I have a question about Proposition 6.5., Chap III (on page 234) from Hartshorne's
Algebraic Geometry.
The statement is:
Let $(X, \mathcal{O}_X$ be ringed space. Suppose there is an exact sequence
$$ ... \to \mathcal{L}_1 \to \mathcal{L}_0 \to \mathcal{F} \to 0$$
in $\mathcal{Mod}(X)$, the category of $\mathcal{O}_X$ modules, where the $\mathcal{L}_i$ are locally free sheaves of finite rank (in this case we say $\mathcal{L}_{\bullet}$. is a locally free resolution of $\mathcal{F}$). Then for any $\mathcal{G} \in \mathcal{Mod}(X)$ we have
$$ \mathcal{Ext}^i(\mathcal{F}, \mathcal{G}) \cong h^i(\mathcal{Hom}(\mathcal{L}_{\bullet}, \mathcal{G})). $$
here the $\mathcal{Ext}^i(\mathcal{F}, -)$ are the right derived functors of $\mathcal{Hom}(\mathcal{F}, -)$ and the right hand side the cohomology sheaves of complex $\mathcal{Hom}(\mathcal{L}_{\bullet}, \mathcal{G})$.
The proof says that both sides are $\delta$-functors in $\mathcal{G}$ from $\mathcal{Mod}(X)$ to $\mathcal{Mod}(X)$. For $i = 0$ they are equal, because then $\mathcal{Hom}(-,\mathcal{G})$ is contravariant and left exact. Both sides vanish for $i > 0$ when is $\mathcal{F}$ injective, because then $\mathcal{Hom}(-,\mathcal{G})$ is exact. So by (1.3A) they are equal.
Theorem 1.3A. on page 206 states: Let $T = (T^i)_{i \ge 0}$ be a covariant $\delta$-functor from $\mathcal{A}$ to $\mathcal{B}$ ($\mathcal{A}$, $\mathcal{B}$ are abelian categories). If $T^i$ is effaceable for each $i > 0$, then $T$ is universal. The definitions of effaceable and universal $\delta$-functor can be looked up on the sage page.
My question is: Why is it neccessary for the proof to assume that
the $\mathcal{L}_i$ are locally free? I see no reason why the same proof
shouldn't go through verbatim if we start with an arbitrary resolution
of $\mathcal{F}$, ie a family $(\mathcal{A}_i)_{i \ge 0}$ in
$\mathcal{Mod}(X)$ there is an exact sequence
$... \to \mathcal{A}_1 \to \mathcal{A}_0 \to \mathcal{F} \to 0$ of
$\mathcal{F}$.