At the end of section A.1.4 of the book "higher topos theory," there is a formula $X\otimes (C\otimes D)\simeq (X\otimes C)\otimes D$ which means the action property of tensoring in enriched categories. I suspect that $X\otimes (C\otimes D)\simeq (X\otimes D)\otimes C$ is right since, in this notation (i.e. $(A\otimes -) \dashv {}^A(-)$ in monoidal categories), the right adjoint of $(A\otimes B)\otimes -$ is ${}^{A\otimes B}(-) \cong ^{B}(^{A}(-))$.

So I think we should denote, in monoidal categories, the right adjoint of $A\otimes -$ (resp. $-\otimes A$) by $(-)^A$ (resp. $^A(-)$). But I found no one writing in this way (Joyal and Tierney are also using Lurie's order, using $(-)/A$ instead of $(-)^A$). I understand that it makes no difference when we are dealing with symmetric monoidal categories, but I wonder why everyone is using the opposite one. Is there any drawback using my notation?

  • $\begingroup$ I think it's just a historical artifact, since that's how exponentiation of numbers is always written. $\endgroup$ – Kevin Carlson May 3 '18 at 21:29
  • $\begingroup$ But then I think we should fix tensor to be a left action. Maybe because of the intuition of “division from one side,” but I feel this might cause some mistake when the monoidal product is not symmetric… $\endgroup$ – Naruki Masuda May 4 '18 at 7:04

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