On the nLab page for sieves and elsewhere it is asserted that a fully faithful functor is 'equivalent' to the inclusion functor of a full subcategory -- what is this intended to mean, explicitly?

Are there equivalences in the $2$-category of functors, natural transformations and modifications between fully faithful functors and inclusions of full subcategories? Is there an equivalence of categories between the domain category of a fully faithful functor and a full subcategory of the codomain category? Concretely:

What does it mean non-heuristically that a fully faithful functor can be thought of as a full subcategory?

I am trying to understand sieves and all the literature I've found defines them as fully faithful discrete fibrations then immediately begins discussing them as full subcategories closed under precomposition, and I'm missing the exact connection between these two interpretations.

It's clear that for any functor $F:\mathcal{C}\to\mathcal{D}$ we have $F(\mathcal{C})$ as a subcategory of $\mathcal{D}$ and $F(\mathcal{C})$ is a full subcategory iff $F$ is full; is this related to the intended meaning? This is incorrect, thanks to Jendrik for catching the error in the comments.

  • 2
    $\begingroup$ If $F \colon \mathcal{C} \to \mathcal{D}$ is a functor then its image $F(\mathcal{C})$ need not be a subcategory of $\mathcal{D}$ since it is not necessarily closed under composition of morphisms. This can fail even if $F$ is full. (See this question for a counterexample.) $\endgroup$ May 11, 2019 at 21:49
  • $\begingroup$ @JendrikStelzner Interesting, thank you for the clarification -- does faithfulness guarantee that $F(\mathcal{C})$ is a subcategory then? $\endgroup$
    – Alec Rhea
    May 11, 2019 at 22:20
  • $\begingroup$ @JendrikStelzner Was just about to say I worked through the linked example and found what you mentioned, thanks haha. Unfortunately this doesn't shed much light on the main question above for me, and perhaps even clouds it a bit since a fully faithful functor doesn't even necessarily yield a category as its image. $\endgroup$
    – Alec Rhea
    May 11, 2019 at 22:32
  • 1
    $\begingroup$ My previous comment was wrong, if $F$ is full then $F(\mathcal{C})$ is a category: The only possible problem is closure under composition. If $f \colon X \to Y$, $g \colon Y \to Z$ are in $F(\mathcal{C})$ then they come from $f' \colon X' \to Y'_1$, $g' \colon Y'_2 \to Z$ in $\mathcal{C}$ with $F(Y'_1) = F(Y'_2) = Y$. Then $\mathrm{id}_Y$ lifts to some $h' \colon Y'_1 \to Y'_2$ since $F$ is full. Now $g' h' f'$ lifts $gf$, so $gf$ is in $F(\mathcal{C})$. (I mistakenly thought the linked counterexample were full. If $F$ is injective then already $Y'_1 = Y'_2$ and then $g'f'$ lifts $gf$.) $\endgroup$ May 12, 2019 at 0:45
  • $\begingroup$ @JendrikStelzner Good catch, thanks for correcting it. $\endgroup$
    – Alec Rhea
    May 12, 2019 at 1:50

1 Answer 1


If $F:C\to D$ is a fully faithful functor and $D_0$ is the full subcategory of $D$ on the objects in the image of $F$, then $F$ restricts to an equivalence of categories $C\to D_0$. Conversely, the inclusion functor $D_0\to D$ is fully faithful, and thus so is its composition with any equivalence of categories $C\to D_0$. So a functor is fully faithful iff it is the composition of an equivalence of categories followed by the inclusion of a full subcategory. For most purposes (more precisely, any purposes that are unaffected by precomposing with an equivalence of categories), this means you can assume any fully faithful functor actually is just the inclusion of a full subcategory.

  • $\begingroup$ This is exactly what I was looking for, much appreciated Eric. $\endgroup$
    – Alec Rhea
    May 11, 2019 at 23:01
  • $\begingroup$ Could we use isomorphic instead of equivalent here? Say, 'a functor is fully faithful iff it is the composition of an isomorphic of categories followed by the inclusion of a full subcategory.' Is that true as well? $\endgroup$
    – Hydrogen
    Dec 11, 2021 at 17:51
  • $\begingroup$ @Hydrogen: No: a fully faithful functor may not be injective on objects. $\endgroup$ Dec 11, 2021 at 18:24

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .