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


j_*:\pi_1(S^n)\to \pi_1(\mathbb{R}^n-\{0\})

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.

Proof

Homotopy is well defined.

Consider H:(\mathbb{R}^n-\{0\})\times I\to \mathbb{R}^n-\{0\}.

Given (x,t)\mapsto tx+(1-t)\frac{x}{\|x\|}.

Note that (t-\frac{1-t}{\|x\|})x=0\implies t=0\land t=1.

So this map is well defined.

Base point is fixed.

On point (1,0) (or anything on the sphere), H(x,0)=x.

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.)

Example of deformation retract

S^1 is a deformation retract of \mathbb{R}^2-\{0\}

Theorem for Deformation Retract

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

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