# The Unapologetic Mathematician

## Properties of Ab-Categories

There are a number of things we can say right off about the $\mathbf{Ab}$-categories we defined last time. As is common practice, we’ll blur the distinction between an abelian group and its underlying set.

First of all, any $\mathbf{Ab}$-category $\mathcal{C}$ has zero morphisms. That is, there’s a special morphism between any two objects that when composed with any other morphism gives the special morphism in the appropriate hom-set. In fact, since each hom-set is an abelian group it has an additive identity $0\in\hom_\mathcal{C}(A,B)$. Then for any $f\in\hom_\mathcal{C}(B,C)$ we have $f\otimes0=0\in\hom_\mathcal{C}(B,C)\otimes\hom_\mathcal{C}(A,B)$, which composition must send to $0\in\hom_\mathcal{C}(A,C)$. The zero morphisms are exactly the zero morphisms!

Given any object $C\in\mathcal{C}$ the hom-set $\hom_\mathcal{C}(C,C)$ is already an abelian group. But the composition puts the structure of a monoid onto this set as well, and the linearity condition says these two are compatible, making the endomorphism monoid into an endomorphism ring. In fact, every ring is an endomorphism ring. Way back when we first defined categories we noted that a category with one object was the exact same thing as a monoid. And a ring is just an abelian group with a compatible monoid structure on it. So an $\mathbf{Ab}$-category with a single object is the exact same thing as a ring! In fact, a lot of the study of $\mathbf{Ab}$-categories can be seen as extending ring theory from that special case to the more general one. Incidentally, you should see right off that when we consider rings $R$ and $S$ as categories like this, a ring homomorphism from $R$ to $S$ is the same thing as an $\mathbf{Ab}$-functor between the categories.

Remember when we talked about direct sums of modules over a given ring? Well the same thing happens here. We define the “biproduct” $\bigoplus\limits_{i=1}^n A_i$ of the finite collection of objects $A_i$ to be an object along with two families of arrows:

• $\pi_i:\bigoplus\limits_{i=1}^n A_i\rightarrow A_i$
• $\iota_i:A_i\rightarrow\bigoplus\limits_{i=1}^n A_i$

satisfying the relations

• $\pi_i\circ\iota_j=0:A_j\rightarrow A_i$ if $i\neq j$
• $\pi_i\circ\iota_i=1_{A_i}:A_i\rightarrow A_i$
• $\sum\limits_{i=1}^n\iota_i\circ\pi_i=1_{\bigoplus\limits_{i=1}^n A_i}$

From the same arguments as in our coverage of direct sums we see that a biproduct satisfies the universal properties of both a categorical product and coproduct, and conversely that a categorical product or coproduct implies the existence of the biproduct arrows. Note that we’re making no statement whatsoever that such a biproduct actually exists in our category, but when it does it’s both a product and a coproduct.

As a special case, we can consider the biproduct of an empty collection of objects. This will be both a product and a coproduct of an empty collection of objects, if it exists, and will thus be a zero object. Of course, it may or may not exist.

Even if there is no zero object in our category, we still have the above zero morphisms, and so we can still talk about kernels and cokernels. The kernel $\mathrm{Ker}(f)$ of a morphism $f:A\rightarrow B$ is the equalizer $\mathrm{Equ}(f,0)$, and its cokernel $\mathrm{Cok}(f)$ is the coequalizer $\mathrm{Coequ}(f,0)$. In fact, life is even better now that we’re enriched over $\mathbf{Ab}$: every equalizer is a kernel and every coequalizer is a cokernel. Indeed, $\mathrm{Equ}(f,g)=\mathrm{Ker}(f-g)$ and similarly for coequalizers. Again, we’re saying nothing about whether such kernels or cokernels actually exist.

Together, these facts say a lot about the behavior of limits in $\mathbf{Ab}$-categories. Biproducts tell us about finite products and coproducts, while kernels of morphisms tell us about all different equalizers. And then The Existence Theorem for Limits tells us that every finite limit can be constructed from finite products and equalizers, while every finite colimit can be constructed from finite coproducts and coequalizers. So if our $\mathbf{Ab}$-category has all biproducts, all kernels, and all cokernels, then it has all finite limits whatsoever!

Let’s add one more little property that will simplify our life. We know that kernels are monomorphisms, and that cokernels are epimorphisms. If we assume on top of having all biproducts, kernels, and cokernels that every monomorphism is actually the kernel of some arrow in our category, and that every epimorphism is actually the cokernel of some arrow, then we will call our $\mathbf{Ab}$-category an abelian category.

You should verify that given any ring $R$ the category $R\mathbf{-mod}$ of all left $R$-modules satisfies all these properties, and thus is an abelian category. These are the abelian categories that started the whole theory of homological algebra, which is to a large extent the study of general abelian categories.

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Posted by John Armstrong | Category theory

## 10 Comments	»

1. [...] on Kernels and Cokernels The best-known abelian categories are categories of modules over various rings. And as modules, these objects are structured sets. [...]

Pingback by | September 18, 2007 | Reply

2. [...] And it turns out we already know a lot about these sorts of categories! Specifically, they’re abelian categories. In fact, since we’re working over a field (which is a commutative ring) the properties of [...]

Pingback by | May 19, 2008 | Reply

3. [...] Splitting Lemma Evidently I never did this one when I was talking about abelian categories. Looks like I have to go back and patch this [...]

Pingback by | June 25, 2008 | Reply

4. [...] already said that (also ) is a abelian category. Now in any abelian category we have the first isomorphism [...]

Pingback by | June 27, 2008 | Reply

5. [...] remember that in abelian categories we’ve got diagrams to chase and exact sequences to play with. And these have something to say [...]

Pingback by | July 23, 2008 | Reply

6. [...] so we have a direct sum of representations. Is it a biproduct? Luckily, we don’t have to bother with universal conditions here, because a biproduct can be [...]

Pingback by | December 9, 2008 | Reply

7. [...] The Category of Representations is Abelian We’ve been considering the category of representations of an algebra , and we’re just about done showing that is abelian. [...]

Pingback by | December 15, 2008 | Reply

8. [...] Of course, given any inner product space we can “forget” the inner product and get the underlying vector space. This is a forgetful functor, and the usual abstract nonsense can be used to show that it creates limits. And from there it’s straightforward to check that the category of inner product spaces is abelian. [...]

Pingback by | May 6, 2009 | Reply

9. [...] like to see that the category of Lie algebras is Abelian. Unfortunately, it isn’t, but we can come close. It should be clear that it’s an [...]

Pingback by | August 14, 2012 | Reply

10. [...] as the quotient of the range by the image. All we need to see that the category of -modules is abelian is to show that every epimorphism is actually a quotient, but we know this is already true for the [...]

Pingback by | September 12, 2012 | Reply

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