# Isomorphisms in enriched categories

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 $$f^*_Y:1\to C(X,Y)^{C(X',Y)}$$.

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.