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Stream: theory: category theory

Topic: When a Kan = Comma?

Beppe Metere (Jul 21 2026 at 13:02):

When is the canonical comparison associated with a right Kan extension an equivalence?

Let

$$U:E\to S, \qquad U':E\to T$$

be functors, and suppose that the right Kan extension

$$V={\rm Ran}_U U'$$

exists.

Moreover, assume that $U$ admits a fully faithful left adjoint

$$F\dashv U,$$

so that

$$V\cong U'F.$$

Writing

$$\epsilon:FU\to 1_E$$

for the counit, there is a canonical comparison functor

$$K:E\to (V\downarrow 1_T),$$

which sends an object $X$ of $E$ to the morphism

$$U'(\epsilon_X): V(UX)=U'FUX\to U'X.$$

In some cases, this comparison is an equivalence:

$$E\simeq (V\downarrow 1_T).$$

Thus the same functor $V$ plays two different universal roles:

1. it is the right Kan extension of $U'$ along $U$;
2. it classifies the objects of $E$ through a comma category.

My question is the following.

Has this phenomenon been studied in general? More precisely, are there known categorical conditions on

$$U:E\to S, \qquad U':E\to T$$

ensuring that the canonical comparison

$$K:E\to (({\rm Ran}_U U')\downarrow 1_T)$$

is an equivalence?

Thanks for your answers, and please be indulgent if there is something obvious or very well-known that I am missing!

Kevin Carlson (Jul 21 2026 at 15:55):

The only case I can think of is when $S$ is terminal and $U'$ is an equivalence. Do you know any other examples?

Beppe Metere (Jul 22 2026 at 08:22):

Yes, for instance: let $C$ be any category and

$$U=dom:C^2\to C, \qquad U'=cod:C^2\to C.$$

One has

$$U\vdash F: C\to C^2, \qquad A\mapsto 1_A,$$

so that $1_C=Ran_{U}(U').$

On the other hand

$$C^2\cong (1_C\downarrow 1_C).$$

Other examples can be produced similarly...

Notification Bot (Jul 23 2026 at 12:19):

This topic was moved to #learning: questions > When a Kan = Comma? by John Baez.