I'm mainly following references such as Kelly, Loregian and the nLab, and it seems customary there to generalize (co)ends to the enriched context (over a symmetric monoidal category $\mathcal{V}$) by first defining the $\mathcal{V}$-valued (co)ends, for $\mathcal V$-functors
$$P\colon \mathcal C^\mathsf{op}\boxtimes\mathcal{C}\longrightarrow\mathcal{V},$$
and then define the general case of - let's say end - for a $\mathcal V$-functor $T\colon\mathcal{C}^\mathsf{op}\boxtimes\mathcal{C}\to\mathcal{D}$ by the object $\int\_cT(c,c)$ of $\mathcal D$ together with its extranatural transformation $\{ p\_a\colon\int\_cT(c,c)\to T(a,a) \}$ (which actually is $\{p\_a\colon \mathbb{1}\to\mathcal{D}(\int\_cT(c,c),T(a,a))\}$) exhibiting via $\mathcal{D}(d,p\_a)$, for every $d$ in $\mathcal{D}$, the end of the $\mathcal V$-functor
$$\mathcal D(d,T(-,-))\colon\mathcal{C}^\mathsf{op}\boxtimes\mathcal{C}\to\mathcal{V}.$$


Now, this makes sense to me, and it clearly gives by definition the formula $\int\_c\mathcal{D}(d,T(c,c))\cong\mathcal{D}(d,\int\_cT(c,c))$, but why do we do that? Is this step (passing through the $\mathcal V$-valued case) necessary for some deep reason I'm not seeing or it's just convenience? My feeling would be that we can generalize verbatim the definition of (co)ends to the enriched case (as a initial/terminal enriched extranatural transformation), just remembering what we actually mean when talking about morphisms in an enriched category. But then I struggle to prove the above continuity formula. Could anyone enlighten me about that?


Thank you so much in advance.



Why are enriched (co)ends defined like that?
The $\newcommand{\V}{\mathcal{V}}\newcommand{\D}{\mathcal{D}}\newcommand{\C}{\mathcal{C}}$universal property of being an “initial/terminal cowedge” — even strengthened to “$\V$-initial/-terminal” — is weaker than the standard definition, and too weak for many purposes. In particular, it’s not strong enough to imply any universal property of the form $\D(\int^\C P, X) \cong \ldots$


The difference shows up already in (co)limits — even in (co)products. For a counterexample, take $\V = \mathrm{Set^2}$, and $\C$ to be the $\V$-category with four objects $a,b,p,q$, generated by “global arrows” from $a$ and $b$ to $p$, and “half-present arrows” from $a$ and $b$ to $q$; so explicitly, $\C(x,x) = (1,1)$ for each $x$, $\C(a,p) = \C(b,p) = (1,1)$, $\C(a,q) = \C(b,q) = (1,0)$, and $\C(x,y) = (0,0)$ otherwise. Then the arrows $a,b \to p$ form a $\V$-initial cospan under $a,b$, since it’s the only such cospan and has no non-trivial endomorphisms. But $p$ is *not* a $\V$-coproduct for $a$ and $b$, since we have $\C(p,q) = (0,0)$, not $(1,0)$ as a $\V$-coproduct should have.


The problem is that the $\V$-category of (co)cones/wedges under a diagram only contains the *global* (co)cones/wedges as its objects — in the example, it can’t see the “half-present” cocone to $q$. This could be correcting by considering it as a $\V$-indexed category, but that requires a whole extra layer of technology.


When a (co)limit/end for a diagram exists, it will be an initial/terminal (co)cone/wedge, and so then any initial/terminal (co)cone/wedge will be isomorphic to it, and hence will be a $\V$-(co)limit/end. In particular, in a (co)-complete category, the weaker property of being an initial/terminal (co)cone/wedge does imply the full universal property. But outside the (co)complete setting, they diverge, as the counterexample shows, and the full universal property is what one generally wants.


