Prove that quotient map is covering map











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I'm self-studying algebraic topology and need help with the following problem (I'm only at part a.)



enter image description here



The relevant definitions are as follows.




Definition: Let F be a discrete space and X be any space. Then X $times$ F is a disjoint union of copies of X, indexed by F. The projection $pi:$ X $times$ F $to$ X is called a trivial covering.



Definition: A map $p:tilde{X} to X$ is a covering map if it is locally a trivial covering. That is, if X has an open cover ${N_alpha: alpha in A}$ by trivializing neighborhoods for p, i.e. there exists discrete $F_alpha$, and a homeomorphism $varphi_alpha: p^{-1}(N_alpha) to N_alpha times F_alpha$ such that $p = pi circ varphi_alpha$ on $p^{-1}(N_alpha)$.




In terms of diagram, my interpretation of the problem is as follows.



enter image description here



To show that $p$ is a covering map, I guess I'd have to find the discrete space $F$ and the homeomorphism $varphi_alpha$, but I don't know how to proceed. Also, I understand that $mathbb{R}P^n$ is the space obtained by identifying antipodal points of $S^n$, but I can't figure out what its open sets look like.










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    up vote
    1
    down vote

    favorite
    1












    I'm self-studying algebraic topology and need help with the following problem (I'm only at part a.)



    enter image description here



    The relevant definitions are as follows.




    Definition: Let F be a discrete space and X be any space. Then X $times$ F is a disjoint union of copies of X, indexed by F. The projection $pi:$ X $times$ F $to$ X is called a trivial covering.



    Definition: A map $p:tilde{X} to X$ is a covering map if it is locally a trivial covering. That is, if X has an open cover ${N_alpha: alpha in A}$ by trivializing neighborhoods for p, i.e. there exists discrete $F_alpha$, and a homeomorphism $varphi_alpha: p^{-1}(N_alpha) to N_alpha times F_alpha$ such that $p = pi circ varphi_alpha$ on $p^{-1}(N_alpha)$.




    In terms of diagram, my interpretation of the problem is as follows.



    enter image description here



    To show that $p$ is a covering map, I guess I'd have to find the discrete space $F$ and the homeomorphism $varphi_alpha$, but I don't know how to proceed. Also, I understand that $mathbb{R}P^n$ is the space obtained by identifying antipodal points of $S^n$, but I can't figure out what its open sets look like.










    share|cite|improve this question


























      up vote
      1
      down vote

      favorite
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      up vote
      1
      down vote

      favorite
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      1





      I'm self-studying algebraic topology and need help with the following problem (I'm only at part a.)



      enter image description here



      The relevant definitions are as follows.




      Definition: Let F be a discrete space and X be any space. Then X $times$ F is a disjoint union of copies of X, indexed by F. The projection $pi:$ X $times$ F $to$ X is called a trivial covering.



      Definition: A map $p:tilde{X} to X$ is a covering map if it is locally a trivial covering. That is, if X has an open cover ${N_alpha: alpha in A}$ by trivializing neighborhoods for p, i.e. there exists discrete $F_alpha$, and a homeomorphism $varphi_alpha: p^{-1}(N_alpha) to N_alpha times F_alpha$ such that $p = pi circ varphi_alpha$ on $p^{-1}(N_alpha)$.




      In terms of diagram, my interpretation of the problem is as follows.



      enter image description here



      To show that $p$ is a covering map, I guess I'd have to find the discrete space $F$ and the homeomorphism $varphi_alpha$, but I don't know how to proceed. Also, I understand that $mathbb{R}P^n$ is the space obtained by identifying antipodal points of $S^n$, but I can't figure out what its open sets look like.










      share|cite|improve this question















      I'm self-studying algebraic topology and need help with the following problem (I'm only at part a.)



      enter image description here



      The relevant definitions are as follows.




      Definition: Let F be a discrete space and X be any space. Then X $times$ F is a disjoint union of copies of X, indexed by F. The projection $pi:$ X $times$ F $to$ X is called a trivial covering.



