diff --git a/content/Math4202/Math4202_L6.md b/content/Math4202/Math4202_L6.md index e173b04..dc988eb 100644 --- a/content/Math4202/Math4202_L6.md +++ b/content/Math4202/Math4202_L6.md @@ -1,4 +1,4 @@ -# Math4202 Topology II (Lecture 5) +# Math4202 Topology II (Lecture 6) ## Manifolds diff --git a/content/Math4202/Math4202_L7.md b/content/Math4202/Math4202_L7.md new file mode 100644 index 0000000..e8b32db --- /dev/null +++ b/content/Math4202/Math4202_L7.md @@ -0,0 +1,63 @@ +# Math4202 Topology II (Lecture 6) + +## Algebraic Topology + +Classify 2-dimensional topological manifolds (connected) up to homeomorphism/homotopy equivalence. + +Use fundamental groups. + +We want to show that: + +1. The fundamental group is invariant under the equivalence relation. +2. develop some methods to compute the groups. +3. 2-dimensional topological spaces with the same fundamental group are equivalent (homeomorphism). + +### Homotopy of paths + +#### Definition of path + +If $f$ and $f'$ are two continuous maps from $X$ to $Y$, where $X$ and $Y$ are topological spaces. Then we say that $f$ is homotopic to $f'$ if there exists a continuous map $F:X\times [0,1]\to Y$ such that $F(x,0)=f(x)$ and $F(x,1)=f'(x)$ for all $x\in X$. + +The map $F$ is called a homotopy between $f$ and $f'$. + +We use $f\simeq f'$ to mean that $f$ is homotopic to $f'$. + +#### Definition of homotopic equivalence map + +Let $f:X\to Y$ and $g:Y\to X$ be two continuous maps. If $f\circ g:Y\to Y$ and $g\circ f:X\to X$ are homotopic to the identity maps $\operatorname{id}_Y$ and $\operatorname{id}_X$, then $f$ and $g$ are homotopic equivalence maps. And the two spaces $X$ and $Y$ are homotopy equivalent. + +> [!NOTE] +> +> This condition is weaker than homeomorphism. (In homeomorphism, let $g=f^{-1}$, we require $g\circ f=\operatorname{id}_X$ and $f\circ g=\operatorname{id}_Y$.) + +
+Example of homotopy equivalence maps + +Let $X=\{a\}$ and $Y=[0,1]$ with standard topology. + +Consider $f:X\to Y$ by $f(a)=0$ and $g:Y\to X$ by $g(y)=a$, where $y\in [0,1]$. + +$g\circ f=\operatorname{id}_X$ and $f\circ g=[0,1]\mapsto 0$. + +$g\circ f\simeq \operatorname{id}_X$ + +and $f\circ g\simeq \operatorname{id}_Y$. + +Consider $F:X\times [0,1]\to Y$ by $F(a,0)=0$ and $F(a,t)=(1-t)y$. $F$ is continuous and homotopy between $f\circ g$ and $\operatorname{id}_Y$. + +This gives example of homotopy but not homeomorphism. + +
+ +#### Definition of null homology + +If $f:X\to Y$ is homotopy to a constant map. $f$ is called null homotopy. + +#### Definition of path homotopy + +Let $f,f':I\to X$ be a continuous maps from an interval $I=[0,1]$ to a topological space $X$. + +Two pathes $f$ and $f'$ are path homotopic if + +- there exists a continuous map $F:I\times [0,1]\to X$ such that $F(i,0)=f(i)$ and $F(i,1)=f'(i)$ for all $i\in I$. +- $f(0)=f'(0)$ and $f(1)=f'(1)$. \ No newline at end of file diff --git a/content/Math4202/_meta.js b/content/Math4202/_meta.js index 55468de..349f627 100644 --- a/content/Math4202/_meta.js +++ b/content/Math4202/_meta.js @@ -9,4 +9,5 @@ export