Map of Čech cohomology induced by refinement

Question

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?

Answer 1

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.