In Lemma III 7.4 of Hartshorne we have $X$ is a closed subscheme of co-dim $r$ in $P=\mathbb P_k^N$. $\mathscr F$ is a coherent sheaf on $X$. Then $\mathscr F$ is naturally an $\mathcal O_P$ module. We have to show
$$ Hom_X (\mathscr F, \omega_X)=Ext ^r_P(\mathscr F,\omega_P) $$
Hartshorne takes an injective resolution of $\omega_P$ as an $\mathcal O_P$ module $$0\rightarrow\omega_P\rightarrow \mathcal {\mathcal I}^{\bullet}$$
Then Hartshorne says since $\mathscr F$ is an $\mathcal O_X$ module any morhpism $\mathscr F \rightarrow \mathcal I^i$ factors through $\mathscr I^i :=\mathscr {Hom} _P(\mathcal O_X, \mathcal I^i)$
I have a problem at this stage.
I know $\mathscr F \simeq \mathscr F \otimes _{\mathcal O_P} \mathcal O_X$ as $\mathcal O_P$ modules and hence we have via the adjunction formula
$$Hom_P(\mathscr F, \mathcal I^i)\simeq Hom_P(\mathscr F, \mathscr {Hom}(O_X, \mathcal I^i))$$
Is it because of this reason. It is a little vague to me so I shall be highly obliged if someone spells out the details to me.
Best Answer
I think your reasoning is okay. Alternatively, let $\mathcal{J}$ denote the sheaf of ideals picking out $X$. Since the image of any morphism from a $\mathcal{J}$-torsion sheaf is again $\mathcal{J}$-torsion and $\mathcal{O}_X$ is $\mathcal{J}$-torsion, all we need to do is to identify the $\mathcal{J}$-torsion of $\mathcal{I}^i$ and then we know that our morphism factors through that (via the inclusion of the torsion in to $\mathcal{I}^i$). But this torsion is easy to figure out - the $\mathcal{J}$ torsion is just $\mathcal{Hom}_P(\mathcal{O}_P/\mathcal{J},\mathcal{I}^i)=\mathcal{Hom}_P(\mathcal{O}_X,\mathcal{I}^i)$, so the image of our map factors through this as Hartshorne says.