Suppose that $q: Xto Z$ and that $p: Xto Y$ are covering spaces. Suppose there is a continuous function $r:...












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$begingroup$


Let $X,Y,Z$ be arc-connected and locally arc-connected spaces. Suppose that $q: Xto Z$ and that $p: Xto Y$ are covering spaces. Suppose there is a continuous function $r: Yto Z$ such that $rcirc p=q$. Prove that $r$ is also a covering space.



We know that $r$ is continuous, so we can only verify that $r$ is surjective and that if $cin Z$ then there is an open $U$ of $Z$ such that $r^{-1}(U)=sqcup_{alphain A}V_{alpha}$ where $V_{alpha}$ are open in $Y$ for all $alphain A$ and $r|_{V_{alpha}}: V_{alpha}to U$.



To see the overjection, let's take $cin Z$, therefore there is a $ain X$ such that $q(a)=c$ since $q$ is surjective it because it is a covering function. Then $r(p(a))=c$ and as $p(a)in Y$, then $p(a)$ is a preimage of $c$ and so $r$ is onto.



Let $cin Z$, since $q$ is a covering application, there is an open $U$ of $Z$ such that $cin U$ and $q^{-1}(U)=sqcup_{alphain A}V_{alpha}$ and $q|_{V_{alpha}}: V_{alpha}to U$ is a homeomorphism. Consider the family ${p(V_{alpha})}_{alphain A}$, let's see that $r^{-1}(U)=sqcup_{alphain A}p(V_{alpha})$ and that $r|_{p(V_{alpha})}: p(V_{alpha})to U$ is a homeomorphism. In effect, as $rcirc q$, then $p(q^{-1}(U))=p(sqcup_{alphain A}V_{alpha})=sqcup_{alphain A}p(V_{alpha})$ but $r^{-1}(U)=p(q^{-1}(U))=sqcup_{alphain A}p(V_{alpha})$.



How do I prove that $r|_{p(V_{alpha})}: p(V_{alpha})to U$ is a homeomorphism? Thank you very much.










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$endgroup$

















    3












    $begingroup$


    Let $X,Y,Z$ be arc-connected and locally arc-connected spaces. Suppose that $q: Xto Z$ and that $p: Xto Y$ are covering spaces. Suppose there is a continuous function $r: Yto Z$ such that $rcirc p=q$. Prove that $r$ is also a covering space.



    We know that $r$ is continuous, so we can only verify that $r$ is surjective and that if $cin Z$ then there is an open $U$ of $Z$ such that $r^{-1}(U)=sqcup_{alphain A}V_{alpha}$ where $V_{alpha}$ are open in $Y$ for all $alphain A$ and $r|_{V_{alpha}}: V_{alpha}to U$.



    To see the overjection, let's take $cin Z$, therefore there is a $ain X$ such that $q(a)=c$ since $q$ is surjective it because it is a covering function. Then $r(p(a))=c$ and as $p(a)in Y$, then $p(a)$ is a preimage of $c$ and so $r$ is onto.



    Let $cin Z$, since $q$ is a covering application, there is an open $U$ of $Z$ such that $cin U$ and $q^{-1}(U)=sqcup_{alphain A}V_{alpha}$ and $q|_{V_{alpha}}: V_{alpha}to U$ is a homeomorphism. Consider the family ${p(V_{alpha})}_{alphain A}$, let's see that $r^{-1}(U)=sqcup_{alphain A}p(V_{alpha})$ and that $r|_{p(V_{alpha})}: p(V_{alpha})to U$ is a homeomorphism. In effect, as $rcirc q$, then $p(q^{-1}(U))=p(sqcup_{alphain A}V_{alpha})=sqcup_{alphain A}p(V_{alpha})$ but $r^{-1}(U)=p(q^{-1}(U))=sqcup_{alphain A}p(V_{alpha})$.



    How do I prove that $r|_{p(V_{alpha})}: p(V_{alpha})to U$ is a homeomorphism? Thank you very much.










    share|cite|improve this question











    $endgroup$















      3












      3








      3


      2



      $begingroup$


      Let $X,Y,Z$ be arc-connected and locally arc-connected spaces. Suppose that $q: Xto Z$ and that $p: Xto Y$ are covering spaces. Suppose there is a continuous function $r: Yto Z$ such that $rcirc p=q$. Prove that $r$ is also a covering space.



