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Sorry for this silly question but I can't find a reference.

Let $\mathcal C$ be a preadditive category and $A\in \mathcal C$ an object. Under what conditions (on $\mathcal C$) can one say that the functor $\textrm{Hom}_{\mathcal C}(-,A)$ is left exact? Is it true very often or is it specific to categories like module categories?

In particular, I would like to know if $\textrm{Hom}_{S}(-,X)$ is left exact in the category of $S$-schemes, where $X$ is a fixed $S$-scheme.

If $X$ and $S$ are both affine, say $X=\textrm{Spec}\, A$ and $S=\textrm{Spec}\, B$, then \begin{equation} \textrm{Hom}_S(-,X)\cong \textrm{Hom}_B(A,-), \end{equation} and the latter is left exact because $A$ is a $B$-module. But for arbitrary $S$ and $X$?

Thank you for any help.

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What is the group structure on $Hom_S(-,X)$? How do you define an exact sequence in the category of $S$-schemes? Note that $Hom_B(A,-)$ is not the group of $B$-module homs, its is the set of $B$-algebra homs. - If $C$ is abelian, then $Hom_{\mathcal C}(A, -)$ (and $Hom_{\mathcal C}(-,A)$) is always left exact, e.g. you can use the Freyd–Mitchell embedding thm and proove it in the category of $R$-modules for a suitable ring $R$. –  finite Jul 5 '12 at 22:41
    
I never know what kind of statement one can or cannot prove by using Mitchell Theorem. But probably in this situation one can use it. So you claim also that Sch/$S$ is not even preadditive, right? One can define the product of $f:Y\to X$ and $f':Y'\to X$ to be $f^\ast Y'$, but then I can see no "inverses". Can we already conclude that $\textrm{Hom}_S(-,X)$ is not left exact? –  Brenin Jul 5 '12 at 23:19
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Hom is always left exact, for any category whatsoever. To be precise, $\textrm{Hom}(-, X)$ maps all colimits to the corresponding limits. This is straightforward abstract nonsense and follows from the definition of colimit/limit. –  Zhen Lin Jul 5 '12 at 23:33
    
Thank you, Zhen Lin. And preserving limits (or turning colimits into limits) implies left exact for any category? Perhaps is it even equivalent? –  Brenin Jul 6 '12 at 11:23
    
@ZhenLin Please consider converting your comment into an answer, so that this question gets removed from the unanswered tab. If you do so, it is helpful to post it to this chat room to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see here, here or here. –  Julian Kuelshammer Jun 9 '13 at 21:09
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1 Answer

As Zhen Lin points out:

Hom is always left exact, for any category whatsoever. To be precise, Hom(−,X) maps all colimits to the corresponding limits. This is straightforward abstract nonsense and follows from the definition of colimit/limit.


In response to OP's subsequent comment:

Indeed, the two are equivalent, as mentioned on Wikipedia.

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