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Where we use assumption that resolution of $\mathcal{G}$ must be locally free of finite rank? It seems natural to me, that $h^0(\mathcal{Hom(L.,G))}$ must be equal to $\mathcal{Hom(F,G)}$ without that assumption (thats because $\mathcal{Hom(?,G)}$ is left exact contravariant functor), and $\mathcal{Hom(?,I)}$ must be exact whenever $\mathcal{I}$ is injective object without that assumption too (thats just another definition of injective object). I am confused, hope for your help!

Hartshorne page screen

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If $\mathscr{L}_.$ is not a complex of locally free sheaves, then $h^i(\mathscr{H}om(\mathscr{L}_.,\mathscr{G}))$ is not a $\delta$-functor in $\mathscr{G}$. Indeed, given a short exact sequence $0\rightarrow\mathscr{G}'\rightarrow\mathscr{G}\rightarrow\mathscr{G}''\rightarrow 0$, the sequence $$0\rightarrow\mathscr{H}om(\mathscr{L}_i,\mathscr{G}')\rightarrow\mathscr{H}om(\mathscr{L}_i,\mathscr{G})\rightarrow\mathscr{H}om(\mathscr{L}_i,\mathscr{G}'')\rightarrow 0$$ must be exact for every $i$. Otherwise, there won't be any connecting homomorphism.

If the $\mathscr{L}_i$ are locally free, then $\mathscr{H}om(\mathscr{L}_i,.)$ are exact functors, so the right hand side is indeed a $\delta$-functor.

By the way, if you take the trivial resolution of $\mathscr{F}$, namely $0\rightarrow\mathscr{F}\rightarrow\mathscr{F}\rightarrow 0$, then $h^i(\mathscr{H}om(\mathscr{L}_.,\mathscr{G}))$ would be $0$ for $i>0$. Clearly this is not right, it can't be isomorphic to $\mathscr{E}xt^i(\mathscr{F,G})$ otherwise $\mathscr{E}xt^i$ would not be interesting at all.

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    Thank you! It means that we can use that argument for any resolution $\mathcal{L. \to G}$ for which functor $\mathcal{Hom(L.,?)}$ is exact? – 2017-02-27
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    @kp9r4d Yes but in practice you know only that $\mathscr{H}om(\mathscr{L},.)$ is exact when $\mathscr{L}$ is a locally free sheaf. If $\mathscr{L}$ is coherent, I think this is equivalent, because on an affine subset it will be projective hence locally free. – 2017-02-27
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I was wondering about this only this morning! Could somebody be so kind as to read my interpretation and tell me if it is correct or not?

First, $\mathcal{Hom}(\mathcal L, \ . \ )$ is an exact functor if $\mathcal L$ is locally free. This is because $\mathcal{Hom}(\mathcal L, \ . \ )$ locally looks the same as $\mathcal{Hom}(\mathcal O, \ . \ )$ - just restrict to a trivialising neighbourhood.

So given a short exact sequence, $$ 0 \to \mathcal F \to \mathcal G \to \mathcal H \to 0,$$ we get a short exact sequence of complexes, $$ 0 \to \mathcal{Hom} (\mathcal L^\bullet, \mathcal F) \to \mathcal{Hom} (\mathcal L^\bullet, \mathcal G)\to \mathcal{Hom} (\mathcal L^\bullet, \mathcal H) \to 0.$$

This short exact sequence induces a long exact sequence by the snake lemma (which I believe is valid in all abelian categories): $$ \dots \to h^i (\mathcal{Hom} (\mathcal L^\bullet, \mathcal F) ) \to h^i (\mathcal{Hom} (\mathcal L^\bullet, \mathcal G) ) \to h^i ( \mathcal{Hom} (\mathcal L^\bullet, \mathcal H) ) \to h^{i+1} (\mathcal{Hom} (\mathcal L^\bullet, \mathcal F) ) \to \dots $$ I believe the existence of this long exact sequence is what Hartshorne is refering to when he says that $h^i(\mathcal {Hom}, . \ )$ is a $\delta$-functor.

Like I said, I would be very grateful indeed if somebody could check this - even if there is a better answer to the question.

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    You beat me to it ;) And yes you are completely right, the locally free hypothesis is what makes the right hand side a $\delta$-functor by the procedure you wrote. – 2017-02-26
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    And what about the snake lemma? Is this valid here? My worry is that I only learned how to prove the snake lemma for modules in class. We used diagram chasing. You can't do that in a general abelian category. – 2017-02-26
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    Yes, the snake lemma is valid in any abelian category. There are direct proofs, but there is also the Freyd–Mitchell theorem which says that we can do diagram chasing in any abelian categories. – 2017-02-26
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    Oh wow, that's really nice! Thank you for looking through my answer - I feel much better now. – 2017-02-26
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    @Roland May I trouble you with one more question? Hartshorne often proves that functors are the same by showing that they are "coeffaceable". I have no idea what this means - his definition is impenetrable. Whenever I see this word, I ignore it and construct my own argument. – 2017-02-27
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    My argument goes like this: (i) I show the "$H^0$"'s of the two functions agree, (ii) I show that both functors give LESs, (iii) I show that for every sheaf $\mathcal F$, there is an injective morphism $\mathcal F \to \mathcal M$, where $\mathcal M$ is a sheaf such that $H^1, H^2, H^3, \dots$ vanish for both functors. Then I apply the obvious inductive argument based on the two LESs. Is this equivalaent to what Hartshorne means by "coeffaceability"? – 2017-02-27
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    oops, I meant to say "$H^0$s of the two functors" – 2017-02-27
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    An effaceable functor is a functor $F$ such that for each $A$, there is a monomorphism $A\rightarrow M$ with $F(M)=0$. But what you wrote is the standard trick to prove that a $\delta$-functor $(T^i)$ is isomorphic to $R^iT^0$. (By the way, coeffaceable is for epimorphism $L\rightarrow A$ and are used for left-derived functor) – 2017-02-27
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    Thank you! "I was wondering about this only this morning!" - glad to see that not only me suffering. :3 – 2017-02-27
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    @Roland That makes perfect sense now. Thank you yet again! – 2017-02-27