Local degree of $z\mapsto z^n$ I'm currently computing the local degree of $f:S^1\to S^1$ by $z\mapsto z^n$ with $n>0$. To do this, I tried to compute the local degree $\deg f|x_i$ where $x_i\in f^{-1}(1)$. Since $f$ is a local homeomorphism, $\deg f|x_i =\pm 1$ for each $i$. But from this, how can I show $\deg f|x_i = 1$?
In Hatcher Example 2.32, it states that by stretching each open sets by $k$ factor then by proper rotation, $f$ is homotopic to identity map. Why this is true? In the domain, one point in $S^1$ is deleted and so is the range. It's identity but not a map from $S^1\to S^1$.
 A: You have to consider all $n$-th roots of unity $\zeta_k = e^{2k\pi i/n}$ with $k = 0,\dots,n-1$.
$V = \{e^{it} \mid -1 < t < 1\}$ is an open neigborhood of $1$. Then $U_k = \{ e^{it} \mid (2k\pi - 1)/n < t < (2k\pi + 1)/n\}$ is an open neigborhood of $\zeta_k$ which is mapped by $f$ homeomorphically onto $V$. What Hatcher really wants to say is that $f : U_k \to V$ is the restriction of a homeomorphism $h_k : S^1 \to S^1$ which is homotopic to the identity. Let us prove it.
We begin with $k = 0$. Define
$$\phi_0 : [-\pi ,\pi] \to [-\pi,\pi], \phi_0(t) = \begin{cases} nt &  -1/n \leq t \leq 1/n \\ \phantom{-}1 + \frac{\pi -1}{\pi - 1/n}(t - 1/n) & t \ge 1/n \\ -1 + \frac{\pi -1}{\pi - 1/n}(t + 1/n) & t \leq -1/n  \end{cases}$$
This is a homeomorphism which is homotopic to the identity rel. $\{-\pi, \pi \}$. Via the quotient map $p : [-\pi,\pi] \to S^1, p(t) = e^{it}$,  it induces a homeomorphism $h_0 : S^1 \to S^1$ which is homotopic to the identity. For $e^{it} \in U_0$ we have $h_0(e^{it}) = h_0(p(t)) = p(\phi_0(t)) = p(nt) = e^{int} = (e^{it})^n = f(e^{it})$, i.e. the restriction of $h_0$ to $U_0$ is $f : U_0 \to V$. This is Hatcher's stretching.
The map $\rho_j : S^1 \to S^1, \rho_j(z) = \zeta_j z$, is a rotation by the angle $2j\pi /n$. It is homotopic to the identity. Now define $h_k = h_0 \circ \rho_{n-k}$. For $t \in U_k$ we have $h_k(e^{it}) = h_0(e^{i(t+2(n-k)\pi /n)}) = h_0(e^{i(t - 2k\pi/n)}) = e^{i(nt - 2k\pi)} = e^{int} = (e^{it})^n = f(e^{it})$.
Now consider the following commutative diagram:
$\require{AMScd}$
\begin{CD}
(S^1,S^1 - \zeta_k) @<{\supset}<< (U_k,U_k- \zeta_k)  @>{f}>> (V,V - 1) @>{\subset}>> (S^1,S^1 - 1)\\
@V{h_k}VV  @V{h_k}VV @V{id}VV  @V{id}VV \\
(S^1,S^1 - 1) @<{\supset}<< (V,V - 1) @>{id}>> (V,V - 1) @>{\subset}>> (S^1,S^1 - 1) \end{CD}
Applying $H_1$ we get
$\require{AMScd}$
\begin{CD}
H_1(S^1) @>{\approx}>> H_1(S^1,S^1 - \zeta_k) @<{\approx}<< H_1(U_k,U_k- \zeta_k)  @>{f}>> H_1(V,V - 1) @>{\approx}>> H_1(S^1,S^1 - 1) @<{\approx}<< H_1(S^1)  \\
@V{(h_k)_*}VV  @V{(h_k)_*}VV  @V{(h_k)_*}VV @V{id}VV  @V{id}VV  @V{id}VV \\
H_1(S^1) @>{\approx}>> H_1(S^1,S^1 - 1) @<{\approx}<< H_1(V,V - 1) @>{id}>> H_1(V,V - 1) @>{\approx}>> H_1(S^1,S^1 - 1) @<{\approx}<< H_1(S^1) \end{CD}
But $(h_k)_* = id : H_1(S^1) \to H_1(S^1)$. This shows $\deg f \mid_{\zeta_k} = +1$.
