I follow Kelly [*Basic concepts of enriched category theory*].

**Q1.** In general, no; in practice, yes.
By definition, we have an equaliser diagram
$$\int_{i : \mathcal{I}} \mathcal{A} (F (i), G (i)) \rightarrow \prod_{i \in \operatorname{ob} \mathcal{I}} \mathcal{A} (F (i), G (i)) \rightrightarrows \prod_{(i, j) \in (\operatorname{ob} \mathcal{I})^2} \mathcal{V} (\mathcal{I} (i, j), \mathcal{A} (F (i), G (j)))$$
and applying the underlying set functor preserves limits, so we get an equaliser diagram of sets:
$$\left( \int_{i : \mathcal{I}} \mathcal{A} (F (i), G (i)) \right)_0 \rightarrow \prod_{i \in \operatorname{ob} \mathcal{I}} \mathcal{A}_0 (F (i), G (i)) \rightrightarrows \prod_{(i, j) \in (\operatorname{ob} \mathcal{I})^2} \mathcal{V}_0 (\mathcal{I} (i, j), \mathcal{A} (F (i), G (j)))$$
This is almost – *but not quite* – the definition of $\int_{i : \mathcal{I}_0} \mathcal{A}_0 (F (i), G (i))$.
For the latter we would have $\prod_{(i, j) \in (\operatorname{ob} \mathcal{I})^2} \textbf{Set} (\mathcal{I}_0 (i, j), \mathcal{A}_0 (F (i), G (j)))$ instead.
For every $V$ and $W$ in $\mathcal{V}$, the action of the underlying set functor $\mathcal{V}_0 \to \textbf{Set}$ is a map
$$\mathcal{V}_0 (V, W) \longrightarrow \textbf{Set} (V_0, W_0)$$
so we get a canonical map:
$$\prod_{(i, j) \in (\operatorname{ob} \mathcal{I})^2} \mathcal{V}_0 (\mathcal{I} (i, j), \mathcal{A} (F (i), G (j))) \longrightarrow \prod_{(i, j) \in (\operatorname{ob} \mathcal{I})^2} \textbf{Set} (\mathcal{I}_0 (i, j), \mathcal{A}_0 (F (i), G (j))) \tag{†}$$
This induces a comparison map
$$\left( \int_{i : \mathcal{I}} \mathcal{A} (F (i), G (i)) \right)_0 \longrightarrow \int_{i : \mathcal{I}_0} \mathcal{A}_0 (F (i), G (i)) \tag{‡}$$
and it is easy to see that (‡) is a bijection if (†) is an injection.
This happens if the underlying set functor $\mathcal{V}_0 \to \textbf{Set}$ is faithful, which is usually the case in practice.
(The main exception is when $\mathcal{V}$ is a category of multisorted structures, e.g. simplicial sets or chain complexes.)

**Q2.**
Yes.
When $V$ is a coproduct of $X$ copies of the monoidal unit, then the underlying set functor gives a bijection $\mathcal{V}_0 (V, W) \to \textbf{Set} (X, W)$.
Thus, if $\mathcal{I}$ is the free $\mathcal{V}$-category on an ordinary category, then (†) is a bijection, so (‡) is also a bijection.
(But we could have deduced this directly from the universal property of free $\mathcal{V}$-categories!)

**Q3.**
Yes.
Suppose $F$ and $G$ are constant functors, with values $A$ and $B$ respectively.
You can directly verify that the diagonal morphism $\mathcal{A} (A, B) \to \prod_{i \in \operatorname{ob} \mathcal{I}} \mathcal{A} (F (i), G (i))$ equalises $\prod_{i \in \operatorname{ob} \mathcal{I}} \mathcal{A} (F (i), G (i)) \rightrightarrows \prod_{(i, j) \in (\operatorname{ob} \mathcal{I})^2} \mathcal{V} (\mathcal{I} (i, j), \mathcal{A} (F (i), G (j)))$ and so we have the desired morphism $\mathcal{A} (A, B) \to \int_{i : \mathcal{I}} \mathcal{A} (F (i), G (i))$.
(Again, you could instead use the universal property of free $\mathcal{V}$-categories here.
The point is that a $\mathcal{V}$-functor $\mathcal{A} \to [\mathcal{I}, \mathcal{A}]$ is essentially the same thing as a $\mathcal{V}$-functor $\mathcal{I} \to [\mathcal{A}, \mathcal{A}]$, and when $\mathcal{I}$ is free we can just take the constant functor with value $\textrm{id}_\mathcal{A}$.)

## Best Answer

I can think of three characterizations of products in a $Cost$-enriched category $X$ (the case of coproducts is dual):

adjointness: $X$ has binary products if the functor $\Delta_X : X \to X\otimes X$ has a right adjoint$$\_\odot\_ : X\otimes X \to X.$$ The object $X\otimes X$ is the product metric space, which has the correct universal property to make $Cost$-categories a monoidal category. Note that this is

slightlymore than what you asked for, because this way you define whenallproducts exist, not just for a single pair $(x,y)$.universal property: given $x,y\in X$, there exists a point $p=x\odot y$ with the following property: let $\alpha_x, \alpha_y$ be respectively the distances $d(p,x), d(p,y)\in [0,\infty]$; then, for every other $z\in X$, the distance $d(z,p)$ is terminal among all real numbers such that $$ \begin{cases} h + \alpha_x \ge d(z,x) \\ h + \alpha_y \ge d(z,y) \end{cases} $$ all in all this means that $d(z,p)=\max\{d(z,x)-d(p,x),d(z,y)-d(p,y)\}$. (I think it's just a matter of unwinding the universal property of an adjoint to see that $p = x\odot y$).representability: observe that a generic category admits the binary product of two objects $x,y$ when the functor $a\mapsto X(a,x)\times X(a,y)$ is representable, i.e. when (taking into account that $X$ is enriched over $[0,\infty]^\text{op}$)$$X(a,x\odot y) = X(a,x)\lor X(a,y)$$

i.e. when (taking into account that the hom-object $X(u,v)$ is the real number $d(u,v)$) for every $a\in X$ one has $d(a,x\odot y) = \max\{d(a,x), d(a,y)\}$.

Certainly the three definitions of $x\odot y$ are equivalent; I think I understand the first only in terms of the third. Also, the third generalises easily (well, not so easily if you want your metric to be finite...) to the case of $\kappa$-ary products: given a family $\{x_i\mid i\in\kappa\}$ of elements of $X$, $\bigodot_i x_i$ is a point of $X$ such that for every $a\in X$,

$$\textstyle d(a,\bigodot_ix) = \sup_i d(a,x_i)$$