Hatcher 0.6

Hatcher, Algebraic Topology, Chapter 0

6. (a) Let X be the subspace of \mathbb{R}^2 consisting of the horizontal segment [0,1]\times\{0\} together with the vertical segments \{r\}\times[0,1-r] for r a rational number in [0,1]. Show that X deformation retracts to any point in the segment [0,1]\times\{0\}, but not to any other point (see the preceding problem).

(b) Let Y be the subspace of \mathbb{R}^2 that is the union of an infinite number of copies of X arranged as in the figure below. Show that Y is contractible but does not deformation retract onto any point.

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

Hatcher 0.6
Figure 1
The spaces mentioned above

Proof. (a) Let X be the subspace of \mathbb{R}^2 shown on the far left of Figure 1 consisting of the horizontal segment [0,1]\times\{0\} together with the vertical segments \{r\}\times[0,1-r] for r a rational number in [0,1]. First, note that the segment J=[0,1]\times\{0\}\cong[0,1] deformation retracts onto any point x_0\in J by way of the straight-line homotopy r_t which fixes x_0 for all t\in I and which linearly “squeezes” the intervals [0,x_0) and (x_0,1] towards their right and left endpoints, respectively. Algebraically, then, r_t is the family of maps for which

[0,x_0)\mapsto [(1-t)x_0,x_0) and (x_0,1]\mapsto (x_0,x_0+t(1-x_0)].

Next, note that the space X deformation retracts onto the space J\cap\mathbb{Q} in a fashion similar to the one above, namely by way of the family of maps which, for each rational r\in\mathbb{Q}, maps the segment \{r\}\times[0,1-r] to the point (r,0) continuously with respect to t as t ranges through I=[0,1]. Denoting this family of maps s_t, it follows that the space X deformation retracts to any point x_0\in J by way of the composition \Gamma_t, where

\Gamma_t = \left\{       \begin{array}{rl}         r_{1-2t}, & 0\leq t\leq 1/2\\         s_{2t-1}, & 1/2\leq t\leq 1       \end{array}     \right..

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

Note that X is path-connected1 but that X\setminus J clearly isn’t path-connected. In particular, suppose U is a neighborhood of x in X\setminus J. Then U can be thought of as an open ball in \mathbb{R}^2, disjoint from J, and intersecting a countably infinite number of line segments, all of which are disjoint. Thus, U fails to be path-connected. For the same reason, any neighborhood V\subset U containing x must also be path-disconnected, whereby it follows that V cannot contract to a point. Hence, \iota:V\to U cannot be homotopic to the constant map, contradicting the result of 0.5.

 

(b) Let Y be the subspace of \mathbb{R}^2 shown in the rightmost part of Figure 1 consisting of an infinite number of copies of X arranged as above. The fact that Y 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 y\in Y, there exists neighborhood U of y in Y that is necessarily path-disconnected. This claim follows immediately from (a) if y fails to lie on the zigzag subspace Z\subset Y (see part (c)). In the case that y\in Z, this can be heuristically described in cases as follows.

Suppose y\in Z fails to be one of the “vertex points”. Any neighborhood U\ni y in Y intersects two “fundamentally different” countably infinite disjoint collections of line segments of Y, one collection which is directed “bottom-left-to-top-right” – call it C_1 – and one which is directed “top-left-to-bottom-right” – call it C_2. In this case, any point on a segment in C_1 cannot be connected to any point in C_2 because of the fact that Z intersects only one of C_i, i=1,2.

The argument for y a “vertex point” of Z is similar with the exception that an open neighborhood U of y now intersects three fundamentally different collections of line segments, two of which – say D_1 and D_2 – are directed in a parallel fashion (either “bottom-left-to-top-right” or “top-left-to-bottom-right”) with the third, D_3, directed in the other fashion. In this case, any point in D_i, i=1,2, can be connected to any point in D_3 by a path due to the fact that Z intersects D_3 and one of D_i, i=1,2. On the other hand, Z intersections precisely one of D_1 or D_2, meaning that no path in Y exists connecting points in D_1 to D_2.

Now, it suffices to show that Y is contractible. Note, first, that Z\cong\mathbb{R} is clearly contractible. Next, note that the space Y can be retracted (not deformation retracted) onto Z by way of a modified version of the family r_t given in part (a). More precisely, the map r:Y\to Z is a retraction, where r is the constant map on Z and where r maps any “straw” \{r\}_i\times[0,1-r]_i onto the point (0,r)_i\in Z_i, where the subscript _i here denotes the elements in the i^\text{th} copy of X.3 Therefore, it’s been shown that Y retracts onto the zigzag Z which is a contractible subspace, thereby proving that Y itself is contractible.

 

(c) To begin, note that there is clearly no (strong) deformation retract of Y onto Z, else the composition with the deformation retraction of Z\cong\mathbb{R} to a point as noted in (b) would yield a deformation retraction of Y to a point. This contradicts the result of (b). As such, it’s enough to show that the “contraction” of Y onto the zigzag subspace Z\subset Y is a (weak) deformation retraction. To that end, let s_t be the family which fixes the space Z and takes each “leaf” \{r\}_i\times[0,1-r]_i and shrinks its length by a factor of t, i.e.

s_t:\{r\}_i\times[0,1-r]_i\mapsto \{r\}_i\times[0,t(1-r)]_i for all t\in I,i\in\mathbb{Z} (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.     \square


   1. The path-connectedness of X is easily shown by an analysis of cases for arbitrary points x\neq y\in X. If x,y\in J, the result is immediate. If one of the pair is in J and the other isn’t (WLOG, x\in J and y\in\{r\}\times[0,1-r]), then the path y\mapsto (r,0)\mapsto x works. Finally, if x\in\{r_1\}\times[0,1-r_1] and y\in\{r_2\}\times[0,1-r_2], then the path y\mapsto (r_2,0)\mapsto(r_1,0)\mapsto x 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 Y with the union \cup_{i\in\mathbb{Z}}X_i where X_i=X for all i\in\mathbb{Z}. In particular, the _i notation is a sloppy, lazy way of designating which copy X_i of X we’re considering.

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8 thoughts on “Hatcher 0.6

  1. Your map in (b) isn’t continuous, as it keeps the points in Z 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.

      • 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 really want 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 Z are kept fixed while the straws are moved, perhaps a solution would be shrink Z at the same time the straws are being shrunk? Imagine unfolding the zig-zag space into a copy of \mathbb{R} with “straws” periodically placed in alternating fashion (some pointing upward, some facing downward), and imagine using the same time parameter to retract the \mathbb{R} part and the straws onto the \mathbb{R} part? Not sure; try this, and if it works, please do comment with any/all details!

  2. 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.

  3. 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.

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