= Solution
A <Möbius transformation> is $M(z)=(az+b)/(cz+d)$ with $ad-bc\ne0$, extended to the <Riemann sphere> at its <pole> and at infinity. Wherever its finite coordinate is defined,
$$
M'(z)=\frac{ad-bc}{(cz+d)^2}\ne0.
$$
At a <pole>, the ordinary complex <derivative> is undefined, rather than zero; using the target coordinate $1/M$ gives a nonzero <derivative> there as well. Thus the source's nowhere-zero assertion is interpreted in local sphere coordinates, or on the finite domain with the <pole> excluded.
Composition corresponds to multiplication of the invertible <matrices> $\begin{pmatrix}a&b\\c&d\end{pmatrix}$, with scalar multiples giving the same map. The identity <matrix> gives the identity, and the inverse <matrix> gives an inverse Möbius map. This proves the group property on the <Riemann sphere>.
The two boundary circles meet at $\pm i$. Put
$$
\boxed{w=M(z)=\frac{z-i}{z+i},\qquad z=i\frac{1+w}{1-w}.}
$$
For $w=u+iv$, direct substitution gives
$$
|z+1|^2-2=\frac{4(u-v)}{|1-w|^2},\qquad
|z-1|^2-2=\frac{4(u+v)}{|1-w|^2}.
$$
This gives a <lens-to-half-plane map through the intersection points>. Both lens inequalities hold exactly when $u<-|v|$, the open sector with continuous argument between $3\pi/4$ and $5\pi/4$. This proves bijectivity, since the inverse is defined throughout that sector and the excluded endpoint $-i$ is not in the lens.
Rotate the sector through $-3\pi/4$ and square it, obtaining a bijection onto the <upper half-plane>. Therefore a <conformal map> is
$$
\boxed{F(z)=i\left(\frac{z-i}{z+i}\right)^2.}
$$
Squaring is injective on this sector because it contains no pair of opposite points. Its <derivative> and the Möbius <derivative> are nonzero in the interior, so the bijection is conformal.
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