most recent 30 from http://mathoverflow.net 2013-05-19T05:38:03Z http://mathoverflow.net/feeds/question/827 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/827/embedding-abelian-categories-to-have-enough-projectives Akhil Mathew 2009-10-17T02:06:55Z 2009-10-17T16:47:36Z

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

http://mathoverflow.net/questions/827/embedding-abelian-categories-to-have-enough-projectives/845#845 Eric Wofsey 2009-10-17T07:09:28Z 2009-10-17T16:35:23Z

<p>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.</p>

http://mathoverflow.net/questions/827/embedding-abelian-categories-to-have-enough-projectives/892#892 Abdó Roig-Maranges 2009-10-17T16:47:36Z 2009-10-17T16:47:36Z

<p>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.</p> <p>I don't know of any universal way of adding enough projectives.</p>