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Given a category ${\cal C}$, a functor $F:{\cal C}\rightarrow {\rm Sets}$, an object $A$ in ${\cal C}$ and an element $x\in F(A)$, one may consider the smallest subfunctor $F_x$ of $F$ that contains $x$. Explicitly, $F_x(B) = \{ F(f)(x) \}_{f\in {\rm Hom}(A,B)}$ and $F_x(f) = F(f)|_{F_x(B)}$ for $f:B\rightarrow B'$.

While functors of the form $F_x$ are not necessarily representable, I believe a bit of the Yoneda lemma survives. Namely given any set valued functor $G:{\cal C}\rightarrow {\rm Sets}$, and natural transformations $T,T'$ taking $F_x$ to $G$, $(T(id_A))(x)= (T'(id_A))(x)$ (in $G(A)$) implies $T=T'$. Thus one can still identify the set of all natural transformations from $F_x$ to $G$ with a subset of $G(A)$. The proof works just like Yoneda.

Can one give an intrinsic characterization of functors of the form $F_x$ solely in terms of ${\cal C}$?

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Up to isomorphism, they are precisely quotients of representables. Indeed, these $F_x$ arise as image (epi-mono) factorizations of the classifying map $\theta_x: \hom(A, -) \to F$ of $x \in F(A)$:

$$\theta_x = (\hom(A, -) \stackrel{epi}{\to} F_x \stackrel{mono}{\to} F)$$

since the image of the map $\theta_x$ is by definition the smallest subobject of $F$ through which it factors. (Note that epi-mono factorizations in the presheaf topos are computed pointwise. Note also that if $Q$ is a quotient of a representable, $\theta: \hom(A, -) \to Q$, then $Q \cong Q_x$ where $x \in Q(A)$ is the element $\theta_A(1_A)$.)

This explains your modified Yoneda lemma in terms of the usual Yoneda lemma: to say $\theta: \hom(A, -) \to Q$ is epi is to say that for any two $T, T': Q \to G$, that $T \circ \theta = T' \circ \theta$ implies $T = T'$. But following the Yoneda lemma, $T \circ \theta$ is the unique map classified by the element $(T \circ \theta)_A(1_A)$, and your result follows.

Edit: In my comment below in response to David, I referred to the end calculation of the set of natural transformations $F \to G$ between functors $F, G: C \to Set$ (where $C$ is small). In terms of vanilla products and equalizers, this set $Nat(F, G)$ is the equalizer of a pair of maps of the form

$$\prod_{A \in Ob(C)} G(A)^{F(A)} \stackrel{\to}{\to} \prod_{A, B} G(B)^{F(A) \times \hom(A, B)}$$

I am almost tempted to leave the description of these maps to the reader. One takes a tuple of maps $\theta_A: F(A) \to G(A)$ to the tuple whose component at $(A, B)$ is the evident composite $F(A) \times \hom(A, B) \to F(B) \to G(B)$; the other takes the same tuple to the evident composite $F(A) \times \hom(A, B) \to G(A) \times \hom(A, B) \to G(B)$. The condition that the maps are equalized corresponds exactly to the naturality condition on $\theta_A: F(A) \to G(A)$.

A nice exercise is to see the definition of sheaf over a space in terms of this construction. Let $U_i$ be an open cover of $U$, and let $F = \bigcup_i \hom(-, U_i) \hookrightarrow \hom(-, U)$ be the corresponding covering sieve. Then a presheaf $G$ is a sheaf iff the restriction map

$$G(U) \cong Nat(\hom(-, U), G) \to Nat(F, G)$$

is an isomorphism for every such covering sieve.

Edit 2: This answer was downvoted for unspecified reasons. If there are good reasons, they should be made explicit.

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  • $\begingroup$ Thanks, Todd. My original aim was understanding the class of all natural transformations between two arbitrary set-valued functors. So natural transformations from $F$ to $G$ restrict to natural transformation from the various sub-functors $F_x$ to G. Then one wants to cover $F$ with the various $F_x$, keep track of how things patch together, and ultimately identify the set of natural transformations with some kind of possibly large limit of subsets of various sets $G(A)$. Does this make any sense, and can I find it worked out somewhere? I would ask a whole new MO question if so advised. $\endgroup$ Commented Dec 29, 2010 at 1:31
  • $\begingroup$ It makes a lot of sense! But also this stuff is well worked out. In outline, each presheaf $F$ is a colimit of representables, in fact a union (a special type of filtered colimit) of quotients $F_x$ of representables as you are saying. (There is another way of expressing this via the calculus of coends; see for instance ncatlab.org/nlab/show/co-Yoneda+lemma) This means $Nat(F, G)$ is a limit of sets $Nat(\hom(A, -), G)$, in other words a limit of sets $G(A)$ or: an end $\int_A G(A)^{F(A)}$ where the exponential is the $F(A)$-indexed product of copies of $F(A)$. (cont.) $\endgroup$ Commented Dec 29, 2010 at 2:35
  • $\begingroup$ It is slightly scandalous that I cannot find an immediate online link for this (classic) material, but there is brief reference to this in the nLab at ncatlab.org/nlab/show/end#enriched_functor_categories_39. I might recommend reading about ends in Categories for the Working Mathematician, and then working out the meaning of the end formula for $Nat(F, G)$ (in the Set-valued case). You'll probably be able to work out the sheaf-y patching formulas for the end from these hints, but I'd be happy to supply them as an edit to this answer as well. $\endgroup$ Commented Dec 29, 2010 at 2:40
  • $\begingroup$ @Todd I've been saddled since graduate school with the misconception that ${\rm Nat}(F,G)$ was somehow deeply intractable even just at the conceptual level. Do there exist criteria (e.g. interesting necessary condition or sufficient conditions weaker than Yoneda) to determine when ${\rm Nat}(F,G)$ forms a set (rather than a proper class)? Can one describe, say $Nat(H_3(\ast,Z),\pi_5(\ast))$, or at least reduce it to a purely topological calculation? $\endgroup$ Commented Dec 29, 2010 at 5:19
  • $\begingroup$ You're right: much of the time $Nat(F, G)$ is hard-to-understand, and you're right also that it's often not even a set if the domain $C$ is large. Basically it's a set if $F: C \to Set$ ($Set$ can be replaced by other bases of hom-enrichment) can be presented as a small colimit of representables. I don't know off the bat if $H_3(\ast, Z)$ is so presentable (say if $C$ is the homotopy category of CW complexes), but that would be something to determine in these calculations. $\endgroup$ Commented Dec 29, 2010 at 14:11

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