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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!

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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
9  
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
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
2  
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  
I think t.b.'s argument in the locally small case can be made to work in general by being stingier with the subcategories $\mathcal B_i$. Instead of taking full subcategories at every step, take only the morphisms you need in order to witness the relevant definitions (e.g., the map that "is" the kernel of another, and the factorizations witnessing that it's a kernel, etc.). –  Andreas Blass Nov 7 '12 at 18:41
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