Induced group homormophism $f_{*} : \pi_1( S^1,p) \rightarrow \pi_1( S^1,f(p))$ due to $z \mapsto z^2$.

Question

Let $f: S^1 \rightarrow S^1$ s.t. $z \mapsto z^2$. This map induces a group homormophism $f_{*} : \pi_1( S^1,p) \rightarrow \pi_1( S^1,f(p))$ . How can we explicitly write this map? I know that the mapping is that a loop gets mapped to 2 loops. But I am not sure how to write it?

Answer 1

Let's consider $\mathbb{S}^1 = \{ z \in \mathbb{C} \mid \lvert z \rvert = 1 \}$. If you'd rather think of the circle living in $\mathbb{R}^2$, just use the standard map $z \mapsto (\Re(z), \Im(z))$, under which $re^{i\theta} \mapsto (r \cos \theta, r \sin \theta)$.

Fix the isomorphism: $\pi_1(\mathbb{S}^1, 1) \to \mathbb{Z}$ with $[\alpha] \mapsto 1$, where $\alpha: \bigl( [0, 1], \{0, 1\} \bigr) \to (\mathbb{S}^1, 1)$ is given by the parametrization $t \mapsto e^{2\pi i t}$. Notice that the endpoints of the interval map to the basepoint $1 \in \mathbb{S}^1 \subset \mathbb{C}$, as required.

Now, the map $f: (\mathbb{S}^1, 1) \to (\mathbb{S}^1, 1)$, $z \mapsto z^2$ produces the parametrization $f \circ \alpha: \bigl( [0, 1], \{0, 1\} \bigr) \to (\mathbb{S}^1, 1)$ given by the formula $t \mapsto \smash[t]{\bigl( e^{2\pi i t} \bigr)^2} = e^{4\pi i t}$. This is a trajectory that moves around the circle at double speed, completing two complete loops as $t$ varies over $[0, 1]$.

What's interesting to observe is that this double-speed loop is precisely the same loop as the one produced by adding $1 + 1 = 2$ in the fundamental group. All that's needed is that they represent the same homotopy class, but in this case they already match as parametrized loops.

Note: generally the fundamental group of a pointed space is non-abelian, so we use multiplicative notation. But in this case, it is abelian, and it's more standard to use the additive notation of the integers. If you insist on multiplicative notation, you could consider a generator $g$ of the group $G$ under the isomorphism $1 \mapsto g^1$, where $m + n \mapsto g^m g^n$ more generally.

Back to the double-speed loop. The way that two classes are combined ("added" in this abelian fundamental group) is by running double speed over the parametrizing intervals for each of the loops, then concatenating the two loops. Recall, if $\beta, \gamma: \bigl( [0, 1], \{0, 1\} \bigr) \to (\mathbb{S}^1, 1)$ are any two loops representing classes in the fundamental group, then $\beta + \gamma: \bigl( [0, 1], \{0, 1\} \bigr) \to (\mathbb{S}^1, 1)$ is defined piecewise by $$ (\beta + \gamma)(t) = \begin{cases} \beta(2t), & 0 \leq t \leq \tfrac{1}{2} \\ \gamma(2t-1), & \tfrac{1}{2} \leq t \leq 1 \end{cases} $$ and we define $[\beta] + [\gamma] = [\beta + \gamma] \in \pi_1(\mathbb{S}^1, 1)$.

But with $\beta = \gamma = \alpha$ as defined above, the parametrization $\alpha + \alpha$ is given by $$ \alpha(2t) = e^{2 \pi (2t)} = e^{4 \pi t} \qquad \bigl( 0 \leq t \leq \tfrac{1}{2} \bigr) $$ and $$ \alpha(2t-1) = e^{2 \pi (2t-1)} = e^{4 \pi t - 2\pi} = e^{4 \pi t} \qquad \bigl( 0 \leq t \leq \tfrac{1}{2} \bigr). $$ Since these are the same formula, you can glue them together without breaking the domain into two pieces. This is the parametrization of the loop representing $2 \in \mathbb{Z}$.

Therefore, $[f \circ \alpha] = [\alpha] + [\alpha] = 2[\alpha] \mapsto 2 \in \mathbb{Z}$.

Addendum: this generalizes to show that for $d \in \mathbb{Z}$, $f_d(z) = z^d$ induces the map $n \mapsto dn \in \mathbb{Z}$ on the fundamental group of $\mathbb{S}^1$.