Kan extensions and coends in the enriched case

I have been thinking about the nerve-realization theorem. See here:

https://ncatlab.org/nlab/show/nerve+and+realization

This blog post is about my attempts to think through it.

The theorem is just about the usual coend formula for the left Kan extension, except that it studies a complicated case where some of the categories are enriched and some are not; on the other hand it is simpler in that one of the functors is fixed to be some variant of the Yoneda embedding.  In their notation, $\mathcal{C}$ is enriched but $S$ may not be. For example, a simplicial Abelian group or a simplicial topological space are functors from an unenriched category to a category that is enriched ($\mathbf{Ab}$) or at least somewhat enriched (hom sets in $\mathbf{Top}$ can be equipped with the compact open topology)

Let $M$ be a small category. Let $\mathcal{A}$ be a co-complete category. Let $T: M\to \mathcal{A}$ be a functor.

Theorem:
  Let $R : \mathcal{A}\to \hat{M}=[M^{op},\mathbf{Sets}]$ be defined as $R(A) =
  \operatorname{Hom}(T(-),A)$. Then $R$ has a left adjoint, $L$, and $L$ is the pointwise left Kan extension
  of $T$ along the Yoneda embedding $y : M\to \hat{M}$.

$L$ is given by the standard coend formula for the left Kan extension
\begin{equation}
\label{eq:6}
L(X)= \int^m \operatorname{Hom}(y(m),X)\cdot T(m)
\end{equation}
Here, $\cdot$ denotes the ``copower'' - the coproduct $\coprod_{x\in \operatorname{Hom}(y(m),X)}T(m)$.  In the case in question, by the Yoneda lemma $L(X)$ simplifies to
\begin{equation}
\label{eq:3}
L(X)= \int^m X(m)\cdot T(m)
\end{equation}
Now I have noticed that a similar construction is often given in a slightly different case, where instead of the two categories being $[M^{op};\mathbf{Sets}]$ and $\mathcal{A}$, we have $[M^{op};\mathcal{A}]$ and $\mathcal{C}$, where $\mathcal{C}$ is enriched over $\mathcal{A}$, and this is the situation that the n-lab covers in their page. For the sake of this blog post, I am going to consider the case $\mathcal{C}=\mathcal{A}$, so that our category is enriched over itself, as I prefer not to complicate things right now by introducing a proliferation of categories.

If $\mathcal{A}$ is a monoidal category, then for any $X: M^{op}\to \mathcal{A}$ we can form
\begin{equation}
\label{eq:7}
L'(X) = \int^m X(m)\otimes T(m)
\end{equation}
This is sometimes referred to as the tensor product of $X$ with $T$. I want to know, if $T$ is fixed, does this functor have any similar properties to the left Kan extension?

I am interested in the specific case where $\mathcal{A}$ is concrete over $\mathbf{Sets}$, say by a forgetful functor $U:\mathcal{A}\to \mathbf{Sets}$. I am most interested in the case where $U$ is faithful, but I don't require this. I ask that $U$ have a left adjoint, $F$. Note as well that this gives rise to an adjunction between $[M^{op},\mathbf{Sets}]$ and $[M^{op},\mathcal{A}]$. Here I am thinking of $\mathcal{A}$ Abelian groups or topological spaces.

Thus, I assume $\mathcal{A}$ is equipped with a monoidal product $\otimes$. The most interesting case is where $\otimes : \mathcal{A}\times \mathcal{A}\to \mathcal{A}$ is an extension (up to natural isomorphism) of the copower functor $\mathbf{Sets}\times \mathcal{A}\to \mathcal{A}$ along $F : \mathbf{Sets}\to \mathcal{A}$. In this case, the induced tensor product functor
\begin{equation}
\label{eq:8}
L'(X) =\int^mX(m)\cdot T(M)
\end{equation}
should be an extension of $L : [M^{op};\mathbf{Sets}]$ along $F : [M^{op}, \mathbf{Sets}]\to [M^{op}, \mathcal{A}]$. Consider here the examples of the product of topological spaces, or the tensor product of Abelian groups, which both satisfy this property.

