Interchange in 2-category. Consider a (strict) 2-category $\textbf{A}$, with vertical composition denoted $\circ$, and horizontal composition denoted $\ast$. Let $f,g: A \to B$ be $1$-cells, and $\alpha: g \to f$ a $2$-cell. If $f$ and $g$ are invertible, does it follow that 
$g^{-1} \ast \alpha \ast f^{-1} = f^{-1} \ast \alpha \ast g^{-1}$?
I can prove this in the case that $\alpha$ is invertible since in this case both $g^{-1} \ast \alpha \ast f^{-1}$ and $f^{-1} \ast \alpha \ast g^{-1}$ are inverse to $\alpha^{-1}$ and so must be the same. I havn't been able to work out whether this is true in general.
 A: I believe it is not true in general. My counterexample is in the $2$-category $\mathbf{CAT}$ of categories, functors, and natural transformations.
First, we define the category $A = B$. Its objects are integers $\mathbb Z$. Morphisms are given as:


*

*identities

*for all $n \in \mathbb Z$, there are two morphisms $t_n, b_n : n \to n+1$

*for all $n \in \mathbb Z$ and $i \in \mathbb N$, there is one ("trivial") morphism $z_{n,i} : n \to n + i + 2$


There is only one way to define the composition of morphisms. That is, the category $A$ looks like this:
$$
\cdots \underset{b_{-2}}{\overset{t_{-2}}{\rightrightarrows}}
-1 \underset{b_{-1}}{\overset{t_{-1}}{\rightrightarrows}}
0 \underset{b_0}{\overset{t_0}{\rightrightarrows}}
1 \underset{b_1}{\overset{t_1}{\rightrightarrows}}
\cdots
$$
where any composition of length $\geq 2$ gives the appropriate morphism $z$.
We define the functor $g$ to be the identity. We define $f$ as $fn = n-1$ on objects, while $ft_n = t_{n-1}$, $fb_n = b_{n-1}$, etc. on morphisms. That is, $f$ moves everything one notch "leftwards". It has an obvious inverse $f^{-1}n = n + 1$.
We define a natural transformation $\alpha : \mathrm{Id} \to f$ as
$$\alpha_{2k} = t_{2k} \qquad \alpha_{2k+1} = b_{2k+1} \qquad \text{for all } k \in \mathbb Z$$
That is, we select $t$ for even objects and $b$ for odd objects. This is trivially a natural transformation, as the condition for natural transformations involves compositions of length $2$, which are always equal in the category $A$.
Now, we need to show that in general $\alpha * f^{-1} \neq f^{-1} * \alpha$. For example:
$$(\alpha * f^{-1})_0 = \alpha_{f^{-1}0}= \alpha_1 = b_1 \neq t_1 = f^{-1}t_0 = f^{-1}\alpha_0 = (f^{-1} * \alpha)_0$$
