Let $S^1 = \{ z \in \mathbb{C} : |z| = 1 \}$. Consider the map $\mathrm{Sq} : S^1 \longrightarrow S^1$,
$$\operatorname{Sq}(z) = z^2$$
Show that there does not exist a continuous map $\mathrm{Sqrt} : S^1 \longrightarrow S^1$ such that $\mathrm{Sq} \circ \mathrm{Sqrt} = Id_{S^1}$. (That is, $(\operatorname{Sqrt}(w))^2 = w$.) (Hint: If such a map existed, show that there would be a bijective continuous map $S^1 \times \{1, -1\} \longrightarrow S^1$.)