# Universal covering map from $\mathcal{H}$ to $\mathbb{C}\setminus \mathbb{Z}\oplus i\mathbb{Z}$ (the countably punctured complex plane)

It's a consequence of the uniformization theorem for simply connected Riemann surfaces that the universal cover of $\mathbb{C}\setminus(\mathbb{Z}\oplus i\mathbb{Z})$ ($\mathbb{C}$ punctured at all the integral lattice points) is the upper half plane $\mathcal{H}$.

How should I think about this map? How does the map behave near the missing lattice points?

A related question is this: $\mathcal{H}$ is also the universal cover for a punctured torus, whose fundamental group is $F_2$, the free group on two generators. By comparing the punctured torus to the wedge of two circles, I feel like the universal cover for the punctured torus, ie $\mathcal{H}$, ought to be deformation-retractable to an infinite 4-regular tree. Ie, the infinite 4-regular tree ought to be able to be embedded in $\mathcal{H}$ such that the vertices of the tree all lie on the boundary of $\mathcal{H}$. What does this tree look like in $\mathcal{H}$?

• I actually mean $\mathbb{C}$ minus all the lattice points. Sep 5, 2013 at 20:24
• Ah, sorry, I should have read more carefully. Sep 5, 2013 at 20:58
• I did not understand the second question. The universal cover of the punctured torus is the open unit disc. Sep 5, 2013 at 22:24
• Alexandre Eremenko: The Riemann mapping theorem shows that the open unit disc is conformally equivalent to the upper half plane. Sep 5, 2013 at 22:59
• The infinite 4-regular tree embeds nicely in the unit disk as the dual graph to the tesselation by regular ideal quarilaterals described by Prof. Eremenko below. This is because the puntured torus is covered by your space $\mathbb{C} \setminus (\mathbb{Z} + i \mathbb{Z})$. Of course, you can lift any deformation-retraction of the punctured-torus onto the wedge of two circles to a deformation-retraction of the unit disk onto the infinite tree. Sep 7, 2013 at 0:17