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.


share|improve this question
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
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
@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
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
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

Your Answer


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

Browse other questions tagged or ask your own question.