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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.

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    $\begingroup$ I take it $\mathbf{Set}_*$ is the category of pointed sets? $\endgroup$ Commented Feb 8, 2019 at 23:11
  • $\begingroup$ @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. $\endgroup$
    – qualcuno
    Commented Feb 8, 2019 at 23:59
  • $\begingroup$ 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. $\endgroup$ Commented Feb 9, 2019 at 0:04
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    $\begingroup$ 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 $\endgroup$ Commented Feb 9, 2019 at 10:47

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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 $1 \rightarrow X \times A \rightarrow X \wedge A$.

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  • $\begingroup$ I understood the abstraction process. Thanks a lot! I can award my bounty yet, will do so when available. Cheers $\endgroup$
    – qualcuno
    Commented Feb 11, 2019 at 20:01

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