# Why doesn't the “zig-zag” comb deformation retract onto a point, even though it's contractible?

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: 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!

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 $U$.

• 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

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 $z$.