# Covering Surfaces

Compact orientable surfaces are classified (up to homeomorphism) by its genus. The genus is just a natural number. Thus we can enumerate them: The genus zero surface is the sphere $S^2$, the genus one surface is the torus $T^2$, the genus two surface can be constructed by taking two toruses, cutting a little disc out of each torus, and then gluing the two boundaries that we created when cutting the holes. In this way we can construct the genus $n$ compact orientable surface for each $n$. The operation of cutting a little disc out of two (connected) surfaces and gluing them by the boundary is called the connected sum. For two surfaces $A$ and $B$ their connected sum is usually denoted by $A \# B$. For $n \geq 0$ we denote the genus $n$ surface by $n T^2 = T^2 \# \ldots \# T^2$. When $n$ is zero we define the connected sum of zero surfaces to be the sphere $S^2$.

Problem 1: Prove that $n T^2$ covers $2 T^2$ for any $n>1$.

The classification of compact non-orientable surfaces is similar. If $P^2$ is the projective plane then a compact non-orientable surface is homeomorphic to $n P^2$ for some positive natural number $n$.

Problem 2: Prove that $(n-1)T^2$ covers $n P^2$ for any positive natural $n$.

Problem 3: Notice that $T^2 \# P^2$ is a compact surface. Describe it.

Problem 4: Prove that $n P^2$ covers $3 P^2$ for any $n\geq 3$.

A nice corollary of this exercise is that, since composition of finite coverings is again a covering, the fundamental group of any compact surface is a subgroup of the fundamental group of $3 P^2$, except maybe for $T^2$ and $2 P^2$.