I ask as well that $\mathcal{A}$ be equipped with an exponential bifunctor $\mathcal{A}^{op}\times \mathcal{A}\to \mathcal{A}$ sending $(A,B)$ to an object I denote $B^A$. This exponential must satisfy $U(B^A)\cong \operatorname{Hom}_{\mathcal{A}}(A,B)$, naturally in $A, B$. I ask that for any object $m\in M$, there is a natural isomorphism
\begin{equation}
\label{eq:5}\operatorname{Hom}(A\otimes T(m), B)\cong \operatorname{Hom}(A, B^{T(m)})
\end{equation}
I don't ask that in general, the exponential be adjoint to the monoidal product. I am only interested in the exponential correspondence in the case where the ``pivot'' object is $T(m)$. (The important example here is $\mathbf{Top}$, where you can express any hom-set with the compact-open topology, and it's bifunctorial, but it doesn't function as adjoint unless the ``pivot'' object is locally compact Hausdorff. Since simplices are certainly LCH, this is enough.)

I define a functor $R' : \mathcal{A}\to [M^{op};\mathcal{A}]$ sending $A$ to the functor $m\mapsto A^{T(m)}$. By the defining assumption of our exponential, $U\circ R' = R$, as $U(m\mapsto A^{T(m)}) = m\mapsto U(A^{T(m)}= m\mapsto \operatorname{Hom}(T(m),A)$.

For convenience I introduce the abbreviation $K: M\to [M^{op}, \mathcal{A}]$ for $F\circ y : M\to [M^{op},\mathbf{Sets}]\to [M^{op};\mathcal{A}]$.

We will show that there is a natural bijection, for any functor $S : [M^{op};\mathcal{A}]\to \mathcal{A}$
\begin{equation}
\label{eq:9}\operatorname{Nat}(L',S)\cong \operatorname{Nat}(T,S\circ K)
\end{equation}
as long as $S$ satisfies a certain property; thus $L'$ is the representing object for the functor $\operatorname{Nat}(T, -\circ K)$ on the category of all $S$ satisfying the desired property.  The property is somewhat ad hoc, as it simply arises from what is necessary in the proof; $L'$ may perhaps satisfy a weaker universal property and thus be initial among a larger class of functors $S$. Let us proceed.

Fix an object $m\in M$. By the Yoneda lemma,

\begin{equation}
\label{eq:10}\operatorname{Hom}(Tm, SKm)\cong\operatorname{Nat}(\operatorname{Hom}_{[M^{op},\mathcal{A}]}(Km,-),\operatorname{Hom}_{\mathcal{A}}(Tm,S-))
\end{equation}.
By the adjunction between $[M^{op},\mathbf{Sets}]$ and $[M^{op},\mathcal{A}]$, $\operatorname{Hom}_{[M^{op},\mathcal{A}]}(Km,-)\cong \operatorname{Hom}_{[M^{op};\mathbf{Sets}]}(y(m),U(-)) = U(-)_m$. By this last $U(-)_m$ I mean the functor which, given $X:M^{op}\to \mathcal{A}$, returns $U(X(m))$, the ``underlying set'' of $X(m)$.

By our assumption about the behavior of $U$ on exponential objects, $\operatorname{Hom}_{\mathcal{A}}(Tm,S-)$ can be identified with $U(S(-)^{Tm})$. Therefore we have
\begin{equation}
\label{eq:11}\operatorname{Hom}(Tm,SKm) \cong \operatorname{Nat}(U(-)_m,U(S(-)^{Tm}))
\end{equation}
Now any natural transformation $(-)_m\to S(-)^{Tm}$ (i.e. a family of maps $X_m\to S(X)^{Tm}$ in $\mathcal{A}$, natural in $X$) gives by precomposition with $U$, a natural transformation $U(-)_m\to U(S(-)^{Tm})$. This leads us to the assumption we need on $S$ for the proof to go through:

Property: The map 
\begin{equation}
\label{eq:12}\operatorname{Nat}((-)_m, S(-)^{Tm})\xrightarrow{U\circ -} \operatorname{Nat}(U(-)_m, U(S(-)^{Tm}))
\end{equation}
is bijective for all $m$ in $M$.

