I need to prove the following proposition:
Let $X$ be a path connected topological space. Let $\varphi \colon \mathbb{S}^1 \subset \mathbb{R}^2 \to X$ be a continuous function and $\psi \colon [0, 1] \to X,\ s \mapsto \varphi(\cos2\pi s, \sin2\pi s)$. Then $\varphi$ is homotopic to a constant map if, and only if, $\psi$ is path homotopic to a constant path.
Here is what I have so far:
Denote $x_0 := \varphi(1, 0) \in X$. It is convenient to introduce the maps $\ell$ and $\alpha_0$, in addition to those given in the statement:
\begin{align*}
\ell &\colon [0, 1] \to \mathbb{S}^1,\ s \mapsto (\cos2\pi s, \sin2\pi s) \\
\alpha_0 &\colon \mathbb{S}^1 \to X,\ (a, b) \mapsto x_0 \\
\varphi &\colon \mathbb{S}^1 \to X,\ (a, b) \mapsto \varphi(a, b) \\
\psi = \varphi \circ \ell &\colon [0, 1] \to X,\ s \mapsto \varphi(\cos2\pi s, \sin2\pi s)
\end{align*}
being all continuous maps.
- ($\Rightarrow$) Suppose $\varphi$ is homotopic to a constant map $\alpha'_0 \colon \mathbb{S}^1 \to X,\ (a, b) \mapsto x'_0$ for some $x'_0 \in X$. Then the constant maps $\alpha'_0$ and $\alpha_0$ are homotopic. By hypothesis $\varphi \simeq \alpha'_0$ and so by transitivity $\varphi \simeq \alpha_0$. Now, using the fact that $\ell \simeq \ell$, we have that $\varphi \circ \ell \simeq \alpha_0 \circ \ell$. But the left-hand side map is just $\varphi \circ \ell = \psi$ while the right-hand side correspond to $(\alpha_0 \circ \ell)(s) = \alpha_0(\ell(s)) = x_0$. Therefore, $\psi$ is homotopic to the constant map $\alpha_0$. Moreover,
\begin{align*}
\psi(0) = (\varphi \circ \ell)(0) = \varphi(\ell(0)) \varphi(1, 0) = x_0 \quad \text{and} \quad (\alpha_0 \circ \ell)(0) = \alpha_0(\ell(0)) = x_0 \\
\psi(1) = (\varphi \circ \ell)(1) = \varphi(\ell(1)) \varphi(1, 0) = x_0 \quad \text{and} \quad (\alpha_0 \circ \ell)(1) = \alpha_0(\ell(1)) = x_0
\end{align*}
so both paths are path homotopic.
But I'm stuck in the ($\Leftarrow$)-direction. Can you help me with hints on how to proceed in this direction?
Best Answer
Suppose $\psi$ is path homotopic to a constant path $\gamma_0 \colon [0, 1] \to X,\ s \mapsto x_0$ for $x_0 \in X$, i.e. there exists a homotopy $H \colon [0, 1] \times [0, 1] \to X$ such that $H(s, 0) = \psi(s)$ and $H(s, 1) = x_0$. Note that the map $\ell$ is continuous, surjective, and closed, so it is a quotient map. Consider now the function \begin{equation} \ell \times \text{Id} \colon [0, 1] \times [0, 1] \to \mathbb{S}^1 \times [0, 1],\ (s, t) \mapsto (\ell(s), t) \end{equation} which also corresponds to a quotient map. Indeed:
Now we are going to use Theorem $22.2$ [Topology, J.~R.~Munkres (2nd Edition)]. Let let us analyze the fiber of each element $\mathbb{S}^1 \times [0, 1]$ under $\ell \times \text{Id}$. We have $2$ cases:
We claim that the function $F$ corresponds to a homotopy between $\varphi$ and the constant map $(a, b) \mapsto x_0$. Indeed, first note that, explicitly, we have \begin{equation} H(s, t) = F\bigl((a(s), b(s)), t\bigr) \end{equation} where $s \in [0, 1]$ is such that $\ell(s) = (a, b)$ (note this equation is well-defined even for $s \in \{0, 1\}$). Then,
Therefore, $F$ corresponds to a homotopy between $\varphi$ and $s \mapsto x_0$.