1. Topological Spaces

2. Basis of Topology

3. The Metric Topology

3. The Subspace topology

4. The Disjoint Union Topology

5. Continuity and Homeomorphism

6. The Product Topology

7. The Quotient Topology

Exercises

Find a homeomorphism $\varphi:\mathbb{R}^n\to\mathbb{R}^n$ satisfies:

$\varphi=1_{\mathbb{R}^n\setminus\mathring{D^n}}$.
$\varphi(0)=p$, $p\in\mathring{D^n}$.

Where $D^n$ denotes the set $\{x\in\mathbb{R}^n\colon\ |x|\leq 1\}$.

Suppose that $\varphi\colon S^1 \to S^1$ is a homeomorphism. Prove that there exists a homeomorphism $\psi\colon D^2 \to D^2$, satisfies $\psi|_{S^1}=\varphi$.

(Hopf Link)

$L=\\{(x,y)\in\mathbb{C}^2\colon\ x^2+y^2=0, |x|^2+|y|^2=1\\}$

Prove that $L\cong S^1\sqcup S^1$.
Describe the image of the stereographic projection of $L$.
For $(x,y)\in L$，we have
$\begin{aligned} &x^2+y^2 =0 \implies(y-ix)(y+ix)=0 \implies y=\pm{i}x \\\\ &|x|^2+|y|^2 =1 \implies |x|=|y|=\frac{\sqrt{2}}{2} \end{aligned}$
Let:
$\begin{aligned} U &=\\{(x,ix)\in\mathbb{C}^2\colon\ |x|=\frac{\sqrt{2}}{2}\\} \\\\ V &=\\{(x,-ix)\in\mathbb{C}^2\colon\ |x|=\frac{\sqrt{2}}{2}\\} \end{aligned}$
we have $L=U\sqcup V$ and we can simply construct homeomorphisms $(x,ix)\mapsto (\sqrt{2}\operatorname{Re}(x),\sqrt{2}\operatorname{Im}(x))$, and $(x,-ix)\mapsto (\sqrt{2}\operatorname{Re}(x),\sqrt{2}\operatorname{Im}(x))$ from $U\to S^1$ and $V\to S^1$ respectively, so we have the conclusion $L=U\sqcup V \cong S^1\sqcup S^1$.
Consider the unit 3-sphere $\{(x,y)\in\mathbb{C}^2\colon\ |x|^2+|y|^2=1\}$, the stereographic projection $P$ of the sphere from the point $(0,1)$ maps each point $(x,y)$ differs from $(0,1)$ of the sphere to the point of intersection of the real line joining $(x,y)$ and $(0,1)$ with the 3-space $\{(p,q)\in\mathbb{C}^2\colon\ \operatorname{Re}(q)=0\}$. Expicitly, when $\operatorname{Re}(y)\neq 1$:
$P\colon\quad (x,y)\mapsto (\frac{\operatorname{Re}(x)}{1-\operatorname{Re}(y)},\frac{\operatorname{Im}(x)}{1-\operatorname{Re}(y)},\frac{\operatorname{Im}(y)}{1-\operatorname{Re}(y)})$
Since for each points $(x,y)\in L$, $\operatorname{Re}(y)\neq 1$, we can find the image for $U$ and $V$ separately from the formula above:
$P(U)= \begin{cases} u=w\\\\ 2u^2+(v+1)^2=2 \end{cases}\quad P(V)= \begin{cases} u=-w\\\\ 2u^2+(v-1)^2=2 \end{cases}$
Equivalently:
$P(U)= \begin{cases} u=w\\\\ u^2+(v+1)^2+w^2=2 \end{cases}\quad P(V)= \begin{cases} u=-w\\\\ u^2+(v-1)^2+w^2=2 \end{cases}$
where $P(U)$ and $P(V)$ are two circles on two intersecting planes respectively (in this case the two planes are perpendicular), and both two centers are located on the line of the intersection of the two planes. Since the distance between centers which is $2$ is less than the sum of radii that is $2\sqrt{2}$, the two circles are going to be linked together, in the sense that neither can be continuously shrunk to a point without pcissing through the other one (see below).

If $(X,d)$ is a metric space and $A$ is a non-empty closed set of $X$, prove that $f(x)=\inf_{a\in A}\{d(x,a)\}$ is a continuous function on $X$ and also, $f(x)=0$ if and only if $x\in A$.

Let $\mathbb{Z}$ denotes the set of integers, $U\subseteq\mathbb{Z}$ is said to be symmetrical if $n\in U$ and $-n\in U$ are equivalent for $\forall n\in \mathbb{Z}$. We define a topology $\mathcal{T}_{\mathbb{Z}}$ on $\mathbb{Z}$, that the set of $\mathbb{Z}$ is open if it is symmetrical.

Prove that $\mathcal{T}_{\mathbb{Z}}$ is actually a topology.
If $A=\{-1,0,1,2\}\subseteq \mathbb{Z}$, determine $\mathring{A}$, $\bar{A}$ and $A’$.
Is $A$ an open set on $\mathbb{Z}$? Or is $A$ closed?

Prove that $(\mathbb{R}^{n+1}\setminus\{0\})\cong (S^n\times\mathbb{R}_{>0})$, where $S^n=\{x\in\mathbb{R}^{n+1}\colon\ |x|=1\}$.

(Zariski Topology) Subset $\mathbb{Z}$ of $\mathbb{C}^n$ is said to be Zariski closed if there exists polynomials $(Pi){i\in I}$ in $\mathbb{C}[x_1,x_2,\dots,x_n]$, such that $Z=\{z\in\mathbb{C}^n\colon\ P_i(z)=0, \forall i\in I\}$. If the complement of $U\subseteq \mathbb{C}^n$ is Zariski closed, we say $U$ is Zariski open. Prove the following:

All Zariski open sets in $\mathbb{C}^n$ forms a topology on $\mathbb{C}^n$, which is called Zariski topology.
$U_f=\{x\in\mathbb{C}^n\colon f(x)\neq{0}\}$ where $f\in\mathbb{C}[x_1,x_2,\dots,x_n]$, forms a topology basis of Zariski topology on $\mathbb{C}^n$.
Suppose that $f_1,f_2,\dots,f_m$ are polynomials of $\mathbb{C}[x_1,x_2,\dots,x_n]$. Prove the map: $(x_1,x_2,\dots,x_n)\mapsto(f_1(x),f_2(x),\dots,f_m(x))$ is continuous if we consider the Zariski topology on both $\mathbb{C}^n$ and $\mathbb{C}^m$.