Under these circumstances, we can employ the end formula for natural transformations, and write
\begin{equation}
\label{eq:13}\operatorname{Hom}(Tm,SKm)\cong \int_X \operatorname{Hom}{\mathcal{A}}(X_m,S(X)^{Tm})
\end{equation}
(where the right hand side of \ref{eq:13} is identified with the left hand side of \ref{eq:12} via the end formula)

Then the usual argument in the calculus of coends goes through to prove $\operatorname{Nat}(L',S)\cong \operatorname{Nat}(T,SK)$ (see Categories for the Working Mathematician for details), which gives the universal property of $L'$, as follows:
\begin{align}
  \operatorname{Nat}(L',S) &\cong \int_{X:M^{op}\to \mathcal{A}} \operatorname{Hom}(L'X,SX)\\
             &\cong \int_{X} \operatorname{Hom}\left( \int^{m:M}X_m\otimes T(m),SX\right)\\
             &\cong \int_{X}\int_m \operatorname{Hom}\left( X_m\otimes T(m),SX\right)\\
             &\cong \int_{X}\int_m \operatorname{Hom}\left( X_m,SX^{Tm}\right)\\
             &\cong \int_m\int_X \operatorname{Hom}\left( X_m,SX^{Tm}\right)\\
             &\cong \int_m\operatorname{Nat}_{\mathcal{A}}\left(  (-)_m,S(-)^{Tm}\right)\\
             &\cong \int_m \operatorname{Hom}(Tm, SKm)\\
             &\cong \operatorname{Nat}(T, SK)
\end{align}

Let me now sketch lightly an idea of how such an $S$ may arise, satisfying the given property, so that we have an idea of what kind of categories of functors from $[M^{op};\mathcal{A}]$ and $\mathcal{A}$ will have $L'$ as a representing object and serve as the Kan extension.

Suppose that $\mathcal{A}$ is complete, as well as co-complete. Then the end construction of the object natural transformations can be carried through in $\mathcal{A}$, so it makes sense to define, for any two objects $X, Y : M^{op}\to \mathcal{A}$, an $\mathcal{A}$-object of natural transformations $Y^X$, by
\begin{equation}
\label{eq:14}
Y^X = \int_mY(m)^{X(m)}
\end{equation}
where the exponentials $Y(m)^{X(m)}$ are computed in $\mathcal{A}$. In the cases of $\mathbf{Top}$ or $\mathbf{Ab}$, these are just the subspace/subgroup of $\prod Y(m)^{X(m)}$ of all the families of maps that define natural transformations, equipped with the subspace topology or inheriting the obvious subgroup structure. Thus it makes sense to talk about a homotopy of natural transformations $X\to Y$ in $\mathbf{sTop}$ as a path in $Y^X$, etc.

Because $U$ is a right adjoint, it commutes with limits and ends, so
\begin{equation}
\label{eq:15}
U(Y^X) = \int_mU(Y(m)^{X(m)} = \int_m \operatorname{Hom}(X(m),Y(m)) = \operatorname{Nat}(X,Y)
\end{equation}

Suppose for a moment for each $m$ that there is a unique map, natural in $X$

$S_{X,m}: X_m\to S(X)^{S(Km)}$ which codes the behavior of the functor $S$ on morphisms $K(m)\to X$ for varying $m$, or more precisely that
\begin{equation}
\label{eq:17}
U(S_{X,m}) : U(X_m) = \operatorname{Hom}(Km,X) \to \operatorname{Hom}(S(Km), SX)
\end{equation}
Given a natural transformation $U(-)_m\to U(S(-)^{Tm})$, regard it by Yoneda as a morphism $\sigma: Tm\to SKm$ by \ref{eq:11}. Then simply because exponents are bifunctorial, $\sigma$ determines a map $-\circ \sigma: S(X)^{SKm}\to S(X)^{Tm}$, natural in $X$. Combining these with the maps $S_{X,m}$ we get the desired natural transformation $X_m\to S(X)^{Tm}$ which is clearly over the original one via $U$.

How can we come by such a natural map $X_m\to S(X)^{S(Km)}$?

