Complement of tubular neighborhood The 2019 Stack Overflow Developer Survey Results Are InFinding a metric on a tubular neighborhood of an embedded surfaceTangent bundle of a noncompact surfaceTubular neighborhood with an additional projectionA non orientable closed surface cannot be embedded into $mathbbR^3$Normal variation of embedded surfacesDerivative of projectionTubular neighborhoods of embedding of manifolds and vector bundles$epsilon$ function of a tubular neighborhoodExistence of certain surfaces in flat riemannian 3-manifoldDifferential of inverse function to a tubular neighborhood
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Complement of tubular neighborhood
The 2019 Stack Overflow Developer Survey Results Are InFinding a metric on a tubular neighborhood of an embedded surfaceTangent bundle of a noncompact surfaceTubular neighborhood with an additional projectionA non orientable closed surface cannot be embedded into $mathbbR^3$Normal variation of embedded surfacesDerivative of projectionTubular neighborhoods of embedding of manifolds and vector bundles$epsilon$ function of a tubular neighborhoodExistence of certain surfaces in flat riemannian 3-manifoldDifferential of inverse function to a tubular neighborhood
$begingroup$
Let $M$ be a closed, connected, orientable and embedded surface inside the unit 3-sphere $mathbbS^3$ and consider a small tubular neighborhood $U$ of $M$:
$$U = x in mathbbS^3 : d(x, M) leq varepsilon ,$$
(for small $varepsilon > 0$). I know that $U$ has the same homotopy type of $M$. Is it true that $mathbbS^3 setminus U$ has the same homotopy type of $mathbbS^3 setminus M$?
differential-geometry differential-topology smooth-manifolds geometric-topology
asked Apr 8 at 1:52
Eduardo LongaEduardo Longa
1,8802719
$endgroup$
add a comment |
$begingroup$
Let $M$ be a closed, connected, orientable and embedded surface inside the unit 3-sphere $mathbbS^3$ and consider a small tubular neighborhood $U$ of $M$:
$$U = x in mathbbS^3 : d(x, M) leq varepsilon ,$$
(for small $varepsilon > 0$). I know that $U$ has the same homotopy type of $M$. Is it true that $mathbbS^3 setminus U$ has the same homotopy type of $mathbbS^3 setminus M$?
differential-geometry differential-topology smooth-manifolds geometric-topology
asked Apr 8 at 1:52
Eduardo LongaEduardo Longa
1,8802719
$endgroup$
add a comment |
$begingroup$
Let $M$ be a closed, connected, orientable and embedded surface inside the unit 3-sphere $mathbbS^3$ and consider a small tubular neighborhood $U$ of $M$:
$$U = x in mathbbS^3 : d(x, M) leq varepsilon ,$$
(for small $varepsilon > 0$). I know that $U$ has the same homotopy type of $M$. Is it true that $mathbbS^3 setminus U$ has the same homotopy type of $mathbbS^3 setminus M$?
differential-geometry differential-topology smooth-manifolds geometric-topology
asked Apr 8 at 1:52
Eduardo LongaEduardo Longa
1,8802719
$endgroup$
Let $M$ be a closed, connected, orientable and embedded surface inside the unit 3-sphere $mathbbS^3$ and consider a small tubular neighborhood $U$ of $M$:
$$U = x in mathbbS^3 : d(x, M) leq varepsilon ,$$
(for small $varepsilon > 0$). I know that $U$ has the same homotopy type of $M$. Is it true that $mathbbS^3 setminus U$ has the same homotopy type of $mathbbS^3 setminus M$?
differential-geometry differential-topology smooth-manifolds geometric-topology
differential-geometry differential-topology smooth-manifolds geometric-topology
asked Apr 8 at 1:52
Eduardo LongaEduardo Longa
1,8802719
asked Apr 8 at 1:52
Eduardo LongaEduardo Longa
1,8802719
asked Apr 8 at 1:52
Eduardo LongaEduardo Longa
1,8802719
asked Apr 8 at 1:52
Eduardo LongaEduardo Longa
1,8802719
asked Apr 8 at 1:52
Eduardo LongaEduardo Longa
1,8802719
1,8802719
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
A tubular neighborhood $U$ of $M$ has a homeomorphism $Mtimes (-1,1)to U$, where $M$ is the image of $Mtimes 0$. By, say, restricting $(-1,1)$ to a closed interval and going to a smaller tubular neighborhood, we can assume we have a homeomorphism $Mtimes [-1,1]to overlineU$, where $overlineU$ is the closure of $U$ in $M$. The homeomorphism can be used to deformation retract $overlineU-M$ to $partial overlineU$, the image of $Mtimes-1,1$. This extends to a deformation retract of $S^3-M$ onto $S^3-U$. Thus $S^3-U hookrightarrow S^3-M$ is a homotopy equivalence.
answered Apr 9 at 0:18
Kyle MillerKyle Miller
10.1k930
$endgroup$
$begingroup$
Why does it extend?
$endgroup$
– Eduardo Longa
Apr 9 at 0:20
$begingroup$
@EduardoLonga Extend each step of the deformation retract by $operatornameid_M-U$. The maps agree on $partial overlineU$, so there exists such a continuous map. (This is the "Pasting Lemma", Theorem 18.3 of Munkres.)
