Proving torus is a topological manifold
$begingroup$
I want to prove that the torus is a topological variety of dimension 2, by definition $T= frac{[0,1]times[0,1]}{(0,y)R(1,y), (x,0)R(x,1)}$
Then, I need to show that for every $xin T$ there is an open set containing $x$ such that is homeomorphic to $mathbb{R}^2$. For every point inside $(0,1)times(0,1)$ there is an open ball inside de open square and we are done. Then we only need to show that this is true for the points in the borders of the square.
If $x=(0,y)$ ($y neq 1,0$) then an open set cointaining $x$ is of the form $A=([0,a)times I)cup ((b,1]times I)$ where $I in (0,1)$ is open.
As we know that $(b,1]$ is homeomorphic to $(-a,0]$
Then $A$ is homeomorphic to $(-a,a)times I$ which is homeomorphic to $mathbb{R}^2$.
I'm not sure if this is correct and I would apreciate if someone can check it.
Thanks in advance.
general-topology proof-verification
edited Nov 30 '18 at 19:17
Rob Arthan
29.1k42866
asked Nov 30 '18 at 18:36
JohannaJohanna
599
$endgroup$
add a comment |
$begingroup$
I want to prove that the torus is a topological variety of dimension 2, by definition $T= frac{[0,1]times[0,1]}{(0,y)R(1,y), (x,0)R(x,1)}$
Then, I need to show that for every $xin T$ there is an open set containing $x$ such that is homeomorphic to $mathbb{R}^2$. For every point inside $(0,1)times(0,1)$ there is an open ball inside de open square and we are done. Then we only need to show that this is true for the points in the borders of the square.
If $x=(0,y)$ ($y neq 1,0$) then an open set cointaining $x$ is of the form $A=([0,a)times I)cup ((b,1]times I)$ where $I in (0,1)$ is open.
As we know that $(b,1]$ is homeomorphic to $(-a,0]$
Then $A$ is homeomorphic to $(-a,a)times I$ which is homeomorphic to $mathbb{R}^2$.
I'm not sure if this is correct and I would apreciate if someone can check it.
Thanks in advance.
general-topology proof-verification
edited Nov 30 '18 at 19:17
Rob Arthan
29.1k42866
asked Nov 30 '18 at 18:36
JohannaJohanna
599
$endgroup$
$begingroup$
Your argument doesn't cover the point represented by the corners of the square (where, to follow your line of thinking, you would need to consider a neighbourhood made of 4 squares). Another possibility is to think about the local homeomorphism $Bbb{R}^2 to T$ that maps $(x, y)$ to (the point represented by) $(x mathop{mbox{mod}} 1, y mathop{mbox{mod}} 1)$.
$endgroup$
– Rob Arthan
Nov 30 '18 at 19:24
add a comment |
$begingroup$
I want to prove that the torus is a topological variety of dimension 2, by definition $T= frac{[0,1]times[0,1]}{(0,y)R(1,y), (x,0)R(x,1)}$
Then, I need to show that for every $xin T$ there is an open set containing $x$ such that is homeomorphic to $mathbb{R}^2$. For every point inside $(0,1)times(0,1)$ there is an open ball inside de open square and we are done. Then we only need to show that this is true for the points in the borders of the square.
If $x=(0,y)$ ($y neq 1,0$) then an open set cointaining $x$ is of the form $A=([0,a)times I)cup ((b,1]times I)$ where $I in (0,1)$ is open.
As we know that $(b,1]$ is homeomorphic to $(-a,0]$
Then $A$ is homeomorphic to $(-a,a)times I$ which is homeomorphic to $mathbb{R}^2$.
I'm not sure if this is correct and I would apreciate if someone can check it.