      We know that $r$ is continuous, so we can only verify that $r$ is surjective and that if $cin Z$ then there is an open $U$ of $Z$ such that $r^{-1}(U)=sqcup_{alphain A}V_{alpha}$ where $V_{alpha}$ are open in $Y$ for all $alphain A$ and $r|_{V_{alpha}}: V_{alpha}to U$.



      To see the overjection, let's take $cin Z$, therefore there is a $ain X$ such that $q(a)=c$ since $q$ is surjective it because it is a covering function. Then $r(p(a))=c$ and as $p(a)in Y$, then $p(a)$ is a preimage of $c$ and so $r$ is onto.



      Let $cin Z$, since $q$ is a covering application, there is an open $U$ of $Z$ such that $cin U$ and $q^{-1}(U)=sqcup_{alphain A}V_{alpha}$ and $q|_{V_{alpha}}: V_{alpha}to U$ is a homeomorphism. Consider the family ${p(V_{alpha})}_{alphain A}$, let's see that $r^{-1}(U)=sqcup_{alphain A}p(V_{alpha})$ and that $r|_{p(V_{alpha})}: p(V_{alpha})to U$ is a homeomorphism. In effect, as $rcirc q$, then $p(q^{-1}(U))=p(sqcup_{alphain A}V_{alpha})=sqcup_{alphain A}p(V_{alpha})$ but $r^{-1}(U)=p(q^{-1}(U))=sqcup_{alphain A}p(V_{alpha})$.



      How do I prove that $r|_{p(V_{alpha})}: p(V_{alpha})to U$ is a homeomorphism? Thank you very much.










      share|cite|improve this question











      $endgroup$




      Let $X,Y,Z$ be arc-connected and locally arc-connected spaces. Suppose that $q: Xto Z$ and that $p: Xto Y$ are covering spaces. Suppose there is a continuous function $r: Yto Z$ such that $rcirc p=q$. Prove that $r$ is also a covering space.



      We know that $r$ is continuous, so we can only verify that $r$ is surjective and that if $cin Z$ then there is an open $U$ of $Z$ such that $r^{-1}(U)=sqcup_{alphain A}V_{alpha}$ where $V_{alpha}$ are open in $Y$ for all $alphain A$ and $r|_{V_{alpha}}: V_{alpha}to U$.



      To see the overjection, let's take $cin Z$, therefore there is a $ain X$ such that $q(a)=c$ since $q$ is surjective it because it is a covering function. Then $r(p(a))=c$ and as $p(a)in Y$, then $p(a)$ is a preimage of $c$ and so $r$ is onto.



      Let $cin Z$, since $q$ is a covering application, there is an open $U$ of $Z$ such that $cin U$ and $q^{-1}(U)=sqcup_{alphain A}V_{alpha}$ and $q|_{V_{alpha}}: V_{alpha}to U$ is a homeomorphism. Consider the family ${p(V_{alpha})}_{alphain A}$, let's see that $r^{-1}(U)=sqcup_{alphain A}p(V_{alpha})$ and that $r|_{p(V_{alpha})}: p(V_{alpha})to U$ is a homeomorphism. In effect, as $rcirc q$, then $p(q^{-1}(U))=p(sqcup_{alphain A}V_{alpha})=sqcup_{alphain A}p(V_{alpha})$ but $r^{-1}(U)=p(q^{-1}(U))=sqcup_{alphain A}p(V_{alpha})$.



      How do I prove that $r|_{p(V_{alpha})}: p(V_{alpha})to U$ is a homeomorphism? Thank you very much.







      general-topology proof-verification algebraic-topology






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      edited Dec 29 '18 at 14:15









      Paul Frost

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      asked Apr 8 '18 at 16:52









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          $begingroup$

          We shall use some elementary fact about functions.



          (a) Let $f : A to B, g : B to C$ be two functions. If $g circ f$ is injective (surjective), then $f$ is injective ($g$ is surjective).



          (b) $f : A to B$ is surjective iff $f(f^{-1}(M)) = M$ for all $M subset B$.



          Let us show that $r$ is a covering map.



          (1) $r$ is a surjective open map.



          We know that $p, q$ are surjective open maps. Then by (a) $r$ is surjective and by (b) we get $q(p^{-1}(M)) = rp(p^{-1}(M)) = r(M)$ for any $M subset Y$. This shows that $r$ is an open map.