Assume that for each $X,Y$ in $[M^{op}, \mathcal{A}]$ there exists a unique morphism $S_{XY}: Y^X\to S(X)^{S(Y)}$ such that $U(S_{XY}): \operatorname{Hom}(X,Y)\to \operatorname{Hom}(S(X),S(Y))$ gives the behavior of $S$ on morphisms. Then we only have to construct a map $X_m\to X^{K_m}$, as we can use the map $S_{X,K_m}:X^{Km}\to S(X)^{S(Km)}$.

Let us see what is necessary to construct a map $X_m\to X^{K(m)}$. Since $X^{K(m)}$ is an end, it suffices to give a family of maps $\omega_n$ $X_m\to X(n)^{K(m)(n)}$ indexed by the objects of $M$ such that for any $f: n\to n'$ in $M$ we have a commutative square
\begin{equation}
\left[-\circ K(m)(f)\right]\circ \omega_n = \left[ X(n)(f)\circ -\right]\circ \omega_{n'}
\end{equation}

Under very basic assumptions about the exponent and the monoidal product of $\mathcal{A}$ we can give these maps $\omega_n$. For example, suppose that $F$ applied to the singleton set in $\mathbf{Sets}$ returns the unit $1$ of the monoidal product, and that $A^1\cong A$. And suppose that the exponential behaves reasonably with products, so that $Y^{\coprod X_i}\cong \prod Y^{X_i}$ and $\left( \prod Y_i \right)^X \cong \prod \left( Y_i^X \right)$.  Then $K(m)(n) =F(\operatorname{Hom}_M(n,m)) = \coprod_{f:n\to m}1$ (because $F$ is a left adjoint and so preserves coproducts) and so $X(n)^{K(m)(n)}\cong \prod_{f:n\to m}X(n)$. Then the map $X(m)\to \prod_fX(n)$ is trivial, it is the product of the maps $X(f)$ (recall that $X$ is contravariant).

Alternatively we could suppose that the exponential functor is also adjoint to the monoidal product on ``free'' objects in $\mathcal{A}$, i.e. those in the image of the functor $F$.

Let me discuss an example of this that has been bugging me.

Let $\Delta$ denote the simplex category of nonempty finite ordinals and order preserving maps between them. Let $F: \Delta \to \mathbf{Top}$ denote the usual embedding which sends $[n]=\left\{ 0,\dots, n \right\}$ to the $n$-simplex $\Delta^n$. Let $y : \Delta\to \mathbf{SSets}$ be the Yoneda embedding. Let $R: \mathbf{Top}\to \mathbf{SSets}$ be the functor
\begin{equation}
\label{eq:1}
R(X)  = \operatorname{Hom}_{\mathbf{Top}}(F(-),X)
\end{equation}
Here, $R$ is exactly the functor which sends a space $X$ to its simplicial set of singular
simplices.

The coend formula we have mentioned earlier gives the left adjoint to $R$, the geometric realization of a simplicial set.

Here the coend formula should take the form, for $S$ a simplicial set,
\begin{equation}
\label{eq:2}
L(S) = \int^{[m]}S_m\cdot \Delta^m
\end{equation}

Now, there is a natural embedding (fully faithful) of $\mathbf{sSet}$ into $\mathbf{sTop}$, the category of simplicial topological spaces.

Let $G : \mathbf{Sets}\to \mathbf{Top}$ denote the left adjoint to the forgetful functor which equips every set with the discrete topology; then what I mean is that given a simplicial set $X$, we can postcompose it with $G$ and we get a functor $G\circ X: \Delta^{op}\to \mathbf{Top}$, which by abuse of notation I will also call $G$.

Then the functor $L$ extends along $G$ in a natural way. For an arbitrary simplicial space $S$ we can define
\begin{equation}
\label{eq:4}
L'(S) = \int^{[m]}S_m \times \Delta^m
\end{equation}
It is clear that $L' \cong L\circ G$ simply because of the fact that $X_m\cdot \Delta^m = X_m\times \Delta^m$ when $X_m$ is regarded on the left as a set, and on the right as a topological space.

The functor $L'$ is important in homotopy theory. It is also referred to as the geometric realization of a simplicial space.

