The double of a disk is a sphere

For $x\mathrm{\ne }0$, write $x=ru$ with $0<r\le 1$ and define $F([x,\pm ])=(\sin (\pi r\mathrm{/}2)u,\pm \cos (\pi r\mathrm{/}2))$, while $F([0,\pm ])=(0,\dots ,0,\pm 1)$. At $r=1$ both formulas give $(u,0)$, so [L1] makes $F$ well defined on the double.

On either disk the first component is $\frac{\sin (\pi \mathrm{\mid }x\mathrm{\mid }\mathrm{/}2)}{\mathrm{\mid }x\mathrm{\mid }}x$, with its value at $x=0$ defined by the smooth even power series, and the last component is $\pm \cos (\pi \mathrm{\mid }x\mathrm{\mid }\mathrm{/}2)$, also a smooth function of $\mathrm{\mid }x{\mathrm{\mid }}^2$. Thus the restrictions are smooth at the two poles.

In the signed seam coordinate $s$, where $x=(1-\mathrm{\mid }s\mathrm{\mid })u$ and the sign of $s$ records the label, [L2] and [L3] rewrite the map as $F(u,s)=(\cos (\pi s\mathrm{/}2)u,\sin (\pi s\mathrm{/}2))$. This is a smooth local diffeomorphism across $s=0$. The formula is bijective because the last coordinate selects the hemisphere and its absolute value determines $r$; its inverse is smooth in the pole charts and in these seam charts. Hence $F$ is a diffeomorphism from the labelled double to $S^n$.