Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 122 / 6 / a / Solution

Because the direction of a lax monoidal functor compares the target tensor with the image of the source tensor, the given maps ${\phi }_{X,Y}:X\bullet Y\to X\diamond Y$ have the correct direction. They need not be invertible unless a strong monoidal functor is intended.

Their associativity condition is the following commutative diagram, with the associators of the two structures labelled explicitly:The two unit conditions, since ${\phi }_0=1_I$, are the unit commutative diagramsTogether with the assumed naturality, these are exactly the axioms for the requested monoidal structure. There are no further coherence conditions.

The faithful strict monoidal functor $U$ allows all three diagrams to be checked after applying it. Both tensor structures then have the same underlying tensor and constraints in $\mathcal{V}$. In particular, the unit conditions are equivalent toThe associativity square becomes the displayed diagram with every tensor replaced by $\otimes$, both associators by ${\alpha }^{\mathcal{V}}$, and every $\phi$ by $U\phi$. Faithfulness reflects equality of the two resulting composites. One should not identify ${\phi }_{I,X}$ itself with an identity in $\mathcal{C}$: its domain and codomain can be distinct objects with the same image under $U$. Intrinsically it is $({\lambda }_X^{\diamond }{)}^{-1}{\lambda }_X^{\bullet }$, and similarly on the right.