Let $\mathbf{B}$ and $\mathbf{B'}$ be strict bicategories and $F: \mathbf{B} \to \mathbf{B'}$ a weak functor which preserves horizontal composition strictly (i.e. $Ff \* Fg = F(f \* g)$ natural in f and g.)


Does this imply $F$ preserves identities strictly, i.e. $1\_{F\_a} = F(1\_a)$?


With the unit axiom for weak functors, the strict preservation of $\*$ and the strictness of $\mathbf{B}$ and $\mathbf{B'}$ it follows that


$$ Ff \* 1\_{Fa} = Ff = F(f \* 1\_a) = Ff \* F1\_a, $$


for arbitrary f. But I don't see how this implies $1\_{F\_a} = F(1\_a).$


But if this isn't true, the claim that a cubical functor (see for instance definition 2 in <https://arxiv.org/pdf/1409.2148.pdf> ) automatically strictly preserves units would be false. (I've seen this claim as an aside now on at least 3 different occasions, so I might just miss something really simple here.)



Is a weak functor which strictly preserves horizontal composition and which runs between strict bicategories automatically strict?
It seems to me that the simplest possible counterexample works. Restrict first to the case when $\mathbf{B}$ and $\mathbf{B}'$ have one object, so we are talking about a strong monoidal functor $F:C\to D$ between strict monoidal categories. Now let $C$ be the terminal strict monoidal category, with one object $I$, only its identity morphism, and $I\otimes I = I$. And let $D$ be the strict monoidal codiscrete category corresponding to the monoid $\{J,E\}$ where $J$ is the unit object and $E$ an idempotent, $E\otimes E = E$; codiscrete means that we have a unique isomorphism $J\cong E$, so that $D$ is equivalent to $C$ as a category. Let $F:C\to D$ be defined by $F(I) = E$. Then $F$ preserves binary tensors strictly, since $F(I\otimes I) = F(I) = E = E\otimes E = F(I)\otimes F(I)$, but it doesn't preserve the identity strictly, $F(I) \cong J$ coherently but $F(I)\neq J$.


