Are contractible open subsets always evenly covered?











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Let $p:Erightarrow X$ be a covering map.



Let $U$ be a contractible open subset of $X$.



Is $U$ necessarily an evenly covered open subset, i.e. $p^{-1}(U)$ a disjoint union of open subsets of $E$ homeomorphic to $U$?



I know that if $X$ is path connected and locally path connected, then $U$ is an evenly covered open set. This can be shown by lifting the inclusion $Uhookrightarrow X$ to maps $U hookrightarrow E$.



What if $X$ is just a general topological space, not necessarily locally path connected and path connected?










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  • Generally, covering spaces only behave well under some local conditions on the space, e.g. locally path connected and semilocally simply connected. For instance, if your space is locally path-connected, being semilocally simply connected is necessary and sufficient for the existence of a universal cover.
    – Joshua Mundinger
    Nov 17 at 20:17










  • @JoshuaMundinger You're completely right, but what you said is somehow off-topic. Under some local conditions, the space does behave well, as I have mentioned in the question. However, I'm looking for a proof or disproof of the statement without those local conditions. ps. I don't think it has something to do with universal cover.
    – Y. Hu
    Nov 18 at 13:44

















up vote
0
down vote

favorite












Let $p:Erightarrow X$ be a covering map.



Let $U$ be a contractible open subset of $X$.



Is $U$ necessarily an evenly covered open subset, i.e. $p^{-1}(U)$ a disjoint union of open subsets of $E$ homeomorphic to $U$?



I know that if $X$ is path connected and locally path connected, then $U$ is an evenly covered open set. This can be shown by lifting the inclusion $Uhookrightarrow X$ to maps $U hookrightarrow E$.



What if $X$ is just a general topological space, not necessarily locally path connected and path connected?










share|cite|improve this question
























  • Generally, covering spaces only behave well under some local conditions on the space, e.g. locally path connected and semilocally simply connected. For instance, if your space is locally path-connected, being semilocally simply connected is necessary and sufficient for the existence of a universal cover.
    – Joshua Mundinger
    Nov 17 at 20:17










  • @JoshuaMundinger You're completely right, but what you said is somehow off-topic. Under some local conditions, the space does behave well, as I have mentioned in the question. However, I'm looking for a proof or disproof of the statement without those local conditions. ps. I don't think it has something to do with universal cover.
    – Y. Hu
    Nov 18 at 13:44















up vote
0
down vote

favorite









up vote
0
down vote

favorite











Let $p:Erightarrow X$ be a covering map.



Let $U$ be a contractible open subset of $X$.



Is $U$ necessarily an evenly covered open subset, i.e. $p^{-1}(U)$ a disjoint union of open subsets of $E$ homeomorphic to $U$?



I know that if $X$ is path connected and locally path connected, then $U$ is an evenly covered open set. This can be shown by lifting the inclusion $Uhookrightarrow X$ to maps $U hookrightarrow E$.



What if $X$ is just a general topological space, not necessarily locally path connected and path connected?










share|cite|improve this question















Let $p:Erightarrow X$ be a covering map.



Let $U$ be a contractible open subset of $X$.



Is $U$ necessarily an evenly covered open subset, i.e. $p^{-1}(U)$ a disjoint union of open subsets of $E$ homeomorphic to $U$?



I know that if $X$ is path connected and locally path connected, then $U$ is an evenly covered open set. This can be shown by lifting the inclusion $Uhookrightarrow X$ to maps $U hookrightarrow E$.



What if $X$ is just a general topological space, not necessarily locally path connected and path connected?







general-topology algebraic-topology covering-spaces






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edited Nov 15 at 15:39









David C. Ullrich

57.4k43791




57.4k43791










asked Nov 15 at 13:28









Y. Hu

28719




28719












  • Generally, covering spaces only behave well under some local conditions on the space, e.g. locally path connected and semilocally simply connected. For instance, if your space is locally path-connected, being semilocally simply connected is necessary and sufficient for the existence of a universal cover.
    – Joshua Mundinger
    Nov 17 at 20:17










