# Is a weak functor which strictly preserves horizontal composition and which runs between strict bicategories automatically strict?

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.)

• When you say "strict bicategories", do you mean 2-categories? – Maxime Lucas Jul 31 '18 at 16:07
• @ Maxime Lucas, yes. – Peter Guthmann Jul 31 '18 at 16:21

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 $F(I)\neq J$.
• Thanks! Your example can also easily be modified to an example of a cubical functor, which doesn't preserve units strictly, by replacing $C$ with $C \times C$. I guess this means one should add the strict preservation of units as an requirement into the definition of a cubical functor? – Peter Guthmann Aug 3 '18 at 9:49