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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 ${\mathbf{Met}}_b$.

A categorical limit of $D:\mathcal{J}\to \mathcal{C}$ is a terminal cone: it consists of an object $L$ and compatible maps $L\to D(j)$ through which every other cone factors uniquely. For a finite diagram, take the product $P={\prod }_{j}D(j)$ and the product $Q={\prod }_{u:i\to j}D(j)$. There are two maps $P\rightrightarrows Q$: on the $u:i\to j$ coordinate one uses respectively $D(u)$ after projection to $D(i)$ and direct projection to $D(j)$. Their equalizer is exactly the compatible-cone object. Hence finite products and equalizers construct every finite limit.

In the category of metric spaces and non-expansive maps, give $X\times Y$ the maximum metricThe projections are non-expansive, and a pair of non-expansive maps into $X$ and $Y$ induces a non-expansive map into this product, proving the universal property.

Let two maps $1\rightrightarrows X$ select $a,b\in X$. On the set quotient identifying $a$ and $b$, define the quotient metric by shortest paths that may jump from $a$ to $b$ at zero cost. Explicitly,This is the largest metric making the quotient map non-expansive. A map $X\to Z$ equalizing $a$ and $b$ factors through the set quotient and remains non-expansive by the path formula, so this is the coequalizer. Its underlying set is the set-theoretic coequalizer.

For $X=\{0,1,5,6\}$ with $0\sim 6$, write $\ast =\{0,6\}$. The quotient hasThus in $Q\times Y$, with $Y=\{0,3\}$,After first taking $X\times Y$, the product of the parallel pair identifies $(0,y)$ with $(6,y)$ separately for $y=0,3$. In its quotient metric, every path from $(1,0)$ to $(5,3)$ has length at least $4$, and length $4$ is attained either directly or via one of those identifications. The canonical bijection from this coequalizer to $Q\times Y$ is therefore not an isometry. Hence $-\times Y$ does not preserve this coequalizer. In a Cartesian closed category, $-\times Y$ is a left adjoint and preserves all colimits, so $\mathbf{Met}$ is not cartesian closed.