2.3 KiB
Math4202 Topology II (Lecture 19)
Exam announcement
Cover from first lecture to the fundamental group of circle.
Algebraic Topology
Retraction and fixed point
Definition of retraction
If A\subseteq X, a retraction of X onto A is a continuous map r:X\to A such that r|_A is the identity map of A.
When such a retraction r exists, A is called a retract of X.
Example
Identity map is a retraction of X onto X.
X=\mathbb{R}^2, A=\{0\}, the constant map that maps all points to (0,0) is a retraction of X onto A.
This can be generalized to any topological space, take A as any one point set in X.
Let X=\mathbb{R}^2, A=\mathbb{R}, the projection map that maps all points to the first coordinate is a retraction of X onto A.
Can we retract
\mathbb{R}^2to a circle?
Let \mathbb{R}^2\to S^1
This can be done in punctured plane. \mathbb{R}^2\setminus\{0\}\to S^1. by \vec{x}\mapsto \vec{x}/\|x\|.
But
Lemma for retraction
If A is a retract of X, the homomorphism of fundamental groups induced by the inclusion map j:A\to X, with induced j_*:\pi_1(A,x_0)\to \pi_1(X,x_0) is injective.
Proof
Let r:X\to A be a retraction. Consider j:A\to X, r:X\to A. Then r\circ j(a)=r(a)=a. Therefore r\circ j=Id_A.
Then r_*\circ j_*=Id_{\pi_1(A,x_0)}.
\forall f\in \ker j_*, j_*f=0. r_*\circ j_*f=Id_{f}=f, therefore f=0.
So \ker j_*=\{0\}.
So it is injective.
Consider the \mathbb{R}^2\to S^1 example, if such retraction exists, j_*:\pi_1(S^1,x_0)\to \pi_1(\mathbb{R}^2,x_0) is injective. But the fundamental group of circle is \mathbb{Z} whereas the fundamental group of plane is 1. That cannot be injective.
Corollary for lemma of retraction
There is no retraction from \mathbb{R}^2, B_1(0)\subseteq \mathbb{R}^2 (unit ball in \mathbb{R}^2), to S^1.
Lemma
Let h:S^1\to X be a continuous map. The following are equivalent:
his null-homotopic (his homotopic to a constant map).hextends to a continuous map fromB_1(0)\to X.h_*is the trivial group homomorphism of fundamental groups (Image of\pi_1(S^1,x_0)\to \pi_1(X,x_0)is trivial group, identity).