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$begingroup$

Let $p: X' rightarrow X$ be an n-sheeted cover of $X$. I proved that $X$ being compact implies that $X'$ is compact in the standard way. I started with an open cover of $X'$, projected it under $p$, used compactification to get a finite subset that covers $X$, and then took the inverse image of those subsets to get a finite sub collection of the original open cover.

How can I use this local information in conjunction with compactness of $X'$ and $X$ to show that $X'$ does not have a boundary?

Also, how can I show that $E(X')=nE(X)$ where $E$ is the Euler characteristic.

general-topology algebraic-topology geometric-topology low-dimensional-topology

share|cite|improve this question

edited Mar 22 at 23:34

Mathematical Mushroom

asked Mar 22 at 16:38

Mathematical MushroomMathematical Mushroom

1078

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why should you have $E(X') = 6 E(X)$? Take for example $n = 1$ in which case $p$ will be a homeomorphism.
  $endgroup$
  – Paul Frost
  Mar 22 at 18:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  oops was supposed to be n.. Thanks!
  $endgroup$
  – Mathematical Mushroom
  Mar 22 at 23:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You should tell us which theorems you are allowed to use.
  $endgroup$
  – Paul Frost
  Mar 23 at 10:52
  

  
  
  
  

add a comment | 

0

1

$begingroup$

Let $p: X' rightarrow X$ be an n-sheeted cover of $X$. I proved that $X$ being compact implies that $X'$ is compact in the standard way. I started with an open cover of $X'$, projected it under $p$, used compactification to get a finite subset that covers $X$, and then took the inverse image of those subsets to get a finite sub collection of the original open cover.

How can I use this local information in conjunction with compactness of $X'$ and $X$ to show that $X'$ does not have a boundary?

Also, how can I show that $E(X')=nE(X)$ where $E$ is the Euler characteristic.

general-topology algebraic-topology geometric-topology low-dimensional-topology

share|cite|improve this question

edited Mar 22 at 23:34

Mathematical Mushroom

asked Mar 22 at 16:38

Mathematical MushroomMathematical Mushroom

1078

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why should you have $E(X') = 6 E(X)$? Take for example $n = 1$ in which case $p$ will be a homeomorphism.
  $endgroup$
  – Paul Frost
  Mar 22 at 18:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  oops was supposed to be n.. Thanks!
  $endgroup$
  – Mathematical Mushroom
  Mar 22 at 23:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You should tell us which theorems you are allowed to use.
  $endgroup$
  – Paul Frost
  Mar 23 at 10:52
  

  
  
  
  

add a comment | 

$begingroup$

Let $p: X' rightarrow X$ be an n-sheeted cover of $X$. I proved that $X$ being compact implies that $X'$ is compact in the standard way. I started with an open cover of $X'$, projected it under $p$, used compactification to get a finite subset that covers $X$, and then took the inverse image of those subsets to get a finite sub collection of the original open cover.

How can I use this local information in conjunction with compactness of $X'$ and $X$ to show that $X'$ does not have a boundary?

Also, how can I show that $E(X')=nE(X)$ where $E$ is the Euler characteristic.

general-topology algebraic-topology geometric-topology low-dimensional-topology

share|cite|improve this question

edited Mar 22 at 23:34

Mathematical Mushroom

asked Mar 22 at 16:38

Mathematical MushroomMathematical Mushroom

1078

$endgroup$

share|cite|improve this question

edited Mar 22 at 23:34

Mathematical Mushroom

asked Mar 22 at 16:38

Mathematical MushroomMathematical Mushroom

1078

share|cite|improve this question

edited Mar 22 at 23:34

Mathematical Mushroom

asked Mar 22 at 16:38

Mathematical MushroomMathematical Mushroom

1078

asked Mar 22 at 16:38

Mathematical MushroomMathematical Mushroom

1078

asked Mar 22 at 16:38

Mathematical MushroomMathematical Mushroom

1078

1 Answer
1

active

oldest

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1

$begingroup$

Let $x'in X'$. Write $p(x')=x$. Let $U$ an open subset which is a chart of $X$ and which contains $x$ such that $p^-1(U)=V_1,...,V_n$ such that the restriction $p:V_irightarrow U$ is a diffeomorphism. Suppose $x'in V_i_0$. Since $U$ is an open subset of $mathbbR^2$ (the domain of a chart) so is $V_i_0$, we deduce that $x'$ has an open neighborhood diffeomorphic to an open subset of $mathbbR^2$ and that the boundary of $X'$ is empty.

share|cite|improve this answer

answered Mar 22 at 16:45

Tsemo AristideTsemo Aristide

60.4k11446

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  When you say that $U$ is a chart of X you mean that there is some map that maps $U$ homeomorphically onto a neighborhood of X? So you are then identifying this neighborhood with $U$ itself, which allows the make $p^-1(U)$ to make sense, but what is really going on is $p^-1(alpha(U))$ Where $alpha$ parameterizes a neighborhood in X, correct? This question is from my algebraic topology class where we haven't talked about charts at all. But this answer makes sense, thank you.
  $endgroup$
  – Mathematical Mushroom
  Mar 22 at 23:39
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  A chart means a neighborhood homeomorphic to an open subset of $mathbbR$.
  $endgroup$
  – Tsemo Aristide
  Mar 23 at 0:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Wouldn't it depend on what type of manifold you are working with? If you were using an atlas that locally parameterizes a smooth manifold then the charts would have to be diffeomorphisms.
  $endgroup$
  – Mathematical Mushroom
  Mar 24 at 13:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  But what about showing that the Euler Characteristic of the cover space is n times the Euler characteristic of the base?
  $endgroup$
  – Mathematical Mushroom
  Mar 24 at 14:00
  

