# Freyd-Mitchell's embedding theorem

Freyd-Mitchell's embedding theorem states that: if A is a small abelian category, then there exists a ring R and a full, faithful and exact functor F: A → R-Mod.

I have been trying to find a proof which does not rely on so many technicalities as the ones I have found. I have leafed through:

Freyd's Abelian Categories says that the text, excepting the exercises, tries to be a geodesic leading to the theorem. If you take out the exercises, probably the text is 120 pages long...

Mitchell's Theory of Categories is very hard to read, and also to prove the theorem you have tons of definitions and propositions and lemmas to prove. For example, the study of AB-5 categories (C3 in Mitchell's notation) is fairly tedious.

Weibel's An Introduction to Homological Algebra redirects me to Swan, The Theory of Sheaves, a book which is unavailable in my university's library. I've leafed through Swan's Algebraic K-Theory: the theorem is proved, but it is also long, hard and painful to read, and assumes a lot of knowledge I don't have (I had never seen a weakly effaceable functor, or a Serre subcategory; and it certainly is not well known to me that the category of additive functors from a small abelian category to the category of abelian groups is well-powered, right complete, and has injective envelopes!)

Maybe there are more modern proofs which require less heavy machinery and technicalities?

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Then again, there might not be a royal road to the embedding theorem. –  José Figueroa-O'Farrill Nov 30 '10 at 3:22
I am pretty sure that Rotman's Homological algebra proves this quite simply, but I could be remembering incorrectly. –  B. Bischof Nov 30 '10 at 5:23
Freyd's book isn't bad at all. A lot of those pages are just developing standard category theory from scratch. If I remember correctly, the real heart of the proof is contained in just one or two chapters. –  Eric Wofsey Nov 30 '10 at 5:49
Perhaps you should accept that some homological algebra has to be used. Serre subcategories, effaceable functors, injective envelopes etc. are all basic stuff which are, of course, not only important for the embedding Theorem. Perhaps you should delay the proof for some time. –  Martin Brandenburg Nov 30 '10 at 9:01
If you're interested in the abelian categories other than module categories that arise in practice (which I imagine you are, since you're asking about Freyd-Mitchell), then Serre subcategories are lurking in the background. You can construct a new abelian category by taking an existing category, and "modding out" by a subcategory. The condition needed to make this work is that of a Serre subcategory. For example, sheaves of abelian groups are presheaves "modulo" the presheaves that go to zero under sheafification. –  arsmath Nov 30 '10 at 14:54
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Think of the category of functors $\scrA \to \Ab$ as $\scrA$-modules, that's why the notation $\amod$ is fairly common. The Yoneda embedding $A \mapsto \Hom(A,{-})$ even yields a fully faithful contravariant functor $y\colon \scrA\to\amod$.

The category $\amod$ inherits a bunch of nice properties of the category $\Ab$ of abelian groups:

• It is abelian.
• It is complete and cocomplete ((co-)limits can be computed pointwise on objects)
• The functors $\Hom(A,{-})$ are injective and $\prod_{A \in\scrA} \Hom{(A,{-})}$ is an injective cogenerator.
• Since there is an injective cogenerator, the category $\amod$ is well-powered.
• etc.

However, the Yoneda embedding is not exact: If $0 \to A' \to A \to A'' \to 0$ is a short exact sequence, we only have an exact sequence

$$0 \to \Hom(A',{-}) \to \Hom(A,{-}) \to \Hom(A'',{-})$$

in $\amod$.

It turns out that the functor $Q = \coker{(\Hom(A,{-}) \to \Hom(A'',{-}))}$ is “weakly effaceable”, so we would want it to be zero in order to get an exact functor. How can we achieve this? Well, just force them to be zero: say a morphism $f\colon F \to G$ in $\amod$ is an isomorphism if both its kernel and its cokernel are weakly effaceable. If this works then a weakly effaceable functor $E$ is isomorphic to zero because $E \to 0$ has $E$ as kernel. Now the full subcategory $\scrE$ of weakly effaceable functors is a Serre subcategory, so we may form the Gabriel quotient $\amod/\scrE$. By its construction, isomorphisms in the Gabriel quotient have precisely the description above.

On the other hand, the category $\Lex(\scrA,\Ab)$ of left exact functors $\scrA \to \Ab$ is abelian. This is far from obvious when you start from the definitions. However, $\Lex(\scrA,\Ab)$ sits comfortably inside the abelian category $\amod$. The inclusion has an exact left adjoint (!) (= “sheafification”), so again ${\bf Lex}({\scr A}, \Ab)$ inherits many useful properties from $\amod$. Moreover, the kernel of the left adjoint can be identified with the weakly effaceable functors, and that's why $\Lex{(\scrA, \Ab)} = \amod/\scrE$.

All this work shows that $A \mapsto \Hom{(A,{-})}$ is a fully faithful and exact embedding of $\scrA$ into ${\Lex}{(\scrA, \Ab)}$, so it remains to show that the latter can be embedded into a category of modules. This is well described in Weibel's or Swan's books, so I won't elaborate on that point and content myself by saying that you simply need to look at the endomorphism ring of an injective cogenerator.

As for references, I think you can't do much better than Freyd's book. Don't be too intimidated by Swan's exposition in his K-theory book. If you're really interested in understanding this proof, I think it's worth reading the two expositions (first Freyd, then Swan). There also is a proof in volume 2 of Borceux's Handbook of categorical algebra with a more “hands on” approach.

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The named properties of $Lex(A,Ab)$ and the embedding of $A$ into it are presented in Gabriel's thesis "Des categories abeliennes". –  Martin Brandenburg Nov 30 '10 at 9:03
Oh sure, somehow I missed to mention this article. It can be found here: archive.numdam.org/article/BSMF_1962__90__323_0.pdf –  Theo Buehler Nov 30 '10 at 10:41
I really like this answer. –  Sheikraisinrollbank Nov 30 '10 at 12:03
@Sheikraisinrollbank: thank you. –  Theo Buehler Dec 1 '10 at 17:47
@Axel: Yes, you're right, thanks for catching that typo. There are a few more. I will edit soon. –  Theo Buehler Dec 5 '12 at 21:43