The Kan construction, profunctors, and Kan extensions

It's been a long time since I tried to understand the deep meaning of the "Kan construction", or "nerve-realization" adjunction $$\text{Lan}_y F \dashv N_F = \hom(F,1)$$ that exists among the left Kan extension of $F\colon \mathcal{A}\to \mathbf{D}$ ($\cal A$ small, $\bf D$ cocomplete) along the Yoneda embedding $y\colon {\cal A} \to \hat{\cal A}$. It can be expressed as $$\text{Lan}_y F \dashv \text{Lan}_F y,$$ and this property seems pretty peculiar; especially if I think about the definition of a "Yoneda structure"[here, several comments in the discussion are mine].

1. Is there a reason why this is true?
2. What are other examples of a span of functors ${\bf C} \xleftarrow{G} {\cal A} \xrightarrow{F} {\bf B}$ such that $\text{Lan}_GF\dashv \text{Lan}_FG$?
3. My sensation is that this question acquires a (more?) meaning plunging the 2-category $\bf Cat$ in $\bf Prof$ in the usual way. The functor $N_F = \hom(F,1)$ is the image of $F$ via the canonical 2-functor $\varphi^{(-)} : {\bf Cat}^\text{co} \to \bf Prof$, and $N_F$ has a right adjoint $\hom(1,F)$; the Kan construction amounts to say that this extends to a triple of adjoints $$\text{Lan}_yF = \varphi_F^! \dashv \varphi^F\dashv \varphi_F.$$ What's the meaning of this extension, and its universal property, in $\bf Prof$?
• Feel free to improve the question as you like; or did you already edit it? – Fosco Oct 18 '16 at 21:59
• Nice! I can't believe I never noticed the F-nerve was $\mathrm{Lan}_F y$. – Omar Antolín-Camarena Nov 20 '18 at 22:42

I will try to answer the second question.

Prop 1. Let $${\bf C} \xleftarrow{i} {\bf A} \xrightarrow{f} {\bf B}$$ be a span where $$i$$ is dense and fully faithful. Moreover $$\text{lan}_if$$ is pointwise. Then, the following are equivalent.

1. $$\text{lan}_if \dashv \text{lan}_fi$$.
2. $$f$$ is the $$i$$-relative left adjoint of $$\text{lan}_fi$$, i.e. $${\bf C}(i, \text{lan}_fi) \cong {\bf B}(f, - ).$$
3. $$f = \text{lift}_{\text{lan}_fi}i$$ and the lift is absolute.

If $$i$$ is only fully faithful $$1 \Rightarrow 2$$, if $$i$$ is only dense $$2 \Rightarrow 1$$.

Proof.

$$1 \Rightarrow 2$$) $${\bf B}(f, -) \stackrel{i \text{ is ff.}}{\cong} {\bf B}((\text{lan}_if) i, -) \stackrel{1}{\cong} {\bf C}(i, \text{lan}_fi).$$

$$2 \Rightarrow 1$$) $${\bf B}(\text{lan}_if, -) \stackrel{\text{point.}}{\cong} \text{lan}_i{\bf B}(f, -) \stackrel{2}{\cong} \text{lan}_i{\bf C}(i, \text{lan}_fi) \stackrel{\text{point.}}{\cong} {\bf C}(\text{lan}_ii, \text{lan}_fi) \stackrel{i \text{ is dense}}{\cong} {\bf C}(-, \text{lan}_fi).$$

$$3$$ is just a rewriting of $$2$$.

Now we study a very special setting.

Let $${\bf C} \xleftarrow{i} {\bf A} \xrightarrow{f} {\bf B}$$ be a span where $$i$$ is dense and fully faithful. Moreover $$\text{lan}_if$$ is pointwise, $${\bf A}$$ is small, $${\bf C}$$ and $${\bf B}$$ are cocomplete.

In this setting $${\bf C}$$ is a reflective subcategory $$V: {\bf C} \leftrightarrows \text{Set}^{{\bf A}^\circ} : L$$ of $$\text{Set}^{{\bf A}^\circ}$$ via the nerve $$V = \text{lan}_i(y_{{\bf A}})$$ (V is the right adjoint).

Prop 2. Let $${\bf C} \xleftarrow{i} {\bf A} \xrightarrow{f} {\bf B}$$ be a span where $$i$$ is dense and fully faithful. $${\bf A}$$ is small, $${\bf C}$$ and $${\bf B}$$ are cocomplete. Then, the following are equivalent.

1. $$\text{lan}_if \dashv \text{lan}_fi$$.
2. V preserves $$\text{lan}_fi$$.
3. $${\bf C}(i,\text{lan}_fi) \cong \text{lan}_f{\bf C}(i,i)$$.

Proof. Recall that $$\text{lan}_if$$ is pointwise,because $${\bf B}$$ is cocomplete.

$$1 \Rightarrow 2)$$.

Using Prop 1. we know that $${\bf C}(i, \text{lan}_fi) \cong {\bf B}(f, - )$$. Since the presheaf construction is a Yoneda structure, we have that $$\text{Set}^{{\bf A}^\circ}(y_A, \text{lan}_fy_A) \cong {\bf B}(f, -)$$.

