Internal hom takes coends to ends

Question

I know that this is a very general fact about limits and colimits, but I would like to prove it directly for ends and coends. If $\mathcal V$ is a closed braided monoidal category, $V$ an object in $\mathcal V$ and $P\colon\mathcal M^\mathsf{op}\otimes\mathcal M\to \mathcal V$ is a $\mathcal V$-functor, I can define its coend, whether it exists, as the initial $\mathcal V$-extranatural transformation $i\colon P(-,-)\overset{\cdot\cdot}{\Rightarrow}\int^MP(M,M)$. This, as far as I understand, is a necessary step in enriched category theory in order to define the coends "representably" for a functor valued in any other $\mathcal V$-category. Dually, I can define the end $\int_MP(M,M)\overset{\cdot\cdot}{\Rightarrow}P(-,-)$, and one should have $$\int_M\mathcal V(P(M,M),V)\cong\mathcal V(\int^MP(M,M),V).$$

The representable definition will be, for a $\mathcal V$-functor $P\colon\mathcal M^\mathsf{op}\otimes\mathcal M\to \mathcal D$, the object $\int^MP(M,M)$ of $\mathcal D$, together with a $\mathcal V$-extranatural transformation $i\colon P(-,-)\overset{\cdot\cdot}{\Rightarrow}\int^MP(M,M)$ such that $\mathcal D(i,D)\colon\mathcal D(\int^MP(M,M),D)\overset{\cdot\cdot}{\Rightarrow}\mathcal D(P(-,-),D)$ exhibit an end for the functor $D(P(-,-),D)\colon \mathcal M^\mathsf{op}\otimes\mathcal M\to \mathcal V$. This will immediately give the desired result of cocontinuity $\int_M\mathcal D(P(M,M),D)\cong\mathcal D(\int^MP(M,M),D).$ However, in order for it to be consistent with the case where the target is $\mathcal V$ itself, one should be able to prove it also in this case.

The strategy that seems to make more sense to me (in order to prove that the right hand side above is indeed the desired end) is to consider the functor $\mathcal V(\mathbb 1,-)\colon \mathcal V\to \mathrm{Set}$, which takes components of a test extranatural transformation $k_N\colon Y\to\mathcal V(P(N,N),V)$ to a function $k_N^*\colon\mathrm{Hom}_\mathcal V(\mathbb 1,Y)\to\mathrm{Hom}_\mathcal V(\mathbb 1,\mathcal V(P(N,N),V))$. Then, whenever $y\colon \mathbb 1\to Y$ is a map in $\mathcal V$ I can find its image $k_N^*(y)$ which provide a map $P(N,N)\to V$, collectively giving a ($\mathcal V$-)extranatural transformation. These maps now induce morphisms $\tilde k^y\colon\int^MP(M,M)\to V$ by the property of coend, which I would like to use in order to find the desired map $\tilde k\colon Y\to\mathcal V(\int^MP(M,M),V)$.

My problem seems to be that what I just defined is a set map $y\mapsto \tilde k^y$ going $$\mathrm{Hom}_\mathcal V(\mathbb 1,Y)\longrightarrow\mathrm{Hom}_\mathcal V(\int^MP(M,M),V)\cong\mathrm{Hom}_\mathcal V(\mathbb 1,\mathcal V(\int^MP(M,M),V))$$ which I'm not able to lift back since in general $\mathrm{Hom}_\mathcal V(\mathbb 1,-)$ is not a faithful functor (right?)

Thank you so much in advance for any suggestion!

