Showing continuity of partially defined map

There is a theorem in Note on Cofibrations by Arne Strøm. It says

Let $A$ be a closed subspace of a topological space $X$. Then $(X,A)$ has the HEP if and only if there are
(i) a neighborhood $U$ of $A$ which is deformable in $X$ to $A$ rel$A$, i.e. there is an $H:U\times I\to X$ s.t. $H(x,0)=x,\ H(x,1)\in A,\ H(a,t)=a$
for all $a\in A,x\in X,t\in I.$
(ii) a map $\phi:X\to I$ s.t. $A=\phi^{-1}(0)$ and $\phi(x)=1$ for all $x\notin U$

To show that $(X,A)$ has the HEP, one constructs a retraction $X\times I\to A\times I\cup X\times\{0\}$. Such a one is given in the proof, namely:

$$r(x,t)=\begin{cases} (x,\ 0), &\text{ if }\phi(x)=1\\ (H(x,2(1-\phi(x))t),\ 0), &\text{ if }1/2\le\phi(x)<1\\ (H(x,t/(2\phi(x))),\ 0), &\text{ if }0<\phi(x)\le1/2,\ t\le2\phi(x)\\ (H(x,1),\ t-2\phi(x)), &\text{ if }0\le\phi(x)\le1/2,\ t\ge2\phi(x) \end{cases}$$

Now, these partial functions could easily be glued to a single continuous function if the domains were closed, but they are not. So, in order to show continuity, especially on the union of the first two domains, that is on $\phi^{-1}\left[\left[\frac12,1\right]\right]$, one takes an element $(x,t)$ and shows that the function is continuous at that point. Since $\phi^{-1}[[0,1)]$ is an open subset of $U$ and the homotopy $H$ is only used for $\phi(x)<1$ in the construction of $r$, we can assume that $U=\phi^{-1}[[0,1)]$ and is open.

Let us define the map $$s:\{x\in X\mid\phi(x)\ge1/2\}\times I\to X\times I\\ (x,t)\mapsto(x,2(1-\phi(x))t)$$ The image of $s$ is the set $S=\{(x,t)\mid t\le2(1-\phi(x))\le1\}$. Now, $r(x,t)$ can written as $(\tilde Hs(x,t),0)$ where $\tilde H:S\to X$ is defined by $$\tilde H(x,t)=\begin{cases} H(x,t) &\text{ if }x\in U\\ x, &\text{ if }t=0 \end{cases}$$ The map $s$ is continuous, so it would be enough to show continuity of $\tilde H$.
This is only nontrivial for $x\in\partial U.$ Then $x\notin U$ and $\phi(x)=1$, so $t=0$. I take an open neighborhood $V$ about $x=\tilde H(x,0)$ and try to find a neighborhood $W$ of $x$ such that $H[W\times I\cap S]\subseteq V$. But how can I find this $W$? Clearly, it must be a subset of $V$.

Solution 1:

Here's another attempt for the tricky case. Your particular concern is to show that the function $r$ extends continuously from $U$ to its boundary $\partial U$, since all the other cases are obvious. The problem is that $H$ is defined only on $U$.

So define: $$U' = \phi^{-1}\left(\left[0, \frac{9}{10}\right)\right);$$ $$\phi': X \rightarrow [0,1], \quad x \mapsto \min \left(\frac{10}{9}\phi(x), 1\right);$$ $$H' = H \upharpoonright U' \times [0,1].$$

Then $(A, X, U', H', \phi')$ satisfies the hypotheses of the required result, and now we know that $H'$ extends continuously to $\partial U'$ and beyond. So, as this was the only oustanding problem, we can perform the desired construction of $r$ for $(A, X)$ continuously without loss of generality. Phew!