## Embedding abelian categories to have enough projectives

Is it true that the pro-objects of an abelian category form a category with enough projectives?

In general, given an abelian category A, is there a canonical way to embed it a bigger abelian category A' with enough projectives (or injectives) and such that A' is universal with respect to this property?

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 Matthew Emerton's answer sbseminar.wordpress.com/2009/09/10/20-questions/… to question 14 would be worth looking at. – Greg Stevenson Oct 17 2009 at 7:26 Thanks! Do you have a reference for the proof? – Akhil Mathew Oct 18 2009 at 3:40

It seems that Pro(A) does not have enough projectives in general. In Kashiwara-Schapira's book "Categories and Sheaves" they prove (corollary 15.1.3) that Ind(k-Mod) does not have enough injectives. This means, taking opposite categories, that Pro(k-Mod^{op}) does not have enough projectives.

I don't know of any universal way of adding enough projectives.

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I'm not sure this is quite what you're looking for, but if A is small, you can consider the (contravariant) Yoneda embedding of A into the category of left-exact functors from A to Ab. This is an exact full embedding, and the product of all representable functors is an injective cogenerator (this is nontrivial; it is not even obvious that left-exact functors form an abelian category). This is proven in Freyd's book Abelian Categories, and is a key part of his proof of the Mitchell Embedding Theorem. I don't know about any universal properties of this, but it is canonical.

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 Is this just secretly taking the ind-category? It seems true and I have seen it somewhere I can't remember I think, but I might be being dumb. – Greg Stevenson Oct 17 2009 at 7:31