Is the dualizing functor $\mathcal{Hom}( \cdot, \mathcal{O}_{X})$ exact?

In Hartshorne's Algebraic Geometry II.8.20.1 (page 182), he takes the dual of Euler sequence

$$0 \rightarrow \Omega_{X/k} \rightarrow \mathcal{O}_{X}(-1)^{n+1} \rightarrow \mathcal{O}_{X} \rightarrow 0,$$

where $X = \mathbb{P}_{k}^{n}$, and get

$$0 \rightarrow \mathcal{O}_{X} \rightarrow \mathcal{O}_{X}(1)^{n+1} \rightarrow \mathscr{T}_{X} \rightarrow 0,$$

where $\mathscr{T}_{X} = \mathcal{Hom}(\Omega_{X/k}, \mathcal{O}_{X})$ is the tangent sheaf of $X$. But, is the dualizing functor $\mathcal{Hom}( \cdot, \mathcal{O}_{X})$ exact? Is not it only a left exact contravariant functor? Why in this case we have exactness of the sequence?

Solution 1:

If you have a short exact sequence

$0\rightarrow\mathscr{F}\rightarrow\mathscr{G}\rightarrow\mathscr{H}\rightarrow 0$

of finite locally free sheaves on a scheme $X$, then the sequence

$0\rightarrow\mathscr{H}^\vee\rightarrow\mathscr{G}^\vee\rightarrow\mathscr{F}^\vee\rightarrow 0$

is exact. The reason is that exactness can be checked on stalks, and because the sheaves in the original sequence are finitely presented, taking stalks commutes with taking $\mathcal{H}om$ sheaves, so the sequence of stalks of the second sequence is

$0\rightarrow\mathrm{Hom}_{\mathscr{O}_{X,x}}(\mathscr{H}_x,\mathscr{O}_{X,x}) \rightarrow\mathrm{Hom}_{\mathscr{O}_{X,x}}(\mathscr{G}_x,\mathscr{O}_{X,x}) \rightarrow\mathrm{Hom}_{\mathscr{O}_{X,x}}(\mathscr{F}_x,\mathscr{O}_{X,x})\rightarrow 0$

which is exact because the functor $\mathrm{Hom}_{\mathscr{O}_{X,x}}(-,\mathscr{O}_{X,x})$ is exact on short exact sequences of finite free $\mathscr{O}_{X,x}$-modules.

I'm not sure what happens when the sheaves in the sequence are not finite locally free. There is a long exact sequence of $\mathcal{E}xt$ sheaves. Note that the sheaves in the OP's original sequence are all finite locally free.

EDIT: Incidentally, this has nothing to do with schemes, and works for arbitrary locally ringed spaces.