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


1 Answer 1


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.

  • $\begingroup$ 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 $\endgroup$ Commented Jun 5, 2022 at 18:28
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    $\begingroup$ @Ezio Greggio Can you construct an inverse equivalence? $\endgroup$ Commented Jun 5, 2022 at 19:57
  • $\begingroup$ Ah yes the disjoint union is the quasi-inverse of the inclusion $\endgroup$ Commented Jun 5, 2022 at 21:22

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