Thanks in advance.
general-topology proof-verification
edited Nov 30 '18 at 19:17
Rob Arthan
29.1k42866
asked Nov 30 '18 at 18:36
JohannaJohanna
599
$endgroup$
I want to prove that the torus is a topological variety of dimension 2, by definition $T= frac{[0,1]times[0,1]}{(0,y)R(1,y), (x,0)R(x,1)}$
Then, I need to show that for every $xin T$ there is an open set containing $x$ such that is homeomorphic to $mathbb{R}^2$. For every point inside $(0,1)times(0,1)$ there is an open ball inside de open square and we are done. Then we only need to show that this is true for the points in the borders of the square.
If $x=(0,y)$ ($y neq 1,0$) then an open set cointaining $x$ is of the form $A=([0,a)times I)cup ((b,1]times I)$ where $I in (0,1)$ is open.
As we know that $(b,1]$ is homeomorphic to $(-a,0]$
Then $A$ is homeomorphic to $(-a,a)times I$ which is homeomorphic to $mathbb{R}^2$.
I'm not sure if this is correct and I would apreciate if someone can check it.
Thanks in advance.
general-topology proof-verification
general-topology proof-verification
edited Nov 30 '18 at 19:17
Rob Arthan
29.1k42866
asked Nov 30 '18 at 18:36
JohannaJohanna
599
edited Nov 30 '18 at 19:17
Rob Arthan
29.1k42866
asked Nov 30 '18 at 18:36
JohannaJohanna
599
edited Nov 30 '18 at 19:17
Rob Arthan
29.1k42866
edited Nov 30 '18 at 19:17
Rob Arthan
29.1k42866
edited Nov 30 '18 at 19:17
Rob Arthan
29.1k42866
29.1k42866
asked Nov 30 '18 at 18:36
JohannaJohanna
599
asked Nov 30 '18 at 18:36
JohannaJohanna
599
asked Nov 30 '18 at 18:36
JohannaJohanna
599
599
$begingroup$
Your argument doesn't cover the point represented by the corners of the square (where, to follow your line of thinking, you would need to consider a neighbourhood made of 4 squares). Another possibility is to think about the local homeomorphism $Bbb{R}^2 to T$ that maps $(x, y)$ to (the point represented by) $(x mathop{mbox{mod}} 1, y mathop{mbox{mod}} 1)$.
$endgroup$
– Rob Arthan
Nov 30 '18 at 19:24
add a comment |
$begingroup$
Your argument doesn't cover the point represented by the corners of the square (where, to follow your line of thinking, you would need to consider a neighbourhood made of 4 squares). Another possibility is to think about the local homeomorphism $Bbb{R}^2 to T$ that maps $(x, y)$ to (the point represented by) $(x mathop{mbox{mod}} 1, y mathop{mbox{mod}} 1)$.
$endgroup$
– Rob Arthan
Nov 30 '18 at 19:24
$begingroup$
Your argument doesn't cover the point represented by the corners of the square (where, to follow your line of thinking, you would need to consider a neighbourhood made of 4 squares). Another possibility is to think about the local homeomorphism $Bbb{R}^2 to T$ that maps $(x, y)$ to (the point represented by) $(x mathop{mbox{mod}} 1, y mathop{mbox{mod}} 1)$.
$endgroup$
– Rob Arthan
Nov 30 '18 at 19:24
$begingroup$
Your argument doesn't cover the point represented by the corners of the square (where, to follow your line of thinking, you would need to consider a neighbourhood made of 4 squares). Another possibility is to think about the local homeomorphism $Bbb{R}^2 to T$ that maps $(x, y)$ to (the point represented by) $(x mathop{mbox{mod}} 1, y mathop{mbox{mod}} 1)$.
$endgroup$
– Rob Arthan
Nov 30 '18 at 19:24
add a comment |
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$begingroup$
Your argument doesn't cover the point represented by the corners of the square (where, to follow your line of thinking, you would need to consider a neighbourhood made of 4 squares). Another possibility is to think about the local homeomorphism $Bbb{R}^2 to T$ that maps $(x, y)$ to (the point represented by) $(x mathop{mbox{mod}} 1, y mathop{mbox{mod}} 1)$.
$endgroup$
– Rob Arthan
Nov 30 '18 at 19:24