# 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 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 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 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 $g$.
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.