# Does the nerve have a right adjoint, sometimes?

Let $F : G\to PA$ be a functor ($G$ a small category, $PA = [A°,Set]$ the presheaf category of $A$). Let $N_F$ be the functor $PA\to PG$ defined by left Kan-extending the Yoneda embedding $y_G : G\to PG$ along $F$: The following procedure seems to exhibit a right adjoint for $N_F$; this is strange, as $N_F$ is in general only a right adjoint. Can you spot the error (if any) in my argument?

Consider the extended diagram where the functor $PG \to PA$ is the left extension of $y_A$ along $N_F y_A$, that can easily shown to be equal to the left extension of the identity along $N_F$. And this latter functor shall be the right adjoint to $N_F$: counit and unit are determined by the suitable universal properties.

Many things can be true at this point:

1. The functor does not exist.
2. The functor exists, but fails to be a right adjoint to $N_F$ (the extension is not absolute/not preserved by $N_F$).
3. The functor is indeed a right adjoint to $N_F$; I didn't know it had one in the special case where its domain is itself a free cocompletion.

Note that taking the right extension of $y_A$ along $N_Fy_A$ does not give a left adjoint to $N_F$: if such a left adjoint exists, it is unique and must exhbit the universal property of the left extension of $F$ along $y_A$: $$\text{Lan}_{y_A}F \dashv \text{Lan}_Fy_A$$ (that's an instance of the $F$-nerve and $F$-oidal realization yoga). Indeed, a key step in the above argument is that $\text{Lan}_{N_F y_A}y_A\cong \text{Lan}_{N_F}(\text{Lan}_{y_A}y_A) \cong \text{Lan}_{N_F}1$, as $y_A$ is dense. But it's not codense: in fact, $\varphi \to \text{Ran}_{y_A}(y_A)(\varphi)$ is the $\varphi$-component of the unit of Isbell adjunction.

• please correct mistype $PA\to GP$ as $PA\to PG$; very minor but unpleasant. :)) – Evgeny Kuznetsov Apr 8 '18 at 9:58
• I have no words to express how sorry I am for having hurt you. :-) – Fosco Loregian Apr 8 '18 at 22:03
• In terms of the profunctor $\Phi_F:G^\circ\times A\to\mathbf{Set}$ with $\Phi_F(g,a)=\hom_{PA}(F(g),\hom(-,a))$, is not $N_F$ isomorphic to $\Phi_F\otimes_A-$? – მამუკა ჯიბლაძე Apr 8 '18 at 23:12
• Hi prof. Jibladze! So are you saying that in this particular case $N_F \dashv \{\Phi_F,-\}$ (or whatever other notation for the "mean cotensor")? That would save my day, up to the fact that I'm insisting in this style-of-proof because my argument must live in a generic 2-category $\cal K$. – Fosco Loregian Apr 9 '18 at 6:47
• @მამუკაჯიბლაძე I guess the answer is no: $$\Phi_F\otimes Q(g) \cong \int^a Qa\times PA(Fg, y_A(a))$$ whereas $$N_F(Q)(g) \cong \int^{g'}PA(Fg',Q)\times G(g,g') \cong PA(Fg,Q)$$ how do you compare the first with the secnd formula, if $F$ takes values in non-tiny objects? – Fosco Loregian Apr 9 '18 at 11:49

$PG$ and $PA$ are locally presentable categories, so $N_F : PA \to PG$ has a right adjoint if and only if it preserves small colimits.

From the formula $$N_F(Q)(g) \cong \hom_{PA}(F(g), Q)$$ the condition that $N_F$ preserves colimits is equivalent to natural isomorphisms $$\hom_{PA}(F(g), \operatorname{colim}_j Q_j) \cong \operatorname{colim}_j N_F(Q_j)(g) \cong \operatorname{colim}_j \hom_{PA}(F(g), Q_j)$$

In other words, $F(g)$ must be a tiny object for every $g$.

But in a presheaf category, the tiny objects are precisely the retracts of representables. If the domain category is idempotent complete, this simplifies to the tiny objects being precisely the representables.

In other words, $N_F$ has a right adjoint if and only if $F$ can be written as a composite of a functor $G \to \operatorname{Idem}(A)$ followed by the yoneda embedding.

The following is a long comment.

Rmk.1

If $F$ is dense then $$\text{Lan}_{N_Fy_A}(y_A) = \text{Lan}_{N_F}(\text{id}) = \text{Lan}_{N_F}(\text{Lan}_{F}(F)) = \text{Lan}_{N_F F}(F) = \text{Lan}_{y_G}(F),$$

which would be a left adjoint for $N_F$. By a 2-out-of-three argument about Kan extensions, this is a characterization of dense functors $G \to P(A)$, namely:

$G \to P(A)$ is dense if and only if $\text{Lan}_{N_Fy_A}(y_A)$ provides a left adjoint for $N_F$.

Rmk.2

$$\text{Lan}_{N_F}(\text{id})= \text{Lan}_{\text{Lan}_{y_A}(N_Fy_A)}(\text{id}).$$

Thus, if everybody is absolute we are requiring that $N_F = \text{Lan}_{y_G}(N_Fy_G).$ This looks to me as a very strong and unlikely request of cocontinuity for $N_F$.