Some questions about Enrichement Definition in Category Theory

Here is the definition of enrichment captured from Borceux.

My questions: It seems to me we cannot define enrichment over any monoidal category, because:

First, take the 3rd requirement, the assignment of objects of the monoidal category to each pair of objects should be wise in the way they respect the composition definition. i.e. $\mathcal{C}(A, C)$ cannot be any random object of $\mathcal{V}$, it should be the object as the result of $\mathcal{C}(A, B) \otimes \mathcal{C}(B, C)$.

Second, it needs to have enough morphisms from $I$ to every object. Because how do we know for every object there is such a map?

No one requires that $\mathcal{C}(A,C)$ be isomorphic to $\mathcal{C}(A,B) \otimes \mathcal{C}(B,C)$, just consider the fact that every category is enriched over the category of sets.
The existence of a morphism $I \to \mathcal{C}(A,A)$ for all objects $A$ is part of the data of an enriched category; no one requires that such a map exists, let alone uniquely, for every object in $\mathcal{V}$. For example, in an ordinary category one can/must specify the identity in the set $Hom(A,A)$ by the function $\{*\} \to Hom(A,A)$ given by $* \mapsto Id_A$. However, no map $\{*\} \to \emptyset$ exists.
To convince yourself that all the axioms are okay, first prove that every category is indeed enriched over $\mathbf{Set}$, and then maybe prove that the category $\mathbf{Vect}_k$ of vector spaces over a field $k$ is enriched over itself.
• Thanks. I tried to check axioms for the example you said: every category is enriched over Sets, but failed! Imagine I assign the set $S_1$ to the pair $(A, B)$ and the set of $S_2$ to $(B, C)$, and $\emptyset$ to $(A, C)$. $$c_{A, B, C}: S_1 \times S_2 \cong \{(s_1, s_2)| s_1 \in S_1 and s_2 \in S_2\} \longrightarrow \emptyset$$ But this composition does not exist, so this assignment does not seem to be a wise assignment! – Mobius Knot Sep 18 '18 at 11:28
• @MobiusKnot There's an obvious assignment of a set to a pair of objects $(A,B)$... $Hom_{\mathcal{C}}(A,B)$. – leibnewtz Sep 18 '18 at 20:08