          Let $z in Z$. Choose an open path connected neighborhood $W$ of $z$ which is evenly covered by $q$. Write $q^{-1}(W) = bigcup_alpha U_alpha$, where the $U_alpha$ are open in $X$ and the $q_alpha = q : U_alpha to W$ are homeomorphisms.



          Let $p_alpha = p : U_alpha to p(U_alpha)$ and $r_alpha = r : p(U_alpha) to W$. Each $p(U_alpha)$ is open in $Y$ and the $p_alpha, r_alpha$ are continuous open maps.



          (2) The $p_alpha$ and the $r_alpha$ are homeomorphisms.



          It remains to show that they are bijective. We have $r_alpha p_alpha = q_alpha$. Hence $p_alpha$ is bijectve (use (a)). But this implies that also $r_alpha$ is bijectve.



          Let $V = p(q^{-1}(W)) = bigcup_alpha p(U_alpha)$ which is open in $Y$.



          (3) $r^{-1}(W) = V$ and $p^{-1}(V) = q^{-1}(W)$.



          We have $r^{-1}(W) = p(p^{-1}(r^{-1}(W))) = p(q^{-1}(W)) = V$ and $p^{-1}(V) = p^{-1}(r^{-1}(W)) = q^{-1}(W)$.



          Her comes the crucial point which uses the fact that $W$ is path connected.



          (4) Either $p(U_{alpha_1}) cap p(U_{alpha_2}) = emptyset$ or $p(U_{alpha_1}) = p(U_{alpha_2})$.



          So let $p(U_{alpha_1}) cap p(U_{alpha_2}) ne emptyset$. Since the $p(U_{alpha_i})$ are path connected, so is their union. Consider $y_i in p(U_{alpha_i})$ and let $x_i in U_{alpha_i}$ such that $p(x_i) = y_i$. Let $u$ be a path in $p(U_{alpha_1}) cup p(U_{alpha_2}) subset V$ connecting $y_1$ and $y_2$. There are lifts $u_i$ of $u$ such that $u_1(0) = x_1$ and $u_2(1) = x_2$. The paths $u_i$ are contained in $p^{-1}(V) = q^{-1}(W)$. Since $q^{-1}(W)$ is partitioned into the open pairwise disjoint $U_alpha$, we conclude that $u_i$ is contained in $U_{alpha_i}$. But then $u = pu_1 = pu_2$ is contained in both $p(U_{alpha_1}), p(U_{alpha_2})$. This means $y_1,y_2 in p(U_{alpha_1}) cap p(U_{alpha_2})$. We conclude $p(U_{alpha_1}), p(U_{alpha_2}) subset p(U_{alpha_1}) cap p(U_{alpha_2})$ which implies $p(U_{alpha_1}) = p(U_{alpha_2})$.



          This shows that $r^{-1}(W)$ is the disjoint union of open sets having the form $p(U_{alpha})$. Now (2) implies that $W$ is evenly covered by $r$.






          share|cite|improve this answer











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            $begingroup$

            We shall use some elementary fact about functions.



            (a) Let $f : A to B, g : B to C$ be two functions. If $g circ f$ is injective (surjective), then $f$ is injective ($g$ is surjective).



            (b) $f : A to B$ is surjective iff $f(f^{-1}(M)) = M$ for all $M subset B$.



            Let us show that $r$ is a covering map.



            (1) $r$ is a surjective open map.



            We know that $p, q$ are surjective open maps. Then by (a) $r$ is surjective and by (b) we get $q(p^{-1}(M)) = rp(p^{-1}(M)) = r(M)$ for any $M subset Y$. This shows that $r$ is an open map.



            Let $z in Z$. Choose an open path connected neighborhood $W$ of $z$ which is evenly covered by $q$. Write $q^{-1}(W) = bigcup_alpha U_alpha$, where the $U_alpha$ are open in $X$ and the $q_alpha = q : U_alpha to W$ are homeomorphisms.



            Let $p_alpha = p : U_alpha to p(U_alpha)$ and $r_alpha = r : p(U_alpha) to W$. Each $p(U_alpha)$ is open in $Y$ and the $p_alpha, r_alpha$ are continuous open maps.



            (2) The $p_alpha$ and the $r_alpha$ are homeomorphisms.



            It remains to show that they are bijective. We have $r_alpha p_alpha = q_alpha$. Hence $p_alpha$ is bijectve (use (a)). But this implies that also $r_alpha$ is bijectve.



