Math4202 Topology II (Lecture 24)

Algebraic Topology

Deformation Retracts and Homotopy Type

Recall from last lecture, let $h,k:(X,x_0)\to (Y,y_0)$ be continuous maps. If there exists a homotopy of $h,y$ such that $H:X\times I\to Y$ that $H(x_0,t)=y_0$.

Then $h_*=k_*:\pi_1(X,x_0)\to \pi_1(Y,y_0)$.

We can prove this by showing that all the loop $f:I\to X$ based at $x_0$, $h_*([f])=k_*([f])$.

That is $[h\circ f]=[k\circ f]$.

This is a function $I\times I \to Y$ by $(s,t)\mapsto H(f(s),t)$.

We need to show that this is a homotopy between $h\circ f$ and $k\circ f$.

Theorem

The Inclusion map $j:S^n\to \mathbb{R}^n-\{0\}$ induces on isomorphism of fundamental groups

The function is injective.

Recall we showed that $S^1\to \mathbb{R}-\{0\}$ is injective by $x\mapsto \frac{x}{|x|}$.

We want to show that $j_*\circ r_*=id_{\pi_1(S^n)}\quad r_*\circ j_*=id_{\pi_1(\mathbb{R}^n-\{0\})}$, then $r_*$, $j_*$ are isomorphism.

Definition of deformation retract

Let $A$ be a subspace of $X$, we say that $A$ is a deformation retract of $X$ if the identity map of $X$ is homotopic to a map that carries all $X$ to $A$ such that each point of $A$ remains fixed during the homotopy.

Equivalently, there exists a homotopy $H:X\times I\to X$ such that:

• $H(x,0)=x$ forall $x\in X$
• $H(a,t)=a$ for all $a\in A$, $t\in I$
• $H(x,1)\in A$ for all $x\in X$

Equivalently,

$r:H(x,1):X\to A$ is a retract.

If we let $j:A\to X$ be the inclusion map, then $r\circ j=id_A$, and $j\circ r\sim id_X$ (with $A$ fixed.)

Theorem for Deformation Retract

If $A$ is a deformation retract of $X$, then $A$ and $X$ have the same fundamental group.