  • @JoshuaMundinger You're completely right, but what you said is somehow off-topic. Under some local conditions, the space does behave well, as I have mentioned in the question. However, I'm looking for a proof or disproof of the statement without those local conditions. ps. I don't think it has something to do with universal cover.
    – Y. Hu
    Nov 18 at 13:44




















  • Generally, covering spaces only behave well under some local conditions on the space, e.g. locally path connected and semilocally simply connected. For instance, if your space is locally path-connected, being semilocally simply connected is necessary and sufficient for the existence of a universal cover.
    – Joshua Mundinger
    Nov 17 at 20:17










  • @JoshuaMundinger You're completely right, but what you said is somehow off-topic. Under some local conditions, the space does behave well, as I have mentioned in the question. However, I'm looking for a proof or disproof of the statement without those local conditions. ps. I don't think it has something to do with universal cover.
    – Y. Hu
    Nov 18 at 13:44


















Generally, covering spaces only behave well under some local conditions on the space, e.g. locally path connected and semilocally simply connected. For instance, if your space is locally path-connected, being semilocally simply connected is necessary and sufficient for the existence of a universal cover.
– Joshua Mundinger
Nov 17 at 20:17




Generally, covering spaces only behave well under some local conditions on the space, e.g. locally path connected and semilocally simply connected. For instance, if your space is locally path-connected, being semilocally simply connected is necessary and sufficient for the existence of a universal cover.
– Joshua Mundinger
Nov 17 at 20:17












@JoshuaMundinger You're completely right, but what you said is somehow off-topic. Under some local conditions, the space does behave well, as I have mentioned in the question. However, I'm looking for a proof or disproof of the statement without those local conditions. ps. I don't think it has something to do with universal cover.
– Y. Hu
Nov 18 at 13:44






@JoshuaMundinger You're completely right, but what you said is somehow off-topic. Under some local conditions, the space does behave well, as I have mentioned in the question. However, I'm looking for a proof or disproof of the statement without those local conditions. ps. I don't think it has something to do with universal cover.
– Y. Hu
Nov 18 at 13:44












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The answer is "yes". To see this, let us recall the concept of a pullback. Given a pair of maps $(p : E to X,f : Y to X)$, a pullback of $(p,f)$ is given by a pair of maps $(p^* : E^* to Y,f^*: E^* to E)$ such that



(1) $fp^* = pf^*$



(2) For any pair of maps $(u : Z to Y, v : Z to E)$ such that $fu = pv$ there exists a unique map $w : Y to E^*$ such that $p^*w = u$ and $f^*w = v$.



This universal property implies that pullbacks are unique up to canonical homeomorphism. Here is an explicit construction:
$$E^* = { (e,y) in E times Y mid p(e) = f(y) }, p^*(e,y) = y, f^*(e,y) = e .$$
If $f$ is the inclusion map of a subspace $Y subset X$, then we may also take $E^* = p^{-1}(Y) subset E, p^*(e) = p(e), f^*(e) = e$.



It is well-known that pullbacks of covering projections are covering projections. This means that if $p$ is a covering projection, then so is $p^*$. See Pullback of a covering map.



Lemma : Let $p : E to X$ be a covering projection and $f_0, f_1 : Y to X$ be homotopic maps. Then the pullback covering projections $p^*_k : E^*_k to Y$ along $f_k$ are equivalent which means that there exists a homeomorphism $h : E^*_0 to E^*_1$ such that $p^*_1h = p^*_0$.



Proof. Let $H : Y times I to X$ be a homotopy from $f_0$ to $f_1$ and for $k = 0,1$ let $i_k : Y times { k } hookrightarrow Y times I$ denote inclusion. Form the pullback $(p^* : E^* to Y times I, H^* : E^* to E)$ of $(p,H)$ and the pullbacks $(pi_k : E^*_k to Y times { k }, i^*_k : E^*_k hookrightarrow E^*)$ of $(p^*,i_k)$, where $E^*_k = (p^*)^{-1}(Y times { k }) $. Then we may assume that $p^*_k = r_k pi_k : E^*_k to Y$, where $r_k : Y times { k } to Y, r_k(y,k) = y$.