            Let $V = p(q^{-1}(W)) = bigcup_alpha p(U_alpha)$ which is open in $Y$.



            (3) $r^{-1}(W) = V$ and $p^{-1}(V) = q^{-1}(W)$.



            We have $r^{-1}(W) = p(p^{-1}(r^{-1}(W))) = p(q^{-1}(W)) = V$ and $p^{-1}(V) = p^{-1}(r^{-1}(W)) = q^{-1}(W)$.



            Her comes the crucial point which uses the fact that $W$ is path connected.



            (4) Either $p(U_{alpha_1}) cap p(U_{alpha_2}) = emptyset$ or $p(U_{alpha_1}) = p(U_{alpha_2})$.



            So let $p(U_{alpha_1}) cap p(U_{alpha_2}) ne emptyset$. Since the $p(U_{alpha_i})$ are path connected, so is their union. Consider $y_i in p(U_{alpha_i})$ and let $x_i in U_{alpha_i}$ such that $p(x_i) = y_i$. Let $u$ be a path in $p(U_{alpha_1}) cup p(U_{alpha_2}) subset V$ connecting $y_1$ and $y_2$. There are lifts $u_i$ of $u$ such that $u_1(0) = x_1$ and $u_2(1) = x_2$. The paths $u_i$ are contained in $p^{-1}(V) = q^{-1}(W)$. Since $q^{-1}(W)$ is partitioned into the open pairwise disjoint $U_alpha$, we conclude that $u_i$ is contained in $U_{alpha_i}$. But then $u = pu_1 = pu_2$ is contained in both $p(U_{alpha_1}), p(U_{alpha_2})$. This means $y_1,y_2 in p(U_{alpha_1}) cap p(U_{alpha_2})$. We conclude $p(U_{alpha_1}), p(U_{alpha_2}) subset p(U_{alpha_1}) cap p(U_{alpha_2})$ which implies $p(U_{alpha_1}) = p(U_{alpha_2})$.



            This shows that $r^{-1}(W)$ is the disjoint union of open sets having the form $p(U_{alpha})$. Now (2) implies that $W$ is evenly covered by $r$.






            share|cite|improve this answer











            $endgroup$


















              2












              $begingroup$

              We shall use some elementary fact about functions.



              (a) Let $f : A to B, g : B to C$ be two functions. If $g circ f$ is injective (surjective), then $f$ is injective ($g$ is surjective).



              (b) $f : A to B$ is surjective iff $f(f^{-1}(M)) = M$ for all $M subset B$.



              Let us show that $r$ is a covering map.



              (1) $r$ is a surjective open map.



              We know that $p, q$ are surjective open maps. Then by (a) $r$ is surjective and by (b) we get $q(p^{-1}(M)) = rp(p^{-1}(M)) = r(M)$ for any $M subset Y$. This shows that $r$ is an open map.



              Let $z in Z$. Choose an open path connected neighborhood $W$ of $z$ which is evenly covered by $q$. Write $q^{-1}(W) = bigcup_alpha U_alpha$, where the $U_alpha$ are open in $X$ and the $q_alpha = q : U_alpha to W$ are homeomorphisms.



              Let $p_alpha = p : U_alpha to p(U_alpha)$ and $r_alpha = r : p(U_alpha) to W$. Each $p(U_alpha)$ is open in $Y$ and the $p_alpha, r_alpha$ are continuous open maps.



              (2) The $p_alpha$ and the $r_alpha$ are homeomorphisms.



              It remains to show that they are bijective. We have $r_alpha p_alpha = q_alpha$. Hence $p_alpha$ is bijectve (use (a)). But this implies that also $r_alpha$ is bijectve.



              Let $V = p(q^{-1}(W)) = bigcup_alpha p(U_alpha)$ which is open in $Y$.



              (3) $r^{-1}(W) = V$ and $p^{-1}(V) = q^{-1}(W)$.



              We have $r^{-1}(W) = p(p^{-1}(r^{-1}(W))) = p(q^{-1}(W)) = V$ and $p^{-1}(V) = p^{-1}(r^{-1}(W)) = q^{-1}(W)$.



              Her comes the crucial point which uses the fact that $W$ is path connected.



              (4) Either $p(U_{alpha_1}) cap p(U_{alpha_2}) = emptyset$ or $p(U_{alpha_1}) = p(U_{alpha_2})$.