Thus, $${\bf C}(i, \text{lan}_fi) \cong {\bf B}(f, -) \cong \text{Set}^{{\bf A}^\circ}(y_A, \text{lan}_fy_A) \cong \text{Set}^{{\bf A}^\circ}(y_A, \text{lan}_fVi)$$

Observe also that $${\bf C}(i, \text{lan}_fi) \cong {\bf C}(Ly_A, \text{lan}_fi) \stackrel{L \dashv V}{\cong} \text{Set}^{{\bf A}^\circ}(y_A, V\text{lan}_fi),$$ putting the last two equation together, one gets: $$\text{Set}^{{\bf A}^\circ}(y_A, \text{lan}_fVi) \cong \text{Set}^{{\bf A}^\circ}(y_A, V\text{lan}_fi).$$ By Yoneda Lemma the two functors on the right have to coincide.

$$2 \Rightarrow 1).$$

Using Prop 1. it is enough to prove that $${\bf C}(i, \text{lan}_fi) \cong {\bf B}(f, - )$$. Since the presheaf construction is a Yoneda structure, we have that $$\text{Set}^{{\bf A}^\circ}(y_A, \text{lan}_fy_A) \cong {\bf B}(f, -)$$. Now, $${\bf C}(i, \text{lan}_fi) \cong {\bf C}(Ly_A, \text{lan}_fi) \stackrel{L \dashv V}{\cong} \text{Set}^{{\bf A}^\circ}(y_A, V\text{lan}_fi) \stackrel{2}{\cong} \text{Set}^{{\bf A}^\circ}(y_A, \text{lan}_fVi) \cong \text{Set}^{{\bf A}^\circ}(y_A, \text{lan}_fy_A) \cong {\bf B}(f, -).$$

$$3$$ is just a rewriting of $$2$$.

Cor. 1 (Quite surprising). Let $${\bf Set}^{A^\circ} \xleftarrow{y} {\bf A} \xrightarrow{f} {\bf B}$$ be a span where $$y$$ is the Yoneda embedding of a small category and $${\bf B}$$ is cocomplete. Then $$\text{lan}_yf \dashv \text{lan}_fy$$.

Proof 1. In Prop 1 the condition 2 is verified and is essentially a rewriting of the Yoneda Lemma. Moreover $$\text{lan}_yf$$ is pointwise because $${\bf B}$$ is cocomplete.

Proof 2. In Prop 2 the condition 2 is trivially verified because $$V$$ is the identity.

I shall conclude by saying that the adjunction cannot verify too often.

Rem. 4 Let $$y: {\bf A} \to {\bf C}$$ be a dense and fully faithful functor. Then the following are equivalent.

1. For every map $$f: {\bf A} \to {\bf B}$$, where $${\bf B}$$ is cocomplete, the Kan extensions $$\text{lan}_yf$$ and $$\text{lan}_fy$$ exist and are adjoint.
2. $$y$$ is the Yoneda embedding.

Proof. One implications is Cor 1. The other implication can be read as follows, if 1 is verified, then for every functor $$f: {\bf A} \to {\bf B}$$, there is a unique cocontinous extension ($$\text{lan}_yf$$), this is the characterization of the free cocompletion under colimits, that is the (small) presheaf construction.

• This looks correct to me. – Fosco Nov 17 '18 at 15:36
• Hopefully, one never knows. – Ivan Di Liberti Nov 17 '18 at 15:41

Edit. The construction of $\eta$ below is not correct. See the comments.

Ad 2. I think that indeed $\mathrm{Lan}_G(F):\mathbf{C} \to \mathbf{B}$ is left adjoint to $\mathrm{Lan}_F(G):\mathbf{B} \to \mathbf{C}$ when $G:A \to \mathbf{C}$ is fully faithful.

Below, I will construct two morphisms of functors $\eta : \mathrm{id}_\mathbf{C} \to\mathrm{Lan}_F(G) \circ \mathrm{Lan}_G(F)$ and $\varepsilon : \mathrm{Lan}_G(F) \circ \mathrm{Lan}_G(F) \to \mathrm{id}_\mathbf{B}$. I haven't verified the triangle identities yet, so that this answer should be taken with a grain of salt. What really confuses me is that I cannot write down a natural bijection $$\hom(\mathrm{Lan}_G(F)(X),Y) \cong \hom(X,\mathrm{Lan}_F(G)(Y))$$ directly, without using $\eta$ and $\varepsilon$, basically because $\hom(X,-)$ does not interchange with coends.

