The image of a functor need not be a subcategory

Question

Warning 1.2.19 gives an example when the image of a functor is not a subcategory:

But I'm confused: the author defines a functor $F$ right away without saying what the codomain category is. This causes the question: the image of that functor is not subcategory of which category? If the codomain is the category depicted on the right (which is the same as the image of $F$), then that is a subcategory of itself.

Possible duplicate of Can a Functor F:A -> B not map to a subcategory of B? — qualcuno, Jun 24 '19 at 21:46
@GuidoA.I don't think it's a duplicate since the OP is not looking for an example but asking about a particular example that (s)he doesn't understand. — Javi, Jun 24 '19 at 22:20

score 5 · Accepted Answer · answered Jun 24 '19 at 21:43

5

The codomain is the category depicted on the right, and the image of $F$ is not a subcategory because it contains the morphisms $p$ and $q$ but not their composition $qp$. This is explained under the diagram.

answered Jun 24 '19 at 21:43

Javi

6,541

The image is not a subcategory because the image is not even a category. Being a category requires the composition to exist. Am I right? – XYSquared Dec 11 '19 at 06:00
@XYSquared Yes, you are. – Javi Dec 11 '19 at 18:08

The image of a functor need not be a subcategory

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