How does Yoneda lemma give that the natural isomorphism $\operatorname{hom}(A,-)\cong\operatorname{hom}(B,-)$ implies $A\cong B$?

Question

I'm trying to work out an element free proof of the associativity of the tensor product, that $$ (M\otimes_A N)\otimes_B P\cong M\otimes_A (N\otimes_B P). $$

Since $\operatorname{hom}$ and $\otimes$ are adjoint functors, I worked out that $$ \operatorname{hom}((M\otimes_A N)\otimes_B P,-)\cong\operatorname{hom}(M\otimes_A(N\otimes_B P),-) $$ is natural.

How does Yoneda lemma finish this? I was looking at the formal statement here, but I don't see how to apply it.

Answer 1

Let $(\mathcal{C}, \bullet, \mathrm{id})$ be your category, and write $\mathrm{Hom}$ for $\mathrm{Hom_\mathcal{C}}$.

Let $\Phi : \mathrm{Hom}(A,-) \to \mathrm{Hom}(B,-)$ be the natural isomorphism. Apply the Yoneda lemma to get that $\Phi = \mathrm{Hom}(f,-) = \bullet f$ for some $f : B \to A$. Apply it to $\Phi^{-1}$ as well to get $g : A \to B$ such that $\Phi^{-1} = \mathrm{Hom}(g,-) = \bullet g$.

Then $\mathrm{id}_{\mathrm{Hom}(B,-)} = \Phi \circ \Phi^{-1} = \mathrm{Hom}(f,-) \circ \mathrm{Hom}(g,-) = (\bullet f) \circ (\bullet g) = \bullet (g \bullet f) = \mathrm{Hom} (g \bullet f, -)$. This implies that $g \bullet f = \mathrm{id}$. Similarly for the other direction.

Hope this helps!