# Question on an exercise in Hartshorne: Equivalence of categories

This is a slight reformulation of exercise II.5.9.(c) in Hartshorne's "Algebraic Geometry" which I don't understand.

Let $K$ be a field and $S=K[X_0,\ldots,X_n]$ a graded ring. Set $X=Proj(S)$ and let $M$ be a graded $S$-module. The functors $\Gamma_*$ definied by $$\Gamma_*(\mathcal{F})=\bigoplus_{n\in\mathbb{Z}} (\mathcal{F}(n))(X)$$ and $~\widetilde{\phantom{\cdot}}~$ (the "graded associated sheaf functor", see Hartshorne II.5. page 116 for a definition) induce an equivalence of categories between the category $\mathcal{A}$ of quasi-finitely generated (i.e. in relation to a finitely generated module) graded $S$-modules modulo a certain equivalence relation $\approx$ and the category $\mathcal{B}$ of coherent $\mathcal{O}_X$-modules. The equivalence relation is: $M\approx N$ if there is an integer $d$ such that $\oplus_{k\geq d}M_k\cong\oplus_{k\geq d}N_k$.

I don't know what an "equivalence of categories" is in this context. Formally an "equivalence of categories" means in particular that there are isomorphisms $$\hom_\mathcal{A}(M,N)\cong \hom_\mathcal{B}(\widetilde{M},\widetilde{N})$$ and $$\hom_\mathcal{B}(Y,Z)\cong \hom_\mathcal{A}(\Gamma_*(Y),\Gamma_*(Z))$$ of sets. This is my problem: How is the sheaf $\mathcal{H}om_\mathcal{B}(Y,Z)$ considered as a set? Perhaps it should be $\Gamma_*(\mathcal{H}om_\mathcal{B}(Y,Z))\cong \hom_\mathcal{A}(\Gamma_*(Y),\Gamma_*(Z))$?

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An equivalence of categories $\mathcal{C}$ and $\mathcal{D}$ consists of a pair of functors, $F:\mathcal{C}\rightarrow\mathcal{D}$ and $G:\mathcal{D}\rightarrow\mathcal{C}$, such that $F\circ G$ is naturally isomorphic to the identity on $\mathcal{D}$ and $G\circ F$ is naturally isomorphic to the identity on $\mathcal{C}$. –  Keenan Kidwell Mar 23 '10 at 12:34
Read Serre's FAC... –  Shizhuo Zhang Mar 23 '10 at 12:45
In the category of coherent $\mathcal{O}_X$ modules, the morphisms from $E$ to $F$ are the global sections of the sheaf $\mathcal{H}om(E,F)$. This object is sometimes denoted $\mathrm{Hom}(E,F)$, to distinguish it from $\mathcal{H}om(E,F)$. –  David Speyer Mar 23 '10 at 13:22
(and to go along with Shizhuo Zhang's comment) ...and it is probably in Euler. –  Mariano Suárez-Alvarez Mar 23 '10 at 15:10

The homomorphisms in the category of sheaves are not sheaves themselves. The hom sheaves have the data of things that are only homomorphisms over open subsets. So if $Y,Z$ are coherent $\mathcal{O}_X$-modules and you are looking for $\mathcal{O}_X$-module homomorphisms, you don't actually get $\mathcal{H}om(X,Y)$, what you actually get are the global sections only, because these are the only homomorphisms that are defined on the whole space, and so the only actual homomorphisms in the category of sheaves.
Thank you, Charles. Are you speaking of the "usual" global sections or do one get in fact an isomorphism of graded modules $\Gamma_*(\mathcal{H}om_\mathcal{B}(Y,Z))\cong\hom_\mathcal{A}(\Gamma_*(Y),\Gam‌​ma_*(Z))$? –  roger123 Mar 23 '10 at 13:35