Compute the induced homomorphisms $g_{*}, h_{*}$ of the infinite cyclic group $\pi_1 (S^1,b_0)$ into itself. (Here we represent $S^1$ as the set of complex numbers $z$ of absolute value 1).
When we have a continous function $h: X \to Y$ such that $h(x_0)=y_0$, for some $x_0 \in X, y_0 \in Y$. ($h:(X,x_0) \to (Y,y_0)$). Then $h$ induces a homomorphism $h_{*}: \pi_1 (X,x_0) \to \pi_1(Y,y_0)$ by the equation $h_{*}([f]) = [h \circ f].$
How can I show that $h$ and $g$ are continous? I also must find a generator of $\pi_1 (S^1,b_0)$, right ? Once I get $[f]$ a generator of $\pi_1 (S^1,b_0)$ I can compute $g_{*}([f]) = [g \circ f]$. I don't know how to finish it.
Ps: I took it from Munkres, and there is a hint: $(\cos \theta + i \sin \theta)^n = \cos n \theta + i \sin n \theta$. But I don't know what to do with this either.
These maps are continuous because they are complex differentiable (the derivative of $z^n$ being $nz^{n-1}$) and this implies continuity.
Now $\pi_1(S^1,1)$ is infinite cyclic, generated by the identity (which is a loop $\mathit{id}:S^1 \to S^1$). The induced map $g_*:\pi_1(S^1,1) \to \pi_1(S^1,1)$ is entirely determined by where it sends the positive generator, so we need just compute the winding number of $g_*([\mathit{id}]) = [g \circ \mathit{id}] = [g]$. Let $p:\mathbb{R} \to S^1$ denote the universal cover. It will be more convenient to treat $g$ as a map $[0,1] \to S^1$ with $$g(\theta) = (\cos2\pi\theta + i\sin2\pi\theta)^n.$$ This $g$ lifts to a path $f:[0,1] \to \mathbb{R}$ defined by $f(\theta) = n\theta$, because $$(p\circ f)(\theta)= p(n\theta) = \cos(2\pi n\theta)+ i\sin(2\pi n\theta) = (\cos 2\pi\theta +i\sin2\pi\theta)^n = g(\theta).$$
This shows that $[g]$ corresponds to $n$ via the isomorphism $\pi_1(S^1,1) \cong \mathbb{Z}$. Therefore, the induced homomorphism $$\mathbb{Z} \cong \pi_1(S^1,1) \xrightarrow{g_*} \pi_1(S^1,1) \cong \mathbb{Z}$$ is just multiplication by $n$. The argument for $h$ is identical.