Upper half-plane $\overline{\mathbb{H}}$ with two boundary punctures Consider $\overline{\mathbb H}$ with two puncture $P_1$ and $P_2$ on the real line, with coordinates $z = x_1$ and $z = x_2$, respectively. Consider another copy of $\overline{\mathbb H}$ with two punctures $P_1$ and $P_2$ on the real line, with coordinates $z = x_1'$ and $z = x_2'$, respectively. Are these two surfaces the same Riemann surface?
Idea. I suspect they are, and it would suffice to exhibit the conformal map that takes the punctures into each other while preserving $\overline{\mathbb H}$. But I am not sure how to do this, could anybody help?

It's enough to consider the case $x_1^{\prime}=0, x_2^{\prime}=\infty$. If $x_1>x_2\in \mathbb{R}$, the map
$$ z\mapsto \frac{z-x_1}{z-x_2} $$
sends $x_1$ to $0$ and $x_2$ to $\infty$, and is conformal because the determinant of the corresponding matrix is $x_1-x_2>0$. And since
$$ \Im\frac{z-x_1}{z-x_2}=\frac{(x_1-x_2)\Im z}{|z-x_2|^2}$$
it follows that the map sends $\mathbb{H}$ into itself and $\mathbb{R}\cup\{\infty\}$ into itself.
If for instance $x_2=\infty$, then the map $z\mapsto z-x_1$ sends $x_1$ to $0$ and $x_2$ to $\infty$, and clearly preserves $\mathbb{H}$ and $\mathbb{R}\cup\{\infty\}$.
