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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.

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    @mixedmath Oops. Good job you spotted that. Sorry.2012-07-04

2 Answers 2

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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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    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.2018-10-29
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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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    @Tyrell This is only a weak deformation retraction, not a true deformation retraction. [Note that is what this question is asking for]2017-07-21