diff --git a/content/Math4202/Math4202_L23.md b/content/Math4202/Math4202_L23.md index 8a953b5..4ff60f8 100644 --- a/content/Math4202/Math4202_L23.md +++ b/content/Math4202/Math4202_L23.md @@ -94,3 +94,48 @@ For this section, we will show that $h_*$ is an isomorphism. #### Lemma for equality of homomorphism Let $h,k: (X,x_0)\to (Y,y_0)$ be continuous maps. If $h$ and $k$ are homotopic, and if **the image of $x_0$ under the homotopy remains $y_0$**. The homomorphism $h_*$ and $k_*$ from $\pi_1(X,x_0)$ to $\pi_1(Y,y_0)$ are equal. + +
+Proof + +Let $H:X\times I\to Y$ be a homotopy from $h$ to $k$ such that +$$ +H(x,0)=h(x), \qquad H(x,1)=k(x), \qquad H(x_0,t)=y_0 \text{ for all } t\in I. +$$ + +To show $h_*=k_*$, let $[f]\in \pi_1(X,x_0)$ be arbitrary, where +$f:I\to X$ is a loop based at $x_0$, so $f(0)=f(1)=x_0$. + +Define +$$ +F:I\times I\to Y,\qquad F(s,t)=H(f(s),t). +$$ +Since $H$ and $f$ are continuous, $F$ is continuous. For each fixed $t\in I$, the map +$$ +s\mapsto F(s,t)=H(f(s),t) +$$ +is a loop based at $y_0$, because +$$ +F(0,t)=H(f(0),t)=H(x_0,t)=y_0 +\quad\text{and}\quad +F(1,t)=H(f(1),t)=H(x_0,t)=y_0. +$$ +Thus $F$ is a based homotopy between the loops $h\circ f$ and $k\circ f$, since +$$ +F(s,0)=H(f(s),0)=h(f(s))=(h\circ f)(s), +$$ +and +$$ +F(s,1)=H(f(s),1)=k(f(s))=(k\circ f)(s). +$$ + +Therefore $h\circ f$ and $k\circ f$ represent the same element of $\pi_1(Y,y_0)$, so +$$ +[h\circ f]=[k\circ f]. +$$ +Hence +$$ +h_*([f])=[h\circ f]=[k\circ f]=k_*([f]). +$$ +Since $[f]$ was arbitrary, it follows that $h_*=k_*$. +
diff --git a/content/Math4202/Math4202_L24.md b/content/Math4202/Math4202_L24.md index 1280e1c..cd89558 100644 --- a/content/Math4202/Math4202_L24.md +++ b/content/Math4202/Math4202_L24.md @@ -70,6 +70,12 @@ If we let $j:A\to X$ be the inclusion map, then $r\circ j=id_A$, and $j\circ r\s $S^1$ is a deformation retract of $\mathbb{R}^2-\{0\}$ +--- + +Consider $\mathbb{R}^2-p=q$, the doubly punctured plane. "The figure 8" space is the deformation retract. + +![Retraction of doubly punctured plane](https://notenextra.trance-0.com/Math4202/Retraction_of_doubly_punctured_plane.jpg) + #### Theorem for Deformation Retract diff --git a/content/Math4202/Math4202_L25.md b/content/Math4202/Math4202_L25.md new file mode 100644 index 0000000..11492f1 --- /dev/null +++ b/content/Math4202/Math4202_L25.md @@ -0,0 +1,89 @@ +# 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 $X$. + +--- + +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](https://notenextra.trance-0.com/Math4202/Math4202_L23/#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 + +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 $g_*$ diff --git a/content/Math4202/_meta.js b/content/Math4202/_meta.js index a12baad..63ad3d7 100644 --- a/content/Math4202/_meta.js +++ b/content/Math4202/_meta.js @@ -30,4 +30,5 @@ export default { Math4202_L22: "Topology II (Lecture 22)", Math4202_L23: "Topology II (Lecture 23)", Math4202_L24: "Topology II (Lecture 24)", + Math4202_L25: "Topology II (Lecture 25)", } diff --git a/public/Math4202/Retraction_of_doubly_punctured_plane.jpg b/public/Math4202/Retraction_of_doubly_punctured_plane.jpg new file mode 100644 index 0000000..8b1315e Binary files /dev/null and b/public/Math4202/Retraction_of_doubly_punctured_plane.jpg differ