Solutions of the problems in the section 1 of GTM 81

Problem $1.1$. (a) We have $\infty\not\in\emptyset$ and $\emptyset$ be open set in $\mathbb{R}^n$ hence $\emptyset$ be open set in $X$, we have also that $X$ be open in $X$ because $\infty \in X$ and $X-X=\emptyset$ be compact set in $\mathbb{R}^n$.
Assume that $U,V$ are open sets in $X$, if $\infty\not\in U$ or $\infty\not\in V$ then $\infty\not\in U\cap V$ and $U\cap V$ is an open set in $\mathbb{R}^n$ therefore $U\cap V$ is open set in $X$. If $\infty\in U$ and $\infty\in V$ then $U=\{\infty\}\cup U',V=\{\infty\}\cup V'$, here $\infty\not\in U'\cup V'$ and $\mathbb{R}^n-U',\mathbb{R}^n-V'$ are compact sets in $\mathbb{R}^n$. We have $X-(U\cap V)=\mathbb{R}^n-(U'\cap V')=(\mathbb{R}^n-U')\cup (\mathbb{R}^n-V')$ is compact set in $\mathbb{R}^n$ and $\infty\in U\cap V$ therefore $U\cap V$ be open set in $X$.
Assume that $\{U_i\}_{i\in I}$ is a family of open sets in $X$ and $\infty\in U_i\forall i \in J,\infty\not\in U_i\forall i\in I-J$ for some $J\subseteq I$ (maybe empty). If $\infty\not\in\cup_{i\in I}U_i$ then $\cup_{i\in I}U_i$ be open in $\mathbb{R}^n$, and so open in $X$. If $\infty\in\cup_{i\in I}U_i$ then $X-\cup_{i\in I}U_i=[\cap_{i\in J}(\mathbb{R}^n-U_i)]\cap [\cap_{i\in I-J}(\mathbb{R}^n-(U_i-\{\infty\}))]$ is closed and bound in $\mathbb{R}^n$, also it is compact set in $\mathbb{R}^n$, therefore $\cup_{i\in I}U_i$ is open set in $X$.
Thanks to all conditions above we have $X$ is a topological space.
Assume $x\not=y$ are elements in $X$. If $\infty \not\in \{x,y\}$ then $x,y\in\mathbb{R}^n$, note that because $\mathbb{R}^n$ is Hausdorff space we can find open sets $U_x,U_y$ in $\mathbb{R}^n$ such that $x\in U_x,y\in U_y$ and $U_x\cap U_y=\emptyset$(Note: $U_x$ and $U_y$ are also open sets in $X$). If $\infty\in \{x,y\}$ , assume that $y=\infty$, denote $U=S(x;1)=\{z\in\mathbb{R}^n|||z-x||<1\},\overline{U}=\{z\in\mathbb{R}^n|||z-x||\leq 1\}$ and $V=\{\infty \}\cup (\mathbb{R}^n-\overline{U})$. Then $U,V$ are open sets in $X$, $x\in U,y\in V$ and $U\cap V=\emptyset$.
Compactness of $X$ is a immediately follows from (b).
(b)We have $\sigma$ be a surjective because $\sigma (0,\cdots,0,1)=\infty$ and $\sigma (x_1,x_2,\cdots,x_{n+1})=(y_1,y_2,\cdots,y_n)\forall (y_i)\in\mathbb{R}^n$, where $x_i=\dfrac{2y_i}{k+1}(i=\overline{1,n}),x_{n+1}=\dfrac{k-1}{k+1},k=\sum_{i=1}^ny_i^2$. But it is easy to see that $\sigma$ is an injective(checking $\sigma (x)=\sigma (y)$ then $x=y$), therefore $\sigma$ is a bijective. By above formular, we have $\sigma$ is a homeomorphism.($x_n\to x_0\Leftrightarrow \sigma (x_n)\to\sigma (x_0)$ in topology of $\mathbb{R}^n$)

Problem $1.2$. We have $D=\{z\in\mathbb{C}|cz+d\not =0\}$ is an open set in $\mathbb{C}$ and $f'(z)=\dfrac{ad-bc}{(cz+d)^2}\forall z\in D$ therefore $f$ is holomorphic on $D$.
If $c=0$ then $ad\not=0$ we definition $f:\mathbb{P}^1\to\mathbb{P}^1$ by setting $f(\infty)=\infty$.
If $c\not =0$ then we definition $f:\mathbb{P}^1\to\mathbb{P}^1$ by setting $f(\infty)=\dfrac{a}{c}$ and $f(-\dfrac{d}{c})=\infty$.
In both cases we have $f$ is continuous over $\mathbb{P}^1$ and $f$ is a bijective, it is easy to find formular of $f^{-1}$ and check $f$ and $f^{-1}$ are holomorphic.

Problem $1.3$.

Problem $1.4$. The problem is wrong, an example $\omega_1=1,\omega_2=i$ and $\omega_1'=\sqrt{2}+i,\omega_2'=1+i\sqrt{2}$ and $A=\left(\begin{matrix}\sqrt{2}&1\\1&\sqrt{2}\end{matrix}\right)$. Then $\left(\begin{matrix}\omega_1'\\ \omega_2'\end{matrix}\right)=A\left(\begin{matrix}\omega_1\\ \omega_2\end{matrix}\right)$ and $\omega_1'\not\in \Gamma$. In my opinion, $A$ must is belong to $\text{SL}(2;\mathbb{Z})$.
Problem $1.5$. (a)Assume that $U$ is open in $\mathbb{C}/\Gamma'$ and $f$ is original map. We have $\pi^{-1}(f^{-1}(U))=\{y\in\mathbb{C}|\pi (y)\in f^{-1}(U)\}=\{y\in\mathbb{C}|\pi'(\alpha y)\in U\}$ and $\pi'^{-1}(U)$ is open in $\mathbb{C}$ therefore $\pi^{-1}(f^{-1}(U))$ is open in $\mathbb{C}$, so $f^{-1}(U)$ is open in $\mathbb{C}/\Gamma$, or $f$ is continuous.
If $\psi_1:U_1\to V_1$ and $\psi_2:U_2\to V_2$ are complex charts of $\mathbb{C}/\Gamma$ and $\mathbb{C}/\Gamma'$, respectively, such that $f(U_1)\subset U_2$. We have $\psi_2\cdot f\cdot \psi_1^{-1}(z)=\psi_2\cdot f ([z]_\Gamma ) =\psi_2 ([\alpha z]_{\Gamma'})=\alpha z\forall z\in V_1$ therefore $f$ is holomorphic.
If $f$ is biholomorphic then $f$ must is bijection and $\alpha \Gamma =\Gamma'$. In fact, if $x\in \Gamma'-\alpha \Gamma$ then $f([\dfrac{1}{\alpha}x])=f([0])$ but $[\dfrac{1}{\alpha}x]\not =[0]$, contradiction!
If $\alpha \Gamma =\Gamma'$ then $f$ is bijective and $\dfrac{1}{\alpha}\Gamma'\subseteq \Gamma$ therefore by above, $f$ and $f^{-1}$ are holomorphic.
(b)We have $\Gamma =\mathbb{Z}\omega_1+\mathbb{Z}\omega_2=\omega_1(\mathbb{Z}+\mathbb{Z}\cdot\dfrac{\pm\omega_2}{\omega_1})$ and use (a).
(c)Use (a) again!