Mixing
Show that the space $X$ is not a surface
$begingroup$
I would like to show that the following space is not a surface:
$X$ is made as an identification space of the unit square $Q={(x,y)mid 0leq x,yleq1}$ with the identifications:
$(0,y)sim(1,y)$ for all $0leq y leq1$
$(x,0)sim(x+frac{1}{2},0)$ for all $0leq x leq frac{1}{2}$
$(x,1)sim(x+frac{1}{2},1)$ for all $0leq x leq frac{1}{2}$
In particular, I would like to find a point on the space which has a neighbourhood that is non-locally Euclidean.
I have so far managed to convince myself that the space cannot be a surface. This is because if we first consider the unit square with the first identification, we get a cylinder, and the second identification then implies that the top-left point of the unit square, as well as the top-centre point both get identified to the top-right point. Likewise, every point between the top-left point and the top-centre point gets identified to a point between the top-centre point and the top-right point. This produces some "overlap". Of course the same applies for the points at the bottom of the square by symmetry.
My difficulty lies in taking, say, the top-centre point, which gets identified to both the top-left and top-right points, and finding a neighborhood about it that is non-locally Euclidean.
I am afraid that my abilities are not quite adequate for me to be able to provide diagrams, but I hope I have made myself understandable verbally.
Note: By surface I mean a compact, connected subspace of $mathbb{R}^3$ that is locally homeomorphic to open subsets of $mathbb{R}^2$.
Any help or input would, as always, be highly appreciated.
general-topology surfaces quotient-spaces
edited Jan 30 at 15:14
valcofadden
566
asked Jan 29 at 21:56
Heinrich WagnerHeinrich Wagner
450211
$endgroup$
add a comment |
$begingroup$
I would like to show that the following space is not a surface:
$X$ is made as an identification space of the unit square $Q={(x,y)mid 0leq x,yleq1}$ with the identifications:
$(0,y)sim(1,y)$ for all $0leq y leq1$
$(x,0)sim(x+frac{1}{2},0)$ for all $0leq x leq frac{1}{2}$
$(x,1)sim(x+frac{1}{2},1)$ for all $0leq x leq frac{1}{2}$
In particular, I would like to find a point on the space which has a neighbourhood that is non-locally Euclidean.
I have so far managed to convince myself that the space cannot be a surface. This is because if we first consider the unit square with the first identification, we get a cylinder, and the second identification then implies that the top-left point of the unit square, as well as the top-centre point both get identified to the top-right point. Likewise, every point between the top-left point and the top-centre point gets identified to a point between the top-centre point and the top-right point. This produces some "overlap". Of course the same applies for the points at the bottom of the square by symmetry.
My difficulty lies in taking, say, the top-centre point, which gets identified to both the top-left and top-right points, and finding a neighborhood about it that is non-locally Euclidean.
I am afraid that my abilities are not quite adequate for me to be able to provide diagrams, but I hope I have made myself understandable verbally.
Note: By surface I mean a compact, connected subspace of $mathbb{R}^3$ that is locally homeomorphic to open subsets of $mathbb{R}^2$.
Any help or input would, as always, be highly appreciated.
general-topology surfaces quotient-spaces
edited Jan 30 at 15:14
valcofadden
566
asked Jan 29 at 21:56
Heinrich WagnerHeinrich Wagner
450211
$endgroup$
$begingroup$
It is a surface in the usual sense but it does not embed in $R^3$ (it is the Klein bottle), hence, it will violate your definition. The standard definition of a surface is 2nd countable Hausdorff topological space locally homeomorphic to $R^2$.
$endgroup$
– Moishe Kohan
Jan 30 at 22:16
$begingroup$
Linked.
$endgroup$
– Alex Ravsky
Feb 2 at 14:58
add a comment |
$begingroup$
I would like to show that the following space is not a surface:
$X$ is made as an identification space of the unit square $Q={(x,y)mid 0leq x,yleq1}$ with the identifications:
$(0,y)sim(1,y)$ for all $0leq y leq1$
$(x,0)sim(x+frac{1}{2},0)$ for all $0leq x leq frac{1}{2}$
$(x,1)sim(x+frac{1}{2},1)$ for all $0leq x leq frac{1}{2}$
In particular, I would like to find a point on the space which has a neighbourhood that is non-locally Euclidean.
I have so far managed to convince myself that the space cannot be a surface. This is because if we first consider the unit square with the first identification, we get a cylinder, and the second identification then implies that the top-left point of the unit square, as well as the top-centre point both get identified to the top-right point. Likewise, every point between the top-left point and the top-centre point gets identified to a point between the top-centre point and the top-right point. This produces some "overlap". Of course the same applies for the points at the bottom of the square by symmetry.
My difficulty lies in taking, say, the top-centre point, which gets identified to both the top-left and top-right points, and finding a neighborhood about it that is non-locally Euclidean.
