# 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

1. Maybe, when summer comes and you have more spare time, you could set up a web page with the (linked) titles of your category theory posts in a table of contents fashion. Unless you are already negotiating the book rights, that is :-).

Comment by estraven | June 13, 2007 | Reply

2. I’ve been pondering a bit of a reorganization. WordPress defaults are a nice interface, but there are some things I’d rather were different. Unfortunately, I don’t really think I’d be that great at tweaking all the CSS on my own.

So, failing a volunteer web designer coming out from the woodwork — with credit given where due, naturally — I’ll be doing what little I can on my own to organize a bit. Also, once I have some web space at Tulane I can set up such a list of topics as you suggest.

Comment by John Armstrong | June 13, 2007 | Reply

3. Hrm. This woodwork?

Oh, and if you want to, I have a wardrobe server where I’d happily host you as well. Either set you up a blog or a wiki or all of it or … well, anything you like, really.

Comment by michiexile | June 14, 2007 | Reply

4. […] our category of vector spaces has a biproduct — the direct sum — and in particular a zero object — the trivial -dimensional vector space . It also has a tensor product, which makes this a […]

Pingback by Linear Algebra « The Unapologetic Mathematician | May 19, 2008 | Reply

5. […] should immediately ask: is this representation a zero object? Suppose we have a representation . Then there is a unique arrow sending every vector to . […]

Pingback by The Zero Representation « The Unapologetic Mathematician | December 8, 2008 | Reply

6. […] because these are the morphisms in the category of -modules. It turns out that this category has kernels and has images. Those two references are pretty technical, so we’ll talk in more […]

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