Proposition: Suppose $(X,A)$ and $(Y,A)$ satisfy the homotopy extension property(HEP) and $f:X \to Y$ is a homotopy equivalence with $f\vert_A=1$ then $f$ is a homotopy equivalance relative $A$.
Proof:
Step1: Construct a homotopy from $g$ to a map $g_1$ with $g_1 \vert_A=1$
Let $h_t:X \to X$ be a homotopy from $gf=h_0$ to $1=h_1$. Since $f\vert_A=1$ we can view $h_t \vert_A$ as a homotopy from $g\vert _A$ to $1$.Since $(Y,A)$ has HEP, we can extend this homotopy to a homotopy $g_t: Y \to X$ from $g=g_0$ to a map $g_1$ with $g_1 \vert_A=1$
Step2: Show $g_1f$ is homotopic to $1$ rel $A$.
It is easy to obtain the homotopy between $g_1f$ and $1$ by using the homotopies of step $1$ and using transitivity. But the hard work is to construct a homotopy which is rel $A$. The proof is given in Hatchers book on page $17$ chapter $0$ but i dont follow the proof. Basically i don't see the motivation of that proof. Can anyone give me 'rough' idea to find a relative homotopy between $g_f$ and $1$
Thank you for all your time