How to calculate rigorously the degree of the map $f: S^1 \to S^1, f(z) = z^n$. This was done in Hatcher's algebraic topology example 2.32. However I do not understand it at all. 
An alternative approach I know of is to show that the degree for homology matches the degree for the fundamental group which requires Hurwitz map. 
For Hatcher's proof, I know that the map is locally a homeomorphism because by inverse function theorem. Then I need to determine the sign of it. How to do this mathematically instead of by words?
Does the sign have anything to do with whether the map is orientation-preserving or not?
 A: If you know that $H^1_{\mathrm{dR}}(S^1)\cong H^1(S^1;\mathbb{R})$, then using the generator $d\theta$ of de Rham cohomology, for $f(\theta)=n\theta$ (on $S^1$ parameterized with $\theta$ modulo $2\pi$) we have $df=n\, d\theta$, hence the induced map $$f^*:H^1_{\mathrm{dR}}(S^1)\to H^1_{\mathrm{dR}}(S^1)$$ is multiplication by $n$.  The de Rham theorem is that there is a natural isomorphism to singular cohomology, so paired with the naturality of the universal coefficient theorem, there is the following commutative diagram:$\require{AMScd}$
\begin{CD}
H^1_{\mathrm{dR}}(S^1) @>\cong>> H^1(S^1;\mathbb{R}) @>\cong>> \hom(H_1(S^1),\mathbb{R})\\
@VVf^*V @VVf^*V @VV(f_*)^*V\\
H^1_{\mathrm{dR}}(S^1) @>\cong>> H^1(S^1;\mathbb{R}) @>\cong>> \hom(H_1(S^1),\mathbb{R})\\
\end{CD}
The UCT isomorphisms come from the fact that the $\operatorname{Ext}^1$ groups are trivial. We know that $H_1(S^1)=\mathbb{Z}$ so $\hom(H_1(S^1),\mathbb{R})=\mathbb{R}$ and $(f_*)^*$ is multiplication by $n$.  This implies $f_*$ itself is multiplication by $n$, and therefore $\deg f=n$.
The Hurwitz map works, too.  The horizontal maps in the following are surjections:
\begin{CD}
\pi_1(S^1) @>>> H_1(S^1) \\
@VVf_*V @VVf_*V \\
\pi_1(S^1) @>>> H_1(S^1)
\end{CD}
Since $\pi_1(S^1)=\mathbb{Z}$ is already abelian, they are of course isomorphisms.  The first $f_*$ is the induced map of the covering map $f:S^1\to S^1$ in the case $n\geq 1$.  From covering space theory, we can see the image is $n\mathbb{Z}$ with $1\mapsto n$, hence $\deg f=n$.  For $n=0$, this is a constant map and $\deg f=0$.  For negative $n$, by composition with $z\mapsto z^{-1}$ (which is degree-$(-1)$, mentioned in Hatcher as property (e) of that chapter), we get a covering map, and by property (d) we get $\deg f=(-1)(-n)=n$.  There are other ways of being careful here.
The Hatcher approach, Proposition 2.30, is to let $x_1,\dots,x_m$ be the preimage of a point through $z\mapsto z^n$, with $m=\lvert n\rvert$ (and assume $n\neq 0$ since constant functions have already been handled).  Concretely, the preimage of $1$ is the $m$th roots of unity, $1,e^{2\pi i/m}, e^{2\pi i\cdot 2/m}, \dots, e^{2\pi i(m-1)/m}$.  Using $\theta$ coordinates, each has a an open neighborhood $U_k=(\pi (2k-1)/m, \pi (2k+1)/m)$ for $k=1,2,\dots,m$, and these are disjoint.  The images of these through $z\mapsto z^n$ map such a neighborhood onto $(-\pi,\pi)$, possibly reversing orientation if $n$ is negative.  The group $H_1(U_k,U_k-e^{2\pi ik/m})$ is $\mathbb{Z}$, generated by a singular $1$-simplex that crosses from one component of punctured $U_k$ to the other.  The induced map
$$f_*:H_1(U_k,U_k-e^{2\pi ik/m})\to H_1((-\pi,\pi),(-\pi,\pi)-0)$$
either sends this generator to a generator in the same direction, or in the opposite direction, depending on the sign of $n$ (and if you want to be careful here, consider the naturality of the long exact sequence of both pairs).  Hence the local degree is $\deg f|e^{2\pi ik/m}=\operatorname{sgn} n$.  Applying the proposition,
$$\deg f=\sum_{k=1}^m \deg|e^{2\pi ik/m} = \sum_{k=1}^m\operatorname{sgn} n=n.$$
A: It is obvious that the degree of the identity map is 1. Then use the fact that the map $x\rightarrow x^n$ is homotopic to the identity map concatenated with itself n times. Lemma 4.60 in Hatcher says that for any homology theory, in particular singular homology, the induced map distributes over this sum, so the degree of $x \rightarrow x^n$ is the integer corresponding to the map $1_\mathbb{Z} + \dots +1_\mathbb{Z}:\mathbb{Z} \rightarrow \mathbb{Z}$ which is n. So the degree is n.