      Definition: A map $p:tilde{X} to X$ is a covering map if it is locally a trivial covering. That is, if X has an open cover ${N_alpha: alpha in A}$ by trivializing neighborhoods for p, i.e. there exists discrete $F_alpha$, and a homeomorphism $varphi_alpha: p^{-1}(N_alpha) to N_alpha times F_alpha$ such that $p = pi circ varphi_alpha$ on $p^{-1}(N_alpha)$.




      In terms of diagram, my interpretation of the problem is as follows.



      enter image description here



      To show that $p$ is a covering map, I guess I'd have to find the discrete space $F$ and the homeomorphism $varphi_alpha$, but I don't know how to proceed. Also, I understand that $mathbb{R}P^n$ is the space obtained by identifying antipodal points of $S^n$, but I can't figure out what its open sets look like.







      algebraic-topology covering-spaces






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      edited Nov 15 at 22:59

























      asked Nov 15 at 20:43









      ensbana

      265113




      265113






















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          Your picture is misleading because it assumes that $p$ is a trivial covering. This is not true, it is only a locally trivial covering. To see that, define
          $U_i^pm = { (x_1,dots,x_{n+1}) in S^n mid (-1)^{pm 1} x_i > 0}$. These set are the intersections of $S^n$ with open half-spaces in $mathbb{R}^{n+1}$, thus open subsets of $S^n$. Note that the $U_i^pm$ cover $S^n$.



          We have $x= (x_1,dots,x_{n+1}) in U_i^+$ if and only if $- x = (-x_1,dots,-x_{n+1}) in U_i^-$. Hence $p(U_i^+) = p(U_i^-)$, and we denote this subset of $mathbb{R}P^n$ by $V_i$.



          $V_i$ is open in $mathbb{R}P^n$ because $p^{-1}(V_i) = U_i^+ cup U_i^-$. Clearly $p_i^pm : U_i^pm stackrel{p}{rightarrow} V_i$ is a bijection. It is even a homeomorphism because it maps open sets to open sets.



          Now let $F = {+1,-1 }$. Then we get a homeomorphism
          $$phi_i : p^{-1}(V_i) to V_i times F, phi_(x) =
          begin{cases}
          (p_i^+(x),+1) & x in U_i^+ \
          (p_i^-(x),-1) & x in U_i^-
          end{cases}
          $$






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            up vote
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            Your picture is misleading because it assumes that $p$ is a trivial covering. This is not true, it is only a locally trivial covering. To see that, define
            $U_i^pm = { (x_1,dots,x_{n+1}) in S^n mid (-1)^{pm 1} x_i > 0}$. These set are the intersections of $S^n$ with open half-spaces in $mathbb{R}^{n+1}$, thus open subsets of $S^n$. Note that the $U_i^pm$ cover $S^n$.



            We have $x= (x_1,dots,x_{n+1}) in U_i^+$ if and only if $- x = (-x_1,dots,-x_{n+1}) in U_i^-$. Hence $p(U_i^+) = p(U_i^-)$, and we denote this subset of $mathbb{R}P^n$ by $V_i$.



            $V_i$ is open in $mathbb{R}P^n$ because $p^{-1}(V_i) = U_i^+ cup U_i^-$. Clearly $p_i^pm : U_i^pm stackrel{p}{rightarrow} V_i$ is a bijection. It is even a homeomorphism because it maps open sets to open sets.



            Now let $F = {+1,-1 }$. Then we get a homeomorphism
            $$phi_i : p^{-1}(V_i) to V_i times F, phi_(x) =
            begin{cases}
            (p_i^+(x),+1) & x in U_i^+ \
            (p_i^-(x),-1) & x in U_i^-
            end{cases}
            $$






            share|cite|improve this answer



























              up vote
              0
              down vote













              Your picture is misleading because it assumes that $p$ is a trivial covering. This is not true, it is only a locally trivial covering. To see that, define
              $U_i^pm = { (x_1,dots,x_{n+1}) in S^n mid (-1)^{pm 1} x_i > 0}$. These set are the intersections of $S^n$ with open half-spaces in $mathbb{R}^{n+1}$, thus open subsets of $S^n$. Note that the $U_i^pm$ cover $S^n$.



