Yoneda map for a composition of a representable functor and an arbitrary functor

Let $$\mathcal{C}$$ and $$\mathcal{D}$$ be categories and let $$T : \mathcal{C} \rightarrow \mathcal{D}$$ be a functor. Suppose that $$F : \mathcal{D}^\mathrm{op} \rightarrow \mathrm{Set}$$ is a functor. (So in the older language I am more used to, $$F$$ is a contravariant functor.) Let $$Y \in \mathcal{D}$$ be an object. Under what conditions on $$T$$ and $$Y$$ is there a bijection $$\mathrm{Nat}\Bigl(\mathrm{Hom}_\mathcal{D}\bigl(T(-), Y\bigr), F\circ T\Bigr) \stackrel{\mathrm{Yon}}{\longrightarrow} F\bigl(Y\bigr)$$ with the properties of the Yoneda embedding?

The rest of my question is background and motivation. I'll first show that there is a unique way to define part of $$\mathrm{Yon}$$ when $$Y$$ is $$T(B)$$ for some object $$B \in \mathcal{C}$$. Suppose that $$\alpha$$ is a natural transformation from the contravariant functor $$\mathrm{Hom}_\mathcal{D}\bigl(T(-), T(B) \bigr) : \mathcal{C} \rightarrow \mathrm{Set}$$ to the contravariant functor $$F \circ T : \mathcal{C} \rightarrow \mathrm{Set}$$. (Again these become ordinary functors replacing $$\mathcal{C}$$ with $$\mathcal{C}^\mathrm{op}$$.) Apply

$$\alpha_{B} : \mathrm{Hom}_{\mathcal{D}}\bigl(T(B),T(B) \bigr) \rightarrow F\bigl(T(B)\bigr)$$

to $$\mathrm{id}_{T(B)}$$ to get $$t = \alpha_B(\mathrm{id}_{T(B)}) \in F(TB) = F(Y)$$. Then, as in the usual proof of the contravariant version of Yoneda's Lemma, a chase around a commutative square shows that if $$f : A \rightarrow B$$ is any morphism in $$\mathcal{C}$$, and so $$Tf : T(A) \rightarrow T(B)$$ is a morphism in $$\mathcal{D}$$, then $$\alpha_A : \mathrm{Hom}_\mathcal{D}\bigl(T(A),T(B)\bigr) \rightarrow F\bigl(T(A)\bigr)$$ satisfies

$$\alpha_A(Tf) = (FTf) \alpha_B(\mathrm{id}_{T(B)}) = (FTf)(t) \in F\bigl(T(A)\bigr).$$

This determines $$\alpha_A$$ on those morphisms of the form $$Tf$$, for an arbitrary object $$A$$ of $$\mathcal{C}$$, and so when $$T$$ is full, we have the required bijection. Restated in terms of $$t$$, my question asks for a necessary and sufficient condition on $$T : \mathcal{C} \rightarrow \mathcal{D}$$ and $$Y \in \mathcal{D}$$ for $$t \in F(Y)$$ to determine $$\alpha$$.

Example 1. Let $$\mathcal{C} = \mathcal{D}$$ be the category with one object $$\star$$ and $$\mathrm{Mor}(\star,\star) = \mathcal{S}$$ where $$\mathcal{S} = \langle \sigma, \tau \rangle$$ is the group of permutations of $$\{1,2,3\}$$, generated by $$\sigma = (1,2,3)$$ and $$\tau = (1,2)$$. Since $$Y = \star = T(\star)$$, the requirement on $$Y$$ for part of $$\mathrm{Yon}$$ to be defined is obviously satisfied. Define $$T : \mathcal{C} \rightarrow \mathcal{C}$$ by abelianization with embedding $$\langle 1, \tau \rangle$$, so

$$T(\mathrm{id}_\star) = T(\sigma) = T(\sigma^2) = \mathrm{id}_\star, \quad T(\tau) = T(\sigma\tau) = T(\sigma^2 \tau) = \tau.$$

The partial definition of $$\alpha_\star$$ requires $$\alpha_\star(Tf) = t(Tf)$$ where $$t = \alpha_\star(\mathrm{id}_\star)$$, for all $$f \in \mathcal{S}$$. There are only two possibilities for $$Tf$$. Taking $$f \in \{\mathrm{id}_\star, \sigma, \sigma^2 \}$$ gives nothing, whereas taking $$f \in \{\tau, \sigma\tau, \sigma^2\tau\}$$ gives $$\alpha_\star(\tau) = t\tau$$. Thus $$t$$ determines $$\alpha_\star(\tau)$$, but not, for instance, $$\alpha_\star(\sigma)$$.

