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Math Help - Exercise with homeomorphism, open and closed: topology

  1. #1
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    Exercise with homeomorphism, open and closed: topology

    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).
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Re: Exercise with homeomorphism, open and closed: topology

    Quote Originally Posted by mameas View Post
    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!
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  3. #3
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    Re: Exercise with homeomorphism, open and closed: topology

    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.
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Re: Exercise with homeomorphism, open and closed: topology

    Quote Originally Posted by mameas View Post
    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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