11.7.3 Epimorphisms of Categories

Let $\mathcal{C}$ and $\mathcal{D}$ be categories.

A functor $F\colon \mathcal{C}\to \mathcal{D}$ is a epimorphism of categories if it is a epimorphism in $\mathsf{Cats}$ (see Unresolved reference, Unresolved reference).

Let $F\colon \mathcal{C}\to \mathcal{D}$ be a functor.

  1. 1.

    Characterisations. The following conditions are equivalent:1

    1. (a)

      The functor $F$ is a epimorphism of categories.

    2. (b)

      For each morphism $f\colon A\to B$ of $\mathcal{D}$, we have a diagram

      in $\mathcal{D}$ satisfying the following conditions:

      1. (i)

        We have $f=\alpha _{0}\circ \phi _{1}$.

      2. (ii)

        We have $f=\psi _{m}\circ \alpha _{2m}$.

      3. (iii)

        For each $0\leq i\leq 2m$, we have $\alpha _{i}\in \operatorname {\mathrm{Mor}}\webleft (\mathrm{Im}\webleft (F\webright )\webright )$.

  2. 2.

    Surjectivity on Objects. If $F$ is an epimorphism of categories, then $F$ is surjective on objects.


  1. 1Further Terminology: This statement is known as Isbell’s zigzag theorem.

Is there a characterisation of functors $F\colon \mathcal{C}\to \mathcal{D}$ such that:

  1. 1.

    For each $\mathcal{X}\in \operatorname {\mathrm{Obj}}\webleft (\mathsf{Cats}\webright )$, the precomposition functor

    \[ F^{*}\colon \mathsf{Fun}\webleft (\mathcal{D},\mathcal{X}\webright )\to \mathsf{Fun}\webleft (\mathcal{C},\mathcal{X}\webright ) \]

    is an epimorphism of categories?

  2. 2.

    For each $\mathcal{X}\in \operatorname {\mathrm{Obj}}\webleft (\mathsf{Cats}\webright )$, the postcomposition functor

    \[ F_{*}\colon \mathsf{Fun}\webleft (\mathcal{X},\mathcal{C}\webright )\to \mathsf{Fun}\webleft (\mathcal{X},\mathcal{D}\webright ) \]

    is an epimorphism of categories?

This question also appears as [Emily, Characterisations of functors $F$ such that $F^*$ or $F_*$ is [property], e.g. faithful, conservative, etc].


Noticed something off, or have any comments? Feel free to reach out!


You can also use the contact form below: