Is the image of a small category small?

Question

Let $C$ be a small category, let $D$ be a locally small category. Given a functor $F:C\to D$, the image of $F$ may not be a category. Now following the nLab, let's instead call the image of $F$ the subcategory of $D$ generated by the images of the objects and morphisms of $C$ under $F$, i.e. close it under composition.

With this notion of image, by construction we have a subcategory of $D$. Is this subcategory again small?

Since $D$ is locally small, the image of $F$ should have a set of objects (the image of the set of objects of $C$, since $C$ is small and you can't reach any object outside that image) and each hom is a set (as a subset of the hom-set of locally-small $D$) so the image is small. — Chessanator, Mar 27 '20 at 03:55
I'm not sure what happens if $D$ isn't locally small, which is probably a more interesting question. — Chessanator, Mar 27 '20 at 03:56

score 4 · Accepted Answer · answered Mar 27 '20 at 05:08

4

Yes, this is trivial. Every morphism in the image of $F$ can be written as a finite composition of morphisms that come from morphisms of $C$, and there is only a small set of such finite compositions.

answered Mar 27 '20 at 05:08

Eric Wofsey

342,377

1

Another way to put this is that the image of $F$ is a quotient of the category freely generated by the underlying graph of $C$, which certainly remains small. – Kevin Carlson Mar 27 '20 at 05:13
@KevinCarlson: I think you mean the underlying graph of the "naive" image of $F$ (which is not necessarily actually a subcategory). – Eric Wofsey Mar 27 '20 at 05:16
Sure, let's say that. – Kevin Carlson Mar 27 '20 at 05:28

Is the image of a small category small?

1 Answers1