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In a complete category that is a locally small category with a coseparator $S$ and all monomorphisms regular, an object $A$ is an exponentiable object exactly when $\mathcal{C}(-\times A,S)$ is a representable functor. A representing object $E$ gives representations $E^I$ for maps into powers $S^I$. An equalizer presentation by powers of a coseparator then constructs a representation for maps into every target. The Yoneda lemma makes these representations functorial and supplies the right adjoint.

A metric space $X$ is interpolating when $d(x,y)=r+s$ implies that some $z$ satisfies $d(x,z)=r$ and $d(z,y)=s$. Interpolation makes the metric exponential candidate satisfy the triangle inequality, so every bounded interpolating metric space is an exponentiable object of ${\mathbf{Met}}_b$.

The terminal object $1$ is an exponentiable object because $-\times 1\cong 1_{\mathcal{C}}$ has the identity as a right adjoint. If $A$ and $B$ are exponentiable objects, compose the two adjunctions to getConsequently exponentiable objects are closed under finite products in a category, including the empty product, and

Under the additional hypotheses, a coseparator $S$ detects unequal parallel arrows by postcomposition: for $u\ne v:X\to B$ there is $t:B\to S$ with $tu\ne tv$. Local smallness makes $I=\mathcal{C}(B,S)$ a set, and completeness forms the product in a category $S^I$. The evaluation arrowis a monomorphism, because $e_{B}u=e_{B}v$ forces $tu=tv$ for every $t$, and the coseparator then forces $u=v$.

By hypothesis this monomorphism is a regular monomorphism, so it is an equalizer of some $r,s:S^I\rightrightarrows Q$. Apply the same evaluation construction to $Q$: $J=\mathcal{C}(Q,S)$ is a set and $e_Q:Q\to S^J$ is monic. Cancelling $e_Q$ shows that equalizing $e_{Q}r,e_{Q}s$ is equivalent to equalizing $r,s$. We have therefore proved the equalizer presentation by powers of a coseparator:The notation $S^I$ here means a product indexed by a set, so its existence does not presuppose categorical exponentiation.

If $A$ is an exponentiable object, $\mathcal{C}(-\times A,S)\cong \mathcal{C}(-,S^A)$ is a representable functor. Conversely, suppose a representing object $E$ and a natural isomorphismare given. For any set $I$, the product in a category property givesThus maps into all powers of $S$ have representations. For an equalizer presentation $B\to S^I\rightrightarrows S^J$ with arrows $r,s$, postcomposition induces natural transformations between these represented functors. The Yoneda lemma identifies them with arrows $r_A,s_A:E^I\rightrightarrows E^J$. Form their equalizer $B_A$. ThenThese bijections are natural in $X$. For a morphism $B\to C$, postcomposition and the Yoneda lemma give the unique arrow $B_A\to C_A$ respecting them; uniqueness proves functoriality. Hence $B\mapsto B_A$ is a right adjoint to $-\times A$. This proves the exponentiability test using a coseparator:

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 ${\mathbf{Met}}_b$.

If $A$ and $B$ are exponentiable objects, thenhas the composite of their exponential functors as a right adjoint. The terminal object supplies the empty product, so exponentiable objects are closed under finite products.