Degree of a covering map is the number of sheets

56 Views Asked by At

Hatcher asks in Exercise 9 of Chapter 3.3 to show that the degree of a $p$-sheeted covering map $f : M \to N$ between closed connected oriented $n$-manifolds is $\pm p$.

Here, the $\textit{degree}$ of a continuous map between closed connected oriented $n$-manifolds is defined to be an integer $d$ such that $f_*[M] = d[N]$, where $[M]$, $[N]$ denote the fundamental classes of $M$, $N$ respectively. (The sign of $d$ depends on the choice of fundamental classes.)

I have managed to solve Exercise 8 of the same chapter, which asks one to show that if $f$ is a continuous map (not necessarily a covering map), and there exists a ball $B\subseteq N$ such that $f^{-1}(B)$ is the disjoint union of balls $B_i$, then the degree of $f$ is the sum $\Sigma_i \epsilon_i$ where $\epsilon_i = \pm 1$ depending on whether $f : B_i \to B$ preserves or reverses local orientations.

Drawing from this problem, I would like to show the following.

Let $B$ be an evenly covered neighborhood of $y$, and let $B_i$ be the sheets above $B$. Let $x_i$ be the preimage of $y$ in $B_i$.

If $f$ preserves (reverses) local orientation at one of the $x_i$'s, then it preserves (reverses) local orientation at every $x_j$.

So my question is, how does one go about proving this claim?