# Why does the “zig-zag comb” weakly deformation retract onto a point?

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:

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.

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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. – mixedmath Jul 4 '12 at 17:15
@mixedmath Oops. Good job you spotted that. Sorry. – Rudy the Reindeer Jul 4 '12 at 18:19

## 2 Answers

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.

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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.

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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 $Z$. – mathreader Aug 31 '14 at 8:39
No, this really works. Perhaps you are forgetting to allow $Z$ to flow as well. – mixedmath Aug 31 '14 at 9:01
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? – mathreader Aug 31 '14 at 9:26
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'. – mathreader Sep 5 '14 at 6:09