              We have $x= (x_1,dots,x_{n+1}) in U_i^+$ if and only if $- x = (-x_1,dots,-x_{n+1}) in U_i^-$. Hence $p(U_i^+) = p(U_i^-)$, and we denote this subset of $mathbb{R}P^n$ by $V_i$.



              $V_i$ is open in $mathbb{R}P^n$ because $p^{-1}(V_i) = U_i^+ cup U_i^-$. Clearly $p_i^pm : U_i^pm stackrel{p}{rightarrow} V_i$ is a bijection. It is even a homeomorphism because it maps open sets to open sets.



              Now let $F = {+1,-1 }$. Then we get a homeomorphism
              $$phi_i : p^{-1}(V_i) to V_i times F, phi_(x) =
              begin{cases}
              (p_i^+(x),+1) & x in U_i^+ \
              (p_i^-(x),-1) & x in U_i^-
              end{cases}
              $$






              share|cite|improve this answer

























                up vote
                0
                down vote










                up vote
                0
                down vote









                Your picture is misleading because it assumes that $p$ is a trivial covering. This is not true, it is only a locally trivial covering. To see that, define
                $U_i^pm = { (x_1,dots,x_{n+1}) in S^n mid (-1)^{pm 1} x_i > 0}$. These set are the intersections of $S^n$ with open half-spaces in $mathbb{R}^{n+1}$, thus open subsets of $S^n$. Note that the $U_i^pm$ cover $S^n$.



                We have $x= (x_1,dots,x_{n+1}) in U_i^+$ if and only if $- x = (-x_1,dots,-x_{n+1}) in U_i^-$. Hence $p(U_i^+) = p(U_i^-)$, and we denote this subset of $mathbb{R}P^n$ by $V_i$.



                $V_i$ is open in $mathbb{R}P^n$ because $p^{-1}(V_i) = U_i^+ cup U_i^-$. Clearly $p_i^pm : U_i^pm stackrel{p}{rightarrow} V_i$ is a bijection. It is even a homeomorphism because it maps open sets to open sets.



                Now let $F = {+1,-1 }$. Then we get a homeomorphism
                $$phi_i : p^{-1}(V_i) to V_i times F, phi_(x) =
                begin{cases}
                (p_i^+(x),+1) & x in U_i^+ \
                (p_i^-(x),-1) & x in U_i^-
                end{cases}
                $$






                share|cite|improve this answer














                Your picture is misleading because it assumes that $p$ is a trivial covering. This is not true, it is only a locally trivial covering. To see that, define
                $U_i^pm = { (x_1,dots,x_{n+1}) in S^n mid (-1)^{pm 1} x_i > 0}$. These set are the intersections of $S^n$ with open half-spaces in $mathbb{R}^{n+1}$, thus open subsets of $S^n$. Note that the $U_i^pm$ cover $S^n$.



                We have $x= (x_1,dots,x_{n+1}) in U_i^+$ if and only if $- x = (-x_1,dots,-x_{n+1}) in U_i^-$. Hence $p(U_i^+) = p(U_i^-)$, and we denote this subset of $mathbb{R}P^n$ by $V_i$.



                $V_i$ is open in $mathbb{R}P^n$ because $p^{-1}(V_i) = U_i^+ cup U_i^-$. Clearly $p_i^pm : U_i^pm stackrel{p}{rightarrow} V_i$ is a bijection. It is even a homeomorphism because it maps open sets to open sets.



                Now let $F = {+1,-1 }$. Then we get a homeomorphism
                $$phi_i : p^{-1}(V_i) to V_i times F, phi_(x) =
                begin{cases}
                (p_i^+(x),+1) & x in U_i^+ \
                (p_i^-(x),-1) & x in U_i^-
                end{cases}
                $$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Nov 16 at 23:25

























                answered Nov 15 at 23:36









                Paul Frost

                7,7141527




                7,7141527






























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