2
$\begingroup$

Consider the strict 2-category of monoidal categories with (lax) monoidal functors as 1-morphisms. There is a general notion of adjunctions and (adjoint) equivalences in (strict) 2-categories. Call these monoidal adjunctions resp. monoidal equivalences in our special case.

Let $L\vdash R$ be a monoidal equivalence. Is it true that $L$ and $R$ are strong monoidal? Note that there is an oplax structure on $F$ given by $$L(X\otimes Y)\longrightarrow L(RLX\otimes RLY)\longrightarrow LR(LX\otimes LY)\longrightarrow LX\otimes LY$$ (in fact, this only requires $L$ to be monoidally left adjoint to $R$), and similarly for $R$.

But I am not able to prove that this is inverse to the lax structure morphism of $L$. Is it even true?

$\endgroup$
  • $\begingroup$ 1) You mean that $F,G$ are strong monoidal, right? 2) I assume that you already know that $F,G$ is part of a monoidal adjunction. Have you tried to use that unit and counit are monoidal transformations? $\endgroup$ – HeinrichD Apr 14 '17 at 16:54
  • $\begingroup$ 1) Yes, you are right. $\endgroup$ – Jakob Werner Apr 14 '17 at 17:03
1
$\begingroup$

It is indeed true that a monoidal adjunction suffices, i.e. the left adjoint of a lax monoidal adjunction is necessarily strong. It seems this also follows from a more general principle called doctrinal adjunction invented by Kelly, but a direct verification is also possible. The diagram pasting goes as follows: Diagram monoidal adjunction Here $\varphi$ denotes both $LX\otimes LY\longrightarrow L(X\otimes Y)$ and $RX\otimes RY\longrightarrow R(X\otimes Y)$.

$\endgroup$
  • $\begingroup$ This would be an "answer" if you share with us the details that you have worked out. :) $\endgroup$ – HeinrichD Apr 15 '17 at 15:13
  • $\begingroup$ @Heinrich Here you go. $\endgroup$ – Jakob Werner Apr 15 '17 at 21:54
  • $\begingroup$ Thank you. Nice diagram :) $\endgroup$ – HeinrichD Apr 15 '17 at 23:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.