 Solution

An exponentiable object $X$ in a category with finite products is one for whichhas a right adjoint $[X,-]$. The terminal object is exponentiable because $-\times 1\cong 1_{\mathcal{C}}$. If $X$ and $Y$ are exponentiable, thenis 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 metricThis is the smallest metric making both projections 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 candidateThe supremum is finite because $Y$ is bounded. For every $x,y$ one has the useful evaluation inequalityIndeed, 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 mapis non-expansive by this inequality. Postcomposition by a non-expansive $k:Y\to Y^{\prime }$ 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. Whenevernon-expansiveness of $h$ forces the latter distance to be at most $d_Z(z,z^{\prime })$. 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

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 $f$.

LetFix 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 $t\le a+b$, choose $r+s=t$ with $r\le a$ and $s\le b$. 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 givesIn 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 $D\le a+b$ for every contributing pair, and taking the supremum givesThus $\bar{d}$ is a metric whenever $X$ is interpolating, and the preceding adjunction proves every bounded interpolating space is exponentiable in ${\mathbf{Met}}_b$.