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I am starting to read Hatcher's book on Algebraic Topology, and I am a little stuck with exercise 6(c) in Chapter $0$. Unfortunately a picture is involved so it doesn't quite make sense for me to repeat it here - but I'll do it anyways (maybe it helps)

Let $Z$ be the zigzag subspace of $Y$ homeomorphic to $\mathbb{R}$ indicated by the heavier line in the picture: enter image description here Show there is a deformation retraction in the weak sense of $Y$ onto $Z$, but no true deformation retraction."

Now the definition for deformation retraction and weak deformation retraction as as follows:

We say that $f: X \to X$ is a deformation retraction of a space $X$ onto a subspace $A \subset X$ if there exists a family of maps $f_t : X \to X$ with $t \in [0,1]$ such that $f_0 = \mathbb{1}$ (the identity) and $f_1 (X) = A$ and also $f_t$ restricts to the identity on $A$ for each $t$.

A weak deformation retraction is almost the same, only that we now relax the conditions $f_1(X) = A$ to $f_1 \subset A$ and, for each $t \in [0,1]$ we require that $f_t(A) \subset A$.

Hence, as far as I understand the question and the concepts involved, I need to show that basically no map can be constructed such that the bold zigzag - line stays put while the thin lines retract to the bold line over a finite time interval $0 \leq t \leq 1$. However, I should be able to show that I can pull the thin lines continously onto the thick line, provided I am allowed to move points on the zigzag line .. is that correct ?

Now, the problem is I can't see why I shouldn't be able to do the former, given that I could find a weak deformation retract. Any help would be great !

This is NOT a homework question as I am a self - learner, so I won't tag it as a homework - question, hope this is the correct way of tagging .. Thanks !

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Great question, thank you harlekin. –  1LiterTears Aug 31 '13 at 17:03

1 Answer 1

up vote 7 down vote accepted

Hint: $Z$, being homeomorphic to $\mathbb{R}$, deformation retracts to a point. Compositions of deformation retractions are deformation retractions (composition in the sense of doing the first deformation retraction for $0\leq t\leq \tfrac{1}{2}$, and doing the second for $\frac{1}{2}\leq t\leq 1$). Thus, if $Y$ deformation retracts to $Z$, it must also deformation retract to a point. Do you see why this is impossible? The argument uses Problem 5 in the same section. The details are in a spoiler box below (put your cursor over it to reveal).

By Problem 5, any neighborhood $U$ of such a point would have to contain an open set $V$ whose inclusion $i:V\hookrightarrow U$ is nullhomotopic; however, this is impossible because any open set $V$ will meet non-path-connected parts of $U$.

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WOW! First, thanks for editing my post and adding the picture! Now I know how to do that so I can post better questions. Secondly - GREAT answer !! It helped a lot working through the hint you gave and then thinking on my own again, with the chance to check against your suggestion in the spoiler box. THANKS !! –  harlekin Mar 30 '12 at 20:22
    
@harlekin: No problem, glad to help! –  Zev Chonoles Apr 1 '12 at 23:03
    
That's very very helpful, thank you Zev. –  1LiterTears Sep 1 '13 at 19:45

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