Hatcher 0.13

Hatcher, Algebraic Topology, Chapter 0

13. Show that any two deformation retractions r_t^0 and r_t^1 of a space X onto a subspace A can be joined by a continuous family of deformation retractions r_t^s, 0\leq s\leq 1, of X onto A, where continuity means that the map X\times I\times I\to X sending (x,s,t) to r_t^s(x) is continuous.

Proof. Given a topological space X and a subspace A with two deformation retractions r_t^0,r_t^1:X\to A, consider the map R:X\times I\times I\to X for which

R(x,s,t)=r_t^s(x)=sr_t^1(x)+(1-s)r_t^0(x) for all x\in X, s,t\in I.

It’s clear that the map R:(x,s,t)\mapsto r_t^s(x) is continuous due to the fact that it’s a straight-line continuous mapping of one continuous map (x,t)\mapsto r_t^0(x) to another (x,t)\mapsto r_t^1(x). In order to show that the map R defines a deformation retraction, note that for arbitrary s\in I, R(x,s,t)=sr_t^1(x)+(1-s)r_t^0(x) is a linear combination of two deformation retractions. In particular, this means that for arbitrary s\in I,

  1. R(x,s,0)=sr_0^1+(1-s)r_0^0 = s1_X+(1-s)1_X=1_X
  2. R(x,s,1)(X)=sA+(1-s)A=A
  3. R(a,s,t)=sa+(1-s)a=a for all a\in A.

Here, not that property 1 above stems from the fact that r_0^0=1_X=r_0^1, property 2 from the fact that r_1^0(X)=A=r_1^1(X), and property 3 from the fact that r_t^0(A)=1_A=r_t^1(A) for all t\in I. Hence, as claimed, R(x,s,t) is a deformation retraction for arbitrary s\in I which connects r_t^0=R(x,0,t) to r_t^1=R(x,1,t), whereby the claim is proved.     \square

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9 thoughts on “Hatcher 0.13

    • You surely are correct, sir. Thanks for taking the time to respond to this; I’ll have an updated solution posted as soon as I can make the time.

      I appreciate your oversight.

      • Anytime. These solutions have helped me studying for the algebraic portion of an Geometry exam I need to take and thought I could give back a little by sharing. I can share you my notes of the Chapter 1 and (especially Chapter 2) stuff when I get around to it (which should be soon). Unfortunately it will mostly be hand written stuff due to time constraints.

      • Anything you have would be hugely appreciated. I messed around and decided to be a topologist so I’m always combining filling in holes in my background with picking up new goodness.

  1. The proof the previous exercise of hatcher implies that image of an element with respect to an homotopy equivalance (between a function and the identity) should stay in the path-component of the element. As both maps are such maps, along the way, the image of $x$ with respect to both deformations lie in the same path component. This observation may help to solve the problem. Concerning your website. Your effort is worth to appreciate. Have a nice mathday.

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