# The Unapologetic Mathematician

## Zero objects, Kernels, and Cokernels

A zero object in a category $\mathcal{C}$ is, simply put, both initial and terminal. Usually we’ll write $\mathbf{0}$ for a zero object, but sometimes $Z$, or even $\mathbf{1}$ in certain circumstances. While initial objects and terminal objects are both nice, zero objects are even nicer.

Since a zero object is terminal, there is a unique morphism in $\hom_\mathcal{C}(X,\mathbf{0})$ for each object $X$. Since it’s initial, there’s a unique morphism in $\hom_\mathcal{C}(\mathbf{0},Y)$ for each object $Y$. Now we can put these together: for any two objects $X$ and $Y$ there is a unique morphism $0:X\rightarrow Y$ which factors through $\mathbf{0}$. Take the unique arrow from $X$ to $\mathbf{0}$, then the unique arrow from $\mathbf{0}$ to $Y$. This picks out a special element of each hom set, just for having a zero object.

We saw that the trivial group $\mathbf{1}$ in the category of groups is both initial and terminal, so it’s a zero object. If we’re just looking in the category $\mathbf{Ab}$ we usually call the trivial abelian group $\mathbf{0}$, and it’s a zero object. The initial and terminal object in $\mathbf{Set}$ are different, so this category does not have a zero object.

It’s often useful to remedy this last case by considering the category of “pointed” sets. A pointed set is a pair $(X,x)$ where $X$ is a set and $x\in X$ is any element of the set. A morphism of pointed sets is a function $f:X\rightarrow Y$ so that $f(x)=y$. The marked point has to go to the marked point. This gives us the category $\mathbf{pSet}$ of pointed sets. It’s easily checked that the pointed set $(\{x\},x)$ is both initial and terminal, so it is a zero object in $\mathbf{pSet}$.

If a category $\mathcal{C}$ has a zero object then we have a special morphism in each hom set, as I noted above. If we have two morphisms between a given pair of objects we can ask about their equalizer and coequalizer. But now we have one for free! So given any arrow $f:X\rightarrow Y$ and the special zero arrow $0:X\rightarrow Y$, we can ask about their equalizer and coequalizer. In this special case we call them the “kernel” and “cokernel” of $f$, respectively. I’ll say more about the kernel, but you should also think about dualizing everything to talk about the cokernel.

Given a morphism $f:X\rightarrow Y$ its kernel is an morphism $k:\mathrm{Ker}(f)\rightarrow X$ so that $f\circ k=0$ as morphisms from $\mathrm{Ker}(f)$ to $Y$. Also, given any other morphism $h:H\rightarrow X$ with $f\circ h=0$ we have a unique morphism $g:H\rightarrow\mathrm{Ker}(f)$ with h=k\circ g\$. This is just the definition of equalizer over again. As with equalizers, the morphism $k$ is monic, so we can view $\mathrm{Ker}(f)$ as a subobject of $X$ and $k$ as the inclusion morphism.

Let’s look at this in the case of groups. If we have a group homomorphism $f:X\rightarrow Y$ the kernel will be group included into $X$ by the monomorphism $k$ — a subgroup. We also have that $f\circ k=0$, so everything that starts out in $\mathrm{Ker}(f)$ gets sent to the identity element in $Y$. If we have any other homomorphism $h:H\rightarrow X$ whose image gets sent to the identity in $Y$, then it factors through $\mathrm{Ker}(f)$, so the image of $h$ lands in the image of $k$. That is, $\mathrm{Ker}(f)$ picks out the whole subgroup of $X$ that gets sent to the identity in $Y$ under the homomorphism $f$. And that’s exactly what we called the kernel of a group homomorphism way back in February.

Often enough we aren’t really interested in all equalizers — just kernels will do. So, if a category has a zero object and if every morphism has a kernel, we say that the category “has kernels”. Dually, we say it “has cokernels”. Having a zero object and having equalizers clearly implies having kernels, but it’s possible to have kernels without having all equalizers.

June 13, 2007 Posted by | Category theory | 6 Comments