The boundary is disjoint from the interior in 2d manifold
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I'm self-studying on Lee's Introduction to Topological Manifolds, and I'm stuck on a problem (8-6). I'm missing something really basic, I fear.
I must prove that the set of the boundary points of a 2-dimensional manifold with boundary is disjoint from the set of interior points. Then show that a 2-manifold is a manifold iff its boundary is empty.
Now, my reasoning is the following. The boundary of a 2d manifold with boundary $M$ must be a 1d manifold. If the manifold is compact, it must be homeomorphic to $S^1$, and $pi_1(partial M,q) simeq pi_1(S^1,1)$ for every $q$.
Given the 2d classification, $Int M$ is homeomorphic to $S^2$, the connected sum of tori, or the connected sum of projected planes.
In the case of the sphere, if $exists q in IntM cap partial M$, I can construct retractions, by using path connection of the sphere, and show that $pi_1(S^1,1)$ must be isomorphic to the trivial fundamental group, which is absurd, since it is infinite cyclic.
For the other cases, I cannot see how to proceed... And what about non-compact manifolds? I would loose the classification theorem (or at least the simple one).
Thank you
manifolds-with-boundary
asked Nov 18 at 8:22
Slz2718
32
add a comment |
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0
down vote
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I'm self-studying on Lee's Introduction to Topological Manifolds, and I'm stuck on a problem (8-6). I'm missing something really basic, I fear.
I must prove that the set of the boundary points of a 2-dimensional manifold with boundary is disjoint from the set of interior points. Then show that a 2-manifold is a manifold iff its boundary is empty.
Now, my reasoning is the following. The boundary of a 2d manifold with boundary $M$ must be a 1d manifold. If the manifold is compact, it must be homeomorphic to $S^1$, and $pi_1(partial M,q) simeq pi_1(S^1,1)$ for every $q$.
Given the 2d classification, $Int M$ is homeomorphic to $S^2$, the connected sum of tori, or the connected sum of projected planes.
In the case of the sphere, if $exists q in IntM cap partial M$, I can construct retractions, by using path connection of the sphere, and show that $pi_1(S^1,1)$ must be isomorphic to the trivial fundamental group, which is absurd, since it is infinite cyclic.
For the other cases, I cannot see how to proceed... And what about non-compact manifolds? I would loose the classification theorem (or at least the simple one).
Thank you
manifolds-with-boundary
asked Nov 18 at 8:22
Slz2718
32
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm self-studying on Lee's Introduction to Topological Manifolds, and I'm stuck on a problem (8-6). I'm missing something really basic, I fear.
I must prove that the set of the boundary points of a 2-dimensional manifold with boundary is disjoint from the set of interior points. Then show that a 2-manifold is a manifold iff its boundary is empty.
Now, my reasoning is the following. The boundary of a 2d manifold with boundary $M$ must be a 1d manifold. If the manifold is compact, it must be homeomorphic to $S^1$, and $pi_1(partial M,q) simeq pi_1(S^1,1)$ for every $q$.
Given the 2d classification, $Int M$ is homeomorphic to $S^2$, the connected sum of tori, or the connected sum of projected planes.
In the case of the sphere, if $exists q in IntM cap partial M$, I can construct retractions, by using path connection of the sphere, and show that $pi_1(S^1,1)$ must be isomorphic to the trivial fundamental group, which is absurd, since it is infinite cyclic.
For the other cases, I cannot see how to proceed... And what about non-compact manifolds? I would loose the classification theorem (or at least the simple one).
Thank you
manifolds-with-boundary
asked Nov 18 at 8:22
Slz2718
32
I'm self-studying on Lee's Introduction to Topological Manifolds, and I'm stuck on a problem (8-6). I'm missing something really basic, I fear.
I must prove that the set of the boundary points of a 2-dimensional manifold with boundary is disjoint from the set of interior points. Then show that a 2-manifold is a manifold iff its boundary is empty.
Now, my reasoning is the following. The boundary of a 2d manifold with boundary $M$ must be a 1d manifold. If the manifold is compact, it must be homeomorphic to $S^1$, and $pi_1(partial M,q) simeq pi_1(S^1,1)$ for every $q$.
Given the 2d classification, $Int M$ is homeomorphic to $S^2$, the connected sum of tori, or the connected sum of projected planes.
In the case of the sphere, if $exists q in IntM cap partial M$, I can construct retractions, by using path connection of the sphere, and show that $pi_1(S^1,1)$ must be isomorphic to the trivial fundamental group, which is absurd, since it is infinite cyclic.
For the other cases, I cannot see how to proceed... And what about non-compact manifolds? I would loose the classification theorem (or at least the simple one).
Thank you
manifolds-with-boundary
manifolds-with-boundary
asked Nov 18 at 8:22
Slz2718
32
asked Nov 18 at 8:22
Slz2718
32
asked Nov 18 at 8:22
Slz2718
32
asked Nov 18 at 8:22
Slz2718
32
asked Nov 18 at 8:22
Slz2718
32
32
add a comment |
add a comment |
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