Example 2. (Edited, since although this was the original motivation, I realised later it doesn't exactly fit the setup above.) Let $$\mathcal{C}$$ be the category of representations of the algebraic group $$\mathrm{GL}_d(\mathbb{C})$$ and let $$\mathcal{D}$$ be the category of bimodules with left $$\mathrm{GL}_d(\mathbb{C})$$ action and right $$S_r$$ action. Let $$T : \mathcal{C} \rightarrow \mathcal{D}$$ be the functor defined by $$T(U) = U^{\otimes r}$$, where the tensor product is regarded as a representation of $$S_r$$ acting on the right by place permutation on tensors. Let $$F$$ be the representable functor

$$\mathrm{Hom}_{\mathcal{D}}\bigl( -, \mathrm{Sp}^\mu \bigr),$$

taking values in $$\mathcal{C}$$, not $$\mathrm{Set}$$ as above. (There is a $$\mathrm{GL}_d(\mathbb{C})$$ action on the hom-set because the $$\mathrm{GL}_d(\mathbb{C})$$ action on each $$T(U)$$ commutes with $$S_r$$; the action on $$\mathrm{Sp}^\mu$$ must be defined somehow to make this module an object in $$\mathcal{D}$$, so take the trivial action.) By Schur–Weyl duality,

$$U^{\otimes r} \cong \bigoplus_\nu \mathrm{\Delta}^\nu(U) \boxtimes \mathrm{Sp}^\lambda,$$

where $$\Delta^\nu$$ is the Schur functor for $$\nu$$. Hence by Schur's Lemma, $$F\bigl(T(U)\bigr) \cong \Delta^\mu(U)$$, naturally in $$U$$. Taking $$Y = \mathrm{Sp}^\lambda$$, we have,

$$\mathrm{Nat}\Bigl(\mathrm{Hom}_\mathcal{D}\bigl((-)^{\otimes r}, \mathrm{Sp}^\lambda\bigr), \mathrm{Hom}_\mathcal{D}\bigl((-)^{\otimes r}, \mathrm{Sp}^\mu\bigr)) \cong \mathrm{Nat}(\Delta^\lambda, \Delta^\mu)$$

and by further applications of Schur's Lemma,

$$\mathrm{Nat}(\Delta^\lambda, \Delta^\mu) \cong \begin{cases} \mathbb{C} & \text{if \lambda=\mu} \\ 0 & \text{otherwise} \end{cases} \cong \mathrm{Hom}_{\mathbb{C}S_r}(\mathrm{Sp}^\lambda, \mathrm{Sp}^\mu) \cong F(\mathrm{Sp}^\lambda)$$

so every natural transformation comes from an element of $$F(\mathrm{Sp}^\lambda) = F(Y)$$, as in the Yoneda embedding, but now into the category $$\mathcal{C}$$ of representations of $$\mathrm{GL}_d(\mathbb{C})$$, rather than $$\mathrm{Set}$$.

This property of the functor $$T$$ is called being a dense functor, or "densely generating functor". The notion was first introduced by Isbell in the case where $$T$$ is fully faithful under the name "adequate subcategory", but that usage has by now disappeared, and "dense" is preferred. It has been occasionally rediscovered -- for example, in the $$\infty$$-categorical context, Lurie calls such a functor "strongly generating", but even there the term "dense" is now preferred, especially because "strong generator" means something else classically.
One of the equivalent ways to say that a functor $$T: C \to D$$ is dense is to say that the induced functor $$T^\ast: \mathrm{Set}^{D^{op}} \to \mathrm{Set}^{C^{op}}$$ defined by $$F \mapsto F \circ T^{op}$$ on categories of presheaves is fully faithful. The condition as you've stated it says that $$T^\ast$$ is fully faithful on the homset from the representable $$\mathrm{Hom}_D(-,Y)$$ to the functor $$F$$, but this is equivalent because every presheaf is a colimit of representables such as $$\mathrm{Hom}_D(-,Y)$$.
• Thank you for the answer. If I understand right, it shows that the existence of the Yoneda map for all objects $Y \in \mathcal{D}$ is equivalent to $T$ being dense. I'd also be interested in conditions that are weaker on the $\mathcal{D}$ objects (e.g. having the embedding for just certain $Y$), even if this means a stronger condition on $T$. – Mark Wildon May 4 at 9:17
• I'm struggling slightly to think of 'dense' in a way that's really different to how I stated the problem, but the presheaf point of view seems helpful. Is there an analogue for presheaves taking values in categories other than $\mathrm{Set}$? – Mark Wildon May 4 at 9:19
• I agree with Philippe Gaucher's point that there is a direct analog of density in enriched category theory, but unfortunately this doesn't seem to be well-documented on the nlab. The definition is "the same": if $T: \mathcal C \to \mathcal D$ is a $\mathcal V$-enriched functor between $\mathcal V$-enriched categories, then $T$ is said to be dense if the induced functor $T^\ast : Fun_{\mathcal V}(\mathcal D^{op},\mathcal V) \to Fun_{\mathcal V}(\mathcal C^{op},\mathcal V)$ is $\mathcal V$-fully faithful; as in the $Set$-enriched case one need not check on every object. – Tim Campion May 4 at 13:31