Round trips, range bounds and continuity for the half-line reparameterisation pair

Let $n$ be a natural number, and let $\psi_n$ and $\varphi_n$ be the published reparameterisation of the half-line $\{x<n+1\}$ onto $\mathbb{R}$ and its explicit inverse.

This record collects the four algebraic and analytic facts about the pair that a later complex lift needs, so that they need not be re-derived.

Round trips. $\varphi_n(\psi_n(x))=x$ for every $x<n+1$, and $\psi_n(\varphi_n(y))=y$ for every $y\in\mathbb{R}$. Writing $u=x-n\in(0,1)$ on the second branch of $\psi_n$, that branch sends $u\mapsto u/(1-u)$; writing $v=y-n>0$ on the second branch of $\varphi_n$, that branch sends $v\mapsto v/(1+v)$. These are literal algebraic inverses, so bijectivity follows with no separate existence-of-a-preimage argument.

Range bounds. Both maps land strictly below the right endpoint: $\psi_n(x)<n+1$ whenever $x<n+1$, and $\varphi_n(y)<n+1$ for every $y$. The first is the statement that the reparameterisation really is defined on its whole domain, and the second is what makes $\varphi_n$ a map into the half-line rather than merely into $\mathbb{R}$.

Continuity of $\psi_n$ is only on the half-line. The function $\psi_n$ is not continuous on all of $\mathbb{R}$: its second branch has a pole at $x=n+1$, precisely the boundary point where $\psi_n$ ceases to be defined. The correct statement, and the one proved here, is that $\psi_n$ is continuous on $\{x:x<n+1\}$.

Continuity of $\varphi_n$ is global. $\varphi_n$ is continuous on all of $\mathbb{R}$, even though its second branch appears to have a pole at $y=n-1$. That pole is harmless because $n-1\le n$, so it lies inside the region where the first branch is selected and is never evaluated. Formally, on the closed ray $\{y\ge n\}$ the denominator $1+y-n$ satisfies $1+y-n\ge 1>0$, so the second branch is a continuous rational function there, and pasting gives continuity everywhere.

The two continuity statements use the pasting lemma for piecewise-continuous functions, with the branches continuous on the closed sets $\{x\le n\}$ and $\{x\ge n\}$; both branches agree at the unique overlap point $x=n$, where each gives $n$.