I've been talking a little about abelian categories these days. That's because I've been going over Weibel's An Introduction to Homological Algebra. It's a book I read before, and I still feel pretty confident about the material. This time, though, I think I'm going to explore a few different paths that I haven't really given much thought to before, such as diagram proofs in abelian categories, group cohomology (more in-depth), and Hochschild homology.

Back to abelian categories. An abelian category is a category with the following properties:

  1. The hom-sets are abelian groups and composition distributes over addition
  2. Every two objects have a product.
  3. There exists an object that is both terminal and initial, denoted by $0$
  4. Every map has a kernel and a cokernel
  5. Every monic is a kernel of its cokernel
  6. Every epic is a cokernel of its kernel

It's quite soothing that these axioms are enough to make small abelian categories exact-embeddable as a full subcategory of a category of modules over a ring. That's the Freyd-Mitchell Embedding. This also means that if you want to prove certain diagram lemmas in an abelian category, it's enough to prove it assuming your diagram is actually a diagram in a module category.

But part of my goal is to get good at proving things just straight from the axioms. In fact, I already started with the proof that every morphism $f:A\to B$ can be factored as $A\to {\rm im}(f)\to B$ where $A\to {\rm im}(f)$ is epic and ${\rm im}(f)\to B$ is monic. That's quite easy in module categories but not at all obvious with abelian categories.

Before I continue, I thought it might be interesting to look at some examples and non-examples of abelian categories.

Category of modules: abelian

The category of left or right $R$-modules over a ring $R$ is an abelian category. That pretty much goes without saying. All the axioms are straightforward to show.

Category of rings: not abelian

The problem with the category of rings is that they are too rigid. The hom-sets are not abelian groups, because ring homomorphisms send the multiplicative identity to the multiplicative identity. However, even if you don't require that, homomorphisms still have to respect multiplication. The multiplicative nature of rings introduces a whole new perspective.

The category of rings does have products. But it doesn't have a zero object: in the category of rings, the initial object is $\Z$ but the terminal object is the zero ring. Kernels don't even exist, though if you consider rings without identity then they do exist, as kernels are ideals.

In short, rings are too rigid to be abelian.

Category of chain complexes: abelian

A chain complex of $R$-modules is a diagram
$$\cdots\to C_n\xrightarrow{d} C_{n-1}\xrightarrow{d} C_{n-2}\to\cdots$$ where the maps $d = d_n:C_n\to C_{n-1}$ satisfy $d^2 = d\circ d = 0$. A map of chain complexes $f:C\to D$ is an indexed sequence of module homomorphisms $\{ f_n:C_n\to D_n \}$ such that $df = fd$.

The category of chain complexes of $R$-modules for a ring $R$ is an abelian category: all the contructions are done pointwise. The zero object is the zero complex, and products, kernels and cokernels are pointwise.

Category of presheaves of abelian groups: abelian

Given a topological space $X$, a presheaf $F$ on $X$ is a contravariant functor from the category of open sets of $X$ to abelian groups (or more generally, $R$-modules) such that $F(\varnothing) = 0$. This category is an abelian category. In fact, chain complexes and (pre)sheaves forming abelian categories is one of the main reasons to actually consider abelian categories in the first place: there are plenty of categories that are abelian and useful in homological algebra that aren't constructed as a category of $R$-modules!

Category of torsion-free abelian groups: not abelian

This is an interesting one. Recall that an abelian group $A$ is torsion-free if there does not exist a nonzero $x\in A$ such that $\sum_{i=1}^n x = 0$ for some $n\geq 1$. So free groups and $\prod_{i=1}^\infty \Z$ are free but $\Z/5$ is not free. The category of torsion-free abelian groups is the full subcategory of abelian groups that are torsion-free.

So, why is this category not abelian? Let's go through the axioms. We know that hom-sets are abelian groups and composition distributes over addition. That's pretty straightforward.

Next, what about products? Well, if $A$ and $B$ are torsion-free, their product $A\times B$ certainly exists in the category of abelian groups. Then it's easy to see that this also must be the product in the category of torsion-free abelian groups: the morphisms $A\times B\to A$ and $A\times B\to B$ still exist here and the universal property is of course satisfied.

There is a zero object, the zero group. Now what about kernels? Let's take a look at the map of abelian groups $\Z\xrightarrow{2}\Z$ given by multiplication by two. In the category of abelian groups, the cokernel of this map is the quotient map $Z\to \Z/2$. Obviously, $\Z/2$ is a torsion abelian group and so this object doesn't exist in the category of torsion-free abelian groups.

Does this mean that cokernels don't exist? Not necessarily. We are in a subcategory and the cokernel could be soemthing else. Let's say for example that the composition $\Z\xrightarrow{2}\Z\xrightarrow{f} X$ is actually the zero map. That means that $f(2) = 0$. But now,
$$0 = f(2) = f(1 + 1) = f(1) + f(1).$$ But $X$ is a torsion-free abelian group! Thus $f(1) = 0$. Based on this construction, there exists a unique map (the zero map, a dotted arrow in the diagram) $0\to X$ such that the following commutes:
What we have shown is that the zero map $\Z\to 0$ is actuall the cokernel of the multiplication-by-two map $\Z\xrightarrow{2}\Z$. Yes, we took the original cokernel out of the equation, but by taking out all torsion abelian groups, we also took out the number of ways you can kill off the multiplication-by-two map by composition with it on the left.

So why is the category of torsion-free abelian groups not abelian? It's because of that funny axiom: every monic is the kernel of its cokernel, and every epic is the cokernel of its kernel. The map $\Z\xrightarrow{2}\Z$ is monic, but it is not the kernel of the map $\Z\to 0$, because the kernel of $\Z\to 0$ is the identity map $\Z\xrightarrow{1}\Z$.

In fact, the axiom that every monic is the kernel of its cokernel (and dually), is one of the most crucial axioms of abelian categories. Here's another consequence of this axiom: a map that is both monic and epic is an isomorphism. It's what gives abelian categories their true flavour.

P.S. Until August 31, 2018, don't forget to vote for my blog for the Canadian Goosy Awards (for best blog)!

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