NoteNextra-origin/content/Math4202/Math4202_L25.md

Math4202 Topology II (Lecture 25)

Algebraic Topology

Deformation Retracts and Homotopy Type

Recall from last lecture, Let A\subseteq X, if there exists a continuous map (deformation retraction) H:X\times I\to X such that

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

then the inclusion map$\pi_1(A,a)\to \pi_1(X,a)$ is an isomorphism.

Example for more deformation retract

Let X=\mathbb{R}^3-\{0,(0,0,1)\}.

Then the two sphere with one point intersect is a deformation retract of X.

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Let X be \mathbb{R}^3-\{(t,0,0)\mid t\in \mathbb{R}\}, then the cyclinder is a deformation retract of X.

Definition of homotopy equivalence

Let f:X\to Y and g:Y\to X be a continuous maps.

Suppose

• the map g\circ f:X\to X is homotopic to the identity map \operatorname{id}_X.
• the map f\circ g:Y\to Y is homotopic to the identity map \operatorname{id}_Y.

Then f and g are homotopy equivalences, and each is said to be the homotopy inverse of the other.

X and Y are said to be homotopy equivalent.

Example

Consider the punctured torus X=S^1\times S^1-\{(0,0)\}.

Then we can do deformation retract of the glued square space to boundary of the square.

After glueing, we left with the figure 8 space.

Then X is homotopy equivalent to the figure 8 space.

Recall the lemma, Lemma for equality of homomorphism

Let f:X\to Y and g:X\to Y, with homotopy H:X\times I\to Y, such that

• H(x,0)=f(x) for all x\in X
• H(x,1)=g(x) for all x\in X
• H(x,t)=y_0 for all t\in I, and y_0\in Y is fixed.

Then f_*=g_*:\pi_1(X,x_0)\to \pi_1(Y,y_0) is an isomorphism.

We wan to know if it is safe to remove the assumption that y_0 is fixed.

Idea of Proof

Let k be any loop in \pi_1(X,x_0).

We can correlate the two fundamental group f\cric k by the function \alpha:I\to Y, and \hat{\alpha}:\pi_1(Y,y_0)\to \pi_1(Y,y_1). (suppose f(x_0)=y_0, g(x_0)=y_1), it is sufficient to show that

f\circ k\simeq \alpha *(g\circ k)*\bar{\alpha}

Lemma of homotopy equivalence

Let f,g:X\to Y be continuous maps. let f(x_0)=y_0 and g(x_0)=y_1. If f and g are homotopic, then there is a path \alpha:I\to Y such that \alpha(0)=y_0 and \alpha(1)=y_1.

Defined as the restriction of the homotopy to \{x_0\}\times I, satisfying \hat{\alpha}\circ f_*=g_*.

Imagine a triangle here:

• \pi_1(X,x_0)\to \pi_1(Y,y_0) by f_*
• \pi_1(Y,y_0)\to \pi_1(Y,y_1) by \hat{\alpha}
• \pi_1(Y,y_1)\to \pi_1(X,x_0) by g_*