 Solution

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: