycle is necessarily alternating – FShrike Apr 11 '24 at 20:29
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Invariance of the choices of restriction map on Cech cohomology indeed follow by a homotopy argument, as Kenny suggests. I'll write up and paraphrase the argument of Godement here.

Let $X$ be a space, $\mathscr{F}$ a presheaf, $\mathfrak{V}\preceq\mathfrak{U}$ open covers. Let $\alpha,\beta$ be two choices of refinement function $\mathfrak{V}\to\mathfrak{U}$. Model Cech cohomology w.r.t a cover as the cohomology of the cosimplicial object $n\mapsto\prod_{(W_0,\cdots,W_n)\in\mathfrak{W}^{n+1}}\mathscr{F}(W_0\cap\cdots\cap W_n)$ with the standard face and degeneracy maps. I impose no ordering or antisymmetry conditions or whatever else; this all disappears in cohomology and is less natural to have in the raw definition, in my opinion.

There are maps of complexes $\alpha^\ast,\beta^\ast:C^\bullet(\mathfrak{U};\mathscr{F})\to C^\bullet(\mathfrak{V};\mathscr{F})$. These are chain homotopic by the following simplicially inspired homotopy: $$h_n:C^n(\mathfrak{U};\mathscr{F})\to C^{n-1}(\mathfrak{V};\mathscr{F})\\\large h_n(\phi)(V_0,V_1,\cdots,V_{n-1}):=\sum_{j=0}^{n-1}(-1)^k\phi_{\alpha(V_0),\alpha(V_1),\cdots,\alpha(V_j),\beta(V_j),\beta(V_{j+1}),\cdots,\beta(V_{n-1})}$$You can directly check this works. But more conceptually $C^\bullet(\mathfrak{W};\mathscr{F})=\hom_k(K^\bullet(\mathfrak{W});\mathscr{F})$ as homomorphisms of $k$-presheaves, if $\mathscr{F}$ is assumed a $k$-presheaf (take $k=\Bbb Z$ if you just want (pre)sheaves of Abelian groups) and ($k$ is just some commutative ring) $K^\bullet(\mathfrak{W})$ is the following simplicial sheaf:

$$K^n(\mathfrak{W}):=\bigoplus_{(W_0,\cdots,W_n)\in\mathfrak{W}^{n+1}}(j_{W_0\cap\cdots\cap W_n})_!\underline{k}$$With obvious face and degeneracy maps, where $(j_W)_!$ is the right adjoint of $j^\ast,j:W\hookrightarrow X$ for presheaves. So explicitly $(j_W)_!$ for a presheaf $\mathscr{G}$ on $W$ is the presheaf on $X$ given by $\Omega\mapsto\begin{cases}0&\Omega\not\subseteq W\\\mathscr{G}(\Omega)&\Omega\subseteq W\end{cases}$ and obvious restrictions. $\underline{k}$ is the constant presheaf with coefficients in $k$ (with convention $\underline{k}(\emptyset):=0$). $\alpha^\ast,\beta^\ast$ just arise from the simplicial maps $K^\bullet(\mathfrak{V})\to K^\bullet(\mathfrak{U})$ which do what you expect.

Thus to show $\alpha^\ast,\beta^\ast$ are chain homotopic, since $\hom_k(-,\mathscr{F})$ is an additive functor, it suffices to show these simplicial maps are chain homotopic. Oh, but for that it suffices to show they are simplicially homotopic.

Well, to be fair I can't quite phrase it using that formalism since the simplicial set $\Delta^1$ isn't a $k$-sheaf in any obvious way or doesn't have an obvious analogue to my eyes. But nevertheless for any $n$-simplex of the interval $\sigma:[n]\to[1]$ we have a fairly reasonable map of sheaves $K^n(\mathfrak{V})\to K^n(\mathfrak{U})$, which directly moves between $\alpha$ and $\beta$, just given componentwise by $(j_{V_0\cap\cdots\cap V_n})_!\underline{k}\to(j_{\gamma_0(V_0),\cdots,\gamma_n(V_n)})_!\underline{k}\hookrightarrow K^n(\mathfrak{U})$ where $\gamma_j$ is either $\alpha$ or $\beta$ depending on the value of the signal $\sigma(j)$. This obviously varies compatibly with the face and degeneracy maps and ranges from $\alpha\to\beta$; we have a "simplicial homotopy" and we get a chain homotopy from that in the canonical way.

FShrike
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