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.