Can functor carry over a monoidal structure?

Question

Suppose we have two categories $C$ and $D$ and a functor $F: C \rightarrow D$, furthermore suppose that the functor $F$ is an equivalence of categories, if the category $C$ is a monoidal category, can we then give a monoidal structure to $D$?

Answer 1

Let $G : D \to C$ be a functor which is inverse to $F$ (meaning quasi-inverse, of course). Then define a monoidal product on $D$ by $$X \otimes Y := F(G(X) \otimes G(Y)),$$ $$1_D := F(1_C).$$ The associativity constraint is induced by the one of $C$: $$X \otimes (Y \otimes Z) = F(G(X) \otimes G(F(G(Y) \otimes G(Z)))) \cong F(G(X) \otimes (G(Y) \otimes G(Z)))$$ $$ \cong F((G(X) \otimes G(Y)) \otimes G(Z)) \cong \dotsc \cong (X \otimes Y) \otimes Z.$$ Also the unit constraint is induced by the one of $C$: $$X \otimes 1_D \cong F(G(X) \otimes 1_C) \cong F(G(X)) \cong X \cong \dotsc \cong 1_D \otimes X.$$ The diagrams which appear in the definition of a monoidal category are commutative, basically because they are commutative in $C$. I omit the details, because it is really straight forward. Also observe that $F$ becomes a monoidal equivalence between $C$ and $D$.

Actually nothing happens here at all. Equivalences of categories are exactly defined in such a way that they transport every structure and property which is defined in the language of category theory.

Finally remark that all this categorifies the following well-known observation: If $f: C \to D$ is a bijection between sets, and $C$ carries the structure of a monoid (or any other algebraic structure, in fact), then also $D$ carries the structure of a monoid via $x \cdot y := f(f^{-1}(x) \cdot f^{-1}(y))$ and $1_D := f(1_C)$, and $f$ becomes an isomorphism of monoids.