Consider $y=-x^2+1$. The distinct real numbers $-1$ and $1$ have the same value $0$: it is not injective on ℝ, but it is strictly decreasing on $[0,\mathrm{+\infty })$ and strictly increasing on $(\mathrm{-\infty },0]$, on each side of its vertex

From $y=-x^2+1$, we obtain $x^2=1-y$, possible only if $y\le 1$

On $[0,\mathrm{+\infty })$, $x$ is nonnegative, so $x=\sqrt{1-y}$; on $(\mathrm{-\infty },0]$, $x$ is nonpositive, so $x=-\sqrt{1-y}$

▶Renaming the output variable $y$ as $x$, the inverse of the restriction to $[0,\mathrm{+\infty })$ is $\sqrt{1-x}$ and that of the restriction to $(\mathrm{-\infty },0]$ is $-\sqrt{1-x}$, defined for $x\le 1$