Proving torus is a topological manifold
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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 2 days ago
Rob Arthan
28.6k42865
asked 2 days ago
Johanna
428
add a comment |
up vote
0
down vote
favorite
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 2 days ago
Rob Arthan
28.6k42865
asked 2 days ago
Johanna
428
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) $(x mathop{mbox{mod}} 1, y mathop{mbox{mod}} 1)$.
– Rob Arthan
2 days ago
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
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 2 days ago
Rob Arthan
28.6k42865
asked 2 days ago
Johanna
428
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 2 days ago
Rob Arthan
28.6k42865
asked 2 days ago
Johanna
428
edited 2 days ago
Rob Arthan
28.6k42865
asked 2 days ago
Johanna
428
edited 2 days ago
Rob Arthan
28.6k42865
edited 2 days ago
Rob Arthan
28.6k42865
edited 2 days ago
Rob Arthan
28.6k42865
28.6k42865
asked 2 days ago
Johanna
428
asked 2 days ago
Johanna
428
asked 2 days ago
Johanna
428
428
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) $(x mathop{mbox{mod}} 1, y mathop{mbox{mod}} 1)$.
– Rob Arthan
2 days ago
add a comment |
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) $(x mathop{mbox{mod}} 1, y mathop{mbox{mod}} 1)$.
– Rob Arthan
2 days ago
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) $(x mathop{mbox{mod}} 1, y mathop{mbox{mod}} 1)$.
– Rob Arthan
2 days ago
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) $(x mathop{mbox{mod}} 1, y mathop{mbox{mod}} 1)$.
– Rob Arthan
2 days ago
add a comment |
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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) $(x mathop{mbox{mod}} 1, y mathop{mbox{mod}} 1)$.
– Rob Arthan
2 days ago