Let:
- $\mathcal{K}$ be a (weak) 2-category
- $A$, $B$ and $C$ be objects of $\mathcal{K}$
- $f_1, f_2 : A \to B$ and $g : B \to C$ be 1-cells
- $\alpha, \beta : f_1 \to f_2$ be 2-cells
Assuming that $\textit{id}_g * \alpha = \textit{id}_g * \beta$ (where $*$ is the horizontal composition), is it true that $\alpha = \beta$? If yes, how to prove it? Can it be generalized, e.g., by replacing $\textit{id}_g$ with an arbitrary 2-cell? If no, can it be specialized, e.g., by replacing $g$ with $\textit{id}_B$?
No. For instance if $\mathcal K=\mathrm{Cat}$ and $C=*$ is the terminal category with $g:B\to *$ the unique functor then $id_g *\alpha= id_g*\beta$ for any $A,B,f_1,f_2$.
If $g=\mathrm{id}_B$, then the claim holds. Indeed, then $\mathrm{id}_g *\alpha:g\circ f_1\to g\circ f_2$ is equal to $\lambda_{f_2}\circ \alpha\circ \lambda_{f_1}^{-1}$, where $\lambda:\mathrm{id}_{\mathcal K(A,B)}\to g\circ (-)$ is the left unitor. Similarly $\mathrm{id}_g *\beta=\lambda_{f_2}\circ \beta \circ \lambda_{f_1}^{-1}$. Thus $\mathrm{id}_g *\alpha=\mathrm{id}_g*\beta$ if and only if $\alpha=\beta$.