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Denote by $\mathbf{Cat}$ the category of small categories, i.e. categories whose class of arrows is a set. I am working with classes but I suppose it works the same if one is working in a given universe following the Grothendieck approach.

My question is whether monomorphisms and epimorphisms in $\mathbf{Cat}$ can easily be classified. An arrow between two small categories is just a functor.

In particular I was wondering if the subcategories of a given small category correspond to the category-theoretic subobjects in $\mathbf{Cat}$. Given a fixed object $X$ in a category I call an isomorphism class of monomorphisms with codomain $X$ a $\textit{subobject}$ of $X$.

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I think that epimorphisms in (Cat) have been discussed before on math.SE. –  Martin Brandenburg Feb 28 at 16:03
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1 Answer 1

up vote 5 down vote accepted

Monomorphisms are embeddings functors (i.e. functors whose both the object-function and the arrow-function are injective).

It easily seen that embeddings are monomorphisms because if $F \colon \mathbf C \to \mathbf D$ is an embedding then for every pair of fucntors $G,H \colon \mathbf X \to \mathbf C$ such that $F \circ G=F \circ H$ then

  • for every $x \in \mathbf X$ we should have $F \circ G(x)=F \circ H(x)$ and since $F$-object part is injective it would follow that $G(x)=H(x)$;

  • in exactly the same way for every $f \colon x \to x'$ in $\mathbf X$ we have that $F \circ G(f)=F\circ H(f)$ and for the injectivity of $F$-arrow part we also have that $G(f)=H(f)$;

so $G=H$.

Now let's suppose $F \colon \mathbf C \to \mathbf D$ is monomorphism.

For every object $c \in \mathbf C$ there's a unique functor $\bar x \colon 1 \to \mathbf C$ (where $1$ is the category with one object ($0$) and one arrow) such that $\bar x(0)=x$.

In similar way for every morphism $f \in \mathbf C(c,c')$ there's a unique functor $\bar f \colon 2 \to \mathbf C$ (where $2$ is the category with two objects, $0$ and $1$, and only one non identity morphism $0 \to 1$) such that $\bar f(0)=c$, $\bar f(1)=c'$ and $\bar f(0 \to 1)=f$.

Then it follows that for every pair of objects $c,c' \in \mathbf C$ if we have $F \circ \bar c(0)=F(c)=F(c')=F \circ \bar c'(0)$ then $F \circ c=F \circ c'$ and so by monomorphism property $\bar c=\bar c'$. From this follows that $$c=\bar c(0)=\bar c'(0)=c'$$ which gives the injectivity of the object part of $F$.

If $f,g \colon c \to c'$ are morphisms of $\mathbf C$ such that $F(f)=F(g)$ it can be proven similarly (using the functor $\bar f$ and $\bar g$) that $f=g$.

From this it follows that also the arrow part of $F$ is injective.

Edit I see that I didn't address the part about epimorphisms in $\mathbf{Cat}$. Let's try to make ammend.

For start it can be shown that every functor that is not sujective on objects cannot be an epimorphism. Indeed if $F \colon \mathbf C \to \mathbf D$ is not surjective we can consider the full subcategory $\mathbf D' \hookrightarrow \mathbf D$ that is spanned by the object in $\text{Im} F$. The consider the pushout of the diagram

pushout

where the morphism from $\mathbf D'$ are the embeddings. Since $F$ factors through $i \colon\mathbf D' \hookrightarrow \mathbf D$ (since it's image is contained in the full subcategory $\mathbf D'$) we have that $G \circ F = H \circ F$.

More in detail there's a functor $\mathcal F \colon \mathbf C \to \mathbf D'$ such that $F=i \circ \mathcal F$ and so

$$G \circ F = G \circ i \circ \mathcal F$$ and $$H \circ F = H \circ i \circ \mathcal F\, .$$

Since $G \circ i= H \circ i$, because of the pushout property, we have that $G \circ F=H \circ F$, nonetheless $G \ne H$.

This basically prove that epimorphism should at least be sujective on objects. Clearly this condition is not sufficient, it's easy to find counter examples.

What can be proven is that $F \colon \mathbf C \to \mathbf D$ is sujective on objects and the graph $\text{Im}F$ contains a set of morphism which are generators for the category $\mathbf D$ then this functor is an epimorphism.

By generators for the category $\mathbf D$ I mean a family of morphisms such that every other morphism in $\mathbf D$ is a composite of these morphisms.

The proof of this is no different from the analog proofs for other categories. Nontheless this is just a sufficient condition as the following example show.

Consider the category $2$ defined as above and $\bar 2$ obtained by $2$ just adding an inverse to the morphism $0 \to 1$.

The embedding $j \colon 2 \to \bar 2$ is an epimorphism since for every $G,H \colon \bar 2 \to X$ such that $G(0\to 1)=G(0 \to 1)$ have to send $1 \to 0$ in *the inverse of $G(0 \to 1)=H(0 \to 1)$.

Of course from this example it becomes clear that one can perfect the previoius said condition by proving that $F \colon \mathbf C \to \mathbf D$ is an epimorphism iff it's surjective on object and every morphism in $\mathbf D$ is a composite of morphism in $\text{Im} F$ and their inverse (if they exist).

Nonetheless I'm not aware if this is a full characterization of epimorphisms in $\mathbf{Cat}$.

Hope this helps.

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I'd voluntarily left some of the details. Hope you don't mind. –  Giorgio Mossa Feb 28 at 14:06
    
Very nice idea! Do you also know about epimorphisms? –  user114885 Feb 28 at 14:56
    
@user114885 I've added some stuff about epis too. Nonetheless I don't know if that can be enough for you. There's a claim (but I've not prove of it) for a possibile characterization of epis in $\mathbf {Cat}$. –  Giorgio Mossa Mar 1 at 15:11
    
I'm afraid I am adding this comment very late. But it might help you. Professor Sobral, Adámek, Bashir and Velebil characterized the lax epimorphisms of Cat (ON FUNCTORS WHICH ARE LAX EPIMORPHISMS). And, there, they claim that it seems that the "correct notion" of epimorphisms in Cat would be that of lax epimorphisms. –  Fernando Jul 9 at 19:16
    
On the other hand, in the strict sense, the best result I am aware is the characterization of "universal epimorphisms": which are precisely the full functors surjective on objects. –  Fernando Jul 9 at 19:17

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