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Is the essential image of a fully faithful functor always a category? If not is there a counterexample?

The answer in Can it happen that the image of a functor is not a category? does not apply in this situation: the functor there is not full. This question is not asking why the image of a functor is not a category, but it is asking when it is a category.

Mark Kamsma
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sirjoe
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  • But isn't the functor in the example then an equivalence of categories? – sirjoe Nov 13 '19 at 23:56
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    I think that example fails to be full as Hom(c,b) is empty but Hom(F(b),F(c)) is non empty since F(b)=F(c)=y – sirjoe Nov 13 '19 at 23:59
  • You are right, I judged too quick. Actually, fullness of the functor should be enough. I'll leave my comment still so people can easily find why fullness is really necessary. – Mark Kamsma Nov 14 '19 at 00:05
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    Ok, so I have deleted my comment now because it seemed to do more harm than good. The question actually got closed as duplicate, probably because of my comment. So I edited the question to clarify the mistake I made and voted to reopen. – Mark Kamsma Nov 14 '19 at 11:33

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If $F: \mathcal{C} \to \mathcal{D}$ is full, then the image of $F$ is always a category. Clearly we have identity arrows on each object, and composition will be associative since it is associative in $\mathcal{D}$. So we only need to check that the image of $F$ is closed under composition (see also this answer to see what could go wrong).

Let $F(f): F(A) \to F(B)$ and $F(g): F(C) \to F(D)$ be arrows in the image of $F$, such that $F(B) = F(C)$. Then we can compose $F(f)$ and $F(g)$ in $\mathcal{D}$ to get $F(g)F(f): F(A) \to F(D)$. Since $F$ is full, we find $h: A \to D$ such that $F(h) = F(g)F(f)$. So $F(g)F(f)$ is in the image of $F$, and so the image of $F$ is closed under composition.

It may also be worth pointing out that if $F$ is injective on objects, its image forms a category. Again you need to check that it's closed under composition, which is an easy exercise.

Mark Kamsma
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