Now the map $i_0 pi_0 : E^*_0 to Y times I$ lifts to $i^*_0 : E^*_0 to E^*$. Hence the homotopy
$$G : E^*_0 times I to Y times I, G(e,t) = (p^*_0(e),t)$$
which satisfies $G_0 = i_0 pi_0$ lifts to a homotopy $G' : E^*_0 times I to E^*$. The map $G'_1 : E^*_0 to E^*$ has the property $G'_1(E^*_0) subset (p^*)^{-1}(Y times { 1 }) = i^*_1(E^*_1)$. Therefore we get a map $h : E^*_0 to E^*_1$ such that $p^*_1h = p^*_0$. By lifting $G' : E^*_1 times I to Y times I, G'(e,t) = (p^*_1(e),1-t)$ we obtain a map $h' : E^*_1 to E^*_0$ such that $p^*_0h' = p^*_1$. It is easy to see that $h' h = id$ (this follows from unique path lifting).



Now let $U subset X$ be open and contractible. This means that the inclusion $i : U to X$ is homotopic to a constant map $c : U to X, c(y) = x_0$. Hence the covering projection $p : p^{-1}(U) to U$ is equivalent to the pullback of $p$ along $c$. But the latter is easily seen to be a trivial covering $U times F to U$ (recall $E^* = { (e,y) in E times U mid p(e) = c(y) = x_0 } = { (e,y) in E times U mid e in p^{-1}(x_0)) } , p^*(e,y) = y$). This means that $U$ is evenly covered.



Note that it suffices to assume that $i : U to X$ is inessential which is weaker than $U$ contractible.






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    The answer is "yes". To see this, let us recall the concept of a pullback. Given a pair of maps $(p : E to X,f : Y to X)$, a pullback of $(p,f)$ is given by a pair of maps $(p^* : E^* to Y,f^*: E^* to E)$ such that



    (1) $fp^* = pf^*$



    (2) For any pair of maps $(u : Z to Y, v : Z to E)$ such that $fu = pv$ there exists a unique map $w : Y to E^*$ such that $p^*w = u$ and $f^*w = v$.



    This universal property implies that pullbacks are unique up to canonical homeomorphism. Here is an explicit construction:
    $$E^* = { (e,y) in E times Y mid p(e) = f(y) }, p^*(e,y) = y, f^*(e,y) = e .$$
    If $f$ is the inclusion map of a subspace $Y subset X$, then we may also take $E^* = p^{-1}(Y) subset E, p^*(e) = p(e), f^*(e) = e$.



    It is well-known that pullbacks of covering projections are covering projections. This means that if $p$ is a covering projection, then so is $p^*$. See Pullback of a covering map.



    Lemma : Let $p : E to X$ be a covering projection and $f_0, f_1 : Y to X$ be homotopic maps. Then the pullback covering projections $p^*_k : E^*_k to Y$ along $f_k$ are equivalent which means that there exists a homeomorphism $h : E^*_0 to E^*_1$ such that $p^*_1h = p^*_0$.



    Proof. Let $H : Y times I to X$ be a homotopy from $f_0$ to $f_1$ and for $k = 0,1$ let $i_k : Y times { k } hookrightarrow Y times I$ denote inclusion. Form the pullback $(p^* : E^* to Y times I, H^* : E^* to E)$ of $(p,H)$ and the pullbacks $(pi_k : E^*_k to Y times { k }, i^*_k : E^*_k hookrightarrow E^*)$ of $(p^*,i_k)$, where $E^*_k = (p^*)^{-1}(Y times { k }) $. Then we may assume that $p^*_k = r_k pi_k : E^*_k to Y$, where $r_k : Y times { k } to Y, r_k(y,k) = y$.



    Now the map $i_0 pi_0 : E^*_0 to Y times I$ lifts to $i^*_0 : E^*_0 to E^*$. Hence the homotopy
    $$G : E^*_0 times I to Y times I, G(e,t) = (p^*_0(e),t)$$
    which satisfies $G_0 = i_0 pi_0$ lifts to a homotopy $G' : E^*_0 times I to E^*$. The map $G'_1 : E^*_0 to E^*$ has the property $G'_1(E^*_0) subset (p^*)^{-1}(Y times { 1 }) = i^*_1(E^*_1)$. Therefore we get a map $h : E^*_0 to E^*_1$ such that $p^*_1h = p^*_0$. By lifting $G' : E^*_1 times I to Y times I, G'(e,t) = (p^*_1(e),1-t)$ we obtain a map $h' : E^*_1 to E^*_0$ such that $p^*_0h' = p^*_1$. It is easy to see that $h' h = id$ (this follows from unique path lifting).



