Is a subfunctor of a representable functor also representable?

Question

I'm trying to learn about representable functors but I'm very new to this; even to categories...

Suppose $G:C^{op}\rightarrow Set$ is a representable functor - as I understand this means that there exists some object $c$ of $C$ such that $G$ and $Hom_{C}(\cdot,c)$ are naturally isomorphic.

According to this page https://ncatlab.org/nlab/show/subfunctor ,

"A subfunctor of a functor $G:C→D$ between categories $C$ and $D$ is a pair $(F,i)$ where $F:C→D$ is a functor and $i:F→G$ is a natural transformation such that its components $i_M:F(M)→G(M)$ are monic."

Question 1: If $G:C^{op}\rightarrow Set$ is representable and $(F,i)$ as above is a subfunctor of $G$, is it true that $F$ is also representable?

I did some research on internet but I couldn't figure out a direct answer to the previous question - so, as it is quite general, I suspect that the answer is NO - but I would like some references to read about it, with examples if possible...

Due to my suspect, I would like also to know if

Question 2: are there some "well-known" criteria (say if $C$ is the category of schemes) to tell whether it could be true for particular examples?

Answer 1

Question 1. A typical example is the functor $\mathrm{Nil} : \mathbf{Ring} \to \mathbf{Set}$ of nilpotent elements in a ring. It is a subfunctor of the forgetful functor, which is represented by $\mathbb{Z}[T]$, but $\mathrm{Nil}$ is not representable: If $u$ is a universal element, then $u^n=0$ for some $n$, which would imply that every nilpotent $x$ in any ring would satisfy $x^n=0$, which is wrong. At least, $\mathrm{Nil}$ is ind-representable, since we have $$\mathrm{Nil} \cong \varinjlim_n \mathrm{Hom}(\mathbb{Z}[T]/T^n,-).$$ Similarly, the functor $\mathrm{Tors} : \mathbf{Grp} \to \mathbf{Set}$ of torsion elements in groups is not representable, but it is ind-representable since $$\mathrm{Tors} \cong \varinjlim_n \mathrm{Hom}(C_n,-).$$

Question 2. The following basic result is proved in EGA I. It can be used, for instance, to prove that fiber products of schemes exist. You can see it as a "gluing process guided by a functor".

Let $\mathbf{Sch}$ be the category of schemes (the statement works just as well for ringed spaces, locally ringed spaces, manifolds etc.). A functor $F : \mathbf{Sch}^{\mathrm{op}} \to \mathbf{Set}$ is represesentable if and only if the following two conditions hold:

• $F$ is a sheaf: For every open covering $U = \bigcup_i U_i$ of a scheme $U$ the diagram $F(U) \to \prod_i F(U_i) \rightrightarrows \prod_{i,j} F(U_i \cap U_j)$ is exact.
• $F$ has an open covering by representable functors.

Here, a subfunctor $G \subseteq F$ is called open when for every scheme $X$ and every element $a \in F(X)$, thus corresponding to a morphism $a : \mathrm{Hom}(-,X) \to F$, the pullback $G \times_F \mathrm{Hom}(-,X) \to \mathrm{Hom}(-,X)$ is isomorphic to $\mathrm{Hom}(-,U) \to \mathrm{Hom}(-,X)$ for some open subscheme $U \subseteq X$. Also, a family of subfunctors $G_i \subseteq F$ covers $F$ when for every scheme $X$ which is the spectrum of a field (not for any scheme!), we have $\bigcup_i G_i(X) = F(X)$.

Answer 2

Here is my answer to your first question. Consider the non trivial category $\mathcal{C}=\{0\circlearrowleft\circlearrowleft\},$ where $0$ has a non-identity morphism $f$ (of order $2$) to itself.

Now consider the representable functor $G:\mathcal{C}^{\text{op}}\to\mathbf{Set}$ given by $G(x)=\operatorname{hom}_{\mathcal{C}}(x, 0).$ This has exactly one proper sub-functor, the empty functor, and it is clearly not representable.

Added: The full description of $G$ is

• $0\mapsto \{\operatorname{id}_0, f\}$
• $\operatorname{id}_0\mapsto\operatorname{id}_{\{\operatorname{id}_0, f\}}$
• $f\mapsto f_*=(\operatorname{id}_0\mapsto f, f\mapsto \operatorname{id}_0)$

and the only proper sub-functor that can make a naturally square is the empty functor.

Answer 3

If you consider $\mathbb Q$ as a poset and thus also naturally a category we can consider the functor category $[\mathbb Q, \textbf{Set}]$.

The set of subfunctors of a representable functor $\text{Hom}(q,-)$ are in bijection with the set $\{r \in \mathbb R \cup \{\infty \}:r \geq q \}$

(this is essentially the statement that every set bounded from above in $\mathbb R$ has a least upper bound)

which is clearly strictly bigger than the set of representable subfunctors of $Hom(q,-)$ which is in bijection with $\{ p \in \mathbb Q \cup \{ \infty \}: p \geq q \}$.