# Pure Quotient and pure sub-object

Let $\mathcal{C}$ be the category of modules over a ring. Let also $\mathcal{F}$ be a class of objects in $\mathcal C$ closed under pure subobject (pure quotient) and direct limit. Is $\mathcal{F}$ closed under pure quotient (pure subobject)?

What can be said about the category of quasi-coherent sheaves?

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What do you mean for "pure subobject" (pure quotient)? –  Buschi Sergio Jul 26 '13 at 21:53
Let $\cal C$ be a category with tensor product (For example the category of modules over a ring, the category of representations of a quiver by modules, the category of sheaves of abelian groups,...). Let $$\varepsilon: 0\to A\to B\to C\to 0$$ be an exact sequence and $D$ be an arbitrary object. $\varepsilon$ is pure if The sequence $D\otimes \varepsilon$ is exact for each $D$. –  Gholam Jul 27 '13 at 10:07
So, your abelian category is assumed to have a monoidal structure too. Do you want to assume any properties of the monoidal structure (e.g. how it interacts with colimits)? I think the question would be improved if you stated your assumptions. You can edit it; that's better than adding comments or leaving answers to your own question. –  Tom Leinster Aug 13 '13 at 18:12
Actually by the category I mean very well known categories such as the category of modules over a ring, the category of sheaves of $\mathcal{O}_X$-modules, the category of quasi-coherent sheaves, the category of representations of a quiver by a modules and etc. All of these categories are Grothendieck categories. –  Gholam Aug 13 '13 at 18:42
I suggest you edit your question to make your assumptions clear. At the moment it doesn't quite make sense, because "pure" doesn't make sense in the absence of a tensor product. As I said, that would be better than adding comments, and will make it more likely that people will answer your question. –  Tom Leinster Aug 13 '13 at 19:08

I think the answer should be yes, because cokerenls are direct limits. Let $\mathcal{G}$ be a pure subsheaf of $\mathcal{H}$ and consider $$0\to \mathcal{G}\to \mathcal{H}\to \frac{\mathcal{H}}{\mathcal{G}}\to 0.$$ Since $\mathcal{G}$ and $\mathcal{H}$ are both in $\mathcal{F}$ and $\frac{\mathcal{H}}{\mathcal{G}}$ can be viewd as a direct limit, all terms in the above short exact sequence are in $\mathcal{F}$.