Any category which is both cartesian closed and cocartesian coclosed is a thin category, though it may not be the terminal category (e.g., any Boolean algebra is such a category).

Proof

Let a category be given which is both cartesian closed and cocartesian coclosed. Cartesian closure tells us that the product of any object with the initial object $0$ will be itself initial (as left adjoints preserve colimits). Furthermore, given any morphism from an object $A$ to $0$, we can pair this with the identity morphism on $A$ to obtain a morphism from $A$ into $A \times 0$ with a left inverse given by projection, thus identifying $A$ as a retract of an initial object, and therefore as an initial object itself. It follows immediately that any two parallel morphisms to $0$ are equal; equivalently, by the dual reasoning, any two parallel morphisms out of $1$ are equal. But this means any two parallel morphisms in general are equal, as maps from $X$ to $Y$ can be identified with maps from $1$ to $Y^X$; accordingly, the category must be a thin category.

Examples

The Kleisli category of a continuation monad$K_R = R^{R^{(-)}}$ on a cartesian closed category with coproduct $\mathbf{C}$ is cocartesian coclosed. In fact $(-) \times R^A \dashv A + (-)$, which we can prove using the cartesian closure of $\mathbf{C}$: