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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?

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The answer depends on how much you know. The next term in the sequence would be an Ext group, but $\mathcal{O}_X$ is free and hence it is $0$. –  Matt Sep 8 '12 at 14:03
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Dear @Matt, If you're thinking about $\mathrm{Ext}$ groups of, say, $R$-modules, then $\mathrm{Ext}_R^i(\cdot,\cdot)$ kills injective modules in the second variable, and $R$ is not necessarily self-injective (there are certain kinds of rings which are, such as fields, or the rings $\mathbf{Z}/p^n\mathbf{Z}$). Also, in the case of sheaves, while you do get a long exact sequence of $\mathcal{E}xt$ sheaves, the contravariant internal $\mathcal{H}om$ functor doesn't usually have derived functors because the category of $\mathscr{O}_X$-modules doesn't have enough projectives. –  Keenan Kidwell Sep 8 '12 at 15:09
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Proposition III.6.3, part (b) of Hartshorne's Algebraic Geometry states that $\mathscr{Ext}^i(\mathscr{O}_X,\mathscr{G})=0$ for $i>0$, which basically says that the sheaf Ext is acting like it 'should' even though we don't have projectives(that is, it effaces free sheaves) –  John Stalfos Sep 8 '12 at 15:43
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I'm sure the point Matt is making is that the next term in the sequence is $\mathscr{E}xt^1(\mathscr{O}_X, \mathscr{O}_X)$, which is zero by the observation. –  Zhen Lin Sep 8 '12 at 16:04
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Ah, of course! I didn't even notice that the last term was $\mathscr{O}_X$ when I wrote my answer\comment. I was just thinking about answering the question of whether or not $\mathcal{H}om_{\mathscr{O}_X}(-,\mathscr{O}_X)$ was always exact. Thank you for pointing this out, and to @Matt, my apologies. –  Keenan Kidwell Sep 8 '12 at 16:14
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1 Answer

up vote 7 down vote accepted

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.

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