Question about manifolds with boundary The Next CEO of Stack OverflowTopological Manifold is...
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Question about manifolds with boundary
The Next CEO of Stack OverflowTopological Manifold is Manifold with Empty BoundaryProving diffeomorphism invariance of boundaryWell definedness of derivative of smooth map between smooth manifoldsOn the dimension of the product of manifoldsGP $1.2.9(d)$: $d(ftimes g)_{(x,y)}=df_xtimes dg_y$Differential of a smooth extension of a mapDiffeomorphism mapping interior point to boundary point of manifolds with boundaryif $f: X rightarrow Y$ is a diffeomorphism of manifolds with boundary, then $partial f$ maps $partial X$ diffeomorphically onto $partial Y$.Manifolds with Boundary and Maximal AtlasA question about diffeomorphism between manifolds with boundary
$begingroup$
Prove that if $f:Xto Y$ is a diffeomorphism of manifolds with boundary, then $f$ maps $partial X$ to $partial Y$ diffeomorphically.
Answer:
Let $Usubset H^k$ be an open subset and let $phi:Urightarrow X$ be a parametrization of $X$, $fcirc phi : Urightarrow Y$ is a parametrization of $Y$. Then $partial Y cap fcirc phi (U) = fcirc phi (partial U)$, thus $partial Y subset f(partial X)$ as $Y$ is covered by such parametrizations. Similarly, $partial X subset f^{-1}(partial Y)$ and thus $f(partial X) = partial Y$.
I don't understand why $partial Y cap fcirc phi (U) = fcirc phi (partial U)$, I understand $f circ phi$ is a diffeomorphism but don't understand why it maps boundary of $U$ to boundary of $Y$. Thanks and appreciate a hint.
manifolds smooth-manifolds manifolds-with-boundary
edited Mar 17 at 7:36
Andrews
1,2812422
asked Mar 17 at 5:58
manifoldedmanifolded
49519
$endgroup$
add a comment |
$begingroup$
Prove that if $f:Xto Y$ is a diffeomorphism of manifolds with boundary, then $f$ maps $partial X$ to $partial Y$ diffeomorphically.
Answer:
Let $Usubset H^k$ be an open subset and let $phi:Urightarrow X$ be a parametrization of $X$, $fcirc phi : Urightarrow Y$ is a parametrization of $Y$. Then $partial Y cap fcirc phi (U) = fcirc phi (partial U)$, thus $partial Y subset f(partial X)$ as $Y$ is covered by such parametrizations. Similarly, $partial X subset f^{-1}(partial Y)$ and thus $f(partial X) = partial Y$.
I don't understand why $partial Y cap fcirc phi (U) = fcirc phi (partial U)$, I understand $f circ phi$ is a diffeomorphism but don't understand why it maps boundary of $U$ to boundary of $Y$. Thanks and appreciate a hint.
manifolds smooth-manifolds manifolds-with-boundary
edited Mar 17 at 7:36
Andrews
1,2812422
asked Mar 17 at 5:58
manifoldedmanifolded
49519
$endgroup$
add a comment |
$begingroup$
Prove that if $f:Xto Y$ is a diffeomorphism of manifolds with boundary, then $f$ maps $partial X$ to $partial Y$ diffeomorphically.
Answer:
Let $Usubset H^k$ be an open subset and let $phi:Urightarrow X$ be a parametrization of $X$, $fcirc phi : Urightarrow Y$ is a parametrization of $Y$. Then $partial Y cap fcirc phi (U) = fcirc phi (partial U)$, thus $partial Y subset f(partial X)$ as $Y$ is covered by such parametrizations. Similarly, $partial X subset f^{-1}(partial Y)$ and thus $f(partial X) = partial Y$.
I don't understand why $partial Y cap fcirc phi (U) = fcirc phi (partial U)$, I understand $f circ phi$ is a diffeomorphism but don't understand why it maps boundary of $U$ to boundary of $Y$. Thanks and appreciate a hint.
manifolds smooth-manifolds manifolds-with-boundary
edited Mar 17 at 7:36
Andrews
1,2812422
asked Mar 17 at 5:58
manifoldedmanifolded
49519
$endgroup$
Prove that if $f:Xto Y$ is a diffeomorphism of manifolds with boundary, then $f$ maps $partial X$ to $partial Y$ diffeomorphically.