              So let $p(U_{alpha_1}) cap p(U_{alpha_2}) ne emptyset$. Since the $p(U_{alpha_i})$ are path connected, so is their union. Consider $y_i in p(U_{alpha_i})$ and let $x_i in U_{alpha_i}$ such that $p(x_i) = y_i$. Let $u$ be a path in $p(U_{alpha_1}) cup p(U_{alpha_2}) subset V$ connecting $y_1$ and $y_2$. There are lifts $u_i$ of $u$ such that $u_1(0) = x_1$ and $u_2(1) = x_2$. The paths $u_i$ are contained in $p^{-1}(V) = q^{-1}(W)$. Since $q^{-1}(W)$ is partitioned into the open pairwise disjoint $U_alpha$, we conclude that $u_i$ is contained in $U_{alpha_i}$. But then $u = pu_1 = pu_2$ is contained in both $p(U_{alpha_1}), p(U_{alpha_2})$. This means $y_1,y_2 in p(U_{alpha_1}) cap p(U_{alpha_2})$. We conclude $p(U_{alpha_1}), p(U_{alpha_2}) subset p(U_{alpha_1}) cap p(U_{alpha_2})$ which implies $p(U_{alpha_1}) = p(U_{alpha_2})$.



              This shows that $r^{-1}(W)$ is the disjoint union of open sets having the form $p(U_{alpha})$. Now (2) implies that $W$ is evenly covered by $r$.






              share|cite|improve this answer











              $endgroup$
















                2












                2








                2





                $begingroup$

                We shall use some elementary fact about functions.



                (a) Let $f : A to B, g : B to C$ be two functions. If $g circ f$ is injective (surjective), then $f$ is injective ($g$ is surjective).



                (b) $f : A to B$ is surjective iff $f(f^{-1}(M)) = M$ for all $M subset B$.



                Let us show that $r$ is a covering map.



                (1) $r$ is a surjective open map.



                We know that $p, q$ are surjective open maps. Then by (a) $r$ is surjective and by (b) we get $q(p^{-1}(M)) = rp(p^{-1}(M)) = r(M)$ for any $M subset Y$. This shows that $r$ is an open map.



                Let $z in Z$. Choose an open path connected neighborhood $W$ of $z$ which is evenly covered by $q$. Write $q^{-1}(W) = bigcup_alpha U_alpha$, where the $U_alpha$ are open in $X$ and the $q_alpha = q : U_alpha to W$ are homeomorphisms.



                Let $p_alpha = p : U_alpha to p(U_alpha)$ and $r_alpha = r : p(U_alpha) to W$. Each $p(U_alpha)$ is open in $Y$ and the $p_alpha, r_alpha$ are continuous open maps.



                (2) The $p_alpha$ and the $r_alpha$ are homeomorphisms.



                It remains to show that they are bijective. We have $r_alpha p_alpha = q_alpha$. Hence $p_alpha$ is bijectve (use (a)). But this implies that also $r_alpha$ is bijectve.



                Let $V = p(q^{-1}(W)) = bigcup_alpha p(U_alpha)$ which is open in $Y$.



                (3) $r^{-1}(W) = V$ and $p^{-1}(V) = q^{-1}(W)$.



                We have $r^{-1}(W) = p(p^{-1}(r^{-1}(W))) = p(q^{-1}(W)) = V$ and $p^{-1}(V) = p^{-1}(r^{-1}(W)) = q^{-1}(W)$.



                Her comes the crucial point which uses the fact that $W$ is path connected.



                (4) Either $p(U_{alpha_1}) cap p(U_{alpha_2}) = emptyset$ or $p(U_{alpha_1}) = p(U_{alpha_2})$.



                So let $p(U_{alpha_1}) cap p(U_{alpha_2}) ne emptyset$. Since the $p(U_{alpha_i})$ are path connected, so is their union. Consider $y_i in p(U_{alpha_i})$ and let $x_i in U_{alpha_i}$ such that $p(x_i) = y_i$. Let $u$ be a path in $p(U_{alpha_1}) cup p(U_{alpha_2}) subset V$ connecting $y_1$ and $y_2$. There are lifts $u_i$ of $u$ such that $u_1(0) = x_1$ and $u_2(1) = x_2$. The paths $u_i$ are contained in $p^{-1}(V) = q^{-1}(W)$. Since $q^{-1}(W)$ is partitioned into the open pairwise disjoint $U_alpha$, we conclude that $u_i$ is contained in $U_{alpha_i}$. But then $u = pu_1 = pu_2$ is contained in both $p(U_{alpha_1}), p(U_{alpha_2})$. This means $y_1,y_2 in p(U_{alpha_1}) cap p(U_{alpha_2})$. We conclude $p(U_{alpha_1}), p(U_{alpha_2}) subset p(U_{alpha_1}) cap p(U_{alpha_2})$ which implies $p(U_{alpha_1}) = p(U_{alpha_2})$.



