Strong NDR Pairs --- A Technical Question

January 16, 2008

Strong NDR Pairs — A Technical Question

Posted by John Baez

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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!

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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 $u: X \to [0,1]$ and a homotopy $h: X \times [0,1] \to X$ such that:

$A = u^{-1}(0)$ .
$h(0,x) = x$ for all $x \in X$ .
$h(a,t) \in A$ for all $(a,t) \in A \times [0,1]$ .
if $u(x) \lt 1$ then $h(x,1) \in A$ .
if $u(x) \lt 1$ then $u(h(x,t)) \lt 1$ .

The first four give the usual concept of NDR pair: intuitively, $A$ has an open neighborhood $U = \{x : u(x) \lt 1\}$ 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: $h(a,t) = a$ for all $(a,t) \in A \times [0,1]$ . See comments below.]

My technical questions:

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

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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

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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?

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

Re: Strong NDR Pairs — A Technical Question

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I answer to your question with 2 (perhaps stupid) questions:

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?
Since (X,A) is a cofibration, it’s possible to extend $Id\cup k: X\times \{0\}\cup A\times I\rightarrow X$ 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

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I wrote:

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.

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. $h(a,t) = a$ 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 $(X,A)$ 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

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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 $A$ .

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.

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

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January 16, 2008