Transposing homeomorphism $S^1 \land S^1 \to S^1 \land S^1$ is a reflection of $S^2$

Question

Consider the smash product $S^1_1 \land S^1_2$ of two copies of circles, and let $f:S_1^1 \land S_2 ^1 \to S_2^1 \land S^1_1$ be the homeomorphism transposing the two factors. Since $S^1 \land S^1$ is homeomorphic to the sphere $S^2$, we can view $f$ as a map of $S^2$. Hatcher says that in this view, $f$ is a reflection of $S^2$ and therefore has degree $-1$. Is there a way to see this?

Answer 1

We get $S^1 \times S^1$ by attaching a 2-cell along the glue pattern $aba^{-1}b^{-1}$. When we transpose the factors, the boundary of the 2-cell gets wrapped $bab^{-1}a^{-1}$ along the wedge of $a$ and $b$. This is the same as if we first reflected the boundary and then rotated by 90 degrees. The map on the entire 2-cell can be described in the same way, each concentric circle is reflected and then rotated by 90 degrees. So when we quotient out by the wedge of $a$ and $b$, the map we get is a reflection followed by a 90 degree rotation. Since a rotation is homotopic to the identity and a reflection across 1-axis is degree -1, the overall degree is $-1$.