Hello,

I solved the following exercise (but i am not sure for the solution).

Can you give me a hand?Thank you.

Given the square $\displaystyle Q$

$\displaystyle I_1 = \{(x, y) \in R ^ 2 | x =- 1, -1 \leq y \leq 1 \} $

$\displaystyle I_2 = \{(x, y) \in R ^ 2 | x = 1, -1 \leq y \leq 1 \} $

$\displaystyle U_1 = \{(x, y) \in R ^ 2 | y = 1, -1 \leq x \leq 1 \} $

$\displaystyle U_2 = \{(x, y) \in R ^ 2 | y =- 1, -1 \leq x \leq 1 \} $

Let $\displaystyle Q = I_1 \cup I_2 \cup U_1 \cup U_2 $ where we define the following equivalence relation:

$\displaystyle \A (x, y), (x ', y') \in Q $ $\displaystyle (x, y) \sim (x ', y') $ where $\displaystyle (x, y) = (x ', y') $ or $\displaystyle (x, y), (x ', y') \in I_1 $

$\displaystyle (x, y), (x ', y') \in I_2 $

a) Show that $\displaystyle Q /\sim $ is isomorphic to $\displaystyle S ^ 1 $.

b) Prove that $\displaystyle \pi: Q\rightarrow Q /\sim $ is closed.

c) Prove that $\displaystyle \pi $ is not open.

Solution:

a) first of all we construct the two following applications

$\displaystyle f_1 (x, y) = \{((1, 0) \;\;\;if\;\;\; (x, y) \in I_1), ((-1, 0)\;\;\; if\;\;\; (x, y) \in I_2) \} $ and $\displaystyle f_2 (x, y) = (x, (y /| y |)sin(arccos (x )))$ where $\displaystyle (x, y) \in U_1, U_2 $.

Then we construct $\displaystyle f (x, y): Q \rightarrow S^1 $ pasting together $\displaystyle f_1 (x, y), f_2 (x, y) $.Since $\displaystyle f_1 $ and $\displaystyle f_2 $ satisfy the conditions of The Pasting Lemma $\displaystyle f $ will be continue. In addition it is bijective and the equivalence relation that induces $\displaystyle f $ coincides with the equivalence relation defined on $\displaystyle Q $.Then $\displaystyle Q /\sim$ is isomorphic to $\displaystyle S ^ 1 $.

b) $\displaystyle Q $ is connected and compact, so also $\displaystyle Q /\sim $ will be connected and compact. We must show at this point that $\displaystyle Q /\sim $ is $\displaystyle T_2 $ (Hausdorff). That is to prove that the set

$\displaystyle F = \{(x, y), (u, v): \pi (x, y) = \pi (u, v) \} = \{(x, y), (u, v): (x, y) \sim (u, v) \} $ is a closed $\displaystyle Q \times Q $.

$\displaystyle C_1 = \{(x, y), (x, y) \in Q \times Q | (x, y) \in Q\} $, $\displaystyle C_2 = \{(-1;y), (-1, y) \in Q \times Q | y \in [-1, 1] \} $

$\displaystyle C_3 = \{(1, y), (1, y) \in Q \times Q | y \in [-1, 1] \} $, $\displaystyle C_4 = \{(M_1, M_4), (M_4, M_1) (M_2, M_3), (M_3, M_2) \} $

where $\displaystyle M_1 = (-1,-1) = M_2 (-1,1), M_3 = (1,-1), M_4 = (1,1) $ are the vertices of the square.

$\displaystyle C_1 $ is closed because $\displaystyle Q $ is $\displaystyle T_2 $. $\displaystyle C_2 $, $\displaystyle C_3 $ are closed because they are closed subsets of a space $\displaystyle T_2 $ and images of the compact $\displaystyle [-1.1] $ of $\displaystyle R $ through continuous applications.$\displaystyle C_4 $ is closed because it is a finite set of points in space $\displaystyle T_2 $. Since $\displaystyle F = C_1 \cup C_2 \cup C_3 \cup C_4 $, then also $\displaystyle F $ will be closed, this means that $\displaystyle Q /\sim $ is $\displaystyle T_2 $. We conclude that $\displaystyle \pi $ is closed.

c) Then we know that if $\displaystyle A $ is an open set in $\displaystyle Q $ (in our case) is said to be an open application if $\displaystyle \pi ^ {-1} (\pi (A)) $ will be open. Then we take $\displaystyle A = A_1 \cup A_2 $ where $\displaystyle = A_1 (A \cap I_1) \cup (A \cap I_2) $ as a general case assuming that none of the parties are empty. Similarly $\displaystyle A_2 = (A \cap U_1) \cup (A \cap U_2) $. At this point we compute $\displaystyle \pi(A) = \pi(A_1) \cup \pi (A_2) = I_1 \cup I_2 \cup A_2 $ and then $\displaystyle \pi ^ {-1} (\pi (A)) = I_1 \cup I_2 \cup A_2 $ and since $\displaystyle I_1 $ and $\displaystyle I_2 $ are closed the application $\displaystyle \pi $ can not be open. In fact, the application will be closed (if $\displaystyle A_2 $ is closed).