# When is the adjoint to a monoidal functor monoidal?

Let $\mathcal C,\mathcal D$ be monoidal categories. Recall that a functor $F : \mathcal C \to \mathcal D$ is lax monoidal if it is equipped with maps $1_{\mathcal D} \to F(1_{\mathcal C})$ and $F(X) \otimes_{\mathcal D} F(Y) \to F(X \otimes_{\mathcal C} Y)$, the latter natural in $X,Y\in \mathcal C$, compatible with associators and unitors. It is oplax monoidal if it is instead equipped with maps in the other direction, and strong monoidal if it is equipped with isomorphisms. For each choice of lax/oplax/strong, there is a bicategory of monoidal categories, lax/oplax/strong monoidal functors, and monoidal natural transformations.

Suppose that $F : \mathcal C \to \mathcal D$ is one of lax/oplax/strong monoidal, and also admits a left adjoint $F^L$ in the bicategory of all functors. Under what circumstances is $F^L$ naturally lax/oplax/strong monoidal? Under what circumstances does this adjunction come from an adjunction in the bicategory of monoidal categories and lax/oplax/strong monoidal functors?

• Of course, the answer will probably be different for different ordered pairs in {lax,oplax,strong}. For example, if $F$ is lax, it is not too hard to believe that $F^L$ is oplax, based on the way that adjoints work (e.g. we do have a canonical map $F^L(1_D) \to 1_C$ corresponding to $1_D \to F(1_C)$). – Theo Johnson-Freyd Jul 21 '15 at 22:24
• If $F$ is lax $F^L = L$ is oplax via $L1 \rightarrow 1$ that you referred to, and the $L(X\otimes Y) \rightarrow L(FLX\otimes FLY) \rightarrow LF(LX\otimes LY) \rightarrow LX\otimes LY$. Nothing else is granted. In particular, if $F$ is strong, then $L$ is only oplax, and this is necessarily the only situation when we have an adjoint pair of oplax functors. – Dimitri Chikhladze Jul 21 '15 at 23:08
• This is a old standard question, this is generalizable also to (op)lax.functors (see a bicategory as a monoidal with many objects). Do you read Italian? (I wrote about..) – Buschi Sergio Jul 21 '15 at 23:38

If $L$ and $R$ are a left and right adjoint, then doctrinal adjunction asserts that $L$ is oplax monoidal iff $R$ is lax monoidal. (I'm being a bit imprecise here, treating monoidality as if it were a property instead of a structure, but hopefully the meaning is clear.)

• Doctrinal adjunction also answers the second question: if $R$ is lax, then its left adjoint $L$ is automatically oplax, and the whole adjunction lifts to an adjunction in the 2-category of lax monoidal functors iff $L$ is actually strong (i.e. its canonically induced oplax structure maps are invertible). – Mike Shulman Jul 30 '15 at 4:23

Given a monoidal functor $^mG=(G, \tilde{G}, G_0) : \textbf{V} \to \textbf{V'}$ with a (simply) left adjoint $F$ with $\eta :\textbf{V'} \Rightarrow G\circ F,\ \varepsilon: F\circ G \Rightarrow \textbf{V}$. Consider the associate morphisms $F_0^\ast: F(I') \to I$, $\tilde{F}_{A, B}^\ast$ of:

$G_0: I' \to G(I)$, $\tilde{G}_{FA,FB}\circ (\eta_A \otimes \eta_B ): A \otimes B \rightarrow GF(A) \otimes GF(B) \to G(F(A) \otimes F(B))$ We can prove that this define a op.monoidal functor ${}^{m.op}F=(F, \tilde{F}^\ast, F_0^\ast) : \textbf{V'}^{op}\to \textbf{V}^{op}$.

