Infinitely many nonhomotopic retractions $S^1 \lor S^1 \to S^1$

Construct infinitely many nonhomotopic retractions $S^1 \lor S^1 \longrightarrow S^1$.

Solution

1. $\pi_1(S^1 \vee S^1) = \ZZ * \ZZ = \langle a, b \rangle$ (free on two generators), and $\pi_1(S^1) = \ZZ$.

standard computation.

2. A retraction $r: S^1 \vee S^1 \to S^1$ induces a homomorphism $r_*: \ZZ * \ZZ \to \ZZ$ that is a retraction of the inclusion $i_*: \ZZ \to \ZZ * \ZZ$ (sending the generator of $\ZZ$ to $a$).

$r \circ i = \id_{S^1}$, so $r_* \circ i_* = \id$.

3. Hence $r_*(a) = 1$ (the generator of $\ZZ$), while $r_*(b)$ can be any integer $n$.

$r_*(a) = 1$ is forced by the retraction condition; $r_*(b) = n$ is arbitrary.

4. For each $n \in \ZZ$, define $r_n: S^1 \vee S^1 \to S^1$ by $r_n(a) = a$ (identity on the first circle) and $r_n(b) = a^n$ (the $n$-fold power on the second circle).

this is a well-defined continuous map (it is the identity on the first $S^1$ and the map $z \mapsto z^n$ on the second $S^1$).

5. Each $r_n$ is a retraction.

$r_n$ restricts to the identity on the first $S^1$ (the target).

6. The $r_n$ are pairwise nonhomotopic.

$r_n$ induces the homomorphism $a \mapsto 1$, $b \mapsto n$ on $\pi_1$; distinct $n$ give distinct homomorphisms, so the $r_n$ are pairwise nonhomotopic (homotopic maps induce equal homomorphisms on $\pi_1$).

7. Hence there are infinitely many nonhomotopic retractions.

<1>4–<1>6.

8. Q.E.D.

Source Collections

Hatcher, Algebraic Topology, problem 17