# Prove\disprove that a category is cartesian closed

In Awodey's Category Theory I saw these two, quite similar, exercises.

Is the category of pointed sets cartesian closed? No, and it should be sufficient to say that, if a pointed category is cartesian closed, the $$\operatorname{Hom}$$-sets must be all trivial; in fact, in any CCC, holds $$\operatorname{Hom}(\top,A^B)\cong \operatorname{Hom}(A,B)$$.

Consider the category $$\mathbf S$$ of sets equipped with a distinguished subset, $$(A, P \subseteq A)$$, with maps $$f : (A, P)\to(B, Q)$$ being those functions $$f : A \to B$$ such that $$a \in P$$ iff $$f(a) \in Q$$. Show this category is cartesian closed by describing it as a category of pairs of sets. Here I'm having troubles. The only thing that I noticed is that $$\mathbf S$$ can be described as a subcategory of $$\mathbf{Set}^\mathbf I$$, where $$\mathbf I$$ is the category with two objects and one non-identity arrow (from one object to the other): the objects are the injections of sets $$Y\hookrightarrow X$$, and the morphisms are those who give raise to a cartesian (and not only commutative) square.

I know that $$\mathbf{Set}^\mathbf I$$ is a CCC but, in order to deduce it for $$\mathbf S$$, shouldn't I prove also that the inclusion $$\mathbf S\hookrightarrow \mathbf{Set}^\mathbf I$$ is a left adjoint, and the unity of the adjunction is a natural iso? I made a few attempts, but it actually doesn't seem the natural path to follow.

I thought of describing $$\mathbf S$$ directly as $$\mathbf{Set}^\mathbf J$$ for some small $$\mathbf J$$, but I really wouldn't know how to catch the property of being injective \ cartesian in this way. Would you give me a hint? Thank you

Yes, in a pointed CCC hom-sets are all trivial because $$\operatorname{Hom}(A,B) = \operatorname{Hom}(A \times 0,B) = \operatorname{Hom}(0,B^A) = *,$$ in particular this holds for pointed sets.
In the second one, I agree with your description of S, but I am not sure why you ignore the hint (or do you want to prove it exactly in the way you describe?). The exercise expects you to note that $$A \subset B$$ can be rewritten as $$A \subset A \coprod (B \setminus A),$$ and that this is compatible with the maps.
• That's the first approach I thought of, but doesn't it give just an inclusion of $\mathbf S$ into $\mathbf {Set}^2$? This last category is CC, but to deduce it for $\mathbf {S}$ shouldn't I prove that the inclusion is left adjoint? Again, I was stuck with this part Jun 5, 2022 at 18:28