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Mitchell's embedding theorem http://en.wikipedia.org/wiki/Mitchell%27s_embedding_theorem 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 $R$.

Now, $V$ being exact is the same as saying that it preserves all finite limits and colimits.

I would be glad to know if Mitchell's embedding theorem could be improved in order to have that $V$ preserves also:

(a) arbitrary products, and (b) filtered colimits.

Or, alternatively,

(c) injective objects.

Or, which conditions on the abelian category ${\cal A}$ would guarantee (a) and (b), or (c)? Are there any results in these directions?

The reason behind my question is the following: I realized that my answer to my previous question vanishing theorems 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.

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Interesting question. What happens when you go through the proof of the theorem and try to check the properties? –  Martin Brandenburg Jul 16 '10 at 16:37
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My advice (based on much experience) is to never ever use the Mitchell embedding theorem. It promotes not paying attention to what is going on, and I have never seen any situation anywhere for which this theorem is of much use anyway. Once you learn the trick of "chasing members" from MacLane's book "Categories for the working..." it ceases to have any purpose at all (as far as I know). Better off to acquire the skill of working in abelian categories by thinking functorially. I would be interested to see even one example where ignoring this fact really creates a problem. –  BCnrd Jul 17 '10 at 3:18
    
@Martin. Have you seen the proof? (For instance, in Borceux's book.) That's what I was trying NOT to do (at least, not before being sure that no one has already done it). :-) –  a.r. Jul 17 '10 at 3:29
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@Martin, I'm offering advice based on experience (which you lack) and seeing students use Mitchell's theorem only to realize later that it doesn't help much, that's all. I also know a thing or two about useful ways to work with sheaf cohomology; the given motivation is not leading in a good direction for that. I don't see any big deal about offering precise advice based on experience in a comment box. I wrote "as far as I know" to make it clear that I am open to hearing examples to the contrary. I never wrote that I have an "important opinion"; don't put words in my mouth. Lighten up, dude. –  BCnrd Jul 17 '10 at 20:10
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Everybody knows that you have a lot of experience. But this does not justify that below every question you don't like you guide the OP towards other more "relevant" questions or problems, without answering at all the actual question. This is also the reason why I won't post any question here anymore. As for my experience, I have tought several algebra-related courses and also dislike the affection of using Mitchell's theorem, but that's not the point at all here. Agusti's question is interesting in it's own right, it does not suggest to actually use the theorem anytime. –  Martin Brandenburg Jul 18 '10 at 9:14

1 Answer 1

EDIT: I'm completely rewriting this in detail now that I think I've worked it out. Lots of possibilities for mistakes here, so stay alert!

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).

Now, for each $x \in R-mod$, consider the index set $I = \{f: x \rightarrow Va \vert a\in A\} = \bigcup_{a \in A} Hom(x, Va)\}$. (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 $R-mod \rightarrow A$.

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.

How does that look?

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Thank you for your interest. It would be nice to have a counter-example at least. But I'm not sure to understand the one you are proposing. As an aside question: $A$ is the category of sheaves over $[0,1]$ with values in which category? But the main point I would like to ask you is the following: are you sure that $A$ is small? That is, its objects form a set? [to be continued] –  a.r. Sep 25 '10 at 2:29
    
[sequel] Notice that this is not the same as being small-complete, as in the hypothesis of Freyd's adjoint functor theorem: unfortunately, "small and complete" is not the same as "small-complete". Otherwise you would have proved that the solution set condition of this theorem is unnecessary, since every functor from a small category verifies it. –  a.r. Sep 25 '10 at 2:30
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Hmmm... well, Mitchell's theorem is for small categories, correct? Now that I think about it, sheaves of abelian groups on $[0,1]$ is probably NOT small! Which means the embedding theorem doesn't apply anyway, right? Meanwhile, perhaps my proof would work if we replaced $A$ by any small, complete (small AND complete!) category without nonzero projectives. –  Dylan Wilson Sep 25 '10 at 3:47
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Okay I think it should work... so long as there is SOME small, complete abelian category with no nonzero projectives. Seems like that should be true, right? –  Dylan Wilson Sep 25 '10 at 4:16
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[sequel] we have at most one morphism between two objects. But in an abelian category we always have at least one morphism- the zero morphism. So every morphism should be the zero morphism. But then we've got a bigger problem... this means the identity morphism and the zero morphism have to be the same, for each object... which leads me to believe that every object is isomorphic to $0$. So small abelian categories with arbitrary products are just... points? I feel as though I've just broken math. What did I do wrong here!? –  Dylan Wilson Sep 26 '10 at 20:43

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