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!

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