# Regarding the pointwise left adjoint of the $\mathrm{Hom}_*$ functor for cartesian closed categories

This question is motivated by a lack of understanding in the following (part of an) exercise of Emily Riehl's Category Theory in Context. More specifically, I do not quite see how to make my solution to the second part of the problem is sufficiently categorical:

• (i) There is a bifunctor $$\newcommand{\cat}[1]{\mathbf{#1}}\newcommand{\op}[0]{^{\mathrm{op}}}\newcommand{\Hom}{\mathrm{Hom}} \cat{Set}\op_* \times \cat{Set}_* \xrightarrow{\Hom_*}\cat{Set}_*$$,where $$\Hom_∗((X, x), (Y, y))$$ is defined to be the set of pointed functions $$(X, x) \to (Y, y)$$, with the constant function at $$y$$ serving as the basepoint. Define a two-variable adjunction determined by this bifunctor, the pointwise left adjoints to $$\Hom_∗((X, x), −)$$.
• (ii) Describe the left adjoint bifunctor $$\cat{Set}_* \times \cat{Set}_* \xrightarrow{\wedge}\cat{Set}_*$$ constructed in (i) in a sufficiently categorical way so that Set can be replaced by any cartesian closed category with pushouts and pullbacks.

For (i), I noted that for a fixed object $$(X,x)$$, functions $$(Z,z) \to (\hom((X,x),(Y,y),c_y)$$ correspond to maps $$f : X \times Z \to Y$$ such that $$f(x,-) = f(-,z) = y$$. That is, $$f$$ has to map $$x \times Z$$ and $$X \times z$$ to $$y$$, the rest is up to choice. This corresponds precisely to a selection of a map $$X \wedge Z \to Y$$ (where $$X \wedge Z$$ is $$(X \times Z)/{\sim}$$ with $$(x,z') \sim (x',z)$$ for all $$x',z'$$) such that the class of $$(x,z)$$ maps to $$y$$.

This defines a functor $$W = W_{(X,x)} : (Z,z) \mapsto (X \wedge Z, [(x,z)])$$ from $$\operatorname{ob} \cat{Set}_*$$ to $$\operatorname{ob} \cat{Set}_*$$. It remains to define an action on arrows and to prove that for each $$X$$ and $$x \in X$$ the functor $$W_{(X,x)}$$ is left adjoint to $$\Hom_*((X,x),-)$$. The former can be constructed as follows: given an arrow $$f : (Z,z) \to (Z',z')$$ the map $$(x,z) \to [(x,f(z))]$$ factors through the quotient $$q_Z : X \times Z \to X \wedge Z$$ via a map which we define to be $$Wf$$. From here (and a bit of diagram chasing), one can see that this assignment is functorial. Finally, we have bijections that send $$h \in \hom((X \wedge Z,[(x,z)]), (Y,y))$$ to $$h_\wedge(w) := h(q(-,w))) \in \hom((Z,z),\hom((X,x),(Y,y)))$$

which (if I have not made a mistake) are natural in $$(Z,z)$$ and $$(Y,y)$$.

Is the above a sound argument? And if so, any hints as to how to generalize it? I suppose that in a category $$C$$ with $$f$$ final, $$X \wedge Z$$ can be constructed as the pullback of $$(X \times Z, (x,z))$$ along the 'embeddings' of $$X$$ as $$X \times z$$ and $$Z$$ as $$x \times Z$$, defined by maps $$1_X : X \to X,\quad X \xrightarrow{\exists!} f \xrightarrow{z} Z$$ in the case of $$X$$ and likewise for $$Z$$.

If correct, is there a 'clean' way of showing naturality and 'left adjointness' of this construction? I'm failing to see where I should use that $$C$$ is cartesian closed. I do use the existence of a final object (to be able to work in $$C_*$$) and finite products, so I suppose the existence of exponentials comes into play when showing functoriality and/or naturality of hom-set bijections.

• I take it $\mathbf{Set}_*$ is the category of pointed sets? Commented Feb 8, 2019 at 23:11
• @MaliceVidrine indeed, this is the notation the author uses throughout the book, I apologize if it is not standard or if something I wrote made it unclear. Commented Feb 8, 2019 at 23:59
• It's quasi-standard, in the sense that it seems to be an emerging standard notation, but I've seen few enough sources use it that I just wanted to be sure. Nothing to apologize for. Commented Feb 9, 2019 at 0:04
• At the end, the pullback will not be what you want, the pullback will give you $\{(x,z)\}$ in $\mathbf{Set}_*$. You want a quotient, that is a colimit. You have 3 subobjects of $Z$ ($X\times z, x\times Z, \{(x,z)\}$) and you want them to collapse to a point Commented Feb 9, 2019 at 10:47

In $$\mathbf{Set}$$, fixing $$X \in \mathbf{Set}$$, we have $$\mathbf{Set}(A,\mathrm{Hom}(X,B)) \cong \mathbf{Set}(X \times A,B).$$ In $$\mathbf{Set_*}$$, fixing $$(X,x) \in \mathbf{Set_*}$$, we use the same correspondence except we insist that $$\forall(m,n) \in X \times A$$ we identify $$(x,n)$$ with $$(m,a)$$, and so $$\mathbf{Set_*} \Big( (A,a),\mathrm{Hom} \big( (X,x),(B,b) \big) \Big) \cong \mathbf{Set_*} \Big( \big( X \times A/\sim,[(x,a)] \big) ,(B,b) \Big).$$ You have made the correct assignment on morphisms to make this into a functor.
In a Cartesian closed category $$\mathscr{A}$$, we have $$\mathscr{A}(A,B^X) \cong \mathscr{A}(X \times A,B).$$ Now we approach $$1/\mathscr{A}$$ (the corresponding category of pointed objects), fixing $$x:1 \rightarrow X \ \in 1/\mathscr{A}$$. Note that the maps $$1 \rightarrow B^X$$ are in bijection with the maps $$X \rightarrow 1 \times X \rightarrow B$$, so the exponential object $$B^X$$ has a canonical point $$c_b : 1 \rightarrow B^X$$ corresponding to $$X \rightarrow 1 \rightarrow B$$ (this is analogous to distinguishing the constant map). As with $$\mathbf{Set}$$ and $$\mathbf{Set_*}$$, we use the same correspondence except we take the coequalizer $$q: X \times A \to X \wedge A$$ of $$(X \rightarrow 1 \rightarrow X, \mathrm{id}_A)$$, $$(\mathrm{id}_X,A \rightarrow 1 \rightarrow A):X \times A \rightarrow X \times A$$ (this is analogous to insisting that we identify $$(x,n)$$ with $$(m,a)$$). The canonical point of $$X \wedge A$$ is $$[(x,a)]$$, or more explicitly $$1 \rightarrow X \times A \rightarrow X \wedge A$$.