Isomorphisms in enriched categories

Question

Let $(M,\otimes,1)$ be closed monoidal category and $C$ an $M$-enriched category. Assume we have $C$-objects $X$ and $X'$ and a morphism $f:1\to C(X,X')$ in $M$. We call $f$ an isomorphism if there is a $g:1\to C(X',X)$ in $M$ such that $$1\cong 1\otimes 1\stackrel{g\otimes f}{\to} C(X',X)\otimes C(X,X')\to C(X,X)$$ equals $\mathrm{id}:1\to C(X,X)$ and the same for $C(X',X')$. Now for each $Y$, $f$ induces a morphism $$f^*_Y:C(X',Y)\cong C(X',Y)\otimes 1\stackrel{\mathrm{id}\otimes f}{\to} C(X',Y)\otimes C(X,X')\to C(X,Y).$$ Assume that $f_Y^*$ is an $M$-isomorphism for all $Y$. As in the basic case ($C$ enriched over $\mathbf{Set}$), I want to conclude that $f$ is an isomorphism. To construct an inverse $g$, we should use $$g:1\stackrel{\mathrm{id}}{\to} C(X,X)\stackrel{(f_X^*)^{-1}}{\to} C(X',X).$$

First of all: Is this correct? At some point, it seems that I have to use the (enriched) naturality of $f^*$, now as a map $f^*_Y:1\to C(X,Y)^{C(X',Y)}$.

Answer 1

Unless I've misunderstood something, I think this is all fine. You would need the enriched naturality of $f^*$ in order to conclude that the collection of $f^*_Y$ was induced by some $f$, but in your setup you already assumed this. It's hard to gauge from your question how much you know about enriched categories, so let me recommend Kelly's book. He uses $\mathcal{V}$ where you use $M$, and then $\mathcal{V}$-naturality in general is (1.7), and in the context of your question is (1.21). That's where he shows that $\mathcal{V}$-naturality can be checked variable-by-variable, and what I stated about $f_Y^*$ inducing $f$.

I would caution against using the name isomorphism for your $f:1\to C(X,X')$, because this is a morphism in $M$, so there's already a notion of isomorphism. If you call this an isomorphism, you should prove it really is one in $M$. Perhaps you could instead say "$f$ induces an isomorphism" if this condition is met, and then you check that the corresponding morphism $X \to X'$ in $C$ is an isomorphism. To get that corresponding morphism, look at (1.33) in Kelly's book. He painstakingly defines a 2-functor $(-)_0: \mathcal{V}-CAT \to CAT$, which would take $C$ to an actual category $C_0$. Under this 2-functor the $\mathcal{V}$-object $C(X,X')$ goes to a set $C_0(X,X')$, and $f$ picks out an element of this set.

Using this approach, it's easy to check that the composites in your setup are the identities on $X$ and $X'$, in $C_0$, and hence are isomorphic to the identities on $X$ and $X'$ in $C$. Again, I refer you to Kelly's book.