Answer 1

$\require{AMScd}$Fix an object $X\in\cal V$. Applying ${\cal V}(-,X)$ to the initial cowedge $\alpha : P(-,-)\overset{..}\Rightarrow \int^MPMM$ you get a wedge ${\cal V}(\alpha,X):{\cal V}(\int^MPMM,X) \overset{..}\Rightarrow {\cal V}(P(-,-),X)$, where it's clear how the functor ${\cal V}(P(-,-),X)$ is defined: $$ \begin{CD} {\cal M}^\text{op}\times{\cal M}@= ({\cal M}^\text{op}\times{\cal M})^\text{op} @>P^\text{op}>> {\cal D}^\text{op} @>{\cal V}(-,X)>> \cal V \end{CD}$$ Now, in order to prove that ${\cal V}(\alpha,X)$ is a terminal wedge, take another object $Z\in \cal V$, and a wedge $Z \overset{..}\Rightarrow {\cal V}(P(-,-),X)$; this corresponds to a cowedge $P(-,-) \overset{..}\Rightarrow {\cal V}(Z,X)$, under the correspondence (true for every $M$) $$ \tag{$\star$}{\cal V}(Z,{\cal V}(PMM,X))\cong {\cal V}(PMM,{\cal V}(Z,X))$$ hence to a unique morphism $$ \begin{CD} \int^M PMM @>>>{\cal V}(Z,X) \end{CD}$$ hence (for the same isomorphism $(\star)$) to a unique morphism $Z\to {\cal V}(\int^M PMM,X)$. So, wedges $Z$ for ${\cal V}(P(-,-),X)$ correspond bijectively to morphisms $Z\to {\cal V}(\int^M PMM,X)$: ${\cal V}(\int^M PMM,X)$ is the end of ${\cal V}(P(-,-),X)$.

Answer 2

$\require{AMScd}\newcommand{\op}{{^\mathsf{op}}}\newcommand{\M}{\mathcal{M}}\newcommand{\V}{\mathscr{V}}\newcommand{\D}{\mathscr{D}}\newcommand{\C}{\mathscr{C}}$Why can you only define (co)ends in the ordinary way for $\V$-valued functors?

Let me say $\V$ is a closed symmetric monoidal category. I stipulate that only because I've developed and read theory of enriched stuff only in the symmetric case and I can't be sure what does and doesn't rely on $\gamma^2=1$ without doing a very thorough review.

Say $F:\underline{\C\op\boxtimes\C}\to\underline{\D}$ is a $\V$-functor. I can define a $\V$-wedge to be an object $\partial\in\D$ with a family of arrows $\omega_\varsigma:\partial\to F(\varsigma,\varsigma)$ for $\varsigma\in\C$ - "arrows" being in the underlying category $(\underline{\D})_0$ - that satisfy: $$\begin{CD}\underline{\C}(\varsigma,\varsigma')@>F^\varsigma_{\varsigma,\varsigma'}>>\underline{\D}(F(\varsigma,\varsigma),F(\varsigma,\varsigma'))\\@VF\op^{\varsigma'}_{\varsigma',\varsigma}VV@VV\underline{\D}(\omega_\varsigma,1)V\\\underline{\D}(F(\varsigma',\varsigma'),F(\varsigma,\varsigma'))@>>\underline{\D}(\omega_{\varsigma'},1)>\underline{\D}(\partial,F(\varsigma,\varsigma'))\end{CD}$$For all $\varsigma,\varsigma'$. I can then define the enriched end $\int_{\underline{\C}} F$ to be a terminal $\V$-wedge to $F$. This definition seems to be used by Kelly and doesn't require the case $\underline{\D}=\underline{\V}$ to be handled separately.

Then it is true by a standard adjunction argument that we have unenriched isomorphisms $\underline{\D}(\partial,\int_{\underline{\C}}F)\cong\int_{\underline{\C}}\underline{\D}(\partial,F)$ if $\underline{\D}$ is strongly tensored over $\V$. By that I mean there is for each $\partial\in\D$ a $\V$-functor $-\odot\partial:\underline{\V}\to\underline{\D}$ which is an enriched left adjoint of $\underline{\D}(\partial,-)$. The only subtlety there is that you have to check the enriched analogues of ordinary statements along the lines of "adjunctions transport wedges to wedge" and this will need some diagram chasing that no one seems to be bothered to do. Wedges $v\to\underline{\D}(\partial,F)$ transpose to wedges $v\odot\partial\to F$ to $v\odot\partial\to\int_{\underline{\C}}F$ to $v\to\underline{\D}(\partial,\int_{\underline{\C}}F)$ in bijection.

What if you wanted more than that? Maybe you want integration to actually be a $\V$-functor, and for $\underline{\D}(\partial,\int_{\underline{\C}}F)\cong\int_{\underline{\C}}\underline{\D}(\partial,F)$ to be $\V$-natural in both $\partial$ and $F$, appropriately formulating both sides as composites of $\V$-functors. To make integration functorial we seem to need $\underline{\D}$ strongly tensored anyway, so there is no loss, and then the enriched naturality ... is true, but more fiddly to check.