Examples of comma categories

There is a basic construction in category theory which I've only just recently become acquainted with, that is the comma category.

It seems to be a quite basic construction for which, however, I've seen really few "real life" examples.

I know the slice, coslice and arrow categories are particular cases of a comma category. This is in MacLane, or in the wikipedia article. In that article there are also the examples of pointed sets or graphs, which are of a more concrete nature.

I'm asking, then, for more examples of this construction in mathematics. Examples of (co)slice categories are also welcome.

Here's one example I've come up with. The completion of a metric space $M$ consists of a pair $(\overline{M},i)$ where $\overline{M}$ is a complete metric space and $i:M\to \overline{M}$ is a uniformly continuous function which satisfies the following universal property: if $N$ is another complete metric space and $g:M\to N$ is a uniformly continuous function, then there exists a unique uniformly continuous function $h:\overline{M}\to N$ such that the following diagram commutes:

I claim this completion is an initial object in a suitable comma category. Consider the functors where $\mathbf{Met_u}$ is the category of metric spaces with uniformly continuous functions and $\mathbf{CompMet_u}$ is the category of complete metric spaces with uniformly continuous functions. The functor $F$ is such that $F(\star)=M$ where $\star$ is the sole object of $\mathbf{1}$, and $U$ is the inclusion functor.

Then an initial object of $(F\downarrow U)$ is exactly a completion of $M$.

Bonus question: is this approach to the completion not interesting/not useful? I ask this because it seems the categorial approach to completions has nothing to do with comma categories. (I can see, though, that the universal property of this completion can also be seen as an adjunction).

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I'm sorry for the "bonus question" part, I know it is not good practice to ask more than one question per question (hmm, that's slightly redundant). If it is a more interesting question than it seemed to me and there is someone willing to give a comprehensive answer that is not suitable for a comment here, then I will happily split it into another question. – Bruno Stonek Apr 21 '12 at 14:45
Have you read the proof of the general adjoint functor theorem? It constructs left adjoints via initial objects of certain comma categories... – Zhen Lin Apr 21 '12 at 16:23
@ZhenLin: I haven't (but plan on doing so relatively shortly). That's very interesting to me, and it can't be just a coincidence that both my example and the one given in the answer by Matt N. are both left adjoints... Thank you. – Bruno Stonek Apr 21 '12 at 18:07
Did you want things that weren't slice/coslice categories? I assume you know that you can think of $R\text{-}\mathbf{Alg}$ as the slice category $(R\downarrow \mathbf{CRing}$. – Alex Youcis Apr 21 '12 at 18:08
@AlexYoucis: no, I didn't mean to say that, I'm also interested in "concrete" examples of (co)slice categories! I didn't know what you mention, perhaps you could post an answer? – Bruno Stonek Apr 21 '12 at 18:11
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2 Answers

One example of a comma category $(F \downarrow G)$, I think of it as "simplified" comma category but I don't think that's a commonly used term, is what you get when you take $F: A \to C$ to be the selection functor (i.e. $A = \textbf{1}$).

More concretely, if $G:\textbf{Group} \to \textbf{Set}$ is the forgetful functor mapping a group to its underlying set and $F: \textbf{1} \to \textbf{Set}$ is the selection functor selecting a set $S$ then you get the category $(S \downarrow G)$ where the objects are pairs $(f,Y)$ where $f:S \to Y$ is a morphism in $\textbf{Set}$ and $Y = G(X)$ is the underlying set of some group $X$. The morphisms $(f,Y) \to (f^\prime, Y^\prime)$ are induced by group homomorphisms $\varphi : X \to X^\prime$ such that $G(\varphi) \circ f = f^\prime$.

You can use this category to define the free group over a set $S$, namely, it is a group $F$ such that $(f,G(F))$ is an initial object in $(S \downarrow G)$.

Hope this serves as an example. As for the bonus question: I'll have to pass on that.

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Yes, it is a nice example, thank you! I'm so used to thinking of the free functor as a left adjoint of the forgetful functor, that I hadn't thought of that triangle as an arrow in a comma category. This example is similar to the one I gave, where also one of the functors was a "selection functor" as you call them; and it could also be seen as an adjoint (see the nlab article). I guess seeing both these constructions as adjoints is nicer because it gets functors (i.e. free functor, completion functor) out of them, not only objects. – Bruno Stonek Apr 21 '12 at 15:04
@BrunoStonek Yes, actually I'd call it the same as what you give in your question. : ) I suspect there are many more "same" examples of this kind: all things defined in terms of universal properties, such as for example free modules. – Matt N. Apr 21 '12 at 15:09

Ok, if you are interested in slice/coslice categories then there are two obviously interesting ones:

It's pretty trivial that the category $\mathbf{Top}_\ast$ of pointed topological spaces is just $(\bullet\downarrow\mathbf{Top})$.

As another example let's find out what $(R\downarrow\mathbf{CRing})$ looks like, when $R$ is come commutative unital ring. We see that, almost by definition, we can think of the objects of $(R\downarrow\mathbf{CRing})$ as being commutative unital rings $A$ with a distinguished ring homomorphism $f:R\to A$. Ok, there's no more "simplification" that can be done there. So, what do the arrows in this comma category look like? Well, if we have two objects $f:R\to A$ and $g:R\to B$, we see that an arrow between them is an arrow $h:A\to B$ (of course, technically it's a pair of arrows, but when the left element of $(-\downarrow-)$ is discrete this is always the identity arrow, and so unimportant) such that $g=h\circ f$. But, this precisely the formulation for commutative algebras over $R$. In other words, we have commutative unital rings $A$ with specified $\mathbf{CRing}$-arrows $R\to A$ and the morphisms between two such objects $(R,A,f)$ and $(R,B,g)$ is just a $\mathbf{CRing}$-arrow $A\to B$ which respects $f$ and $g$. Thus, we see that $\left(R\downarrow\mathbf{CRing}\right)\cong R\text{-}\mathbf{CAlg}$.

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Thank you for your elaboration. – Bruno Stonek Apr 21 '12 at 18:23
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