Let $\mathcal{C}$ be a symmetric monoidal category and let $A$ and $A^*$ be dual in the sense of Definition 2.1 in nLab.
Dold & Puppe (1984) show (Thm 1.3) that the map
$$ \text{Hom}(X, Y \otimes A^{\ast}) \xrightarrow{(-) \otimes A} \text{Hom}(X \otimes A, Y \otimes {A}^{\ast} \otimes A) \xrightarrow{{\textrm{id}}_{Y} \otimes \text{ev}_A \circ (-)} \text{Hom}(X \otimes A, Y) $$
is an isomorphism for all objects $X$, $Y$. Let us refer to this map by $\varphi$.
I am trying to understand how it follows from this observation that
$$ (-) \otimes A \dashv (-) \otimes A^{\ast} $$
as pointed out in Remark 2.15 on nLab.
If the isomorphism $\varphi$ is natural, we automatically get the desired adjunction by definition. How can we know that $\varphi$ is indeed natural?
We will check that $\varphi_{X,Y}$ is natural in $X$ and $Y$ separately, by checking that the same holds for $-\otimes A$ and $(\mathrm{id}_Y\otimes \mathrm{ev}_A)\circ-$. However, since $-\otimes A$ is a functor, naturality in $X$ and $Y$ of the induced map $\mathrm{Hom}(X,Y\otimes A^*)\to\mathrm{Hom}(X\otimes A,Y\otimes A^*\otimes A)$ is just functoriality of $-\otimes A$, more specifically the statement that a functor preserves composition of morphisms.
The assignment $(\mathrm{id}_Y\otimes \mathrm{ev}_A)\circ-\colon\mathrm{Hom}(X\otimes A,Y\otimes A^*\otimes A)\to\mathrm{Hom}(X\otimes A,Y)$ is a postcomposition operation. Naturality in $X$ (which is tested by looking at precompositions) is hence a consequence of the statement that postcomposition and precomposition commute, in the sense that, given three composable morphisms $f$, $g$ and $h$, we have $(hg)f=h(gf)$ (this is of course also simply called associativity of composition).
We hence only have to still check that the assignment $(\mathrm{id}_Y\otimes \mathrm{ev}_A)\circ-\colon\mathrm{Hom}(X\otimes A,Y\otimes A^*\otimes A)\to\mathrm{Hom}(X\otimes A,Y)$ is natural in $Y$. For this, consider a morphism $g\colon Y\to Y'$. We must verify that the square $$ \require{AMScd} \begin{CD} \mathrm{Hom}(X\otimes A,Y\otimes A^*\otimes A) @>{(\mathrm{id}_Y\otimes \mathrm{ev}_A)\circ-}>> \mathrm{Hom}(X\otimes A,Y)\\ @V{(g\otimes\mathrm{id}_{A^*}\otimes\mathrm{id}_A)\circ-}VV @VV{g\circ-}V\\ \mathrm{Hom}(X\otimes A,Y'\otimes A^*\otimes A) @>{(\mathrm{id}_{Y'}\otimes \mathrm{ev}_A)\circ-}>> \mathrm{Hom}(X\otimes A,Y') \end{CD} $$ commutes. Suppose given a morphism $p\colon X\otimes A\to Y\otimes A^*\otimes A$. Going down in the diagram and then to the right, we end up with the morphism $(\mathrm{id}_{Y'}\otimes \mathrm{ev}_A)\circ (g\otimes\mathrm{id}_{A^*}\otimes\mathrm{id}_A)\circ p$, while first going right in the diagram and then going down, we end up with the morphism $g \circ (\mathrm{id}_Y\otimes \mathrm{ev}_A)\circ p$. These two maps are equal, because $g \circ (\mathrm{id}_Y\otimes \mathrm{ev}_A)=g\otimes \mathrm{ev}_A=(\mathrm{id}_{Y'}\otimes \mathrm{ev}_A)\circ(g\otimes\mathrm{id}_{A^*}\otimes\mathrm{id}_A)$ as maps $Y\otimes A^*\otimes A\to Y'\otimes I\cong Y'$ (where I write $I$ for the unit object of the symmetric monoidal structure). This in turn holds because $-\otimes-$ is a bifunctor, meaning that, for any two morphisms $a\colon A\to B$ and $c\colon C\to D$, we have $(a\otimes\mathrm{id}_D)\circ(\mathrm{id}_A\otimes c)= a\otimes c=(\mathrm{id}_B\otimes c)\circ(a\otimes\mathrm{id}_C)$ because the bifunctor respects composition in each input. This means that we have verified naturality of $\varphi$ in both $X$ and $Y$, so we are done.