TITLE: Is a weak functor which strictly preserves horizontal composition and which runs between strict bicategories automatically strict?
QUESTION [6 upvotes]: 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.)

REPLY [3 votes]: 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 $F(I)\neq J$.