I will use the coend formula

$$\forall X \in \mathbf{C}:\quad \mathrm{Lan}_G(F)(X) = \int^{a \in A} \hom(G(a),X) \otimes F(a).$$ Now let's do some co/end fu: $$\begin{eqnarray} &&\mathrm{Lan}_F(G)(\mathrm{Lan}_G(F)(X)) \\ & = & \int^{a \in A} \hom\biggl(F(a),\int^{a' \in A} \hom(G(a'),X) \otimes F(a')\biggr) \otimes G(a) \\ & \leftarrow & \int^{a \in A} \int^{a' \in A} \hom(G(a'),X) \otimes \hom(F(a),F(a')) \otimes G(a) \\ & \cong & \int^{a' \in A} \hom(G(a'),X) \otimes \int^{a \in A} \hom(F(a),F(a')) \otimes G(a) \\ & \leftarrow & \int^{a' \in A} \hom(G(a'),X) \otimes \int^{a \in A} \hom(a,a') \otimes G(a) \\ & \cong & \int^{a' \in A} \hom(G(a'),X) \otimes G(a') \\ & \cong & \int^{X' \in \mathbf{C}} \hom(X',X) \otimes X' \\ & \cong & X\end{eqnarray}$$ This describes $\eta$. We describe $\varepsilon$ as follows: $$\begin{eqnarray} && \mathrm{Lan}_G(F)(\mathrm{Lan}_F(G)(Y)) \\ &= & \int^{a \in A} \hom(G(a),\mathrm{Lan}_F(G)(Y)) \otimes F(a) \\ & \cong & \int^{a \in A} \int^{a' \in A} \hom(F(a'),Y) \otimes \hom(G(a),\mathrm{Lan}_F(G)(F(a'))) \otimes F(a) \\ & \rightarrow & \int^{a \in A} \int^{a' \in A} \hom(F(a'),Y) \otimes \hom(G(a),G(a')) \otimes F(a) \\ & \cong & \int^{a \in A} \int^{a' \in A} \hom(F(a'),Y) \otimes \hom(a,a') \otimes F(a) \\ & \cong & \int^{a' \in A} \hom(F(a'),Y) \otimes \int^{a \in A} \hom(a,a') \otimes F(a) \\ & \cong & \int^{a' \in A} \hom(F(a'),Y) \otimes F(a') \\ & \rightarrow & Y \end{eqnarray}$$

• I don't understand why fully faithfulness of $G$ implies $\int^{a' \in \cal A} \hom(G(a'),X) \otimes G(a') \cong \int^{X' \in \mathbf{C}} \hom(X',X) \otimes X'$, can you expand? – Fosco Oct 19 '16 at 15:31
• This is a mistake, sorry. So probably there won't be an adjunction in this generality. – HeinrichD Oct 19 '16 at 17:55

What follows is kind of preliminary and complementary to Ivan's answer.

Let $$A$$ be a small category, $$\mathcal B$$ and $$\mathcal C$$ be locally presentable categories and $$i\colon A \to \mathcal B$$ and $$j \colon A \to \mathcal C$$ two functors. The left Kan extension $$\operatorname{Lan}_i j \colon \mathcal B \to \mathcal C$$ is a left adjoint functor if and only if it commutes with colimits. By this answer of John Bourke, this happens if, for any object $$a$$ of $$A$$, the object $$i(a)$$ of $$\mathcal B$$ is tiny. Suppose that this is the case and call $$V\colon \mathcal C \to \mathcal B$$ the right adjoint to $$\operatorname{Lan}_i j$$.

If the functor $$i \colon A \to \mathcal B$$ is fully faithful, then for every object $$a$$ of $$A$$ we have $$\operatorname{Lan}_i j \,(i(a)) \cong j(a)$$. Thus for every object $$a$$ of $$A$$ and every object $$Y$$ of $$\mathcal Y$$ we get $$\text{Hom}_{\mathcal C}(\operatorname{Lan}_i j \,(i(a)), Y) = \text{Hom}_{\mathcal C}(j(a), Y) = \text{Hom}_{\mathcal B}(i(a), V(Y))$$ naturally in $$a \in A$$. This is equivalent to say that the two commas $$A/Y$$ and $$A/V(Y)$$ (aka $$j \downarrow Y$$ and $$i\downarrow V(Y)$$) are isomorphic categories.

In the case where $$i$$ is fully faithful, the functor $$V$$ is isomorphic to $$\operatorname{Lan}_j i$$ if and only if for any object $$Y$$ of $$\mathcal C$$ we have that $$V(Y)$$ is the colimit of the functor $$A/Y \to A \to \mathcal B$$. By the previous paragraph, this is isomorphic to the colimit of the functor $$A/V(Y) \to A \to \mathcal B$$. Thus $$V \cong \operatorname{Lan}_j i$$ if and only if every object in the image of $$V$$ is canonically a conical colimit with respect to the functor $$i$$. (The functor $$i$$ is dense in the subcategory of $$\mathcal B$$ spanned by the essential image of $$V$$.)

We obtain a sufficient condition for $$V$$ to be isomorphic to $$\operatorname{Lan}_j i$$ imposing the functor $$i$$ to be dense (and fully faithful). Indeed, in this case by definition we have that $$V(Y)$$ is the colimit of the functor $$A/V(Y) \to A \to \mathcal B$$ for any object $$Y$$ of $$\mathcal C$$ and so we conclude using the preceding paragraph.

• To my knowledge, in John's answer that is not an "if and only if", it is just an "if". – Ivan Di Liberti Jan 16 '19 at 22:39
• You are absolutely right, thanks! – Andrea Gagna Jan 17 '19 at 12:10