Let us consider the scheme $X$ and the coherent sheaf $\cal F$ on it. We consider finitely many affine open covers $U_{\lambda}$ with $\lambda \in \Lambda$. When I calculate Cech cohomology with $U_{\lambda}$'s such that the cardinality $|\Lambda| = n$, I need to put the label on the index set $\Lambda$ like

L: $\{$1,....,n$\}$ $\Leftrightarrow$ $\{\lambda \colon \lambda \in \Lambda\}$.

If the sheaf ${\cal F}$ is quasi-coherent, we have an isomorphism

$\theta_L \colon H_{\mathrm{cech}}^i(X,{\cal F}) \cong H_{\mathrm{zar}}^i(X, {\cal F})$ for $i \geq 0$.

For any $i \geq 0$ any representative $x \in H_{\mathrm{cech}}^i(X, {\cal F})$, if we consider it in the direct sum $\bigoplus \Gamma(U_{\lambda,...,\lambda'},{\cal F})$ over all combinations $\lambda,..,\lambda'$ of $i$ different $\lambda$'s, is $not$ affected by the labelling $L$ because only the differential morphism between Cech complex changes equivariantly. My question is

## Q: Is $\theta_L(x)$ same irrespective of the labelling $L$?

As far as I know once the labelling $L$ is fixed, the proof of $\theta_L$ being bijective is by using double complex $C^{\bullet,\bullet}$ having $\underset{k_1<k_2...<k_i}{\bigoplus} \Gamma(U_{\lambda_{k_1},...,\lambda_{k_i}}, {\cal I}^j)$ as its $(i,j)$-term, where ${\cal I}^{\bullet}$ is the injective resolution of $\cal F$. Because all homology $H^i(C^{\bullet,\bullet})$ are zero for $i > 0$ in both horizontal and vertical directions, the argument shows the single complex $D^{\bullet}$ associated to $C^{\bullet,\bullet}$ turns out

$H_{\mathrm{cech}}^i(X,{\cal F}) \cong H^{i}(D^{\bullet}) \cong H_{\mathrm{zar}}^i(X, {\cal F})$.

However this proof quite not shows the independence of the image of $x \in H_{\mathrm{cech}}^i(X,{\cal F})$ in $H_{\mathrm{zar}}^i(X, {\cal F})$ from the choice of $L$. I heartily wish someone to teach me about this. Many thanks. Pierre

notbe the same group, just 'merely' canonically isomorphic. I think once you write the canonical isomorphism $\rho : H^i_{cech}(\mathfrak{U}_L,\mathcal{F}) \rightarrow H^i_{cech}(\mathfrak{U}_{L'},\mathcal{F})$ then what you have is that $\theta_L = \theta_{L'} \circ \rho$, but that's just my intuition. $\endgroup$ – Joe Berner Apr 3 '14 at 3:00