default { Math4202_L4: "Topology II (Lecture 4)", Math4202_L5: "Topology II (Lecture 5)", Math4202_L6: "Topology II (Lecture 6)", + Math4202_L7: "Topology II (Lecture 7)", } diff --git a/content/Math4302/Math4302_L7.md b/content/Math4302/Math4302_L7.md new file mode 100644 index 0000000..1ceb228 --- /dev/null +++ b/content/Math4302/Math4302_L7.md @@ -0,0 +1,83 @@ +# Math4302 Modern Algebra (Lecture 7) + +## Subgroups + +### Cyclic group + +Last time, let $G$ be a group and $a\in G$. $|\langle a\rangle|=$ smallest positive $n$ such that $a^n=e$. + +$\langle a\rangle=\{a^0,a^1,a^2,\cdots,a^{n-1}\}$. + + +#### Lemma subgroup of cyclic group is cyclic + +Every subgroup of a cyclic group is cyclic. + +$G=\langle a\rangle$. + +
+Proof + +Let $H\leq G$ be a subgroup. + +If $H=\{e\}$, we are done. + +Otherwise, let $m$ be the smallest positive integer such that $a^m\in H$. We claim $H=\langle a^m\rangle$. + +- $\langle a^m\rangle\subseteq H$. trivial since $a^m\in H$ and $H$ is a subgroup. +- $H\subseteq\langle a^m\rangle$. Suppose $a^k\in H$, need to show $a^k\in \langle a^m\rangle$ + Divide $k$ by $m$: $k=qm+r$, $0\leq r\leq m-1$, Then $a^k\in H\implies a^{qm+r}\in H$. Also $a^m\in H$, then $(a^m)^q\in H$, so $a^mq\in H$, $a^-mq\in H$, so $a^{k}a^{-mq}\in H$, so $a^r\in H$, so $r$ has to be zero. + By our choice of $m$, $k=mq$, so $a^k=a^mq\in \langle a^m\rangle$. + +
+ +
+Example + +Every subgroup of $(\mathbb{Z},+)$ is of the form + +like the multiples of $n$: $n\mathbb{Z}=\langle n\rangle$ for some $n\geq 0$. + +In particular, if $n,m\geq 1$ are in $\mathbb{Z}$, then the subgroup $\{nr+ms|r,s\in \mathbb{Z}\}\leq \mathbb{Z}$. + +is equal to $d\mathbb{Z}$ where $d=\operatorname{gcd}(n,m)$. + +
+ +Skip $\operatorname{gcd}$ part, check for Math 4111 notes in this site. + + +#### Lemma for size of cyclic subgroup + +Let $G=\langle a\rangle$, $|G|=n$, and $H=\langle a^m\rangle\subseteq G$. Then $|H|=\frac{n}{d}$ where $d=\operatorname{gcd}(|G|,|H|)$. + +
+Proof + +Recall $|H|$ is the smallest power of $a^m$ which is equal to $e$. + +Let $d=\operatorname{gcd}(m,n)$, so $m=m_1d$, $n=n_1d$. and $\frac{n}{\operatorname{gcd}(m,n)}=n_1$, + +- $(a^m)^{n_1}=a^{mn_1}=a^{m_1dn_1}=a^{m_1n}=(a^n)^{m_1}=e$. +- If $(a^m)^k=e$, the $a^{mk}=e\implies$ $mk$ is a multiple of $n$, + - If $a^\ell=e$, divide $\ell$ by $n$, $\ell=nq+r$, $0\leq r\leq n-1$, then $e=a^\ell=a^{nq+r}=a^r$, $r$ has to be zero, so $a^\ell=a^r=e$. $n|\ell$. +- $n_1d|m_1dk$, but by the definition of smallest common divisor, $m_1,n_1$ should not have common divisor other than $1$. So $n_1|m_1k$, $n_1|k\implies k\geq n_1$. + +
+ +
+Example Applying the lemma + +Let $G=\langle a \rangle$, $|G|=6$, $H=\langle a^4\rangle$. Then $|H|=\frac{6}{d}=3$ where $d=\operatorname{gcd}(6,4)=2$. + +To check this we do enumeration $\langle a^4\rangle=\{e,a^4,a^2\}$. + +--- + +Find generator of $\mathbb{Z}_9$: + +Using the coprime, we have $g=\{1,2,4,5,7,8\}$. + +
+ +Corollary: $\langle a^m\rangle=G\iff |H|=n\iff \frac{n}{d}=n\iff \operatorname{gcd}(m,n)=1$ $m,n$ are coprime.