Answer:
Let $Usubset H^k$ be an open subset and let $phi:Urightarrow X$ be a parametrization of $X$, $fcirc phi : Urightarrow Y$ is a parametrization of $Y$. Then $partial Y cap fcirc phi (U) = fcirc phi (partial U)$, thus $partial Y subset f(partial X)$ as $Y$ is covered by such parametrizations. Similarly, $partial X subset f^{-1}(partial Y)$ and thus $f(partial X) = partial Y$.
I don't understand why $partial Y cap fcirc phi (U) = fcirc phi (partial U)$, I understand $f circ phi$ is a diffeomorphism but don't understand why it maps boundary of $U$ to boundary of $Y$. Thanks and appreciate a hint.
manifolds smooth-manifolds manifolds-with-boundary
manifolds smooth-manifolds manifolds-with-boundary
edited Mar 17 at 7:36
Andrews
1,2812422
asked Mar 17 at 5:58
manifoldedmanifolded
49519
edited Mar 17 at 7:36
Andrews
1,2812422
asked Mar 17 at 5:58
manifoldedmanifolded
49519
edited Mar 17 at 7:36
Andrews
1,2812422
edited Mar 17 at 7:36
Andrews
1,2812422
edited Mar 17 at 7:36
Andrews
1,2812422
1,2812422
asked Mar 17 at 5:58
manifoldedmanifolded
49519
asked Mar 17 at 5:58
manifoldedmanifolded
49519
asked Mar 17 at 5:58
manifoldedmanifolded
49519
49519
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Here's an easier argument. If $f:Xto Y$ is a homeomorphism of manifolds with boundary, then it restricts to a homeomorphism of the boundaries. (Then if $f$ is a diffeomorphism, the restriction will also be a diffeomorphism.)
Proof:
There is a purely topological way to distinguish boundary points from non-boundary points.
Boundary points have contractible punctured neighborhoods, non-boundary points do not have contractible punctured neighborhoods. This is immediate since deleting the center of an open ball in $Bbb{R}^k$ yields a space homotopic to the $k-1$-sphere, whereas deleting the center of an open ball that has been intersected with a half-space passing through the center gives a contractible space (it's star shaped around any point not on the boundary of the half-space).
Since this characterization of boundary points is purely topological, it is preserved by homeomorphisms.
How this addresses your question
Well, I hope it illuminates why $partial Ycap f circ phi(U) = fcirc phi (partial U)$. It's because homeomorphisms (and thus diffeomorphisms) preserve boundary points.
answered Mar 17 at 14:26
jgonjgon
16.2k32143
$endgroup$
$begingroup$
Interesting! May I know how you intuitively know that boundary is a contractible manifold or in general how do you look at an arbitrary manifold and know if it's contractible? I know just know the definition: $X$ is contractible if $id_X $ is homotopic to a constant map i.e. $X$ can be smoothly shrunk to a point.
$endgroup$
– manifolded
Mar 17 at 19:03
$begingroup$
Also, may I know what you mean by purely topological that makes it homeomorphism invariant? Thanks.
$endgroup$
– manifolded
Mar 17 at 19:04
$begingroup$
@manifolded The boundary is not contractible, at least not generally, but any point in the boundary has a contractible punctured open neighborhood. As I mentioned, take a sufficiently small open ball around the boundary point, which is homeomorphic to an ordinary open ball in $Bbb{R}^n$ intersected with a half-space that passes through the center of the open ball. Deleting the center of the open ball leaves us with half an open ball minus the center point, and this remaining set is star shaped around any point not in the hyperplane that defined the half-space, and thus contractible.
$endgroup$
– jgon
Mar 17 at 19:06
$begingroup$
You may want to try proving from the definition that (nonempty) star shaped subsets of $Bbb{R}^n$ are contractible.
$endgroup$
– jgon
Mar 17 at 19:06
$begingroup$
By purely topological, I mean that the existence of a contractible punctured open neighborhood of a point is something that is preserved by homeomorphisms. If this is not obvious, I suggest trying to prove it.
$endgroup$
– jgon
Mar 17 at 19:07
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Here's an easier argument. If $f:Xto Y$ is a homeomorphism of manifolds with boundary, then it restricts to a homeomorphism of the boundaries. (Then if $f$ is a diffeomorphism, the restriction will also be a diffeomorphism.)
