Why is $Ext_{\mathcal{S}}^{j}(H,G) = 0$ if $H$ is a free sheaf and $G$ a flabby sheaf?

Question

So, I am reading Schapira's and Kashiwara's "Sheaves on Manifolds" and in the proof of proposition 2.6.3, it is stated that, for a free $\mathcal{S}$-module $H$ and a flabby $\mathcal{S}$-module $K$, we have $\operatorname{Ext}_{\mathcal{S}}^{j}(H,K) = 0$ for $j \neq 0$. But why is this? I can't seem to incorporate the flabbiness of $K$ in my reasoning effectively, and I know that the fact that $H$ is free (and not necessarily finitely generated) is not enough. Can someone give me a hint?

Answer 1

As in the comments, we may recognize that a free sheaf is by definition isomorphic to $\bigoplus_{i\in I} \mathcal{O}_X$, so $$\operatorname{Ext}^j(H,K)\cong \operatorname{Ext}^j(\bigoplus_{i\in I} \mathcal{O}_X,K)\cong \prod_{i\in I} \operatorname{Ext}^j(\mathcal{O}_X,K)$$ and it suffices to determine $\operatorname{Ext}^j(\mathcal{O}_X,K)$. But $\operatorname{Hom}(\mathcal{O}_X,-)$ is isomorphic as a functor to $\Gamma(X,-)$, the global sections functor, so the derived functors of $\operatorname{Hom}(\mathcal{O}_X,-)$ are the derived functors of $\Gamma(X,-)$, that is, cohomology. So $\operatorname{Ext}^j(\mathcal{O}_X,K)\cong H^j(K)$, and now you may apply the fact that all higher sheaf cohomology vanishes for flasque sheaves.

(Your error in the comments was mis-identifying $\operatorname{Hom}(\mathcal{O}_X,-)$ as an exact functor - it is only left-exact in general. Perhaps you were mixing it up with the hom-sheaf functor, $\mathcal{Hom}(\mathcal{O}_X,-)$, which is actually exact, and in fact isomorphic to the identity functor on $\mathcal{O}_X$-mod.)