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Here's a similar question, but I'm not sure if what I'll ask is already answered there.

Let $(C,\odot), (D,\square)$ be monoidal categories. If $F \colon D \to C$ is left adjoint to $U \colon C \to D$, where $F$ is a monoidal functor ($U$ does not need to be), is it the case that the induced functors $F' \colon Mon(D,\square) \to Mon(C, \odot)$ and $U' \colon Mon(C,\odot) \to Mon(D,\square)$ are adjoint? (EDIT: $F'$ is induced by $F$ in the natural way since $F$ is monoidal; $U'$ is induced by $U$ in the sense that the square with $U, U'$ and the forgetful functors from internal monoids to their monoidal categories commute).

Here's my motivation for the question: if $(D, \square)=(Set, \times)$ (where $\times$ is the cartesian product), $(C, \odot)=(Vect_{\mathbb{K}}, \otimes)$ (where $\otimes$ is the usual tensor product), $U$ is the forgetful functor which takes vector spaces to their underlying sets and linear maps to their underlying functions and $F$ takes a set to a vector space which basis is that set and functions to uniquely determined linear maps, you also have functors $U_V \colon Mon(Vect_{\mathbb{K}}, \otimes) \to Vect_{\mathbb{K}}, U_S \colon Mon(Set, \times) \to Set$ (the usual forgetful functors from associative unital algebras to vector spaces and from monoids to sets, respectively), which by Theorem 2 (Construction of free monoids) in CWM have left adjoints $F_V, F_S$.

It's elementary that $U U_V = U_S U'$. If I could prove that $U'$ is right adjoint to $F'$, then $U_S U'$ would be right adjoint to $F' F_S$. Since $U U_V$ is already a right adjoint to $F_V F$, we would have a natural isomorphism $F_V F \cong F' F_S$, which I wish to prove (without messy calculations).

Hilario Fernandes
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    Is $\times$ necessarily the categorical product ? And how does $U$ induce $U'$ if it is not monoidal ? – Maxime Ramzi Aug 13 '18 at 21:09
  • @Max I apologize for the bad choice there. It doesn't need to be the categorical product, but for my purposes it suffices. Also you're absolutely right, $U$ can't induce $U'$ in a direct way if it's not monoidal. So let's say that $U$ induces $U'$ in the sense that it makes the square with right adjoints and $U'$ to commute. – Hilario Fernandes Aug 13 '18 at 21:36
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    @HilarioFernandes: FYI, \square ($\square$) is a common symbol to use for monoidal operations when you can't use $\otimes$. Some other options are \odot and \star ($\odot$ and $\star$). –  Aug 13 '18 at 21:55
  • @Hurkyl Thank you, I'm going to make an edit. – Hilario Fernandes Aug 13 '18 at 21:58

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Your question is addressed by the linked question, but I'll spell it out so that it's clear how. If $F \dashv G$ is an adjunction between monoidal categories, then colax monoidal structures on $F$ are in bijection with lax monoidal structures on $G$ (this is called doctrinal adjunction). Therefore, if $F$ is strong monoidal, then $F \dashv G$ is a lax monoidal adjunction between monoidal categories. The monoid constructon $\mathbf{Mon}({-})$ is 2-functorial with respect to monoidal categories and lax monoidal functors. Furthermore, 2-functors preserve adjunctions. Therefore, every adjunction $F \dashv G$ between monoidal categories for which $F$ is strong monoidal induces an adjunction $\mathbf{Mon}(F) \dashv \mathbf{Mon}(G)$.

varkor
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