# Equality of maps in a monoidal category

Let $C_1,C_2$ be monoidal categories (aka tensor categories) with tensor bifunctor $$\otimes_i: C_i\times C_i\to C_i$$ and tensor units $1_i$.

Assume I have a monoidal functor $F:C_1\to C_2$, it comes with an isomorphism $F(1_1)\simeq 1_2$ and a natural family $$F(U,V):F(U\otimes_1 V)\to F(U)\otimes_2 F(V).$$

Further, assume that $G:C_2\to C_2$ is a monoidal equivalence of $C_2$.

The map $F(U,V)\otimes_2 id:F(U\otimes_1 V)\otimes_2 F(W)\to (F(U)\otimes_2 F(V))\otimes_2 F(W)$ is a morphism in $C_2$ and thus I can look at $G(F(U,V)\otimes_2 id)$.

Is the last map equal to the map $G(F(U,V))\otimes_2 G(id)=G(F(U,V))\otimes_2 id$?

• Is $F$ lax monoidal (no restrictions on your natural family) or monoidal (meaning the natural family is a natural family of isomorphisms) ? – jeanmfischer Sep 19 '18 at 7:37
• It should be monoidal – user592465 Sep 19 '18 at 7:56
• Actually we don't need the condition on $F$ but rather on $G$, anyways the natural equation is valid even if both $F$ and $G$ are lax, just then you cannot conjugate because you cannot take inverses. – jeanmfischer Sep 19 '18 at 8:03

Consider $f_{U,V} : F(U \otimes_1 V) \to F(U) \otimes_2 F(V)$ to be natural morphisms in $C_2$. Next let's denote by $g_{X,Y} : G(X \otimes _2 Y) \to G(X) \otimes_2 G(Y)$ the strucutral isomorphisms of the monoidal auto-equivalence $G$. Then by naturality arguments you can deduce that, $$g_{F(U) \otimes_2 F(V), F(W)} \circ G(f_{U,V} \otimes_2 id_{F(W)}) = (G(f_{U,V}) \otimes id_{GF(W)}) \circ g_{F(U \otimes_1 V), F(W)}$$ So considering that $g$ are isomorphisms in $C_2$, we get that $G(f_{U,V} \otimes_2 id_F(W))$ is equal $G(f_{U,V})\otimes_2 id_{GF(W)}$ modulo conjugation by natural isomorphisms. But they are not equal unless your structure morphism $g$ is the identity.