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Here is the definition of enrichment captured from Borceux.

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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?

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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.

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  • $\begingroup$ 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! $\endgroup$
    – Mobius Knot
    Sep 18, 2018 at 11:28
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    $\begingroup$ @MobiusKnot It's indeed not wise in the sense that what you describe is not a category. $\endgroup$
    – Arnaud D.
    Sep 18, 2018 at 11:44
  • $\begingroup$ @ArnaudD.Which one is not a category? I mean I just followed the axioms. $\endgroup$
    – Mobius Knot
    Sep 18, 2018 at 15:10
  • $\begingroup$ @MobiusKnot There's an obvious assignment of a set to a pair of objects $(A,B)$... $Hom_{\mathcal{C}}(A,B)$. $\endgroup$
    – Exit path
    Sep 18, 2018 at 20:08
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    $\begingroup$ @MobiusKnot you didn't follow the axioms, because one of the axioms is that the composition map, which doesn't exist in your example, must exist. $\endgroup$
    – Kevin Arlin
    Sep 18, 2018 at 20:36

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