Prove isomorphism of rings using Yoneda lemma I have seen many questions which prove the isomorphism of rings using the Yoneda lemma, for example, this question, this question and this question. The basic idea is that we can prove the two functors (from Ring category to Set category) $\hom(X,*)$ and $\hom(Y,*)$ are isomorphism, then by Yoneda lemma we know $X$ and $Y$ are isomorphic.
For example, the first question proves that $(A/I)\otimes (B/J)\cong (A\otimes B)/(I\otimes 1+1\otimes J)$.
Question: By using category theory, can we see this isomorphism is natural?
What I mean by natural isomorphism is that the isomorphism between $(A/I)\otimes (B/J)$ and $(A\otimes B)/(I\otimes 1+1\otimes J)$ is exactly given by the obvious map $\bar{a}\otimes\bar{b}\rightarrow\overline{a\otimes b}$. I think that using category theory, one can indeed prove that $(A/I)\otimes (B/J)\cong (A\otimes B)/(I\otimes 1+1\otimes J)$, but one can't actually know what the explicit isomorphism is.
I think in commutative algebra and algebraic geometry, the naturalness or canonicalness of an isomorphism is very important, right?
 A: I completely agree knowing the explicit isomorphism is important, and I claim you can recover the isomorphism with the Yoneda lemma proof, you just need to know to know what's going on "inside" the Yoneda lemma. Recall the Yoneda lemma (in one form) says for any object $X$ of a category $\mathscr C$ and a functor $G:\mathscr C\to\mathbf{Set}$, we have a bijection of sets $\operatorname{Nat}(h^X,G)\cong G(X)$ which is natural in $X$. Explicitly this is given as follows: if we have $\varphi:h^X\to G$, then we take $\varphi_X(\mathrm{id}_X)\in G(X)$.
If we are given $X,Y$ in $\mathscr C$ then taking $G=h_Y$, we see that $\operatorname{Nat}(h^X,h^Y)\cong h^Y(X)=\hom(Y,X)$; this is the "fully faithfullness" of the Yoneda embedding, and in particular if we have an isomorphism $h^X\cong h^Y$ then the corresponding morphism $Y\to X$ is our isomorphism, which we see from the above paragraph we can recover as $\varphi_X(\mathcal id_X)$.
Let's apply this to the example you give; let's figure out exactly what is the explicit isomorphism $\hom(A/I\otimes_R B/J,T)\cong \hom((A\otimes_RB)/(I\otimes1+1\otimes J),T)$: I will follow the isomorphisms given in the answer you linked to, so you can compare line by line. Starting with $h\in\hom(A/I\otimes_R B/J,T)$ we have
\begin{align*}
h&\mapsto(\bar a\mapsto h(\bar a\otimes1),\,\bar b\mapsto h(1\otimes\bar b))\\
&\mapsto (a\mapsto h(\bar a\otimes 1),\,b\mapsto h(1\otimes\bar b))\\
&\mapsto (a\otimes b\mapsto h(\bar a\otimes 1)h(1\otimes\bar b)=h((\bar a\otimes 1)(1\otimes\bar b))=h(\bar a\otimes\bar b))\\
&\mapsto (\overline{a\otimes b}\mapsto h(\bar a\otimes\bar b))
\end{align*}
Now taking $h=\mathrm{id}:A/I\otimes_R B/J\to A/I\otimes_R B/J$ we see that the isomorphism $(A\otimes_R B)/(I\otimes 1+1\otimes J)\to A/I\otimes_R B/J$ coming from Yoneda is exactly $\overline{a\otimes b}\mapsto\bar a\otimes\bar b$.
