Proposition 9

Let X and B be topological spaces, let A be a subspace of B, and let $f: A\rightarrow X$ be a continuous mapping.

a) If the pair $(B,A)$ has the homotopy extension property, the same is true of the pair $(X\cup_fB, \alpha_f(X))$.

b) Suppose that the mapping $f$ is injective and strict. If the pair $(X, f(A))$ has the homotopy extension property, the same is true of the pair $(X\cup_fB, \beta_f(B))$.

a) Suppose that the pair $(B,A)$ has the homotopy extension property. Let Z be a topological space, let $u: X\cup_fB\rightarrow Z$ be a continuous mapping, and let $\sigma :\alpha_f(X)\times \mathbf{I}\rightarrow Z$ be a homotopy whose final term is the restriction of $u$ to the subspace $\alpha_f(X)$.

Let us put $v_1=u\circ \alpha_f$ and denote by $\tau_1: X\times \mathbf{I}\rightarrow Z$ the mapping given by $(x, t)\mapsto \sigma (\alpha_f(x), t)$. Put $v_2=u\circ \beta_f$. Since $\beta_f|A =\alpha_f\circ f$, the mapping from $A\times \mathbf{I}$ into Z which maps $(a, t)$ to $\sigma (\alpha_f(f(a)), t)$ for $a\in A$ and $t\in \mathbf{I}$ is a homotopy whose final term is equal to the mapping $v_2|A$. Since the pair $(B,A)$ has the homotopy extension property, there exists a homotopy $\tau_2: B\times \mathbf{I}\rightarrow Z$ whose final term is the mapping $v_2$ and such that $\tau_2(a, t) =\sigma (\alpha_f(f(a)), t)$ for $(a, t)\in A\times \mathbf{I}$.

For $a\in A$ and $t\in \mathbf{I}$, we have $\tau_2(a, t) =\tau_1(f(a), t)$. By Proposition 8, there exists a unique continuous mapping $\sigma : (X\cup_fB)\times \mathbf{I}\rightarrow Z$ such that $\sigma (\alpha_f(x), t) =\sigma_1(x, t)$ and $\sigma (\beta_f(b), t) =\sigma_2(b, t)$, for $x\in X$, $b\in B$ and $t\in \mathbf{I}$. It is a homotopy with final term $u$ which extends $\tau$, hence assertion a).

b) In view of Example 2 (III, p. 249), assertion b) follows from assertion a).