Connecting homomorphism in Cech cohomology

Question

Let $M$ be a smooth manifold and $\mathcal{U}$ be a good open cover of $M$. If I have an exact sequence of sheaves

$$0 \longrightarrow A \stackrel{f}\longrightarrow B \stackrel{g}\longrightarrow C \longrightarrow 0,$$ then there is an exact long sequence from Cech's cohomology under what chances?

$$...\rightarrow \check{H}^{q}(\mathcal{U}, A) \rightarrow \check{H}^{q}(\mathcal{U}, B) \rightarrow \check{H}^{q}(\mathcal{U}, C) \stackrel{\delta^q} \rightarrow \check{H}^{q+1}(\mathcal{U}, A) \rightarrow ...$$

How would connecting homomorphism $\delta^q$ be? Can you recommend any literature that deals with this?

Appreciate.

Answer 1

A short exact sequence of sheaves will give you a sequence of Cech complexes $0\to \mathcal{\check{C}}^\bullet(\mathcal{U}, A)\to \mathcal{\check{C}}^\bullet(\mathcal{U}, B)\to \mathcal{\check{C}}^\bullet(\mathcal{U}, C)\to 0$, which is in general not exact on the right and the connecting homomorphism has to be defined by going to a refinement (see the proof in https://stacks.math.columbia.edu/tag/09V2). However if $H^1(U_{i_0,\ldots, i_n}, A) =0$ for all $n\ge 0$ and all $i_0,\ldots, i_n$ (since you assume that $\mathcal{U}$ is a good cover then this is true if for example $A$ is a locally constant sheaf) then the above sequence of Cech complexes is exact on the right and the connecting homomorphism is the usual one defined by diagram chasing (https://stacks.math.columbia.edu/tag/0111).