In a category $\mathcal{C}$, a generalized element of an object $A$ means a morphism to $A$. It follows from Yoneda lemma that the object $A$ is determined by the collection of generalized of elements of $A$.

In a monoidal category $(\mathcal{C},\otimes)$, it seems more natural to consider generalized elements of a tensor product $A\otimes B$ in the form $a\otimes b$, meaning a tensor product of two morphisms, one to $A$, another to $B$. So, my first questions are

  1. Is the tensor product $A\otimes B$ determined by generalized elements of the form $a\otimes b$?
  2. Let $f,g\colon A\otimes B\to C$ be two morphisms. If for all generalized elements of the form $a\otimes b$, the compositions $f(a\otimes b)=g(a\otimes b)$, is it true that $f=g$?

In nLab, it mentioned that generalized elements of the form $I\to A$, where $I$ is the tensor unit, is important in the theory of enriched category. However, the functor $A\mapsto\mathrm{Hom}(I,A)$ may not be full or faithful. Hence those generalized elements cannot determine the objects or morphisms in general. So

  1. Why generalized elements of the form $I\to A$ important in enriched category theory regardless its application to define underlying categories?

1 Answer 1


If I'm not wrong, the following is an answer to 2.

For any objects $A,B,C$ of $\mathcal{C}$, we have the following natural map \begin{align} \Phi\colon\mathrm{Hom}(A\otimes B,C) &\to \mathrm{Nat}(h_A\boxtimes h_B,h_C\circ\otimes) \\ f &\mapsto f\circ- \end{align} where

  • $h_A\boxtimes h_B$ denotes the functor $\mathcal{C}\times\mathcal{C}\to\mathrm{Set}$ mapping each pair of objects $S,T$ of $\mathcal{C}$ to the set $\mathrm{Hom}(S,A)\times\mathrm{Hom}(T,B)$;
  • $h_C\circ\otimes$ denotes the functor $\mathcal{C}\times\mathcal{C}\to\mathrm{Set}$ mapping each pair of objects $S,T$ of $\mathcal{C}$ to the set $\mathrm{Hom}(S\otimes T,C)$;
  • $f\circ-$ means mapping each pair $a\colon S\to A, b\colon T\to B$ to the composition $f(a\otimes b)$.

On the other hand, by the definition of End, we have (each "$=$" means a canonical isomorphism) \begin{align} \mathrm{Nat}(h_A\boxtimes h_B,h_C\circ\otimes) &= \int_{(S,T)\in\mathrm{ob}\mathcal{C}\times\mathcal{C}}\mathrm{Map}(\mathrm{Hom}(S,A)\times\mathrm{Hom}(T,B),\mathrm{Hom}(S\otimes T,C)) \\ &= \int_{S\in\mathrm{ob}\mathcal{C}}\int_{T\in\mathrm{ob}\mathcal{C}} \mathrm{Map}(\mathrm{Hom}(S,A)\times\mathrm{Hom}(T,B),\mathrm{Hom}(S\otimes T,C)) & (\text{by Fubini theorem})\\ &= \int_{T\in\mathrm{ob}\mathcal{C}}\int_{S\in\mathrm{ob}\mathcal{C}} \mathrm{Map}(\mathrm{Hom}(S,A)\times\mathrm{Hom}(T,B),\mathrm{Hom}(S\otimes T,C)) & (\text{since $\mathrm{Set}$ is symmetric monoidal})\\ &= \int_{T\in\mathrm{ob}\mathcal{C}}\int_{S\in\mathrm{ob}\mathcal{C}} \mathrm{Map}(\mathrm{Hom}(S,A),\mathrm{Map}(\mathrm{Hom}(T,B),\mathrm{Hom}(S\otimes T,C))) \\ &= \int_{T\in\mathrm{ob}\mathcal{C}} \mathrm{Nat}(h_A,\mathrm{Map}(\mathrm{Hom}(T,B),\mathrm{Hom}(-\otimes T,C))) \\ &= \int_{T\in\mathrm{ob}\mathcal{C}} \mathrm{Map}(\mathrm{Hom}(T,B),\mathrm{Hom}(A\otimes T,C)) & (\text{by Yoneda lemma}) \\ &= \mathrm{Nat}(h_B,\mathrm{Hom}(A\otimes -,C)) \\ &= \mathrm{Hom}(A\otimes B,C)& (\text{by Yoneda lemma again}) \end{align} By trace the canonical isomorphism involved, the above gives the inverse of $\Phi$. The bijectivity of $\Phi$ implies 2 as desired.


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