Induced homomorphisms on fundamental group

Question

Define the map $f : S^{1} \times S^{1} \to S^{1}$ with $f(x,y) = xy$ and $g : S^{1} \times S^{1} \to S^{1} \times S^{1}$ with $g(x,y) = (xy,x)$ where $x, y \in \mathbb{C}$ on the unit circle $S^{1}$. How would one compute the induced homomorphisms $f_{*} : \pi_{1}(S^{1} \times S^{1}) \to \pi_{1}(S^{1})$ and $g_{*} : \pi_{1}(S^{1} \times S^{1}) \to \pi_{1}(S^{1} \times S^{1})$?

Answer 1

The obvious map $\pi_1(X) \times \pi_1(Y) \to \pi_1(X\times Y)$ is an isomorphism, and $T^2 = S^1\times S^1$ has $\pi_1(T^2) = \mathbb{Z}^2$, with generators corresponding to (for example) the maps $\lambda, m: S^1 \to S^1\times S^1$ given by $\lambda(\theta) = (\theta, 1)$ and $m(\theta) = (1, \theta)$. It's then easy to write down $f_*\lambda, f_*m, g_*\lambda, g_*m$ explicitly.

Another fact that might be useful here is that $\pi_1 (S^1)$ and $\pi_1(T^2)$ are abelian, so we can do this computation in homology rather than homotopy. Poincare duality shows that the intersection map $H_1(T^2) \otimes H_1(T^2) \to H^2(T^2) = \mathbb{Z}$ (note that $H_*(T^2)$ is torsion-free) is non-degenerate, so we can identify classes in $\pi_1(T^2) = H_1(T^2)$ by computing their intersection numbers with the two generators $\lambda, m$ of $T^2$. More directly, this is done by homotoping a given loop $\gamma$ so that it's transverse to $\lambda$ and $m$, then counting the points of intersection with respect to orientation. (In this particular case, there are obvious choices for the loops that make the orientation clear.)

Answer 2

Just take a look at what happens to the generators of the groups. The fundamental group of the torus is $\mathbb{Z}^2$. Lets call the generators $a$ and $b$. Clearly $f_*$ sends $a$ and $b$ to the same element (the generator $c$ of $\mathbb{Z} = \pi_1(S^1)$).

$$f_*: a \mapsto c, b \mapsto c$$

For the map $g$ the map $g_*$ sends $a \mapsto c_1 c_2$ and $b \mapsto c_1$. Where $c_1$ and $c_2$ are the two generators of $\mathbb{Z}^2$.