$L'$ appears in the notion of homotopy colimit.  If $X$ is a simplicial set and $F: \Pi_1(X)\to \mathbf{Top}$ is a functor from the fundamental category of $X$ (i.e. $F$ is a diagram of spaces indexed by the vertices and edges of $X$ subject to composition coherence laws arising from the 2-simplices) there is a natural way to regard $F$ as a simplicial topological space, which I can call $\overline{F}$; fibred over $X$ by a map $\pi : F\to X$ (here I regard $X$ as a discrete simplicial space, so $\pi$ is a morphism in $\mathbf{sTop}$) To briefly recap, we define $\overline{F}_0 = \coprod_{x\in X_0}F(x)$ as the coproduct of the spaces in the diagram $F$; then $\pi_0$ is the obvious projection, and $\overline{F}_n= \coprod_{x=[v_0,\dots, v_n]\in X_n}F(v_0)$. Almost all the face and degeneracy maps of $\overline{F}_n$ are essentially reindexing maps which delete or duplicate vertices in the indexing simplices, so that if $p$ is a point in $\overline{F}_n$ lying over the $n$-simplex $x = [v_0,\dots, v_n]$, then the face map $\partial_i$ will simply send it to the same point $p$ lying over $[v_0,\dots,\hat{v}_i,\dots, v_n]$ in $\overline{F}_n$, and the degeneracy $s_{i}$ will send it to the same point $p$ lying over $[v_0,\dots,v_i,v_i,\dots, v_n]$ in $\overline{F}_{n+1}$. The exception are the face maps $\partial_0 :\overline{F}_n\to \overline{F}_{n-1}$, which are defined as follows: if $x= [v_0,\dots, v_n]$ is an $n$-simplex in $X_n$, and if by $e$ I denote the edge from $v_0\to v_1$ in $x$, then $F$ associates to $e$ a continuous map $F(v_0)\to F(v_1)$ by definition. Then if $p$ is a point in $\overline{F}_n$ in the fibre over $x=[v_0,\dots, v_n]$ (so more specifically $p$ is a point in the direct summand $F(v_0)$ of the coproduct) then we define $\partial_0(p) = F(e)(p)$ over $[v_1,\dots, v_n]$ in $\overline{F}_{n-1}$.

I have realized recently that the construction of the object $\overline{F}$ as fibred over $X$ depended only on ``pullback'' type operations, and one can prove that all the appropriate commutativity laws of the face and degeneracy maps are satisfied only by appeal to properties that hold in any Grothendieck fibration. In this case, viewing $X$ as a discrete simplicial topological space, i.e. as a functor $\Delta^{op}\to \mathbf{Top}$, I can view the construction of $\overline{F}$ as a certain lift of $X$ along the codomain fibration $\mathbf{Top}^{\to}\to \mathbf{Top}$ subject only to the condition that all degeneracy maps $s_j$ and face maps $\partial_i$ with $i\neq 0$ are Cartesian. I have given a generalization of this construction which shows that if $p : \mathbb{E}\to \mathbb{B}$ is a Grothendieck fibration, then the appropriate definition for an "internal diagram in p indexed by a simplicial object $X : \Delta^{op}\to \mathbb{B}$" (paralleling the definition of an internal diagram in $p$ indexed by an internal category in $\mathbb{B}$, see Chapter 7 of Categorical Logic and Type Theory by Bart Jacobs, or another treatment of internal category theory) is exactly a lift of $X$ along $p$ such that all face maps $\partial_i$ for $i>0$ are Cartesian, and I will gladly make it available upon request.

Anyway - to return to the functor $L'$. Going back to the functor $F$, the homotopy colimit of $F$ is defined to be the geometric realization $L'(\overline{F})$

I have several questions about the functor $L'$. How can we interpret it? Is it the left Kan extension of $F : \Delta^{op}\to \mathbf{Top}$ along $G\circ y$? Can we regard $S_m$, as it appears in \ref{eq:4}, as a kind of enriched Hom of $\operatorname{Hom}(G(y([m]),S)$, and then understand it to be a kind of enriched Kan extension? I think the answer to all of these should be yes but it will take some time to appreciate the precise sense in which this is true.