Count paths of length $2n$ that don't hit the diagonal after starting 
Consider a path on the coordinate plane that can only move up or right. Prove that the number of such paths of length $2n$ that don't ever touch (previously read: "cross") the diagonal is $\binom{2n}n$ by a bijection.

Attempt: Tried to implement a Catalan-style proof by considering the first point it crosses the diagonal. Reflecting across the diagonal doesn't do anything, since the endpoint of a path is not necessarily $(n,n)$. I have no other ideas to attempt... Help! :)
 A: There is a natural bijection between these paths and paths of length $2n$ that start at $\langle 0,0\rangle$, move in up-steps from $\langle x,y\rangle$ to $\langle x+1,y+1\rangle$ and down-steps from $\langle x,y\rangle$ to $\langle x+1,y-1\rangle$, and never drop below the $x$-axis. I find it easier to think in terms of these ‘mountain’ paths, so I’ll work with them.
I’ll describe the bijection that I found a few years ago in another context, but I’ll warn you that it’s pretty ugly. I’ll describe one direction and for now leave it to you to verify that what I’ve described really is a bijection.
Clearly there are $\binom{2n}n$ such paths of length $2n$ that end at $\langle 2n,0\rangle$, so we want a bijection between those paths and the paths of length $2n$ that do not hit the $x$-axis after leaving $\langle 0,0\rangle$. Let $P$ be a path of length $2n$ that ends at $\langle 2n,0\rangle$. If $P$ never drops below the $x$-axis, let $P'=P$. Otherwise, let $-m<0$ be the minimum $y$-coordinate of any point on $P$, and let $\langle h,-m\rangle$ be the first point on $P$ at height $-m$. Construct a new path $P'$ as follows.
Let the $k$-th step of $P$ be to $\langle k,y_k\rangle$ and the $k$-th step of $P'$ be to $\langle k,y_k'\rangle$. The first $n-h$ steps of $P'$ mimic the last $2n-h$ steps of $P$: for $k=1,\ldots,2n-h$, $y_k'=y_{h+k}$. Pictorially speaking, we’ve taken the graph of $P$ to the right of $x=h$ and moved it $h$ units to the left and $m$ units up, so that it starts at the origin. Clearly this segment of $P'$ ends at $\langle 2n-h,m\rangle$. Complete $P'$ by reflecting the part of $P$ to the left of $x=h$ in the line $y=-m$, so that it starts at $\langle 0,-2m\rangle$ and ends at $\langle h,-m\rangle$, raising it $3m$ units to start at $\langle 0,m\rangle$ and end at $\langle h,2m\rangle$, and shifting it $2n-h$ units to the right to start at $\langle 2n-h,m\rangle$ and end at $\langle 2n,2m\rangle$. In other words, for $k=1,\ldots,h$, $y_{2n-h+k}'=m-y_k$. If $y_k'>0$ for $k=1,\ldots,2n$, let $\widehat P=P'$; $\widehat P$ lies entirely above the $x$-axis after it leaves the origin.
Otherwise, let $j$ be minimal such that $y_j'=0$; clearly $y_{j-1}'=1$, since the $j$-th step must have been a down-step. Change that to an up-step and leave the remaining sequence of up- and down-steps as is; this simply replaces $y_k'$ by $y_k'+2$ for $k=j,\ldots,2n$ and results in a path $P''$ from $\langle 0,0\rangle$ to $\langle 2n,2m+2\rangle$ that lies strictly above the $x$-axis after leaving the origin. Finally, let $\widehat P$ be the reflection of this path in the $x$-axis; $\widehat P$ lies strictly below the $x$-axis after leaving the origin and ends at $\langle 2n,-2m-2\rangle$.
You shouldn’t have too much trouble verifying that the mapping $P\mapsto\widehat P$ is injective, but you may have to work a bit to show that every path that never hits the $x$-axis after leaving the origin is $\widehat P$ for some $P$ that ends at $\langle 2n,n\rangle$.
