Product of two NDR pairs This is a question about a lemma in May's Algebraic Topology, asserting that if $(X,A)$ and $(Y,B)$ are NDR pairs, then so is $(X\times Y,X\times B\cup A\times Y)$.
By definition $(X,A)$ is a NDR pair if there exists a map $u:X\to I$ and a homotopy $h:X\times I\to X$ such that $u^{-1}(0)=A$ and $h(x,0)=x$ for all $x\in X$, $h(a,t)=a$ for all $a\in A$ and $t\in I$, and $h(x,1)\in A$ for all $x\in u^{-1}([0,1))$.
Suppose $(h,u)$ and $(j,y)$ represent $(X,A)$ and $(Y,B)$ as NDR-pairs, and define $k:X\times Y\times I\to X\times Y$ by letting
$$k(x,y,t)=\begin{cases}
(h(x,t),j(y,tu(x)/v(y)))&\text{if }v(y)\geq u(x)\\
(h(x,tv(y)/u(x)),j(y,t))&\text{if }u(x)\geq v(y).
\end{cases}
$$
We understand $u(x)/v(y)=1=v(y)/u(x)$ if $u(x)=v(y)=0$. My question is: How can we check continuity of $k$?
 A: You have to show that $k$ is continuous on the subsets $C$ and $D$ of $X\times Y\times I$
defined by $v(y)\ge u(x)$ and $u(x)\ge v(y)$ respectively, and the
definitions agree on $C\cap D$. That suffices, since $C$ and $D$ are closed in
$X\times Y\times I$. It's also clear that on the intersection they match up,
and so all one needs to prove is that $k$ is continuous on $C$ and on $D$.
The proofs for both will be similar, so let's concentrate on $C$. I think it's
clear that $k$ is continuous on all points with $v(y)>0$, so take $P=(x_0,y_0,t_0)$
with $u(x_0)=v(y_0)=0$, that is $x_0\in A$ and $y_0\in B$. Certainly $h(x,t)$
is continuous at $P$, so we ask if $j(y,tu(x)/v(y))$ is also. This will
follow from the continuity of $tu(x)/v(y)$. Note that we take $t_0u(x_0)/v(y_0)$
to be $t_0$.
Let $U$ be a neighbourhood of $t_0$ in $I$. By the continuity
of $u$ and $v$, it suffices to prove that
$$E=\{(r,s,t):0\le r\le s\le1,0\le t\le 1,t(r/s)\in U\}$$
is open in
$$F=\{(r,s,t):0\le r\le s\le1,0\le t\le 1\}.$$
From the convention that $0/0=1$,
$$E=\{(0,0,t):t\in U\}\cup\{(r,s,t)\in F,s>0,rt/s\in U\}$$
which is open in $F$.
