Exercise 3, § 4

Let $I$ denote the interval $[0, 1]$ of $\mathbf{R}$, and let $f$ be a homeomorphism of $I$ onto itself such that $f(0) = 0$ and $f(1) = 1$. Show that there is a continuous mapping $g$ of $I \times I$ into $I$ such that: (i) for each $x \in I$, $g(x, 0) = x$ and $g(x, 1) = f(x)$; (ii) for each $y \in I$ the partial mapping $x \to g(x, y)$ is a homeomorphism of $I$ onto itself such that $g(0, y) = 0$ and $g(1, y) = 1$.