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I am doing a self-study in algebraic topology and have a question about Hatcher's Ex.0.6(b) and contractibility on p18. There have been several other posts concerning why the "zigzag comb" (space Y) is not deformation retractable, and that part is very clear to me. But my question concerns why it IS contractible.

Some of the proofs I have seen use the fact that each "bristle" can be retracted to the heavy zigzag line (Z), and then Z can be contracted to a point. But that is not the definition of contractible, and seems to require that we tear the structure.

By definition, we must show that Y is nullhomotopic, i.e., that the identity map of Y is homotopic to a constant map. I can't see how we do this since all maps have to be continuous, as required in the definitions of homotopic and homotopy. So why don't we run into exactly the same problem that prevented the space from deformation retracting?

My question is similar to articles 495856 and 166648, but answers to both of those questions seemed to deal with an infinite number of copies of the triangular comb starting at the origin. This gives us a "starting point" for the contraction, and I see how the shape would contract in that case. But that is not Hatcher's problem.

Hatcher's problem involves combs extending infinitely in both directions. Thus, it appears to me that we have a problem "getting started" with any contraction algorithm without making tears.

In other words, my confusion concerns how we start the process when there are no ends to the combs where we can get the contraction process started.

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    $\begingroup$ Possible duplicate of exercise 6 chapter 0 Hatcher $\endgroup$ – Najib Idrissi Dec 19 '15 at 10:48
  • $\begingroup$ I have edited my question to explain why it was not answered by earlier posts (see last three paragraphs). $\endgroup$ – PossumP Dec 19 '15 at 21:35
  • $\begingroup$ Sorry to say, but you misunderstood the problem (and it is indeed a duplicate, they are dealing with a comb that extends indefinitely in both directions), and the answer you propose is incorrect. $\endgroup$ – Najib Idrissi Dec 20 '15 at 7:31
  • $\begingroup$ @Najib, what is the "answer" you are refering to that is incorrect? (My proposed answer below, or Balarka's answer below?) $\endgroup$ – PossumP Dec 20 '15 at 16:52
  • $\begingroup$ The answer that you, user54301, wrote below. $\endgroup$ – Najib Idrissi Dec 20 '15 at 19:46
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enter image description here

If $Y$ is the space above, one can define a homotopy $f_t : Y \to Y$ between $f_0 = \text{id}$ and $f_1$, the constant map, as described here. For any $a \in Y$, define path $\sigma_a : [0, \infty) \to Y$ with $\sigma_a(0) = a$ which goes down the bristles $a$ is in, and then moves along $Z$ to the right. Thus, define $f_t(x) = \sigma_x(t)$.

You can imagine this as moving to the right, and as you move, contracting the bristles as you head off to right. So you're not contracting all the space to $Z$ at once and then moving off to right by $Z$, but you're doing both simultaneously. This way, continuity is not destroyed.

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  • $\begingroup$ pls. see my revised question, in which I emphasize my difficulty in finding an initial point for starting the contraction you describe. $\endgroup$ – PossumP Dec 20 '15 at 16:48
  • $\begingroup$ @user54301 I am unsure what you mean by initial point. $\endgroup$ – Balarka Sen Dec 21 '15 at 19:11
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Perhaps I can give a very informal answer/approach to solving my own problem:

In the same sense that the real line is homeomorphic (and thus homotopy equivalent) to the open unit interval, the zigzag comb arrayed all along the real line is similarly homeomorphic (and homotopy eqv.) to a "compressed" copy of itself arrayed only along the open unit interval.

And, since the open unit interval can clearly be contracted to a point (without concern for a "starting point") then by a similar (but somewhat more complicated) process the compressed zigzag comb is also contractable without reference to any starting point.

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