I am afraid that my abilities are not quite adequate for me to be able to provide diagrams, but I hope I have made myself understandable verbally.
Note: By surface I mean a compact, connected subspace of $mathbb{R}^3$ that is locally homeomorphic to open subsets of $mathbb{R}^2$.
Any help or input would, as always, be highly appreciated.
general-topology surfaces quotient-spaces
edited Jan 30 at 15:14
valcofadden
566
asked Jan 29 at 21:56
Heinrich WagnerHeinrich Wagner
450211
$endgroup$
I would like to show that the following space is not a surface:
$X$ is made as an identification space of the unit square $Q={(x,y)mid 0leq x,yleq1}$ with the identifications:
$(0,y)sim(1,y)$ for all $0leq y leq1$
$(x,0)sim(x+frac{1}{2},0)$ for all $0leq x leq frac{1}{2}$
$(x,1)sim(x+frac{1}{2},1)$ for all $0leq x leq frac{1}{2}$
In particular, I would like to find a point on the space which has a neighbourhood that is non-locally Euclidean.
I have so far managed to convince myself that the space cannot be a surface. This is because if we first consider the unit square with the first identification, we get a cylinder, and the second identification then implies that the top-left point of the unit square, as well as the top-centre point both get identified to the top-right point. Likewise, every point between the top-left point and the top-centre point gets identified to a point between the top-centre point and the top-right point. This produces some "overlap". Of course the same applies for the points at the bottom of the square by symmetry.
My difficulty lies in taking, say, the top-centre point, which gets identified to both the top-left and top-right points, and finding a neighborhood about it that is non-locally Euclidean.
I am afraid that my abilities are not quite adequate for me to be able to provide diagrams, but I hope I have made myself understandable verbally.
Note: By surface I mean a compact, connected subspace of $mathbb{R}^3$ that is locally homeomorphic to open subsets of $mathbb{R}^2$.
Any help or input would, as always, be highly appreciated.
general-topology surfaces quotient-spaces
general-topology surfaces quotient-spaces
edited Jan 30 at 15:14
valcofadden
566
asked Jan 29 at 21:56
Heinrich WagnerHeinrich Wagner
450211
edited Jan 30 at 15:14
valcofadden
566
asked Jan 29 at 21:56
Heinrich WagnerHeinrich Wagner
450211
edited Jan 30 at 15:14
valcofadden
566
edited Jan 30 at 15:14
valcofadden
566
edited Jan 30 at 15:14
valcofadden
566
566
asked Jan 29 at 21:56
Heinrich WagnerHeinrich Wagner
450211
asked Jan 29 at 21:56
Heinrich WagnerHeinrich Wagner
450211
asked Jan 29 at 21:56
Heinrich WagnerHeinrich Wagner
450211
450211
$begingroup$
It is a surface in the usual sense but it does not embed in $R^3$ (it is the Klein bottle), hence, it will violate your definition. The standard definition of a surface is 2nd countable Hausdorff topological space locally homeomorphic to $R^2$.
$endgroup$
– Moishe Kohan
Jan 30 at 22:16
$begingroup$
Linked.
$endgroup$
– Alex Ravsky
Feb 2 at 14:58
add a comment |
$begingroup$
It is a surface in the usual sense but it does not embed in $R^3$ (it is the Klein bottle), hence, it will violate your definition. The standard definition of a surface is 2nd countable Hausdorff topological space locally homeomorphic to $R^2$.
$endgroup$
– Moishe Kohan
Jan 30 at 22:16
$begingroup$
Linked.
$endgroup$
– Alex Ravsky
Feb 2 at 14:58
$begingroup$
It is a surface in the usual sense but it does not embed in $R^3$ (it is the Klein bottle), hence, it will violate your definition. The standard definition of a surface is 2nd countable Hausdorff topological space locally homeomorphic to $R^2$.
$endgroup$
– Moishe Kohan
Jan 30 at 22:16
$begingroup$
It is a surface in the usual sense but it does not embed in $R^3$ (it is the Klein bottle), hence, it will violate your definition. The standard definition of a surface is 2nd countable Hausdorff topological space locally homeomorphic to $R^2$.
$endgroup$
– Moishe Kohan
Jan 30 at 22:16
$begingroup$
Linked.
$endgroup$
– Alex Ravsky
Feb 2 at 14:58
$begingroup$
Linked.
$endgroup$
– Alex Ravsky
Feb 2 at 14:58
add a comment |
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$begingroup$
It is a surface in the usual sense but it does not embed in $R^3$ (it is the Klein bottle), hence, it will violate your definition. The standard definition of a surface is 2nd countable Hausdorff topological space locally homeomorphic to $R^2$.
$endgroup$
– Moishe Kohan
Jan 30 at 22:16
$begingroup$
Linked.
$endgroup$
– Alex Ravsky
Feb 2 at 14:58