Describe the homomorphism induced on the fundamental group

Question

Consider the unit circle $S^1$ and consider the continuous application f: $S^1 \to S^1$, f(z)=$\bar{z}^mz^n$, where $\bar{z}$ indicates the conjugate of z and $n,m \in \mathbf{Z}$. The exercise asks me to describe the homomorphism induced on the fundamental group $f_\ast: \pi_1(S^1,1) \to \pi_1(S^1,1) $. My question is what means describe the homomorphism induced on the fundamental group? Does anyone know how I should proceed?

Answer 1

The elements of $\pi_1(S^1, 1)$ are closed paths starting (and ending) at $1$. Let $p \in \pi_1(S^1, 1)$, so that $p: [0,1] \rightarrow S^1$ with $p(0) = p(1) = 1$.

You have several maps $f$, depending on the choices of $m$ and $n$. As written, $f$ is a map from $\mathbb{C}$ to $\mathbb{C}$, but we restrict it to $|z| = 1$ so that the restriction is a map from the complex unit circle to itself.

To allow $f$ to compose with $p$, we must identify the $S^1$ codomain of $p$ with the unit complex circle. $S^1 \cong \{z \in \mathbb{C}: |z| = 1\}$ Having done so, we are ready to ask, what is $f \circ p$?

Note that $f \circ p : [0,1] \rightarrow S^1$, so the $f$-images of each loop in $\pi_1(S^1,1)$ are loops in $S^1$. Since $f \circ p(0) = f(1) = 1$ and $f \circ p(1) = f(1) = 1$ for any $m,n$, the $f$-images of each loop in $\pi_1(S^1,1)$ are loops in $\pi_1(S^1,1)$. So from $f$, a map of points to points, we get another map, $f_*:\pi_1(S^1,1) \rightarrow \pi_1(S^1,1)$, a group homomorphism. This map is the homomorphism induced on the fundamental group.

$$\require{AMScd} \begin{CD} [0,1] @>{p}>> S^1 @. @. \pi_1(S^1,1)\\ @. @V{f}VV @. @V{f_*}VV \\ @. S^1 @. @. \pi_1(S^1,1); \end{CD}$$

Each equivalence class of loops in $\pi_1(S^1,1)$ has a representative, $p_k: x \mapsto \mathrm{e}^{2\pi \mathrm{i} k x}$ for $k \in \mathbb{Z}$. Each of these is a suitable $p$ in the discussion above. The set of these maps $\{p_k : k \in \mathbb{Z}\}$ is a complete set of representatives of the equivalence classes in $\pi_1(S^1,1)$. Now, for each choice of $(m,n)$ study the set $\{f \circ p_k : k \in \mathbb{Z}\}$.

The behaviour is different for different choices of $m$ and $n$. For $m = 0$, $n = 1$, this $f$ is the identity, so $f_*(p_k) = p_k$ and the induced map is an isomorphism. For $m = n = 0$, $f$ is the constant map $z \mapsto 1$, so this $f$ induces the trivial homomorphism. Other choices of $m$ and $n$ will show maps to various subgroups of $\pi(S^1,1)$. It is possible to write down (in a short way) which subgroup is the image of the induced homomorphism as a function of $m$ and $n$ and you are asked to find this short specification. Note that $\pi_1(S^1,1) = \langle p_1 \mid \rangle$ (that is, is infinite cyclic with generator $p_1$), so as soon as you can specify $f_*(p_1)$ as a function of $m$ and $n$, you have completely specified the homomorphism $f_*$.