Mitchell's embedding theorem - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T21:36:22Z http://mathoverflow.net/feeds/question/32173 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/32173/mitchells-embedding-theorem Mitchell's embedding theorem Agusti Roig 2010-07-16T14:04:45Z 2010-09-25T04:15:41Z <p>Mitchell's embedding theorem <a href="http://en.wikipedia.org/wiki/Mitchell%27s_embedding_theorem" rel="nofollow">http://en.wikipedia.org/wiki/Mitchell%27s_embedding_theorem</a> tells us that every small abelian category ${\cal A}$ has a full, faithful and exact embedding $V : {\cal A} \longrightarrow \mathbf{Mod}_R$ in a category of $R$-modules, for some ring $R$.</p> <p>Now, $V$ being exact is the same as saying that it preserves all <em>finite</em> limits and colimits.</p> <p>I would be glad to know if Mitchell's embedding theorem could be improved in order to have that $V$ preserves also:</p> <p>(a) arbitrary products, and (b) filtered colimits.</p> <p>Or, alternatively,</p> <p>(c) injective objects.</p> <p>Or, which conditions on the abelian category ${\cal A}$ would guarantee (a) and (b), or (c)? Are there any results in these directions?</p> <p>The reason behind my question is the following: I realized that my answer to my previous question <a href="http://mathoverflow.net/questions/29883/vanishing-theorems" rel="nofollow">http://mathoverflow.net/questions/29883/vanishing-theorems</a> is wrong: sheaf cohomology is not defined uniquely in terms of exact sequences, so the fact that $V$ is exact doesn't guarantee that $H^n(X; {\cal F}) = H^n(X; V({\cal F}))$ as I claimed. But, if I had (a) and (b), I could say that $V$ preserves Godement resolutions. And if I had (c), $V$ would preserve injective resolutions.</p> http://mathoverflow.net/questions/32173/mitchells-embedding-theorem/39878#39878 Answer by Dylan Wilson for Mitchell's embedding theorem Dylan Wilson 2010-09-24T17:55:48Z 2010-09-25T04:15:41Z <p>EDIT: I'm completely rewriting this in detail now that I <em>think</em> I've worked it out. Lots of possibilities for mistakes here, so stay alert!</p> <p>I claim that the generalization to include $(a)$ fails. My counterexample goes like this... Take $A$ to be any small, complete (that's small AND complete) category with no nonzero projectives. If the embedding into $R-mod$ given by Mitchell preserved arbitrary products, then it would be continuous since $A$ has equalizers and any limits can be built from products and equialisers (where equalisers are preserved by exactness). </p> <p>Now, for each $x \in R-mod$, consider the index <em>set</em> $I = {f: x \rightarrow Va \vert a\in A} = \bigcup_{a \in A} Hom(x, Va)}$. (This is a set since hom-sets are small, and all of $A$ is a set.) Now, any $x \rightarrow Va$ can be realized as $x \rightarrow Va \rightarrow Va$ (with the identity), so we have verified the "solution set condition" of the adjoint functor theorem. If all is good, we may conclude that this embedding has a left adjoint $R-mod \rightarrow A$. </p> <p>Now, left adjoints of exact functors preserve projective objects (see Weibel). Now, if we choose some $b \in A$ that doesn't map to zero under the embedding $V$, and some free module $a \in R-mod$ that maps nontrivially to $b$, then by the bijection on hom-sets we get from the adjunction, we may conclude that $a$ maps to some nonzero element in $A$. But this is a contradiction, as the only projective elements of $A$ are zero.</p> <p>How does that look?</p>