(I am reading Huybrechts's Complex Geometry)
I also looked some classic sources, but I didn't find anything addressing the below.
Proposition $B.0.35$
Let $$0\longrightarrow \mathcal{F}^0\longrightarrow \mathcal{F}^1\longrightarrow \mathcal{F}^2\longrightarrow 0$$ be a short exact sequence of sheaves on M. Then there exists a long exact cohomology sequence $$ 0\to H^0(M,\mathcal{F}^0)\to H^0(M,\mathcal{F}^1)\to H^0(M,\mathcal{F}^2)\to\ldots H^2(M,\mathcal{F}^2)\to\ldots $$
So now I think there is some abuse of notation here, so let me write what I think is happening. For every sheaf $\mathcal{F}^0,\mathcal{F}^1,\mathcal{F}^2$ there exist a resolution on the global sections i.e. For $ \mathcal{F}^{0}$ we have the long sequence
$$0\longrightarrow \mathcal{F}^{0}(M) \longrightarrow \mathcal{F}^{0}(M)^0\longrightarrow \mathcal{F}^{0}(M)^1\longrightarrow \mathcal{F}^{0}(M)^2\longrightarrow \ldots$$
And similarly with $ \mathcal{F}^{1}, \mathcal{F}^{2}$
And when we write $H^1(M,\mathcal{F}^0)$ we actually mean $\dfrac{ \operatorname{Ker}(\phi^1 :\mathcal{F}^{0}(M)^1 \to \mathcal{F}^{0}(M)^2}{\operatorname{Im}(\phi^0 : \mathcal{F}^{0}(M)^0\to \mathcal{F}^{0}(M)^1}$
My question is: how we map these elements in the long exact sequence ?
The obvious way (let's say for the above example) to map $H^1(M,\mathcal{F}^0)\longrightarrow H^1(M,\mathcal{F}^1) $ would be to map $[a]\in H^1(M,\mathcal{F}^0)$ via the map $\psi: \mathcal{F}^0\longrightarrow \mathcal{F}^1$ somehow.
Any help would be appreciated.