Proof:
There is a purely topological way to distinguish boundary points from non-boundary points.
Boundary points have contractible punctured neighborhoods, non-boundary points do not have contractible punctured neighborhoods. This is immediate since deleting the center of an open ball in $Bbb{R}^k$ yields a space homotopic to the $k-1$-sphere, whereas deleting the center of an open ball that has been intersected with a half-space passing through the center gives a contractible space (it's star shaped around any point not on the boundary of the half-space).
Since this characterization of boundary points is purely topological, it is preserved by homeomorphisms.
How this addresses your question
Well, I hope it illuminates why $partial Ycap f circ phi(U) = fcirc phi (partial U)$. It's because homeomorphisms (and thus diffeomorphisms) preserve boundary points.
answered Mar 17 at 14:26
jgonjgon
16.2k32143
$endgroup$
$begingroup$
Interesting! May I know how you intuitively know that boundary is a contractible manifold or in general how do you look at an arbitrary manifold and know if it's contractible? I know just know the definition: $X$ is contractible if $id_X $ is homotopic to a constant map i.e. $X$ can be smoothly shrunk to a point.
$endgroup$
– manifolded
Mar 17 at 19:03
$begingroup$
Also, may I know what you mean by purely topological that makes it homeomorphism invariant? Thanks.
$endgroup$
– manifolded
Mar 17 at 19:04
$begingroup$
@manifolded The boundary is not contractible, at least not generally, but any point in the boundary has a contractible punctured open neighborhood. As I mentioned, take a sufficiently small open ball around the boundary point, which is homeomorphic to an ordinary open ball in $Bbb{R}^n$ intersected with a half-space that passes through the center of the open ball. Deleting the center of the open ball leaves us with half an open ball minus the center point, and this remaining set is star shaped around any point not in the hyperplane that defined the half-space, and thus contractible.
$endgroup$
– jgon
Mar 17 at 19:06
$begingroup$
You may want to try proving from the definition that (nonempty) star shaped subsets of $Bbb{R}^n$ are contractible.
$endgroup$
– jgon
Mar 17 at 19:06
$begingroup$
By purely topological, I mean that the existence of a contractible punctured open neighborhood of a point is something that is preserved by homeomorphisms. If this is not obvious, I suggest trying to prove it.
$endgroup$
– jgon
Mar 17 at 19:07
add a comment |
$begingroup$
Here's an easier argument. If $f:Xto Y$ is a homeomorphism of manifolds with boundary, then it restricts to a homeomorphism of the boundaries. (Then if $f$ is a diffeomorphism, the restriction will also be a diffeomorphism.)
Proof:
There is a purely topological way to distinguish boundary points from non-boundary points.
Boundary points have contractible punctured neighborhoods, non-boundary points do not have contractible punctured neighborhoods. This is immediate since deleting the center of an open ball in $Bbb{R}^k$ yields a space homotopic to the $k-1$-sphere, whereas deleting the center of an open ball that has been intersected with a half-space passing through the center gives a contractible space (it's star shaped around any point not on the boundary of the half-space).
Since this characterization of boundary points is purely topological, it is preserved by homeomorphisms.
How this addresses your question
Well, I hope it illuminates why $partial Ycap f circ phi(U) = fcirc phi (partial U)$. It's because homeomorphisms (and thus diffeomorphisms) preserve boundary points.
answered Mar 17 at 14:26
jgonjgon
16.2k32143
$endgroup$
$begingroup$
Interesting! May I know how you intuitively know that boundary is a contractible manifold or in general how do you look at an arbitrary manifold and know if it's contractible? I know just know the definition: $X$ is contractible if $id_X $ is homotopic to a constant map i.e. $X$ can be smoothly shrunk to a point.
$endgroup$
– manifolded
Mar 17 at 19:03
$begingroup$
Also, may I know what you mean by purely topological that makes it homeomorphism invariant? Thanks.
$endgroup$
– manifolded
Mar 17 at 19:04
$begingroup$
@manifolded The boundary is not contractible, at least not generally, but any point in the boundary has a contractible punctured open neighborhood. As I mentioned, take a sufficiently small open ball around the boundary point, which is homeomorphic to an ordinary open ball in $Bbb{R}^n$ intersected with a half-space that passes through the center of the open ball. Deleting the center of the open ball leaves us with half an open ball minus the center point, and this remaining set is star shaped around any point not in the hyperplane that defined the half-space, and thus contractible.
