Hatcher, *Algebraic Topology*, Chapter 0

**6. (a) Let be the subspace of consisting of the horizontal segment together with the vertical segments for a rational number in . Show that deformation retracts to any point in the segment , but not to any other point (see the preceding problem). **

(b) Let be the subspace of that is the union of an infinite number of copies of arranged as in the figure below. Show that is contractible but does not deformation retract onto any point.

**(c) Let be the zigzag subspace of homeomorphic to indicated by the heavier line. Show that there is a deformation retraction in the weak sense (see Exercise 4) of onto , but no true deformation retraction.**

**Figure 1**

The spaces mentioned above

*Proof.* (a) Let be the subspace of shown on the far left of Figure 1 consisting of the horizontal segment together with the vertical segments for a rational number in . First, note that the segment deformation retracts onto any point by way of the straight-line homotopy which fixes for all and which linearly “squeezes” the intervals and towards their right and left endpoints, respectively. Algebraically, then, is the family of maps for which

and .

Next, note that the space deformation retracts onto the space in a fashion similar to the one above, namely by way of the family of maps which, for each rational , maps the segment to the point continuously with respect to as ranges through . Denoting this family of maps , it follows that the space deformation retracts to any point by way of the composition , where

.

Finally, the goal is to show that fails to deformation retract onto any point *not* on the segment . For a contradiction, suppose that there *does* exist such an for which a deformation retraction exists. By the result of Exercise 0.5, it follows that every neighborhood of in contains a neighborhood of for which the inclusion is null-homotopic. Evidently this is bad; how come?^{2}

Note that is path-connected^{1} but that clearly *isn’t* path-connected. In particular, suppose is a neighborhood of in . Then can be thought of as an open ball in , disjoint from , and intersecting a countably infinite number of line segments, all of which are disjoint. Thus, fails to be path-connected. For the same reason, any neighborhood containing must also be path-disconnected, whereby it follows that cannot contract to a point. Hence, cannot be homotopic to the constant map, contradicting the result of 0.5.

(b) Let be the subspace of shown in the rightmost part of Figure 1 consisting of an infinite number of copies of arranged as above. The fact that fails to deformation retract onto any point is an almost-identical application of Exercise 0.5 to the one mentioned in (a). In particular, given a point , there exists neighborhood of in that is necessarily path-disconnected. This claim follows immediately from (a) if fails to lie on the zigzag subspace (see part (c)). In the case that , this can be heuristically described in cases as follows.

Suppose fails to be one of the “vertex points”. Any neighborhood in intersects two “fundamentally different” countably infinite disjoint collections of line segments of , one collection which is directed “bottom-left-to-top-right” – call it – and one which is directed “top-left-to-bottom-right” – call it . In this case, any point on a segment in cannot be connected to any point in because of the fact that intersects only one of , .

The argument for a “vertex point” of is similar with the exception that an open neighborhood of now intersects *three* fundamentally different collections of line segments, two of which – say and – are directed in a parallel fashion (either “bottom-left-to-top-right” or “top-left-to-bottom-right”) with the third, , directed in the other fashion. In this case, any point in , , *can be* connected to any point in by a path due to the fact that intersects and one of , . On the other hand, intersections *precisely* one of or , meaning that no path in exists connecting points in to .

Now, it suffices to show that is contractible. Note, first, that is clearly contractible. Next, note that the space can be *retracted* (not *deformation retracted*) onto by way of a modified version of the family given in part (a). More precisely, the map is a retraction, where is the constant map on and where maps any “straw” onto the point , where the subscript here denotes the elements in the copy of .^{3} Therefore, it’s been shown that retracts onto the zigzag which is a contractible subspace, thereby proving that itself is contractible.

(c) To begin, note that there is clearly no (strong) deformation retract of onto , else the composition with the deformation retraction of to a point as noted in (b) would yield a deformation retraction of to a point. This contradicts the result of (b). As such, it’s enough to show that the “contraction” of onto the zigzag subspace is a (weak) deformation retraction. To that end, let be the family which fixes the space and takes each “leaf” and shrinks its length by a factor of , i.e.

for all (see [3]).

Continuity here is immediate, and because the three conditions outlined in Exercise 0.4 for weak deformation retractions are clearly met, the result follows.

1. The path-connectedness of is easily shown by an analysis of cases for arbitrary points . If , the result is immediate. If one of the pair is in and the other isn’t (WLOG, and ), then the path works. Finally, if and , then the path works as well.

2. Ideas for the final part of this proof were borrowed from the document found here: http://www.math.uchicago.edu/~may/263/ps1.pdf.

3. Here, we’re identifying with the union where for all . In particular, the notation is a sloppy, lazy way of designating which copy of we’re considering.

Your map in (b) isn’t continuous, as it keeps the points in fixed and moves the straws. In fact, if it were continuous, it would be a deformation retraction in the strong sense, and we know this is not possible.

Hmmmmm. I seem to have overlooked something when I was attempting to understand what needed to happen in this problem originally. That now seems silly.

Would you perhaps be willing to shed some insight on a “not broken” way to proceed with this part of the problem?

I appreciate both your oversight and your insight.

Hi, did you ever figure out this issue in part b)? I am having trouble coming up with a continuous homotopy as well.

I honestly didn’t even think about it after the other comment which pointed out my dumbness. If I had to think about it (which I don’t

reallywant to do but will for 2 seconds because you asked so nicely ^_^), I’d say that because the previous comment indicates badness in the fact that the points in are kept fixed while the straws are moved, perhaps a solution would be shrink at the same time the straws are being shrunk? Imagine unfolding the zig-zag space into a copy of with “straws” periodically placed in alternating fashion (some pointing upward, some facing downward), and imagine using the same time parameter to retract the partandthe straws onto the part? Not sure; try this, and if it works, please do comment with any/all details!in last statement part b) you said that Y retracts on to a zig-zag Z that is contractible then it implies Y is contractible. This is not true in general and it is not clear in this content. It is good idea if you prove (c) first and use previous exercise ( i.e exercise 4) to show Y is homotopy equivalent to Z, it follows that Y is homotopy equivalent to a point.

If Y retracts on to a zig-zag Z that is contractible then it implies there is a continuous function F(deformation retract) from Y * I to into Y such that F(y,0)=y, F(Y,1)=Z for all y in Y.

And since Z is contractible, there is a continuous function H (contraction) from Z*I to Z.

Then for each y in Y, we can construct path product alpha_y by two functions F and H.

If we define new function K:Y*I -> Y such that K(y,t)=alpha_y (t) , then it will be continuous and a contraction of Y

Retractions and deformation retractions are not the same thing. A retraction is simply a continuous map from a space into a subspace that is the identity when restricted to that subspace. Every space retracts to a single point set (by the constant map). By this logic, every space is contractible.

It seems to me that in the fourth paragraph seventh line of part b that the solution should read (r,0) instead of (0,r) because we want the height of the straws to become 0.