    Now let $U subset X$ be open and contractible. This means that the inclusion $i : U to X$ is homotopic to a constant map $c : U to X, c(y) = x_0$. Hence the covering projection $p : p^{-1}(U) to U$ is equivalent to the pullback of $p$ along $c$. But the latter is easily seen to be a trivial covering $U times F to U$ (recall $E^* = { (e,y) in E times U mid p(e) = c(y) = x_0 } = { (e,y) in E times U mid e in p^{-1}(x_0)) } , p^*(e,y) = y$). This means that $U$ is evenly covered.



    Note that it suffices to assume that $i : U to X$ is inessential which is weaker than $U$ contractible.






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      up vote
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      down vote



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      The answer is "yes". To see this, let us recall the concept of a pullback. Given a pair of maps $(p : E to X,f : Y to X)$, a pullback of $(p,f)$ is given by a pair of maps $(p^* : E^* to Y,f^*: E^* to E)$ such that



      (1) $fp^* = pf^*$



      (2) For any pair of maps $(u : Z to Y, v : Z to E)$ such that $fu = pv$ there exists a unique map $w : Y to E^*$ such that $p^*w = u$ and $f^*w = v$.



      This universal property implies that pullbacks are unique up to canonical homeomorphism. Here is an explicit construction:
      $$E^* = { (e,y) in E times Y mid p(e) = f(y) }, p^*(e,y) = y, f^*(e,y) = e .$$
      If $f$ is the inclusion map of a subspace $Y subset X$, then we may also take $E^* = p^{-1}(Y) subset E, p^*(e) = p(e), f^*(e) = e$.



      It is well-known that pullbacks of covering projections are covering projections. This means that if $p$ is a covering projection, then so is $p^*$. See Pullback of a covering map.



      Lemma : Let $p : E to X$ be a covering projection and $f_0, f_1 : Y to X$ be homotopic maps. Then the pullback covering projections $p^*_k : E^*_k to Y$ along $f_k$ are equivalent which means that there exists a homeomorphism $h : E^*_0 to E^*_1$ such that $p^*_1h = p^*_0$.



      Proof. Let $H : Y times I to X$ be a homotopy from $f_0$ to $f_1$ and for $k = 0,1$ let $i_k : Y times { k } hookrightarrow Y times I$ denote inclusion. Form the pullback $(p^* : E^* to Y times I, H^* : E^* to E)$ of $(p,H)$ and the pullbacks $(pi_k : E^*_k to Y times { k }, i^*_k : E^*_k hookrightarrow E^*)$ of $(p^*,i_k)$, where $E^*_k = (p^*)^{-1}(Y times { k }) $. Then we may assume that $p^*_k = r_k pi_k : E^*_k to Y$, where $r_k : Y times { k } to Y, r_k(y,k) = y$.



      Now the map $i_0 pi_0 : E^*_0 to Y times I$ lifts to $i^*_0 : E^*_0 to E^*$. Hence the homotopy
      $$G : E^*_0 times I to Y times I, G(e,t) = (p^*_0(e),t)$$
      which satisfies $G_0 = i_0 pi_0$ lifts to a homotopy $G' : E^*_0 times I to E^*$. The map $G'_1 : E^*_0 to E^*$ has the property $G'_1(E^*_0) subset (p^*)^{-1}(Y times { 1 }) = i^*_1(E^*_1)$. Therefore we get a map $h : E^*_0 to E^*_1$ such that $p^*_1h = p^*_0$. By lifting $G' : E^*_1 times I to Y times I, G'(e,t) = (p^*_1(e),1-t)$ we obtain a map $h' : E^*_1 to E^*_0$ such that $p^*_0h' = p^*_1$. It is easy to see that $h' h = id$ (this follows from unique path lifting).



