The Unapologetic Mathematician

More on Kernels and Cokernels

The best-known abelian categories are categories of modules over various rings. And as modules, these objects are structured sets. Now, even though we’re willing to elide the difference between a hom-group and a hom-set, we would at least like to avoid talking about the objects as sets, and the morphisms as functions. So let’s try to focus on studying the morphisms and use them to understand the properties of the category. Luckily, in an abelian category we know a lot about morphisms.

For one thing, let’s consider an object $C\in\mathcal{C}$ and the set $P_C=\bigcup\limits_{C'\in\mathcal{C}}\hom_\mathcal{C}(C',C)$ of all the arrows coming into $C$. We can put the structure of a preorder $\preceq$ on this set by saying $(f:A\rightarrow C)\preceq(g:B\rightarrow C)$ if there is $h:A\rightarrow B$ with $f=g\circ h$. It’s straightforward to show that this relation is reflexive and transitive. Then, as usual we can symmetrize this relation to get an equivalence relation: $f\equiv g$ if and only if $f\preceq g$ and $g\preceq f$. When we pass to the equivalence classes, the preorder becomes a partial order, which we’ll also call $P_C$. In particular, this includes all monomorphisms, which we know as subobjects.

Similarly, we can construct another preorder $Q^C=\bigcup\limits_{C'\in\mathcal{C}}\hom_\mathcal{C}(C,C')$ of all the arrows leading out of $C$ with $f\succeq g$ if there is an $h$ with $g=h\circ f$. We can also turn this into a partial order, which includes the partial order of all quotient objects of $C$.

Now for every arrow $g$ from $C$ pick a kernel and for every $f$ to $C$ pick a cokernel. Then $f$ and $\mathrm{Ker}(g)$ live in $P_C$, while $\mathrm{Cok}(f)$ and $g$ live in $Q^C$. And we have three equivalent statements:

• $f\preceq\mathrm{Ker}(g)$
• $g\circ f=0$
• $\mathrm{Cok}(f)\succeq g$

Now the partial orders $Q^C$ and $P_C$ can be considered as categories, as any partial order can be. And the maps $\mathrm{Ker}:Q^C\rightarrow P_C$ and $\mathrm{Cok}:P_C\rightarrow Q^C$ reverse these orders, so they are contravariant functors. So let’s flip the order on $P_C$ ($x\geq y$ instead of $x\preceq y$) and instead consider them as covariant functors $\mathrm{Ker}:Q^C\rightarrow P_C^\mathrm{op}$ and $\mathrm{Cok}:P_C^\mathrm{op}\rightarrow Q^C$.

The above equivalence now reads $f\geq\mathrm{Ker}(g)\Leftrightarrow\mathrm{Cok}(f)\succeq g$. That is, we have an adjunction. The identities satisfied by the unit and counit read, in this case:

• $\mathrm{Ker}(\mathrm{Cok}(\mathrm{Ker}(g)))=\mathrm{Ker}(g)$
• $\mathrm{Cok}(\mathrm{Ker}(\mathrm{Cok}(f)))=\mathrm{Cok}(f)$