### Re: Strong NDR Pairs — A Technical Question

If $(X,A)$ and $(A,B)$ are strong NDR pairs, is $(X,B)$ a
strong NDR pair?

I may be starting to see how to do it. I feel sure that the standard proof that the composite of NDR pairs is an NDR pair, applied to a composite of strong NDR pairs, will give a strong NDR pair. So, I just need to reinvent that standard proof — or look it up, if I have time for a trip to the library.

This is the very beginning:

Suppose $(X,A)$ is a strong NDR pair with function $u: X \to [0,1]$ vanishing on $A$ and $h: X \times [0,1] \to X$ a homotopy squashing the neighborhood $U = \{x \in X: u(x) \lt 1\}$ down to $A$.

And suppose $(A,B)$ is a strong NDR pair with function $v: A \to [0,1]$ vanishing on $B$ and $k: B \times [0,1] \to B$ a homotopy squashing the neighborhood $V = \{a \in A: v(a) \lt 1\}$ down to $B$.

We want to show $(X,B)$ is a strong NDR pair with function $w: X \to [0,1]$ vanishing on $B$ and $\ell: X \times [0,1] \to X$ a homotopy squashing the neighborhood $W = \{x \in X: w(x) \lt 1\}$ down to $A$.

How do we get the function $w$, for starters?

First, let’s ‘turn around’ the function $u$, working instead with $1 - u$. This new function equals $1$ on the subset $A$ instead of $0$, and the neighborhood $U$ is where this new function is $\gt 0$ instead of $\lt 1$.

Let’s do the same for $v$, working instead with $1 - v$.

I want to define $w: X \to [0,1]$ by

$1 - w = (1-u)(1-v)$

There’s a problem here, which is that $1-v$ isn’t defined on all of $X$, just on $A$. That won’t be a problem if $1-v \to 0$ as we approach the boundary of $A$. Then we can extend $1-v$ to a continuous function that equals $0$ outside $A$. Let’s assume we can do this, just for now.
Let’s call this extension $v$.

Then $1-u$ is $1$ precisely on $A$ and $1 - v$ is $1$ precisely on $B$ so

$1 - w = (1-u)(1-v)$

is $1$ precisely on $B$, as desired. This is part 1 of the definition of ‘strong NDR pair’.

Let’s jump ahead to condition 5, the one special to ‘strong’ NDR pairs.

I still don’t see how to cook up the homotopy $\ell$ from the homotopies $h$ and $k$. The problem, again, is that $k$ is only defined on $A$, not all of $X$. But, presumably we do something like use $h$ for a while and then use $k$ for a while.

Condition 5 says that at every time $t$, the homotopy $h$ maps $U$ into itself. Similarly, at every time $k$ maps $V$ into itself. Why does $\ell$ map $W$ into itself at every time?

Well, what’s $W$? It’s where $(1-u)(1-v) \gt 0$. But $1-v = 0$ outside $A$, so to find $W$, we need only look inside $A$. Inside $A$, $1-u = 1$. So, we need only look where $1 -v \ge 0$ inside $A$. And that’s the neighborhood $V$.

So, $W = V$.

So, why does $\ell$ map $V$ into itself at every time? This is hard to say, since I don’t know what $\ell$ is. But $\ell$ is built using $h$ and $k$, and $k$ maps $V$ into itself at every time, which is promising. I’d feel confident of success if $h$ *fixed all points of* $A$ at every time, since then it too would map $V$ into itself at every time.

But now, to my horror, I see that condition 3 merely says that $h(a,t) \in A$ for all $a \in A$. So, $h$ maps $A$ into itself at every time. This is much weaker than *fixing all points of* $A$ at every time.

I don’t like this. I see the problem is that $h$ is just a deformation retract, not a strong deformation retract.

So, I have to ask: can every deformation retract be improved to be a strong deformation retract?

### Re: Strong NDR Pairs — A Technical Question

Okay — I know I’m talking to myself, but I’ll keep on, just to prove to all of you that I actually spend a lot of time doing very boring stuff, and normally blog about it only when something interesting occurs. Danny need to settle this question to get a classifying space for a topological 2-group! That sounds interesting. But, it involves some tiresome work.

We were trying to see how two NDR pairs $(X,A)$ and $(A,B)$ give an NDR pair $(X,B)$, so we could see why two *strong* NDR pairs give a *strong* NDR pair.

So, suppose $(X,A)$ is a strong NDR pair with function $u: X \to [0,1]$ vanishing on $A$ and $h: X \times [0,1] \to X$ a homotopy squashing the neighborhood $U = \{x \in X: u(x) \lt 1\}$ down to $A$.

And suppose $(A,B)$ is a strong NDR pair with function $v: A \to [0,1]$ vanishing on $B$ and $k: B \times [0,1] \to B$ a homotopy squashing the neighborhood $V = \{a \in A: v(a) \lt 1\}$ down to $B$.

Then we want to show $(X,B)$ is a strong NDR pair with function $w: X \to [0,1]$ vanishing on $B$ and $\ell: X \times [0,1] \to X$ a homotopy squashing the neighborhood $W = \{x \in X: w(x) \lt 1\}$ down to $A$.

And, Danny has by now told me what seems like a more promising way to get the function $w$. He wrote:

I also had a candidate function $w$:

$w(x) = 1$ if $u(x) = 1$ and $w(x) = max{u(x), v(h(x,1))}$ if $u(x) \lt 1$

I think this is continuous, and I think also that $w^{-1}(0) = B$, and
$W = V$.

Let’s see if I understand this. I just need to think out loud.

**Proof that $w$ is well-defined**

If $u(x) = 1$, $x$ is outside of the open set $U$ which our homotopy $h$
squashes down to $A$. Otherwise $x$ is inside $U$, so $h(x,1) \in A$ and $v(h(x,1))$ is well-defined. So, the formula for $w$ makes sense.

**Proof that $w^{-1}(0) = B$**

For $w$ to be zero we need $x \in U$, but we also need both
$u(x) = 0$ and $v(h(x,1)) = 0$. For $u(x) = 0$ we need $x \in A$,
but for $v(h(x,1)) = 0$ we need $h(x,1) \in B$. However, for
$x \in A$ we have $h(x,1) = x$. So, we need $x \in B$.
Conversely if $x$ is in $B$ we have $u(x) = 0$ (since $x$ is in $A$) and
$v(h(x,1)) = v(x) = 0$ (since $x$ is in $B$), so $w$ is zero.

**Proof that w is continuous**

Each clause in the definition “$w(x) = 1$ if $u(x) = 1$ and $w(x) = max{u(x), v(h(x,1))}$ if $u(x) \lt 1$”
is a continuous function, but the first clause applies only
when $x$ is outside $U$, while the second applies when $x$ is in $U$.

$U$ is open, so for continuity we need only consider a net of points
$x_i \in U$ converging to a point $x$ not in $U$, and check that
$w(x_i)$ converges to $w(x)$. We have $w(x) = 1$ since $x$ is not in $U$.
So, we need to check $w(x_i) \to 1$. Since $w(x_i)$ is defined as a max
of numbers less than or equal to 1, it’s enough to check that one
of these numbers approaches 1. So, check $u(x_i) \to 1$. This
is true just because $u$ is continuous and $x_i \to x$.
Okay, so yes — it’s all true, and it works better than my guess did.

## Re: Strong NDR Pairs — A Technical Question

I doubt there are many NOW, but when I was much younger, there were quite a few. Suggest you post this to Don Davis’ ALG-TOP

blog where such memories may persist.