      Now let $U subset X$ be open and contractible. This means that the inclusion $i : U to X$ is homotopic to a constant map $c : U to X, c(y) = x_0$. Hence the covering projection $p : p^{-1}(U) to U$ is equivalent to the pullback of $p$ along $c$. But the latter is easily seen to be a trivial covering $U times F to U$ (recall $E^* = { (e,y) in E times U mid p(e) = c(y) = x_0 } = { (e,y) in E times U mid e in p^{-1}(x_0)) } , p^*(e,y) = y$). This means that $U$ is evenly covered.



      Note that it suffices to assume that $i : U to X$ is inessential which is weaker than $U$ contractible.






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        The answer is "yes". To see this, let us recall the concept of a pullback. Given a pair of maps $(p : E to X,f : Y to X)$, a pullback of $(p,f)$ is given by a pair of maps $(p^* : E^* to Y,f^*: E^* to E)$ such that



        (1) $fp^* = pf^*$



        (2) For any pair of maps $(u : Z to Y, v : Z to E)$ such that $fu = pv$ there exists a unique map $w : Y to E^*$ such that $p^*w = u$ and $f^*w = v$.



        This universal property implies that pullbacks are unique up to canonical homeomorphism. Here is an explicit construction:
        $$E^* = { (e,y) in E times Y mid p(e) = f(y) }, p^*(e,y) = y, f^*(e,y) = e .$$
        If $f$ is the inclusion map of a subspace $Y subset X$, then we may also take $E^* = p^{-1}(Y) subset E, p^*(e) = p(e), f^*(e) = e$.



        It is well-known that pullbacks of covering projections are covering projections. This means that if $p$ is a covering projection, then so is $p^*$. See Pullback of a covering map.



        Lemma : Let $p : E to X$ be a covering projection and $f_0, f_1 : Y to X$ be homotopic maps. Then the pullback covering projections $p^*_k : E^*_k to Y$ along $f_k$ are equivalent which means that there exists a homeomorphism $h : E^*_0 to E^*_1$ such that $p^*_1h = p^*_0$.



        Proof. Let $H : Y times I to X$ be a homotopy from $f_0$ to $f_1$ and for $k = 0,1$ let $i_k : Y times { k } hookrightarrow Y times I$ denote inclusion. Form the pullback $(p^* : E^* to Y times I, H^* : E^* to E)$ of $(p,H)$ and the pullbacks $(pi_k : E^*_k to Y times { k }, i^*_k : E^*_k hookrightarrow E^*)$ of $(p^*,i_k)$, where $E^*_k = (p^*)^{-1}(Y times { k }) $. Then we may assume that $p^*_k = r_k pi_k : E^*_k to Y$, where $r_k : Y times { k } to Y, r_k(y,k) = y$.



        Now the map $i_0 pi_0 : E^*_0 to Y times I$ lifts to $i^*_0 : E^*_0 to E^*$. Hence the homotopy
        $$G : E^*_0 times I to Y times I, G(e,t) = (p^*_0(e),t)$$
        which satisfies $G_0 = i_0 pi_0$ lifts to a homotopy $G' : E^*_0 times I to E^*$. The map $G'_1 : E^*_0 to E^*$ has the property $G'_1(E^*_0) subset (p^*)^{-1}(Y times { 1 }) = i^*_1(E^*_1)$. Therefore we get a map $h : E^*_0 to E^*_1$ such that $p^*_1h = p^*_0$. By lifting $G' : E^*_1 times I to Y times I, G'(e,t) = (p^*_1(e),1-t)$ we obtain a map $h' : E^*_1 to E^*_0$ such that $p^*_0h' = p^*_1$. It is easy to see that $h' h = id$ (this follows from unique path lifting).



        Now let $U subset X$ be open and contractible. This means that the inclusion $i : U to X$ is homotopic to a constant map $c : U to X, c(y) = x_0$. Hence the covering projection $p : p^{-1}(U) to U$ is equivalent to the pullback of $p$ along $c$. But the latter is easily seen to be a trivial covering $U times F to U$ (recall $E^* = { (e,y) in E times U mid p(e) = c(y) = x_0 } = { (e,y) in E times U mid e in p^{-1}(x_0)) } , p^*(e,y) = y$). This means that $U$ is evenly covered.



