Corollary 1

Let X be a normal topological space and let A be a closed subspace of X. Suppose that there exist a neighbourhood V of A in X and a contraction of V onto A, together with a continuous mapping $f: X\rightarrow \mathbf{R}$ such that $\overset{-1}{f}(0) = A$. Then the pair $(X,A)$ has the homotopy extension property.

Let $\rho : V\times \mathbf{I}\rightarrow V$ be a contraction of V onto A. By definition of a normal space (TG, IX, p. 41, Definition 1), there exists a continuous mapping $g: X\rightarrow \mathbf{I}$ which is 0 on A and 1 at every point of X - V. Let $\varphi : X\rightarrow \mathbf{R}$ be the mapping defined by $\varphi (x) =|f(x)|+g(x)$ for $x\in X$; it is continuous. We have $\overset{-1}{\varphi}(0) = A$ and $\overset{-1}{\varphi}(\mathbf{I})\subset V$. Let $\sigma$ be the mapping from $\overset{-1}{\varphi}(\mathbf{I})\times \mathbf{I}$ into X defined by $\sigma (x, t) =\rho (x,1-t)$ for $x\in \overset{-1}{\varphi}(\mathbf{I})$ and $t\in \mathbf{I}$. For $x\in \overset{-1}{\varphi}(\mathbf{I})$, we have $\sigma (x,1) =\rho (x,0) =x$ and $\sigma (x,0) =\rho (x,1)\in A$.

The mappings $\varphi$ and $\sigma$ satisfy the conditions of assertion (iii) of theorem 1 of III, p. 243; hence the pair $(X,A)$ possesses the homotopy extension property.