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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 $\bf Prof$?
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  • $\begingroup$ I had to read ncatlab.org/nlab/show/nerve+and+realization in order to understand your question. Perhaps others want to follow this link too. $\endgroup$ – HeinrichD Oct 18 '16 at 21:37
  • $\begingroup$ Feel free to improve the question as you like; or did you already edit it? $\endgroup$ – Fosco Loregian Oct 18 '16 at 21:59
  • $\begingroup$ Nice! I can't believe I never noticed the F-nerve was $\mathrm{Lan}_F y$. $\endgroup$ – Omar Antolín-Camarena Nov 20 '18 at 22:42
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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.

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  • $\begingroup$ This looks correct to me. $\endgroup$ – Fosco Loregian Nov 17 '18 at 15:36
  • $\begingroup$ Hopefully, one never knows. $\endgroup$ – Ivan Di Liberti Nov 17 '18 at 15:41
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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: $$\begin{eqnarray} && \mathrm{Lan}_G(F)(\mathrm{Lan}_F(G)(Y)) \\ &= & \int^{a \in A} \hom(G(a),\mathrm{Lan}_F(G)(Y)) \otimes F(a) \\ & \cong & \int^{a \in A} \int^{a' \in A} \hom(F(a'),Y) \otimes \hom(G(a),\mathrm{Lan}_F(G)(F(a'))) \otimes F(a) \\ & \rightarrow & \int^{a \in A} \int^{a' \in A} \hom(F(a'),Y) \otimes \hom(G(a),G(a')) \otimes F(a) \\ & \cong & \int^{a \in A} \int^{a' \in A} \hom(F(a'),Y) \otimes \hom(a,a') \otimes F(a) \\ & \cong & \int^{a' \in A} \hom(F(a'),Y) \otimes \int^{a \in A} \hom(a,a') \otimes F(a) \\ & \cong & \int^{a' \in A} \hom(F(a'),Y) \otimes F(a') \\ & \rightarrow & Y \end{eqnarray}$$

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  • $\begingroup$ 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? $\endgroup$ – Fosco Loregian Oct 19 '16 at 15:31
  • $\begingroup$ This is a mistake, sorry. So probably there won't be an adjunction in this generality. $\endgroup$ – HeinrichD Oct 19 '16 at 17:55
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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.

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

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