        Note that it suffices to assume that $i : U to X$ is inessential which is weaker than $U$ contractible.






        share|cite|improve this answer












        The answer is "yes". To see this, let us recall the concept of a pullback. Given a pair of maps $(p : E to X,f : Y to X)$, a pullback of $(p,f)$ is given by a pair of maps $(p^* : E^* to Y,f^*: E^* to E)$ such that



        (1) $fp^* = pf^*$



        (2) For any pair of maps $(u : Z to Y, v : Z to E)$ such that $fu = pv$ there exists a unique map $w : Y to E^*$ such that $p^*w = u$ and $f^*w = v$.



        This universal property implies that pullbacks are unique up to canonical homeomorphism. Here is an explicit construction:
        $$E^* = { (e,y) in E times Y mid p(e) = f(y) }, p^*(e,y) = y, f^*(e,y) = e .$$
        If $f$ is the inclusion map of a subspace $Y subset X$, then we may also take $E^* = p^{-1}(Y) subset E, p^*(e) = p(e), f^*(e) = e$.



        It is well-known that pullbacks of covering projections are covering projections. This means that if $p$ is a covering projection, then so is $p^*$. See Pullback of a covering map.



        Lemma : Let $p : E to X$ be a covering projection and $f_0, f_1 : Y to X$ be homotopic maps. Then the pullback covering projections $p^*_k : E^*_k to Y$ along $f_k$ are equivalent which means that there exists a homeomorphism $h : E^*_0 to E^*_1$ such that $p^*_1h = p^*_0$.



        Proof. Let $H : Y times I to X$ be a homotopy from $f_0$ to $f_1$ and for $k = 0,1$ let $i_k : Y times { k } hookrightarrow Y times I$ denote inclusion. Form the pullback $(p^* : E^* to Y times I, H^* : E^* to E)$ of $(p,H)$ and the pullbacks $(pi_k : E^*_k to Y times { k }, i^*_k : E^*_k hookrightarrow E^*)$ of $(p^*,i_k)$, where $E^*_k = (p^*)^{-1}(Y times { k }) $. Then we may assume that $p^*_k = r_k pi_k : E^*_k to Y$, where $r_k : Y times { k } to Y, r_k(y,k) = y$.



        Now the map $i_0 pi_0 : E^*_0 to Y times I$ lifts to $i^*_0 : E^*_0 to E^*$. Hence the homotopy
        $$G : E^*_0 times I to Y times I, G(e,t) = (p^*_0(e),t)$$
        which satisfies $G_0 = i_0 pi_0$ lifts to a homotopy $G' : E^*_0 times I to E^*$. The map $G'_1 : E^*_0 to E^*$ has the property $G'_1(E^*_0) subset (p^*)^{-1}(Y times { 1 }) = i^*_1(E^*_1)$. Therefore we get a map $h : E^*_0 to E^*_1$ such that $p^*_1h = p^*_0$. By lifting $G' : E^*_1 times I to Y times I, G'(e,t) = (p^*_1(e),1-t)$ we obtain a map $h' : E^*_1 to E^*_0$ such that $p^*_0h' = p^*_1$. It is easy to see that $h' h = id$ (this follows from unique path lifting).



        Now let $U subset X$ be open and contractible. This means that the inclusion $i : U to X$ is homotopic to a constant map $c : U to X, c(y) = x_0$. Hence the covering projection $p : p^{-1}(U) to U$ is equivalent to the pullback of $p$ along $c$. But the latter is easily seen to be a trivial covering $U times F to U$ (recall $E^* = { (e,y) in E times U mid p(e) = c(y) = x_0 } = { (e,y) in E times U mid e in p^{-1}(x_0)) } , p^*(e,y) = y$). This means that $U$ is evenly covered.



        Note that it suffices to assume that $i : U to X$ is inessential which is weaker than $U$ contractible.







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        answered Nov 25 at 13:55









        Paul Frost

        8,1041527




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