The Unapologetic Mathematician

Mathematics for the interested outsider

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.

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June 13, 2007 - Posted by | Category theory

6 Comments »

  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?

    Tell me what you’d like to accomplish, and I’d help you.

    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 [...]

    Pingback by Images and Kernels « The Unapologetic Mathematician | September 29, 2010 | Reply


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