## January 16, 2008

### Strong NDR Pairs — A Technical Question

#### Posted by John Baez

You may think this blog is technical, but it’s not! We try to focus on broad-brush issues, not the niggling technical details that math research always seems to drag you down into.

But now I’m desperate: I want to finish a paper with Danny Stevenson on the classifying space of a topological 2-group, and the only remaining wrinkle seems to involve ‘strong NDR pairs’. Help!

In Peter May’s book Geometry of Iterated Loop Spaces, he uses the well-known concept of a neighborhood deformation retract pair, but also the concept of a ‘strong’ NDR pair. He says that a topological space $X$ with a subspace $A$ is a strong NDR pair if there exists a map $u:X\to \left[0,1\right]$ and a homotopy $h:X×\left[0,1\right]\to X$ such that:

1. $A={u}^{-1}\left(0\right)$.
2. $h\left(0,x\right)=x$ for all $x\in X$.
3. $h\left(a,t\right)\in A$ for all $\left(a,t\right)\in A×\left[0,1\right]$.
4. if $u\left(x\right)<1$ then $h\left(x,1\right)\in A$.
5. if $u\left(x\right)<1$ then $u\left(h\left(x,t\right)\right)<1$.

The first four give the usual concept of NDR pair: intuitively, $A$ has an open neighborhood $U=\left\{x:u\left(x\right)<1\right\}$ that has $A$ as a deformation retract. The fifth condition makes the NDR pair strong: intuitively, $U$ gets mapped into itself at every stage as we squash it down to $A$.

[Note: I later learned that condition 3 should be: $h\left(a,t\right)=a$ for all $\left(a,t\right)\in A×\left[0,1\right]$. See comments below.]

My technical questions:

If $\left(X,A\right)$ and $\left(A,B\right)$ are strong NDR pairs, is $\left(X,B\right)$ a strong NDR pair? What if $X={\cup }_{n=0}^{\infty }{X}_{n}$, ${X}_{n}\subset {X}_{n+1}$ is closed for every $n$, $X$ has the topology of the union, $\left({X}_{n+1},{X}_{n}\right)$is a strong NDR pair for every $n$ — does this mean that $\left(X,{X}_{n}\right)$ is a strong NDR pair for every n?

I bet the answers to both questions are yes and I bet I could even show this in an hour of uninterrupted hard thought — which alas I don’t have time for today.

I’m also wondering: are there really pairs that are NDR but not strong NDR? Strong NDR gives you a lot more control over the deformation retraction process.

I’m also wondering: how many people think about strong NDR pairs?

Posted at January 16, 2008 7:52 PM UTC

TrackBack URL for this Entry:   http://golem.ph.utexas.edu/cgi-bin/MT-3.0/dxy-tb.fcgi/1573

### 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.

Posted by: jim stasheff on January 17, 2008 1:17 PM | Permalink | Reply to this

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

If $\left(X,A\right)$ and $\left(A,B\right)$ are strong NDR pairs, is $\left(X,B\right)$ 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 $\left(X,A\right)$ is a strong NDR pair with function $u:X\to \left[0,1\right]$ vanishing on $A$ and $h:X×\left[0,1\right]\to X$ a homotopy squashing the neighborhood $U=\left\{x\in X:u\left(x\right)<1\right\}$ down to $A$.

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

We want to show $\left(X,B\right)$ is a strong NDR pair with function $w:X\to \left[0,1\right]$ vanishing on $B$ and $\ell :X×\left[0,1\right]\to X$ a homotopy squashing the neighborhood $W=\left\{x\in X:w\left(x\right)<1\right\}$ 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 $>0$ instead of $<1$.

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

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

$1-w=\left(1-u\right)\left(1-v\right)$

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

$1-w=\left(1-u\right)\left(1-v\right)$

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 $\left(1-u\right)\left(1-v\right)>0$. 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 $h\left(a,t\right)\in A$ 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?

Posted by: John Baez on January 17, 2008 5:21 PM | Permalink | Reply to this

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

1. It’s easy to compose cofibrations. So, would it be a good idea to translate the problem into a problem of cofibrations, and then to come back to NDR pairs?

2. Since (X,A) is a cofibration, it’s possible to extend

$\mathrm{Id}\cup k:X×\left\{0\right\}\cup A×I\to X$

to a homotopy on X. Is this homotopy interesting?

Posted by: Théophile Naïto on January 18, 2008 8:47 AM | Permalink | Reply to this

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

I wrote:

But now, to my horror, I see that condition 3 merely says that $h\left(a,t\right)\in A$ 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.

Danny Stevenson sent me email saying:

Actually to my horror I noticed that I’ve accidentally made a typo in the definition of strong NDR pair: 3 should be

3. $h\left(a,t\right)=a$ for all $a\in A$ and $t\in I$.

Yay! That takes care of that problem. You can tell the tao is with me: when the theorem I want is not true, the definition must have been wrong.

So, if I can figure out how to build the homotopy $\ell$ from the homotopies $h$ and $k$, it’ll be quite likely that this will satisfy condition 5 — the condition that makes an NDR pair ‘strong’.

I know most of you don’t care about this thread. For those of you who’ve vaguely heard about model categories but never have heard about NDR pairs, maybe I should mention that the inclusion of a closed subspace $A$ in a space $X$ is a cofibration if and only if $\left(X,A\right)$ is an NDR pair.

Could there be a young generation of topologists who know about ‘cofibrations’ but have never even heard of ‘NDR pairs’? When I took homotopy theory with G. W. Whitehead, we learned all about NDR pairs… but not these annoying ‘strong’ NDR pairs.

Posted by: John Baez on January 17, 2008 9:49 PM | Permalink | Reply to this

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

NDR pairs are nice from a conceptual point of view - they tell you how cofibrations act without needing an acyclic fibration handy.

Posted by: David Roberts on January 18, 2008 1:16 AM | Permalink | Reply to this

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

The reason might be that they really are the same! Too bad it is not obvious, but if you really need it I can dig for notes (at the moment I don’t even understand what I am saying).

Posted by: Vincent Franjou on January 23, 2008 5:30 PM | Permalink | Reply to this

### 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 $\left(X,A\right)$ and $\left(A,B\right)$ give an NDR pair $\left(X,B\right)$, so we could see why two strong NDR pairs give a strong NDR pair.

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

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

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

$w\left(x\right)=1$ if $u\left(x\right)=1$ and $w\left(x\right)=\mathrm{max}u\left(x\right),v\left(h\left(x,1\right)\right)$ if $u\left(x\right)<1$

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

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

Proof that $w$ is well-defined

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

Proof that ${w}^{-1}\left(0\right)=B$

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

Proof that w is continuous

Each clause in the definition “$w\left(x\right)=1$ if $u\left(x\right)=1$ and $w\left(x\right)=\mathrm{max}u\left(x\right),v\left(h\left(x,1\right)\right)$ if $u\left(x\right)<1$” 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 $w\left({x}_{i}\right)$ converges to $w\left(x\right)$. We have $w\left(x\right)=1$ since $x$ is not in $U$. So, we need to check $w\left({x}_{i}\right)\to 1$. Since $w\left({x}_{i}\right)$ 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 $u\left({x}_{i}\right)\to 1$. 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.

Posted by: John Baez on January 18, 2008 1:45 AM | Permalink | Reply to this

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

Suggest any remaining questions be posted to Don Davis’ ALG-TOP blog.

Posted by: jim stasheff on January 18, 2008 12:48 PM | Permalink | Reply to this

Post a New Comment