  
  
  
  

add a comment | 

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1

$begingroup$

Let $x'in X'$. Write $p(x')=x$. Let $U$ an open subset which is a chart of $X$ and which contains $x$ such that $p^-1(U)=V_1,...,V_n$ such that the restriction $p:V_irightarrow U$ is a diffeomorphism. Suppose $x'in V_i_0$. Since $U$ is an open subset of $mathbbR^2$ (the domain of a chart) so is $V_i_0$, we deduce that $x'$ has an open neighborhood diffeomorphic to an open subset of $mathbbR^2$ and that the boundary of $X'$ is empty.

share|cite|improve this answer

answered Mar 22 at 16:45

Tsemo AristideTsemo Aristide

60.4k11446

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  When you say that $U$ is a chart of X you mean that there is some map that maps $U$ homeomorphically onto a neighborhood of X? So you are then identifying this neighborhood with $U$ itself, which allows the make $p^-1(U)$ to make sense, but what is really going on is $p^-1(alpha(U))$ Where $alpha$ parameterizes a neighborhood in X, correct? This question is from my algebraic topology class where we haven't talked about charts at all. But this answer makes sense, thank you.
  $endgroup$
  – Mathematical Mushroom
  Mar 22 at 23:39
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  A chart means a neighborhood homeomorphic to an open subset of $mathbbR$.
  $endgroup$
  – Tsemo Aristide
  Mar 23 at 0:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Wouldn't it depend on what type of manifold you are working with? If you were using an atlas that locally parameterizes a smooth manifold then the charts would have to be diffeomorphisms.
  $endgroup$
  – Mathematical Mushroom
  Mar 24 at 13:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  But what about showing that the Euler Characteristic of the cover space is n times the Euler characteristic of the base?
  $endgroup$
  – Mathematical Mushroom
  Mar 24 at 14:00
  

  
  
  
  

add a comment | 

1

$begingroup$

Let $x'in X'$. Write $p(x')=x$. Let $U$ an open subset which is a chart of $X$ and which contains $x$ such that $p^-1(U)=V_1,...,V_n$ such that the restriction $p:V_irightarrow U$ is a diffeomorphism. Suppose $x'in V_i_0$. Since $U$ is an open subset of $mathbbR^2$ (the domain of a chart) so is $V_i_0$, we deduce that $x'$ has an open neighborhood diffeomorphic to an open subset of $mathbbR^2$ and that the boundary of $X'$ is empty.

share|cite|improve this answer

answered Mar 22 at 16:45

Tsemo AristideTsemo Aristide

60.4k11446

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  When you say that $U$ is a chart of X you mean that there is some map that maps $U$ homeomorphically onto a neighborhood of X? So you are then identifying this neighborhood with $U$ itself, which allows the make $p^-1(U)$ to make sense, but what is really going on is $p^-1(alpha(U))$ Where $alpha$ parameterizes a neighborhood in X, correct? This question is from my algebraic topology class where we haven't talked about charts at all. But this answer makes sense, thank you.
  $endgroup$
  – Mathematical Mushroom
  Mar 22 at 23:39
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  A chart means a neighborhood homeomorphic to an open subset of $mathbbR$.
  $endgroup$
  – Tsemo Aristide
  Mar 23 at 0:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Wouldn't it depend on what type of manifold you are working with? If you were using an atlas that locally parameterizes a smooth manifold then the charts would have to be diffeomorphisms.
  $endgroup$
  – Mathematical Mushroom
  Mar 24 at 13:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  But what about showing that the Euler Characteristic of the cover space is n times the Euler characteristic of the base?
  $endgroup$
  – Mathematical Mushroom
  Mar 24 at 14:00
  

  
  
  
  

add a comment | 

$begingroup$

Let $x'in X'$. Write $p(x')=x$. Let $U$ an open subset which is a chart of $X$ and which contains $x$ such that $p^-1(U)=V_1,...,V_n$ such that the restriction $p:V_irightarrow U$ is a diffeomorphism. Suppose $x'in V_i_0$. Since $U$ is an open subset of $mathbbR^2$ (the domain of a chart) so is $V_i_0$, we deduce that $x'$ has an open neighborhood diffeomorphic to an open subset of $mathbbR^2$ and that the boundary of $X'$ is empty.

share|cite|improve this answer

answered Mar 22 at 16:45

Tsemo AristideTsemo Aristide

60.4k11446

$endgroup$

share|cite|improve this answer

answered Mar 22 at 16:45

Tsemo AristideTsemo Aristide

60.4k11446

answered Mar 22 at 16:45

Tsemo AristideTsemo Aristide

60.4k11446

answered Mar 22 at 16:45

Tsemo AristideTsemo Aristide

60.4k11446