                This shows that $r^{-1}(W)$ is the disjoint union of open sets having the form $p(U_{alpha})$. Now (2) implies that $W$ is evenly covered by $r$.






                share|cite|improve this answer











                $endgroup$



                We shall use some elementary fact about functions.



                (a) Let $f : A to B, g : B to C$ be two functions. If $g circ f$ is injective (surjective), then $f$ is injective ($g$ is surjective).



                (b) $f : A to B$ is surjective iff $f(f^{-1}(M)) = M$ for all $M subset B$.



                Let us show that $r$ is a covering map.



                (1) $r$ is a surjective open map.



                We know that $p, q$ are surjective open maps. Then by (a) $r$ is surjective and by (b) we get $q(p^{-1}(M)) = rp(p^{-1}(M)) = r(M)$ for any $M subset Y$. This shows that $r$ is an open map.



                Let $z in Z$. Choose an open path connected neighborhood $W$ of $z$ which is evenly covered by $q$. Write $q^{-1}(W) = bigcup_alpha U_alpha$, where the $U_alpha$ are open in $X$ and the $q_alpha = q : U_alpha to W$ are homeomorphisms.



                Let $p_alpha = p : U_alpha to p(U_alpha)$ and $r_alpha = r : p(U_alpha) to W$. Each $p(U_alpha)$ is open in $Y$ and the $p_alpha, r_alpha$ are continuous open maps.



                (2) The $p_alpha$ and the $r_alpha$ are homeomorphisms.



                It remains to show that they are bijective. We have $r_alpha p_alpha = q_alpha$. Hence $p_alpha$ is bijectve (use (a)). But this implies that also $r_alpha$ is bijectve.



                Let $V = p(q^{-1}(W)) = bigcup_alpha p(U_alpha)$ which is open in $Y$.



                (3) $r^{-1}(W) = V$ and $p^{-1}(V) = q^{-1}(W)$.



                We have $r^{-1}(W) = p(p^{-1}(r^{-1}(W))) = p(q^{-1}(W)) = V$ and $p^{-1}(V) = p^{-1}(r^{-1}(W)) = q^{-1}(W)$.



                Her comes the crucial point which uses the fact that $W$ is path connected.



                (4) Either $p(U_{alpha_1}) cap p(U_{alpha_2}) = emptyset$ or $p(U_{alpha_1}) = p(U_{alpha_2})$.



                So let $p(U_{alpha_1}) cap p(U_{alpha_2}) ne emptyset$. Since the $p(U_{alpha_i})$ are path connected, so is their union. Consider $y_i in p(U_{alpha_i})$ and let $x_i in U_{alpha_i}$ such that $p(x_i) = y_i$. Let $u$ be a path in $p(U_{alpha_1}) cup p(U_{alpha_2}) subset V$ connecting $y_1$ and $y_2$. There are lifts $u_i$ of $u$ such that $u_1(0) = x_1$ and $u_2(1) = x_2$. The paths $u_i$ are contained in $p^{-1}(V) = q^{-1}(W)$. Since $q^{-1}(W)$ is partitioned into the open pairwise disjoint $U_alpha$, we conclude that $u_i$ is contained in $U_{alpha_i}$. But then $u = pu_1 = pu_2$ is contained in both $p(U_{alpha_1}), p(U_{alpha_2})$. This means $y_1,y_2 in p(U_{alpha_1}) cap p(U_{alpha_2})$. We conclude $p(U_{alpha_1}), p(U_{alpha_2}) subset p(U_{alpha_1}) cap p(U_{alpha_2})$ which implies $p(U_{alpha_1}) = p(U_{alpha_2})$.



                This shows that $r^{-1}(W)$ is the disjoint union of open sets having the form $p(U_{alpha})$. Now (2) implies that $W$ is evenly covered by $r$.







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                edited Dec 7 '18 at 9:23

























                answered Dec 7 '18 at 0:44









                Paul FrostPaul Frost

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                10.2k3933






























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