Strong deformation retract 
Let $A$ be strong deformation retract of $X$ and $A=\alpha^{-1}(\{0\}) $ for some continuous $\alpha:X\to I$.
If $H:X\times I\to X$ is homotopy between $i\circ r $ and $1_X$ (rel $A$) , where $i:A\to X$ is inclusion and $r:X\to A$ is retraction, we define $\phi:X\times I\to I$ with $\phi(x,t)=$min$\left(1,\frac t{\alpha(x)}\right)$ for all $x\notin A$ and $\phi(x,t)=1$ for all $x\in A$.
Show that $\theta:X\times I\to X$ defined by $\theta(x,t)=H(x,\phi(x,t))$ is continuous.

Any help with this? I really don't know where to start, any help is appreciated.
 A: It is relatively easy to show that $\phi$ is continuous for all $(x,t)\not\in\partial A\times 0$. But it won't be continuous on these critical points unless $A$ is also open. Still, the $\theta$ can be proven to be continuous there. So let $y\in\partial A$ and $t=0$. Since $A$ is closed, we have $\theta(y,0)=y$.
We will need a lemma:

Lemma: Let $X$ be a space, $A\subset X$, and $d:X\times I\to X$ a homotopy such that $d(a,t)=a$ for $a\in A$. If $U$ is open, then there is an open $V\subseteq U$ such that $U\cap A\subseteq V$ and for every $v\in V$ the path $d(v,t)$ is contained in $U$.

Proof: Since $d$ is continuous, $d^{-1}(U)$ is open and contains $U\cap A\times I$. For each $x\in U\cap A$ the set $\{x\}\times I$ is compact, so there is an open $V_x\subseteq U$ with $\{x\}\times I\subseteq V_x\times I\subseteq d^{-1}(U)$. Let $V=\bigcup_{x\in U\cap A}V_x$. Then $A\cap U\times I\subseteq V\times I\subseteq d^{-1}(U)$. This completes the proof. $\Box$
Now let $U$ be an open neighborhood about $\theta(y,0)=y$. By the lemma there is an open $V$ between $A\cap U$ and $U$ with the special property. Let $v\in V$ and $s\in I$ arbitrary. Then $H(v,\phi(v,s))$ is a point in the path of $v$, and this path is contained in $U$ by construction of $V$. We have shown that the desired neighborhood whose image is contained in $U$ is $V\times I$.
