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 $\mathrm{Hom}_D(-,Y)$.

  • 1
    $\begingroup$ More recently, Lurie has been using the name "dense", e.g. in SAG.20.4.1. $\endgroup$ – Alexander Campbell May 3 at 22:55
  • $\begingroup$ 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 $T$. $\endgroup$ – Mark Wildon May 4 at 9:17
  • $\begingroup$ 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 $\mathrm{Set}$? $\endgroup$ – Mark Wildon May 4 at 9:19
  • $\begingroup$ @MarkWildon In the nLab page about dense functor, look at the hyperlink enriched category theory. $\endgroup$ – Philippe Gaucher May 4 at 11:17
  • $\begingroup$ 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. $\endgroup$ – Tim Campion May 4 at 13:31

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