Proposition

Let $F: I \times I \rightarrow X$ be a continuous map, and let $f, g, h$, and $k$ be the paths in $X$ defined by

\[\begin{align*} f(s) &= F(s, 0); \\ g(s) &= F(1, s); \\ h(s) &= F(0, s); \\ k(s) &= F(s, 1). \end{align*}\]

Then $f \cdot g \sim h \cdot k$.

Solution

It makes sense to consider $f \cdot g \sim h \cdot k$ because:

  • $f \cdot g$ is well-defined because $f(1) = F(1, 0) = g(0)$.
  • $h \cdot k$ is well-defined because $h(1) = F(0, 1) = k(0)$.
  • $(f \cdot g)(0) = F(0, 0) = (h \cdot k)(0)$.
  • $(f \cdot g)(1) = F(1, 1) = (h \cdot k)(1)$.

Let $G: I \times I \rightarrow X$ be defined such that

\[\begin{align*} G(s, t) &= \begin{cases} F(2s(1 - t), 2st) & (s \in [0, 1/2]) \\ F(1 - 2(1 - s)t, 1 - 2(1 - s)(1 - t)) & (s \in [1/2, 1]). \end{cases} \end{align*}\]

This formula can be derived by comparing squares. Yes, it was rather tedious.

Squares

$G$ is well-defined because:

  • Let $s \in [0, 1/2], t \in I$.
    • $2s(1 - t) \in I$ because $2s \in I$ and $1 - t \in I$.
  • Let $s \in [1/2, 1], t \in I$.
    • $2(1 - s) \in I$ and $t \in I$, so $2(1 - s)t \in I$. Thus $1 - 2(1 - s)t \in I$.
    • $2(1 - s) \in I$ and $1 - t \in I$. Thus $2(1 - s)(1 - t) \in I$, so $1 - 2(1 - s)(1 - t) \in I$.
  • Let $s = 1/2, t \in I$.
    • $F(2s(1 - t), 2st) = F(1 - t, t)$.
    • $F(1 - 2(1 - s)t, 1 - 2(1 - s)(1 - t)) = F(1 - t, 1 - (1 - t)) = F(1 - t, t)$.

Each of $2s(1 - t), 2st, 1 - 2(1 - s)t, 1 - 2(1 - s)(1 - t)$ is continuous. Compositions of continuous functions are continuous, so $F(2s(1 - t), 2st)$ and $F(1 - 2(1 - s)t, 1 - 2(1 - s)(1 - t))$ are continuous. By the pasting lemma, $G$ is continuous.

We claim that $G$ is a path homotopy from $f \cdot g$ to $h \cdot k$.

  • Let $t = 0$.
    • Let $s \in [0, 1/2]$.
      • $G(s, t) = F(2s(1 - t), 2st) = F(2s, 0) = f(2s) = (f \cdot g)(s)$.
    • Let $s \in [1/2, 1]$.
      • $G(s, t) = F(1 - 2(1 - s)t, 1 - 2(1 - s)(1 - t)) = F(1, 2s - 1) = g(2s - 1) = (f \cdot g)(s)$.
  • Let $t = 1$.
    • Let $s \in [0, 1/2]$.
      • $G(s, t) = F(2s(1 - t), 2st) = F(0, 2s) = h(2s) = (h \cdot k)(s)$.
    • Let $s \in [1/2, 1]$.
      • $G(s, t) = F(1 - 2(1 - s)t, 1 - 2(1 - s)(1 - t)) = F(2s - 1, 1) = k(2s - 1) = (h \cdot k)(s)$.
  • Let $s = 0$.
    • For any $t \in I$, $G(s, t) = F(0, 0)$.
  • Let $s = 1$.
    • For any $t \in I$, $G(s, t) = F(1, 1)$.

Therefore, $G$ is indeed a path homotopy from $f \cdot g$ to $h \cdot k$.