we have that $^mG$ ha a left monoidal adjoint (a left adjoint arrow in the 2-category of monoidal categories) $^mF=(F, \tilde{F}, F_0)$ if and only if it has a simple left adjoint $F$ such that the $F_0^\ast$ and $\tilde{F}^\ast_{A, B}$ (defined as above) are isomorphisms, and $F_0=(F_0^*)^{-1},\ \tilde{F}_{A, B} =(\tilde{F}_{A, B}^\ast)^{-1}$.

we have the commutative diagrams:

\text{ \xy (-10,20)+{(1):};(0,20)+{I'}="1"; (20,20)+{ I'}="2"; (10,10)+{GI}="3"; (20,0)+{GFI'}="4";(50,20)+{ }; {\ar@{>}^{1} "1";"2"};{\ar@{>}_{G_0} "1";"3"}; {\ar@{>}^{\eta_{I'} } "2";"4"}; {\ar@{>}_{GF_0} "3";"4"}; \endxy \xy (0,20)+{I}="1'"; (20,20)+{ I}="2'"; (-10,20)+{(2):}; (10,10)+{FI'}="3'"; (20,0)*+{FGI}="4'"; {\ar@{>}^{1} "1'";"2'"};{\ar@{>}_{F_0} "1'";"3'"}; {\ar@{>}{\epsilon{I} } "4'";"2'"}; {\ar@{>}_{FG_0} "3'";"4'"}; \endxy\ \ }

GIven $^mF$ the associate of $G_0$ is $F_0^*:= \epsilon_{I}\circ FG_0$ and from (1) follow $F_0^*\circ F_0=1$, now we prove that $F_0\circ F_0^*=1$: we show that its associate is $\eta_{I'}$: this associate is give from: $G(F_0)\circ G\epsilon_{I'}\circ GF(G_0)\circ \eta_{I'}$, per naturalità e da (2) questi è $G\epsilon_{FI'}\circ GFG(F_0)\circ GF(G_0)\circ \eta = G\epsilon_{FI'}\circ GF\eta_I\circ \eta_{I'}=\eta_{I'}$. Viceversa we define $F_0:=(F_0^\ast)^{-1}$, then follow the diagrams $(1)$ e $(2)$.

the associate of $\tilde{G}_{FA,FB} \circ (\eta_A \otimes \eta_B )$ is $F^*_{A, B}= \epsilon_{FA \otimes FB}\circ F(\tilde{G}_{FA,FB} )\circ F(\eta_A \otimes \eta_B ): F(A \otimes B) \to F(A) \otimes F(B)$. We have the commutative diagrams:

\text{ \xymatrix{A\otimes B\ar[rr]^1\ar[d]{\eta\otimes \eta}&&A\otimes B\ar[d]^{\eta} GF(A)\otimes GF(B)\ar[r]{\tilde{G}}& G(FA\otimes FB)\ar[r]{G(\tilde{F})}& GF(A\otimes B) } ; \xymatrix{FG(X)\otimes FG(Y)\ar[r]^{\tilde{F}}\ar[d]{\epsilon\otimes \epsilon}& F(G(X)\otimes G(Y))\ar[r]^{F(\tilde{G})}& FG(X\otimes Y)\ar[d]^{\epsilon} X\otimes Y\ar[rr]_1& &X\otimes Y } } acting by $F$ on the first and considering $\tilde{F}_{A,B}\circ \epsilon_{FA\otimes FB} = \epsilon_{F(A\otimes B)}\circ FG(\tilde{F}_{A,B}): FG(FA\otimes FB)\to F(A\otimes B)$ follow that $\tilde{F}_{A, B}\circ \tilde{F}_{A, B}^*=1$. Valuate on $X=F(A),\ Y=F(B)$ the second, from $\tilde{F}_{GFA\otimes GFB}\circ (F\eta_A\otimes F\eta_B)= F(\eta_A\otimes \eta_B)\circ \tilde{F}_{A,B}$ we have that $\tilde{F}_{A, B}^*\circ \tilde{F}_{A, B}=1$. Viceversa define $\tilde{F}_{A, B}:=(\tilde{F}_{A, B}^*)^{-1}$ si we have the commutative diagrams above.