Čech cohomology - independence of covering

Question

Let $X$ be a topological space and let $\mathcal{F}$ be a sheaf over $X$. Suppose we have a open covering $\mathcal{U}$ of $X$. When $\check{\mathrm{H}}^n(X,\mathcal{F})\cong \check{\mathrm{H}}^n(\mathcal{U},\mathcal{F}) $ ? (When Čech cohomology is independent of covering ?)

Answer 1

There is a sufficient condition called Leray Theorem: If the covering $\mathcal{U}$ is acyclic for the sheaf $\mathcal{F}$ in the sense that $$H^{k}(U_{i_1}\cap U_{i_2}\cap...\cap U_{i_p},\mathcal {F})=0 $$ for all $k>0$ and all finite collections $\{i_1,i_2,...,i_p\}$ then $H^{*}(X,\mathcal{F})=H^{*}(\mathcal{U},\mathcal{F}).$ The proof uses spectral sequence mechinary.