Equivalence of Categories does not preserve enrichment

Question

I am learning some category theory currently, and was thinking about the following problem: Suppose $F: C \to D$ is a functor that induces an equivalence of categories and that $C,D$ are enriched over some category $M$. Then does $F$ preserve the enriched structure, that is is $Hom(c,c') \simeq Hom(F(c),F(c'))$, where the isomorphism is in the category $M$. We only know that this isomorphism is an isomorphism of sets, and my intuition tells me that we should not get a natural isomorphism in $M$. However, I tried finding some counter examples with categories enriched over familiar things like topological spaces, and abelian groups, but could not find a counter example. Any ideas?

Answer 1

No, certainly not. For instance, let $R$ and $S$ be rings and $f:R\to S$ be a bijection which preserves multiplication but not addition (for an explicit example of such a bijection, consider $R=S=\mathbb{Z}$ and $f:\mathbb{Z}\to\mathbb{Z}$ which swaps the factors of $2$ and $3$ in the prime factorization of each integer). We can consider $R$ as a one-object category enriched in abelian groups, with the morphisms being elements of $R$, addition of morphisms being addition in $R$, and composition of morphisms being multiplication in $R$. We can similarly consider $S$ as a one-object category enriched in abelian groups. Then $f$ can be considered as an isomorphism between these two categories (since it preserves multiplication), but it does not preserve the enrichment (since it does not preserve addition).

An important case where equivalences of categories do automatically preserve the enrichment is additive categories. Any functor between additive categories which preserves finite products or coproducts (in particular, any equivalence) automatically preserves the enrichment in abelian groups. This is because and addition of morphisms in an additive category can be constructed using direct sums: the sum of $f,g:A\to B$ is the composition of $(f,g):A\to B\oplus B$ with the fold map $B\oplus B\to B$.