If $f:Y \to X$ is a mono in $\textsf{Set}^{\mathbb{C}^{op}}$, then $f$ factors uniquely through $Y \xrightarrow{g} A \stackrel{i}{\hookrightarrow} X$.

Question

I have taken an introductory course in category theory and would like to learn more about presheaves. Currently I am working through "Generic figures and their glueings" by Marie La Palme Reyes, Gonzalo E. Reyes and Houman Zolfaghari (subsequently called RRZ).

Let $\mathbb{C}$ be a small category.

I am stuck with the second part of exercise 1.2.1.1:

If $A \stackrel{i}{\hookrightarrow} X$ is a sub-$\mathbb{C}$-set of $X$, then $i$ is a monomorphism. Conversely, if $f : Y \to X$ is a monomorphism show the existence of a unique sub-$\mathbb{C}$-set $A$ of $X$ and a unique isomorphism $Y \xrightarrow{g} A$ such that $f = ig$.

Since I am not sure how standard this terminology and the notation is, recall the following:

• A $\mathbb{C}$-set is a presheaf.
• A $\mathbb{C}$-set $B$ is a sub-$\mathbb{C}$-set of a $\mathbb{C}$-set $A$, if there exists a morphism of $\mathbb{C}$-sets $i$ (i.e. a natural transformation) such that for each $F \in \textrm{Ob}(\mathbb{C})$, $i_F:B(F) \subset A(F)$ (i.e. $i_F:\textrm{hom}(F, B) \hookrightarrow \textrm{hom}(F, A)$) is the set inclusion.

I find it slightly confusing that RRZ seem to use the same notation for objects in $\mathbb{C}$ and in $\textsf{Set}^{\mathbb{C}^{op}}$. It seems to be sensible for the sake of brevity in the context of that book, but for clarity I reformulate the second part of the exercise using the hom-functor.

Conversely, if $\textrm{hom}(-, f) : \textrm{hom}(-, Y) \to \textrm{hom}(-, X)$ is a monomorphism show the existence of a unique sub-$\mathbb{C}$-set $\textrm{hom}(-, A)$ of $\textrm{hom}(-, X)$ and a unique isomorphism $\textrm{hom}(-, Y) \xrightarrow{\textrm{hom}(-, g)} \textrm{hom}(-, A)$ such that $\textrm{hom}(-, f) = \textrm{hom}(-, i) \circ \textrm{hom}(-, g)$.

We know that in the category of sets, every monomorphism $G \xrightarrow{m} H$ factors uniquely through a bijection $G \xrightarrow{n} G'$ and $j : G' \hookrightarrow H$ such that $m = jn$.

Hence, for every $C \in \textrm{Ob}(\mathbb{C})$, we know that $\textrm{hom}(C, f)$ factors through some bijection $\textrm{hom}(C, Y) \xrightarrow{u} S$ and an inclusion $v : S \hookrightarrow \textrm{hom}(C, X)$ such that $\textrm{hom}(-, f) = v \circ u$.

My problem is now: How can we know that there exists some $A \in \textrm{Ob}(\mathbb{C})$ auch that $S = \textrm{hom}(C, A)$? I don't see how I could use the Yoneda-Lemma at this point, because we don't know whether $S$ is in the codomain of the object function of the Yoneda-embedding. Probably I would have to use the fact that $\textrm{hom}(-, f)$ is a mono somehow - but how?

Answer 1

I think, I managed to solve this exercise. The problem was that I thought of the elements of $\textsf{Set}^{\mathbb{C}^{op}}$ as hom-functors and not as arbitrary functors from $\mathbb{C}^{op}$ to $\textsf{Set}$.

Let $Y \xrightarrow{f} X$ be a mono in $\textsf{Set}^{\mathbb{C}^{op}}$. Then for all $C \in \textrm{Ob}(\mathbb{C})$, $f_C$ is injective and factors uniquely as $f_C = i_C \circ a_C$ where $i_C : A(C) \hookrightarrow X(C)$ is the inclusion of the uniquely determined subset $A(C)$ of $X(C)$ and $a_C : Y(C) \to A(C)$ is a bijection. We can now define the functor $A : \mathbb{C}^{op} \to \textsf{Set}$ to map every $C$ to this uniquely determined set $A(C)$ and every morphism $C \xrightarrow{g} C'$ to the function $A(g) = a_{C'} \circ Y(g) \circ a_{C'}^{-1}$.

All that is left to do is to show that this is actually a functor, which is trivial.