# 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.     $\square$

## 9 thoughts on “Hatcher 0.13”

1. name.

This is not correct in non-convex subspaces of R^n. You cannot do the straight line homotopy for an arbitrary topological space, dummy!

• Indeed, you are correct. Thanks for taking the time to interject your thoughts.

2. Eric

Multiplication of elements of $X$ by a number in $\mathbb R$ isn’t even defined. I don’t see how this function is supposed to work.

• 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.

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.