# Covering of orientable surface (Hatcher)

The following is an exercise from Hatcher, Algebraic Topology, that I'm struggling with (exercise 2.2.23):

Show that if the closed orientable surface $M_g$ of genus $g$ is a covering space of $M_h$, then $g=n(h-1) +1$ for some $n$, namely $n$ is the number of sheets in the covering.

My thoughts on this are the following:

I know that if $Y$ a finite CW-complex and $p: X \rightarrow Y$ a $k$-sheeted covering map, then $\chi(X) = k*\chi(Y)$, where $\chi(\cdot)$ is the Euler-characteristic of a space.

Furthermore, if the CW-complex $Y$ has $a_i$ $i$-cells ($i \geq 0$) then its Euler-characteristic is given by $\chi(Y) = \sum_{i \geq 0} (-1)^i a_i$.

The closed orientable surface of genus $g$ can be endowed with a CW-structure with one $0$-cell, $2g$ $1$-cells, one $2$-cell and no cells in any other dimension (see for example Homology of surface of genus $g$). Therefore we get $\chi(M_g)=-2g$.

Now, if we have an $n$-sheeted covering of $M_h$ by $M_g$, then we obtain by the above arguments that $-2g = n(-2h)$ from which it follows that $g=n h$.

Could someone please point out to me the mistake in the above argumentation.

Any help is much appreciated.

Your argument relating $g$ to $\chi$ is wrong. It's actually $\chi = 2 - 2g$. Check this for a torus: one zero cell, two one-cells, one two-cell: $\chi = 0$. But $g = 1$.
It looks as if you thought that the zero-cell and the 2-cell cancelled, but they should actually both be counted as $+1$.