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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!

Najib Idrissi
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harlekin
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    For future users, here's a question about proving that it does weakly deformation retract onto a point (i.e. it really is contractible). – Najib Idrissi Jan 11 '16 at 14:49
  • @NajibIdrissi Note that following Hatcher's terminology of "deformation retraction in the weak sense", the space above does not weakly deformation retracts to a point, as that would be equivalent to a strong deformation retract in that case. What is true is that there is a deformation retraction to a point in the sense of Wikipedia, i.e., the deformation retraction need not fix the point throughout the homotopy. This is different from the notions in Hatcher.) – PatrickR Jun 28 '22 at 23:26
  • For me it becomes easier when we look to a guy homeomorphic to Y (if I am not wrong on the homeomorphism). Let's construct Y' ~Y that looks nicer: $$\$$ Define X given in item (a) by:

    $$X=[0,1]×{0}∪(⋃_{r\in \mathbb{Q} \cap [0,1]}{r}×[0,1−r])$$

    Then let $T_n:[0,1] \to \mathbb{R}$ such that $T_n(x,y) = (x+n, y(-1)^n)$ for every $n\in \mathbb{Z}$

    Thus $$Y' = \cup_{n\in\mathbb{Z}} T_n (X)$$

    This guy looks like the region between the graph of sin wave and x-axis so we can colapse cleaverly the envelope curve of the spikes to $\mathbb{R}\times{0}$ , which is contractible.

     – BraQuiet Mar 22 '25 at 13:16
  • I've mistyped the domain and range of $T_n$ sorry, I have dyslexia! The correct is:

    $$T_n:[0,1]²\to\mathbb{R}²$$

     – BraQuiet Mar 22 '25 at 14:07

2 Answers2

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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 $U$.

Zev Chonoles
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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 01 '12 at 23:03
  • That's very very helpful, thank you Zev. – 1LiterTears Sep 01 '13 at 19:45
5

enter image description here

The following argument uses the compactness of $I$ in a way different from that used in the 'lemma' of problem 0.5 of the book.

To show that the entire space $Y$ cannot deformation retract to any point $x\in Y$, consider a sequence of points $x_{n}\rightarrow x,$ with each point $x_{n}$ being in a unique distinct strand (the red dots in the figure are supposed to represent the points of the sequence $x_{n}$).

If $Y$ deformation retracts to $x\in Y$, there is a continuous path from $x_{n}$ to $x$, for each $n\geq 0 $. Each such path has to pass through the point $B$ at least once. (if you remove the point $B$ from $Y$, then $x_{n}$ and $x$ are in two distinct path components, hence there cannot exist any path between the two points avoiding $B$).

Given $n$, consider the 'time' $t_{n}$ when $x_{n}$ is first mapped to the point $B$. Then consider the sequence $(x_{n},t_{n})\in Y\times I$. Because $I$ is sequentially compact, we have a converging subsequence $t_{n_{i}}\rightarrow t\in I$. Then $(x_{n_{i}},t_{n_{i}})\rightarrow (x,t)$. Let $F:Y\times I\rightarrow Y$ be the continuous map that gives the deformation retraction. Then the sequence lemma implies $F(x_{n_{i}},t_{n_{i}})\rightarrow F(x,t)$. This is a contradiction as $F(x_{n_{i}},t_{n_{i}})=B \ \forall n_{i}\geq 0$ as constructed, and $F(x,t)=x$, because it is a deformation retraction.

A similar argument can be applied to points $z\in Z$, to show both the facts that there is no def. retraction of $Y$ to $Z$, nor one from $Y$ to $z$.

Aritro Pathak
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