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I've heard that given a ringed topos $(X,\mathcal{O}_X)$, the functor $Hom_{\mathcal{O}_X-\operatorname{Mod}}(\mathcal{O}_X, -)$ often fails to be exact. Is this only the case for the unenriched hom (global sections), i.e. $$Hom_{\mathcal{O}_X-\operatorname{Mod}}(\mathcal{O}_X, -):\mathcal{O}_X-\operatorname{Mod}\to \Gamma(\mathcal{O}_X,X)-\operatorname{Mod},$$ or is it also true in the case of the true Hom-sheaf, i.e. $$Hom_{\mathcal{O}_X-\operatorname{Mod}}(\mathcal{O}_X, -):\mathcal{O}_X-\operatorname{Mod}\to\mathcal{O}_X-\operatorname{Mod}?$$

Morally, it seems like the second case should be exact, if only for the reason that we've done nothing but relativize.

More generally, I've heard that free $\mathcal{O}_X$-modules can often fail to be projective. Is this another case of not considering the "enriched Hom object", but only considering its global sections?

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Dear Harry, It is an exercise to compute sheaf-Hom (what you call the Hom-sheaf) from $\mathcal O_X$ to any $\mathcal O_X$-module $\mathcal F$. Once you do this exercise, you will find that it is indeed exact. – Emerton Oct 31 '10 at 19:24
Dear Emerton, Yes, exactly! The computation is obvious. I wasn't understanding what people meant by "$\mathcal{O}_X$ is not projective", since it clearly is projective in the enriched category of $\mathcal{O}_X$-modules. – Harry Gindi Oct 31 '10 at 19:35
In particular, when people say that a sheaf is projective, they mean to say that it is projective in the $\Gamma(O_X,X)-\operatorname{Mod}$-enriched category of $O_X$-modules. That is, we're "changing the base of enrichment". It exactly generalizes restriction of scalars in commutative algebra (in which case, it's really no surprise that lots of things we would expect to be exact can fail to be exact). – Harry Gindi Oct 31 '10 at 19:42
For example, $\mathbf{F}_2$ is $\mathbf{F}_2$-projective, but not $\mathbf{Z}$-projective. I guess now my only question is: What is the cohomological significance of being $\Gamma(\mathcal{O}_X,X)$-projective instead of being "enriched projective"? – Harry Gindi Oct 31 '10 at 19:47
(Meanwhile, as you can see in Martin's post that he linked below, I did the computation noted by Emerton over two weeks ago). – Harry Gindi Nov 1 '10 at 8:06

Sheaf cohomology deals with the obstruction of the exactness of the global section functor; and we all know that this has been very important in the development of algebraic geometry. This somehow justifies why we should regard sheaves as objects of a usual category. But I think for other purposes it is also very good to regard sheaves as objects of a sheaf-enriched category, and try to imitiate homological algebra in these categories. Then all these properties of commutative algebra carry over to module sheaves. There is no distinction between local and global sections, since, by definition, every notion will be local. For example then $R$ is always projective over $R$, where $R$ is a ring in $Sh(X)$. See also my related question "abelian categories enriched over sheaves".

One of the purposes of this way of thinking about sheaves is that you can actually prove some theorems about them formally by writing down a intuitionistic proof for the corresponding known theorem in commutative algebra. See for example the paper "Intuitionistic algebra and representations of rings" by Christopher Mulvey how this works for Swan's Theorem. When you write down a intuitionistic definition for a free module, you get what is commonly called a locally free module.

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I am not sure what has been asked here but if you take a projective variety $X$, its category of coherent sheaves will have only one projective: the zero sheaf. In particular, $O_X$ is not projective.

For a general scheme, projectivity of $O_X$ would probably imply that $X$ is affine (sorry if this one stinks). Otherwise, $O_X$ would have higher cohomology.

Off course, if you replace HOM by the SHEAF-HOM, the functor SHEAF-HOM($O_X$, ) will be exact. As you say, nothing happens there. But SHEAF-HOM is a strange beast: a category needs HOM-s to be sets, not sheaves. You can probably talk of categories over a category (so your HOM-s are no longer sets but objects of another category), then $O_X$ wil be "projective over another category"...

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Yes, exactly. The idea would be something like: "$\mathcal{O}_X$ is $\mathcal{O}_X$-projective, but not $\Gamma{\mathcal{O}_X,X}$-projective. It's not really surprising if you think of $\Gamma(,X)$ as something like restriction of scalars. The important philosophical point is that morally, the base topos does not matter (this is the philosophy behind relative schemes that motivates things like relative spec and relative proj). – Harry Gindi Oct 31 '10 at 19:28
O'why shalt thou kill the best thing that ever happened to Algebraic Geometry, cohomology of projective varieties? Off course, if there is a big theorem you can prove by doing it, just kill the bastards! – Bugs Bunny Oct 31 '10 at 21:19

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