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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 XX with a subspace AA is a strong NDR pair if there exists a map u:X[0,1]u: X \to [0,1] and a homotopy h:X×[0,1]Xh: X \times [0,1] \to X such that:

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

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

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

My technical questions:

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

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

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

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

How do we get the function ww, for starters?

First, let’s ‘turn around’ the function uu, working instead with 1u1 - u. This new function equals 11 on the subset AA instead of 00, and the neighborhood UU is where this new function is >0\gt 0 instead of <1\lt 1.

Let’s do the same for vv, working instead with 1v1 - v.

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

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

There’s a problem here, which is that 1v1-v isn’t defined on all of XX, just on AA. That won’t be a problem if 1v01-v \to 0 as we approach the boundary of AA. Then we can extend 1v1-v to a continuous function that equals 00 outside AA. Let’s assume we can do this, just for now. Let’s call this extension vv.

Then 1u1-u is 11 precisely on AA and 1v1 - v is 11 precisely on BB so

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

is 11 precisely on BB, 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 hh and kk. The problem, again, is that kk is only defined on AA, not all of XX. But, presumably we do something like use hh for a while and then use kk for a while.

Condition 5 says that at every time tt, the homotopy hh maps UU into itself. Similarly, at every time kk maps VV into itself. Why does \ell map WW into itself at every time?

Well, what’s WW? It’s where (1u)(1v)>0(1-u)(1-v) \gt 0. But 1v=01-v = 0 outside AA, so to find WW, we need only look inside AA. Inside AA, 1u=11-u = 1. So, we need only look where 1v01 -v \ge 0 inside AA. And that’s the neighborhood VV.

So, W=VW = V.

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

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

I don’t like this. I see the problem is that hh 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

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

  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 Idk:X×{0}A×IXId\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

I wrote:

But now, to my horror, I see that condition 3 merely says that h(a,t)Ah(a,t) \in A for all aAa \in A. So, hh maps AA into itself at every time. This is much weaker than fixing all points of AA 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)=ah(a,t) = a for all aAa\in A and tIt\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 hh and kk, 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 AA in a space XX is a cofibration if and only if (X,A)(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

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

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

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

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

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

I also had a candidate function ww:

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

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

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

Proof that ww is well-defined

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

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

For ww to be zero we need xUx \in U, but we also need both u(x)=0u(x) = 0 and v(h(x,1))=0v(h(x,1)) = 0. For u(x)=0u(x) = 0 we need xAx \in A, but for v(h(x,1))=0v(h(x,1)) = 0 we need h(x,1)Bh(x,1) \in B. However, for xAx \in A we have h(x,1)=xh(x,1) = x. So, we need xBx \in B. Conversely if xx is in BB we have u(x)=0u(x) = 0 (since xx is in AA) and v(h(x,1))=v(x)=0v(h(x,1)) = v(x) = 0 (since xx is in BB), so ww is zero.

Proof that w is continuous

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

UU is open, so for continuity we need only consider a net of points x iUx_i \in U converging to a point xx not in UU, and check that w(x i)w(x_i) converges to w(x)w(x). We have w(x)=1w(x) = 1 since xx is not in UU. So, we need to check w(x i)1w(x_i) \to 1. Since w(x i)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)1u(x_i) \to 1. This is true just because uu is continuous and x ixx_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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