8
$\begingroup$

I am starting to read Hatcher's book on Algebraic Topology, and I am a little stuck with exercise 6 in Chapter 0.

Let $Z$ be the zigzag subspace of $Y$ homeomorphic to $\mathbb{R}$ indicated by the heavier line in the picture: enter image description here

(see here for picture and definitions)

Show there is a deformation retraction in the weak sense of $Y$ onto $Z$, but no true deformation retraction.

It's easy to show no true deformation retract is possible, but how does one show that a weak deformation retract is possible? Clearly we must deformation retract onto a disconnected subspace of of $Z$; however, it would appear that all open neighborhoods of every point are disconnected.

$\endgroup$
  • $\begingroup$ This is not a duplicate of that question. Although the source is the same, neither of the other two formulations actually show a weak reatract, but instead show that no def retract is possible. $\endgroup$ – davidlowryduda Jul 4 '12 at 17:15
  • $\begingroup$ @mixedmath Oops. Good job you spotted that. Sorry. $\endgroup$ – Rudy the Reindeer Jul 4 '12 at 18:19
9
$\begingroup$

This is an elaboration of mixedmath's response.

For each point $a$ in Y, there is a natural path leading from $Y$ off to the right. For example, if $a$ is already on the zigzag $Z$, then $a$ just travels rightward along the zigzag. If $a$ is on one of the bristles, then first $a$ travels towards the zigzag, and then subsequently off to the right.

Let $f_a: [0, \infty) \to Y$ be this path, where the point travels at constant speed 1.

Consider the map $H: Y \times I \to Y$ defined by $H(a, t) = f_a(t)$, for $0 \leq t \leq 1$. I claim this is our desired homotopy. The only part that isn't clear is continuity; there are several cases to check here but none of them are hard.

$\endgroup$
  • $\begingroup$ Sorry, I just came across this answer and I would like to know what cases have to be discussed. $\endgroup$ – Javi Oct 27 '18 at 19:37
  • 1
    $\begingroup$ I don't want to spell it out, but bear in mind that the map H is defined by cases (because the maps $f_a$ are defined by cases). This will have to be reflected in a proof that $H$ is continuous. $\endgroup$ – Douglas Ulrich Oct 29 '18 at 4:40
  • $\begingroup$ This answer is so inspiring, which even gives a proof of being contractable part of question (b), by the fact that the concatenation of a weak deformation retraction and a strong deformation retraction is a homotopy. I had been thinking if in question (b) the image point of the constant map stays, then the homotopy should be a strong def ret, a contradiction. However, in this concatenated homotopy, the point did move. $\endgroup$ – Shiyu Liang Aug 4 at 19:47
4
$\begingroup$

HINT

In short, imagine that everything 'flows' to the right (and maybe up or down, depending on where it is), down each of the comb bits.

$\endgroup$
  • $\begingroup$ Could you please elaborate? I cannot see for a few days how $Y$ can weakly deformation retract onto $Z$. If we make things flow, as you suggested, then we get discontinuity right away at the epsilon-neighborhood of $Z$. $\endgroup$ – mathreader Aug 31 '14 at 8:39
  • $\begingroup$ No, this really works. Perhaps you are forgetting to allow $Z$ to flow as well. $\endgroup$ – davidlowryduda Aug 31 '14 at 9:01
  • $\begingroup$ What happens to the long bristles right after the start of the movement? Are they getting 'broken' too, like the segment of zigzag line they were close to? $\endgroup$ – mathreader Aug 31 '14 at 9:26
  • $\begingroup$ Answering my own question: no, the bristles remain straight and move each along itself, decreasing in length until they reach the base. After that the points continue to travel inside zigzag. The situation is like at every rational point of the base there is a small 'hole' through which the bristle is sucked in and moved away inside the zigzag 'tube'. $\endgroup$ – mathreader Sep 5 '14 at 6:09
  • $\begingroup$ You are saying that $Z$ flows too. But how this can happen? Shouldn't it be fixed? $\endgroup$ – Tyrell Jul 21 '17 at 14:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.