In considering the $\mathsf{Hom}$-functor
$$ \mathsf{Hom}_*: \mathsf{Set}_*^{\text{op}} \times \mathsf{Set}_* \to \mathsf{Set}_* $$
it feels somewhat clear to me that the left adjoint is the smash product
$$ (X,x_0) \wedge (Y,y_0) = (X\times Y) / [(X\times \{y_0\}) \cup (\{x_0\} \times Y)]. $$
This feels relatively straightforward to me since if $F$ is the left adjoint, then
$$ \mathsf{Set}_*\left [(X,x_0), \mathsf{Hom}_*\left ((Y,y_0), (Z,z_0)\right ) \right ] \;\; \cong \;\; \mathsf{Set}_*\left [F((X,x_0), (Y,y_0)), (Z, z_0) \right ] $$
which means that $F((X,x_0), (Y,y_0))$ is a pointed set and corresponds precisely to mappings that are constant on all pairs of the form $(x,y_0)$ and $(x_0,y)$ for all $x\in X, y\in Y$.
Now, my question stems from Riehl's book where she claims that this statement can be phrased in a categorical way such that $\mathsf{Set}$ can be replaced by any cartesian closed category.
Can the smash product be described as a pullback? In other words, do we have the following commutative diagram:
$$ \require{AMScd} \begin{CD} (X,x_0)\wedge (Y,y_0) @>>> X \\ @VVV @VVV \\ Y @>>> X\times Y \end{CD} $$
where the first two legs are projection and the next two legs are inclusions of the form $i_{y_0}:X\to X\times Y$ given by $i_{y_0}(x) = (x,y_0)$ and similarly, $i_{x_0}:Y\to X\times Y$ given by $i_{x_0}(y) = (x_0,y)$? Let me know if my logic appears sound, and if not what direction I should take this in. Where
In the category of pointed sets, the point $x_0\in X$ corresponds to a morphism $x_0 : * \to X$. Given two pointed sets you can form the pushout $$ \require{AMScd} \begin{CD} * @>>> X \\ @VVV@VVV \\ Y @>>> X \vee Y \end{CD} $$ where you identify the two points $x_0,y_0$ in a single one in the disjoint union $X\sqcup Y$.
Now, observe that $X\vee Y$ is a (pointed) subset of $X\times Y$: it identifies to the subset $(\{x_0\}\times Y)\times (X\times\{y_0\})$ (simply because the two distinguished points became one point $(x_0,y_0)$ in $X\vee Y$).
What if you took the quotient of $X\times Y$ by $X\vee Y$? This is exactly the smash product of $X$ and $Y$. But at the same time, this is also the pushout $$\require{AMScd} \begin{CD} X\vee Y @>>> X\times Y \\ @VVV@VVV \\ * @>>> X\land Y \end{CD} $$ This works exactly the same in every category that has a zero object and all pushouts: for example in the category of pointed topological spaces.
Interesting fact is that if your category is, say, that of abelian groups, the "smash product" there is the zero object, because $X\vee Y\cong X\times Y=X\oplus Y$.
So, you may think of the smash product as a measure of the failure of products and coproducts to be isomorphic: the bigger the quotient $\frac{X\times Y}{X\vee Y}$, the further the ambient category is from having biproducts.