Exercise with homeomorphism, open and closed: topology

**mameas** · June 27th 2011, 07:18 PM

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 $Q$
$I_1 = \{(x, y) \in R ^ 2 | x =- 1, -1 \leq y \leq 1 \}$
$I_2 = \{(x, y) \in R ^ 2 | x = 1, -1 \leq y \leq 1 \}$
$U_1 = \{(x, y) \in R ^ 2 | y = 1, -1 \leq x \leq 1 \}$
$U_2 = \{(x, y) \in R ^ 2 | y =- 1, -1 \leq x \leq 1 \}$
Let $Q = I_1 \cup I_2 \cup U_1 \cup U_2$ where we define the following equivalence relation:
$\A (x, y), (x ', y') \in Q$ $(x, y) \sim (x ', y')$ where $(x, y) = (x ', y')$ or $(x, y), (x ', y') \in I_1$
$(x, y), (x ', y') \in I_2$
a) Show that $Q /\sim$ is isomorphic to $S ^ 1$ .
b) Prove that $\pi: Q\rightarrow Q /\sim$ is closed.
c) Prove that $\pi$ is not open.
Solution :
a) first of all we construct the two following applications
$f_1 (x, y) = \{((1, 0) \;\;\;if\;\;\; (x, y) \in I_1), ((-1, 0)\;\;\; if\;\;\; (x, y) \in I_2) \}$ and $f_2 (x, y) = (x, (y /| y |)sin(arccos (x )))$ where $(x, y) \in U_1, U_2$ .
Then we construct $f (x, y): Q \rightarrow S^1$ pasting together $f_1 (x, y), f_2 (x, y)$ .Since $f_1$ and $f_2$ satisfy the conditions of The Pasting Lemma $f$ will be continue. In addition it is bijective and the equivalence relation that induces $f$ coincides with the equivalence relation defined on $Q$ .Then $Q /\sim$ is isomorphic to $S ^ 1$ .

b) $Q$ is connected and compact, so also $Q /\sim$ will be connected and compact. We must show at this point that $Q /\sim$ is $T_2$ (Hausdorff). That is to prove that the set
$F = \{(x, y), (u, v): \pi (x, y) = \pi (u, v) \} = \{(x, y), (u, v): (x, y) \sim (u, v) \}$ is a closed $Q \times Q$ .
$C_1 = \{(x, y), (x, y) \in Q \times Q | (x, y) \in Q\}$ , $C_2 = \{(-1;y), (-1, y) \in Q \times Q | y \in [-1, 1] \}$
$C_3 = \{(1, y), (1, y) \in Q \times Q | y \in [-1, 1] \}$ , $C_4 = \{(M_1, M_4), (M_4, M_1) (M_2, M_3), (M_3, M_2) \}$
where $M_1 = (-1,-1) = M_2 (-1,1), M_3 = (1,-1), M_4 = (1,1)$ are the vertices of the square.
$C_1$ is closed because $Q$ is $T_2$ . $C_2$ , $C_3$ are closed because they are closed subsets of a space $T_2$ and images of the compact $[-1.1]$ of $R$ through continuous applications. $C_4$ is closed because it is a finite set of points in space $T_2$ . Since $F = C_1 \cup C_2 \cup C_3 \cup C_4$ , then also $F$ will be closed, this means that $Q /\sim$ is $T_2$ . We conclude that $\pi$ is closed.

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

**Drexel28** · June 27th 2011, 08:14 PM

Originally Posted by mameas

a) first of all we construct the two following applications
$f_1 (x, y) = \{((1, 0) \;\;\;if\;\;\; (x, y) \in I_1), ((-1, 0)\;\;\; if\;\;\; (x, y) \in I_2) \}$ and $f_2 (x, y) = (x, (y /| y |)sin(arccos (x )))$ where $(x, y) \in U_1, U_2$ .
Then we construct $f (x, y): Q \rightarrow S^1$ pasting together $f_1 (x, y), f_2 (x, y)$ .Since $f_1$ and $f_2$ satisfy the conditions of The Pasting Lemma $f$ will be continue. In addition it is bijective and the equivalence relation that induces $f$ coincides with the equivalence relation defined on $Q$ .Then $Q /\sim$ is isomorphic to $S ^ 1$ .

This looks good to me.

b) $Q$ is connected and compact, so also $Q /\sim$ will be connected and compact. We must show at this point that $Q /\sim$ is $T_2$ (Hausdorff). That is to prove that the set
$F = \{(x, y), (u, v): \pi (x, y) = \pi (u, v) \} = \{(x, y), (u, v): (x, y) \sim (u, v) \}$ is a closed $Q \times Q$ .
$C_1 = \{(x, y), (x, y) \in Q \times Q | (x, y) \in Q\}$ , $C_2 = \{(-1;y), (-1, y) \in Q \times Q | y \in [-1, 1] \}$
$C_3 = \{(1, y), (1, y) \in Q \times Q | y \in [-1, 1] \}$ , $C_4 = \{(M_1, M_4), (M_4, M_1) (M_2, M_3), (M_3, M_2) \}$
where $M_1 = (-1,-1) = M_2 (-1,1), M_3 = (1,-1), M_4 = (1,1)$ are the vertices of the square.
$C_1$ is closed because $Q$ is $T_2$ . $C_2$ , $C_3$ are closed because they are closed subsets of a space $T_2$ and images of the compact $[-1.1]$ of $R$ through continuous applications. $C_4$ is closed because it is a finite set of points in space $T_2$ . Since $F = C_1 \cup C_2 \cup C_3 \cup C_4$ , then also $F$ will be closed, this means that $Q /\sim$ is $T_2$ . We conclude that $\pi$ is closed.

Ok, so you're using the theorem that if $Y$ is Hausdorff, $X$ compact and $f:X\to Y$ continuous then $f$ is closed, I assume? But then, I don't understand the problem. You just proved that $Q/\sim\approx \mathbb{S}^1$ and since Hausdorffness is preserved under homeomorphism (in fact, it is preserved under open surjections with closed kernels) you know that $Q/\sim$ is Hasudorff, no?

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

Good, assuming your readership will understand that you are using the fact that since $Q/\sim$ is connected there exists no nontrivial clopen sets.

Good job overall!

**mameas** · June 27th 2011, 08:27 PM

Perhaps, following the rules of the forum I do not respond to my thread before 24 hours, however I want to thank you Drexel28.
About the question b) of the exercise I agree totally with you, but I prefer to answer the differents questions of the exercises independently (when it is possible of course). Thank you again.

**Drexel28** · June 27th 2011, 08:29 PM

Originally Posted by mameas

Perhaps, following the rules of the forum I do not respond to my thread before 24 hours, however I want to thank you Drexel28.
About the question b) of the exercise I agree totally with you, but I prefer to answer the differents questions of the exercises independently (when it is possible of course). Thank you again.

No haha, you can respond as soon as you like! And if you want to thank a user merely press the thank you button in the bottom left corner of their post.

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