$endgroup$
– Kyle Miller
Apr 9 at 0:28
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
A tubular neighborhood $U$ of $M$ has a homeomorphism $Mtimes (-1,1)to U$, where $M$ is the image of $Mtimes 0$. By, say, restricting $(-1,1)$ to a closed interval and going to a smaller tubular neighborhood, we can assume we have a homeomorphism $Mtimes [-1,1]to overlineU$, where $overlineU$ is the closure of $U$ in $M$. The homeomorphism can be used to deformation retract $overlineU-M$ to $partial overlineU$, the image of $Mtimes-1,1$. This extends to a deformation retract of $S^3-M$ onto $S^3-U$. Thus $S^3-U hookrightarrow S^3-M$ is a homotopy equivalence.
answered Apr 9 at 0:18
Kyle MillerKyle Miller
10.1k930
$endgroup$
$begingroup$
Why does it extend?
$endgroup$
– Eduardo Longa
Apr 9 at 0:20
$begingroup$
@EduardoLonga Extend each step of the deformation retract by $operatornameid_M-U$. The maps agree on $partial overlineU$, so there exists such a continuous map. (This is the "Pasting Lemma", Theorem 18.3 of Munkres.)
$endgroup$
– Kyle Miller
Apr 9 at 0:28
add a comment |
$begingroup$
A tubular neighborhood $U$ of $M$ has a homeomorphism $Mtimes (-1,1)to U$, where $M$ is the image of $Mtimes 0$. By, say, restricting $(-1,1)$ to a closed interval and going to a smaller tubular neighborhood, we can assume we have a homeomorphism $Mtimes [-1,1]to overlineU$, where $overlineU$ is the closure of $U$ in $M$. The homeomorphism can be used to deformation retract $overlineU-M$ to $partial overlineU$, the image of $Mtimes-1,1$. This extends to a deformation retract of $S^3-M$ onto $S^3-U$. Thus $S^3-U hookrightarrow S^3-M$ is a homotopy equivalence.
answered Apr 9 at 0:18
Kyle MillerKyle Miller
10.1k930
$endgroup$
$begingroup$
Why does it extend?
$endgroup$
– Eduardo Longa
Apr 9 at 0:20
$begingroup$
@EduardoLonga Extend each step of the deformation retract by $operatornameid_M-U$. The maps agree on $partial overlineU$, so there exists such a continuous map. (This is the "Pasting Lemma", Theorem 18.3 of Munkres.)
$endgroup$
– Kyle Miller
Apr 9 at 0:28
add a comment |
$begingroup$
A tubular neighborhood $U$ of $M$ has a homeomorphism $Mtimes (-1,1)to U$, where $M$ is the image of $Mtimes 0$. By, say, restricting $(-1,1)$ to a closed interval and going to a smaller tubular neighborhood, we can assume we have a homeomorphism $Mtimes [-1,1]to overlineU$, where $overlineU$ is the closure of $U$ in $M$. The homeomorphism can be used to deformation retract $overlineU-M$ to $partial overlineU$, the image of $Mtimes-1,1$. This extends to a deformation retract of $S^3-M$ onto $S^3-U$. Thus $S^3-U hookrightarrow S^3-M$ is a homotopy equivalence.
answered Apr 9 at 0:18
Kyle MillerKyle Miller
10.1k930
$endgroup$
A tubular neighborhood $U$ of $M$ has a homeomorphism $Mtimes (-1,1)to U$, where $M$ is the image of $Mtimes 0$. By, say, restricting $(-1,1)$ to a closed interval and going to a smaller tubular neighborhood, we can assume we have a homeomorphism $Mtimes [-1,1]to overlineU$, where $overlineU$ is the closure of $U$ in $M$. The homeomorphism can be used to deformation retract $overlineU-M$ to $partial overlineU$, the image of $Mtimes-1,1$. This extends to a deformation retract of $S^3-M$ onto $S^3-U$. Thus $S^3-U hookrightarrow S^3-M$ is a homotopy equivalence.
answered Apr 9 at 0:18
Kyle MillerKyle Miller
10.1k930
answered Apr 9 at 0:18
Kyle MillerKyle Miller
10.1k930
answered Apr 9 at 0:18
Kyle MillerKyle Miller
10.1k930
answered Apr 9 at 0:18
Kyle MillerKyle Miller
10.1k930
10.1k930
$begingroup$
Why does it extend?
$endgroup$
– Eduardo Longa
Apr 9 at 0:20
$begingroup$
@EduardoLonga Extend each step of the deformation retract by $operatornameid_M-U$. The maps agree on $partial overlineU$, so there exists such a continuous map. (This is the "Pasting Lemma", Theorem 18.3 of Munkres.)
$endgroup$
– Kyle Miller
Apr 9 at 0:28
add a comment |
$begingroup$
Why does it extend?
$endgroup$
– Eduardo Longa
Apr 9 at 0:20
$begingroup$
@EduardoLonga Extend each step of the deformation retract by $operatornameid_M-U$. The maps agree on $partial overlineU$, so there exists such a continuous map. (This is the "Pasting Lemma", Theorem 18.3 of Munkres.)
$endgroup$
– Kyle Miller
Apr 9 at 0:28
$begingroup$
Why does it extend?
$endgroup$
– Eduardo Longa
Apr 9 at 0:20
$begingroup$
Why does it extend?
$endgroup$
– Eduardo Longa
Apr 9 at 0:20
$begingroup$
@EduardoLonga Extend each step of the deformation retract by $operatornameid_M-U$. The maps agree on $partial overlineU$, so there exists such a continuous map. (This is the "Pasting Lemma", Theorem 18.3 of Munkres.)
$endgroup$
– Kyle Miller
Apr 9 at 0:28
$begingroup$
@EduardoLonga Extend each step of the deformation retract by $operatornameid_M-U$. The maps agree on $partial overlineU$, so there exists such a continuous map. (This is the "Pasting Lemma", Theorem 18.3 of Munkres.)
$endgroup$
– Kyle Miller
Apr 9 at 0:28
add a comment |
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