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 $\mathrm{Set}$.


1 Answer 1


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. $\endgroup$ May 3, 2021 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 $T$. $\endgroup$ May 4, 2021 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 $\mathrm{Set}$? $\endgroup$ May 4, 2021 at 9:19
  • $\begingroup$ @MarkWildon In the nLab page about dense functor, look at the hyperlink enriched category theory. $\endgroup$ May 4, 2021 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, 2021 at 13:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.