Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

One commonly reads that the Freyd-Mitchell's embedding theorem allows proof by diagram chasing in any abelian category.

This is not immediately clear, since only small abelian categories can be embedded into R-mod.

Weibel for example argues that the snake lemma holds in an arbitrary abelian category (p. 12, Introduction to Homological Algebra):

The Snake Lemma also holds in an arbitrary abelian category $\mathcal{C}$. To see this, let $\mathcal{A}$ be the smallest abelian subcategory of $\mathcal{C}$ containing the
objects and morphisms of the diagram. Since $\mathcal{A}$ has a set of objects, the Freyd-Mitchell Embedding Theorem (see 1.6.1) gives an exact, fully faithful
embedding of A into R-mod for some ring $R$.

I am looking for a reference or an explanation as to why we know that the category $\mathcal{A}$, containing the diagram, has to be small.

I fear that this might be a stupid question, but why can't we potentially end up needing the whole category $\mathcal{C}$ to contain the diagram? I don't thoroughly understand how $\mathcal{A}$ will be constructed.

Thanks!

share|improve this question
4  
maybe this is a subtlety in terms of what Weibel means by diagram. People frequently say that a diagram is a functor out of a small category. If this is the definition Weibel is using, then $A$ must be small by definition. –  Sean Tilson Apr 16 '11 at 0:39
10  
Well you need to do this iteratively, but you'll end up with a small category: Take the full subcategory $\mathcal{B}_0$ of $\mathcal{A}$ generated by the objects in the diagram. This is clearly small. Now choose a kernel, an image and a cokernel in $\mathcal{A}$ for each morphism in $\mathcal{B}_0$ and take the full subcategory $\mathcal{B}_1$ generated by that. Repeat this procedure. This clearly yields a small abelian subcategory $\mathcal{B} = \bigcup \mathcal{B}_n$ of $\mathcal{A}$ containing your diagram. Exactness in $\mathcal{B}$ means the same thing as exactness in $\mathcal{A}$. –  t.b. Apr 16 '11 at 8:17
1  
@TheoBuehler Ah, thank you. So to get $\mathcal{B_0}$ you basically just take the objects of your diagram and add morphisms that are still missing? How do we know that $\mathcal{B_1}$ (or any other $\mathcal{B_n}$,$n \neq 0$ for that matter) is small? Maybe I had to add so many morphisms in $\mathcal{B_0}$ that the kernels, cokernels and images don't fit into any set? I mean, couldn't it happen that $\mathcal{B_0}$ might not be locally small and therefore cause problems in the further construction? Sorry, if I am totally off track! –  Felix Hoffmann Apr 16 '11 at 11:44
7  
For me a category is always locally small (in Weibel's book too, by the way), and of course the above argument only works if you're locally small, so I've misread your question a bit. If your category is not locally small, you're certainly out of luck. But you can still prove all the diagram lemmata without ever referring to elements... –  t.b. Apr 16 '11 at 11:53
3  
Yes, definitely. All the metatheorems (such as Freyd-Mitchell) run into foundational difficulties when you're dealing with huge categories. On the other hand, in most concrete situations (and all the ones I needed to consider seriously so far) you can cut the size of things you need to consider down to reasonable by using techniques adapted to the specific situation. I advise you to ignore these issues at the moment because it seems to me that there are more central things for you to learn about at the moment. –  t.b. Apr 16 '11 at 13:58

1 Answer 1

Just to get this out of the unanswered queue: yes, you can diagram chase even in a non-locally small abelian category using Freyd-Mitchell. Let $D:J\to \mathcal{A}$ be a (small) diagram in an arbitrary abelian category, and then work in the abelian category generated by $D(J)$. This is automatically small: you only need homs in the abelian groups generated by morphisms in $D(J)$, as well as by the kernels, cokernels, and finite direct sums of such morphisms. This process of taking kernels and generating morphism groups terminates after countably many steps, so in fact you get an abelian category of cardinality no more than $\aleph_0|J|$, where $|J|$ is the sum of cardinalities of the morphisms and objects of $J$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.