= Solution
An <exponentiable object> $X$ in a category with finite products is one for which
$$
-\times X
$$
has a right adjoint $[X,-]$. The <terminal object> is exponentiable because $-\times1\cong1_{\mathcal C}$. If $X$ and $Y$ are exponentiable, then
$$
-\times(X\times Y)\cong(-\times X)\times Y
$$
is a composite of two left adjoints and therefore has the composite right adjoint $[X,[Y,-]]$. Exponentiable objects are consequently closed under finite products.
In the <category of metric spaces and non-expansive maps>, the terminal object is the one-point space. The product of $X$ and $Y$ has underlying set $X\times Y$ and metric
$$
d((x,y),(x',y'))
=\max\{d_X(x,x'),d_Y(y,y')\}.
$$
This is the smallest metric making both projections <non-expansive map>[non-expansive], and the product pairing of two non-expansive maps is non-expansive. If $X$ and $Y$ are bounded, so is this product. Hence both $\mathbf{Met}$ and the <category of bounded metric spaces and non-expansive maps> $\mathbf{Met}_b$ have finite products.
For bounded $X,Y$, define the <metric exponential candidate>
$$
\bar d(f,g)=sup\{d_Y(fx,gy):d_X(x,y)<d_Y(fx,gy)\}.
$$
The supremum is finite because $Y$ is bounded. For every $x,y$ one has the useful evaluation inequality
$$
d_Y(fx,gy)
\leq\max\{\bar d(f,g),d_X(x,y)\}.
$$
Indeed, if the second term does not already dominate, the pair $(x,y)$ occurs in the defining supremum.
Assume $\bar d$ is a metric. The evaluation map
$$
\operatorname{ev}:[X,Y]\times X\longrightarrow Y,
\qquad(f,x)\longmapsto f(x)
$$
is non-expansive by this inequality. Postcomposition by a non-expansive $k:Y\to Y'$ is non-expansive on function spaces, because every pair contributing to $\bar d(kf,kg)$ also contributes a no-smaller bound to $\bar d(f,g)$. Thus $[X,-]$ is a functor.
If $h:Z\times X\to Y$ is non-expansive, each $h_z(x)=h(z,x)$ is non-expansive. Whenever
$$
d_X(x,y)<d_Y(h_z(x),h_{z'}(y)),
$$
non-expansiveness of $h$ forces the latter distance to be at most $d_Z(z,z')$. Hence $z\mapsto h_z$ is non-expansive into $[X,Y]$. Conversely, a non-expansive $Z\to[X,Y]$ followed by evaluation gives a non-expansive $Z\times X\to Y$. These inverse operations are natural, proving
$$
\mathbf{Met}_b(Z\times X,Y)
\cong\mathbf{Met}_b(Z,[X,Y]).
$$
It remains to obtain the triangle inequality from interpolation. Nonnegativity and symmetry of $\bar d$ are immediate. If $f\ne g$, taking $x=y$ where $f(x)\ne g(x)$ proves separation; and $\bar d(f,f)=0$ follows from non-expansiveness of $f$.
Let
$$
a=\bar d(f,g),\qquad b=\bar d(g,h).
$$
Fix a pair $x,y$ contributing $D=d_Y(fx,hy)$, so $t=d_X(x,y)<D$. Suppose for contradiction that $D>a+b$. If $t\leq a+b$, choose $r+s=t$ with $r\leq a$ and $s\leq b$. If $t>a+b$, choose $r+s=t$ with $r>a$ and $s>b$. Since $X$ is an <interpolating metric space>, there is $z$ with $d(x,z)=r$ and $d(z,y)=s$. Applying the evaluation inequality twice gives
$$
D\leq d_Y(fx,gz)+d_Y(gz,hy)
\leq\max\{a,r\}+\max\{b,s\}.
$$
In the first case the right side is at most $a+b$, and in the second it equals $r+s=t<D$. Both are contradictions. Therefore $D\leq a+b$ for every contributing pair, and taking the supremum gives
$$
\bar d(f,h)\leq\bar d(f,g)+\bar d(g,h).
$$
Thus $\bar d$ is a metric whenever $X$ is interpolating, and the preceding adjunction proves every bounded interpolating space is exponentiable in $\mathbf{Met}_b$.
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