I have some questions about $2$-functors between two $2$-categories and their compatibility with compositions. Also how the stuff extends to $n$-functors between $n$-categories.
Recall that for usual categories $C, D$ a functor $F: C \to D$ respects the compositions and identities: Let $X,Y, Z \in C$ and $f: X \to Y, g: Y \to Z$ in $C$. Then $F(id_X)=id_{F(X)}$ and $F(g \circ f) =F(g) \circ F(f)$ as proper equalities.
If if $C$ and $D$ are now $2$-categories, then a $2$-functor $F: C\to D$ consisits of
- a set map $F: Ob(C) \to Ob(D)$ -for each pair $c, d \in Ob(C) $ the $F_{c,d}:Hom_C(c,d) \to Hom_D(Fc,Fd)$ is a usual functor -"weak composition": Let $f: c \to d, g: d \to e$ in $C$, then in general $F(g \circ f) \neq F(g) \circ F(f)$ in $Hom_D(Fc, Fe)$, but we require only $F(g \circ f) \cong F(g) \circ F(f)$.
And I'm intersted in precise meaning of this "$\cong$".
First of all this means that $F(g \circ f)$ and $F(g) \circ F(f)$ are isomorphic in "weak" sense. What I understood is that this "weak" means that there exist two $2$-morphisms $a: F(g \circ f) \to F(g) \circ F(f)$ and $b: F(g) \circ F(f) \to F(g \circ f)$ with $a \circ b = id_{F(g) \circ F(f)}$ and $b \circ a = id_{F(g \circ f)}$.
If I understood this correctly then then the last identities are proper. The argument why these identities are proper now and not on the level of "weak" isomorphisms seems to be that that $2$-categories are sectretly assumed to be special $\infty$-categories where all $k$-morphisms with $k > 2$ are identities.
My first question is if the reason I tried to explaned before that $a \circ b = id_{F(g) \circ F(f)}$ and $b \circ a = id_{F(g \circ f)}$ should hold as proper identities and not just as "weak" isomorphisms is the correct one?
The next question is how the same question on compatibility of $F(g \circ f) $ and $F(g) \circ F(f)$ generlizes to $n$-functor between $n$-categories $C, D$ with $n >2$.
My naive guess is: First if $C,D$ are $n$-categories and $F: C \to D$ is $n$-functor then we require again $F$ to be a set map $F: Ob(C) \to Ob(D)$ and for each pair $c, d \in Ob(C) $ the $F_{c,d}:Hom_C(c,d) \to Hom_D(Fc,Fd)$ is a $(n-1)$-functor between $(n-1)$-categories $Hom_C(c,d) $ and $Hom_D(Fc,Fd)$ by recursion.
Now the question is what about the relation between $F(g \circ f) $ and $ F(g) \circ F(f)$. It seems naturally also to require $F(g \circ f) \cong F(g) \circ F(f)$ as isomorphisms in weak sense, is it correct?
If that's what I wrote before make sense then the question is how the "$\cong $" in here defined and how it difers from the case of $2$-functor?
Can it be explaned as follows: $F(g \circ f) \cong F(g) \circ F(f)$ is $1$-morphisms means that there exist again two $2$-morphisms $a: F(g \circ f) \to F(g) \circ F(f)$ and $b: F(g) \circ F(f) \to F(g \circ f)$ but now we require $a \circ b \cong id_{F(g) \circ F(f)}$ and $b \circ a \cong id_{F(g \circ f)}$.
And now we play the same game, that is there exist for $a \circ b$ and $ id_{F(g) \circ F(f)}$ two $3$-morphisms $a_1: a \circ b \to id_{F(g) \circ F(f)}$ and $a_2: id_{F(g) \circ F(f)} \to a \circ b$ with $a_1 \circ a_2 \cong id_{id_{F(g) \circ F(f)}}$ and $a_2 \circ a_1 \cong id_{a \circ b}$.
Similary for $b \circ a $ and $ id_{F(g \circ f)}$ and so on and so on. And we do it $n-1$ times. Then since we are working with $n$-categories after $n-1$-steps the "$\circ $" become finally equalities "$=$".
The last question is if that's exactly what "$\cong $" or "weak" isomorphism is in context of $n$-categories?
In an $\infty$-category, you can define an equivalence $f:a\simeq b$ to be a morphism such that there exists a morphism $g:b\to a$ and equivalences $1_a\simeq g\circ f$ and $f\circ g\simeq 1_b$. This coinductive definition unfolds into an infinite tower of higher and higher witnesses to equivalence. If we’re in a 2-category, then since all $n$-morphisms are identities for $n>2$, the tower collapses and a 2-morphism is an equivalence if and only if it’s an isomorphism in the hom-1-categories, just as you describe. The situation is entirely analogous for larger $n$.
As a remark, a 2-functor does not involve just any isomorphisms showing it respects composition, but isomorphisms respecting coherence conditions. These coherence conditions are both very important and become unmanageable to write down explicitly beyond $n=3$.