First Ext group of a sheaf Let $F$ be a quasicoherent sheaf on a scheme $X$, which is supposed to be sufficiently nice.
Does one then have a canonical isomorphism
$Ext^1(F,F) \simeq H^1(X, \underline{End}(F))$,
where with $\underline{End}(F)$ I denote the sheaf of endomorphisms of $F$.
I know that this holds for $F$ locally free, but I read an article where this iso was also used for quasicoherent $F$, but I don't see how to prove it.
 A: There is a spectral sequence $$E_2^{i,j} := H^i(X,\underline{Ext}^j(\mathcal E,\mathcal F)) \implies Ext^{i+j}(\mathcal E,\mathcal F),$$ for two sheaves
$\mathcal E$ and $\mathcal F.$ (Here $\underline{Ext}$ denotes sheaf Ext.)
Looking at the the low degree implications of this, one finds an exact sequence
$$0 \to H^1(X,
\underline{Hom}(\mathcal E,\mathcal F)) \to Ext^1(\mathcal E,\mathcal F) 
\to H^0(X,\underline{Ext}^1(\mathcal E,\mathcal F)) \to H^2(X,\underline{Hom}(\mathcal E,\mathcal F)) \to \cdots.$$
Taking $\mathcal E  = \mathcal F$, the first map is the one you are asking about.
If $\mathcal F$ is locally free than $\underline{Ext}^1$ vanishes, and one
gets the isomorphism that you recalled.  In general, this $\underline{Ext}^1$
need not vanish, the corresponding map in the exact sequence also need not vanish,
and so there will be an injection
$$H^1(X,\underline{End}(\mathcal F)) \hookrightarrow Ext^1(\mathcal F)$$
which is not surjective.
[Added: A typical example would come from taking $\mathcal F$ to be
a skyscraper at a point, with $X$ positive dimensional.]
