Is bijection important in Catalan monotonic path counting? Second proof computes Catalan number by removing bad paths (those which touch the forbidden diagonal) from all monotonic paths to compute the number of permitted paths. The proof seems to say that all bad paths are deflected into a different target square, which makes them easy to count. It however insists that there is a bijection between flipped paths and their preimages. Is it important the we can recover the original bad path or it would suffice that we can just pile up all bad paths together?

The bijcetion is important to ensure that the count is identical. Every path to $(n-1,n+1)$ can be flipped on its first touch on $y=x+1$ to make a Catalan-forbidden path to $(n,n)$ and vice-versa. So the count of paths to $(n-1,n+1)$ is ${2n\choose n+1}$, selecting where the upward steps occur in the sequence, and this allows the calculation shown in the proof.