$endgroup$
– jgon
Mar 17 at 19:06
$begingroup$
You may want to try proving from the definition that (nonempty) star shaped subsets of $Bbb{R}^n$ are contractible.
$endgroup$
– jgon
Mar 17 at 19:06
$begingroup$
By purely topological, I mean that the existence of a contractible punctured open neighborhood of a point is something that is preserved by homeomorphisms. If this is not obvious, I suggest trying to prove it.
$endgroup$
– jgon
Mar 17 at 19:07
add a comment |
$begingroup$
Here's an easier argument. If $f:Xto Y$ is a homeomorphism of manifolds with boundary, then it restricts to a homeomorphism of the boundaries. (Then if $f$ is a diffeomorphism, the restriction will also be a diffeomorphism.)
Proof:
There is a purely topological way to distinguish boundary points from non-boundary points.
Boundary points have contractible punctured neighborhoods, non-boundary points do not have contractible punctured neighborhoods. This is immediate since deleting the center of an open ball in $Bbb{R}^k$ yields a space homotopic to the $k-1$-sphere, whereas deleting the center of an open ball that has been intersected with a half-space passing through the center gives a contractible space (it's star shaped around any point not on the boundary of the half-space).
Since this characterization of boundary points is purely topological, it is preserved by homeomorphisms.
How this addresses your question
Well, I hope it illuminates why $partial Ycap f circ phi(U) = fcirc phi (partial U)$. It's because homeomorphisms (and thus diffeomorphisms) preserve boundary points.
answered Mar 17 at 14:26
jgonjgon
16.2k32143
$endgroup$
Here's an easier argument. If $f:Xto Y$ is a homeomorphism of manifolds with boundary, then it restricts to a homeomorphism of the boundaries. (Then if $f$ is a diffeomorphism, the restriction will also be a diffeomorphism.)
Proof:
There is a purely topological way to distinguish boundary points from non-boundary points.
Boundary points have contractible punctured neighborhoods, non-boundary points do not have contractible punctured neighborhoods. This is immediate since deleting the center of an open ball in $Bbb{R}^k$ yields a space homotopic to the $k-1$-sphere, whereas deleting the center of an open ball that has been intersected with a half-space passing through the center gives a contractible space (it's star shaped around any point not on the boundary of the half-space).
Since this characterization of boundary points is purely topological, it is preserved by homeomorphisms.
How this addresses your question
Well, I hope it illuminates why $partial Ycap f circ phi(U) = fcirc phi (partial U)$. It's because homeomorphisms (and thus diffeomorphisms) preserve boundary points.
answered Mar 17 at 14:26
jgonjgon
16.2k32143
answered Mar 17 at 14:26
jgonjgon
16.2k32143
answered Mar 17 at 14:26
jgonjgon
16.2k32143
answered Mar 17 at 14:26
jgonjgon
16.2k32143
16.2k32143
$begingroup$
Interesting! May I know how you intuitively know that boundary is a contractible manifold or in general how do you look at an arbitrary manifold and know if it's contractible? I know just know the definition: $X$ is contractible if $id_X $ is homotopic to a constant map i.e. $X$ can be smoothly shrunk to a point.
$endgroup$
– manifolded
Mar 17 at 19:03
$begingroup$
Also, may I know what you mean by purely topological that makes it homeomorphism invariant? Thanks.
$endgroup$
– manifolded
Mar 17 at 19:04
$begingroup$
@manifolded The boundary is not contractible, at least not generally, but any point in the boundary has a contractible punctured open neighborhood. As I mentioned, take a sufficiently small open ball around the boundary point, which is homeomorphic to an ordinary open ball in $Bbb{R}^n$ intersected with a half-space that passes through the center of the open ball. Deleting the center of the open ball leaves us with half an open ball minus the center point, and this remaining set is star shaped around any point not in the hyperplane that defined the half-space, and thus contractible.
$endgroup$
– jgon
Mar 17 at 19:06
$begingroup$
You may want to try proving from the definition that (nonempty) star shaped subsets of $Bbb{R}^n$ are contractible.
$endgroup$
– jgon
Mar 17 at 19:06
$begingroup$
By purely topological, I mean that the existence of a contractible punctured open neighborhood of a point is something that is preserved by homeomorphisms. If this is not obvious, I suggest trying to prove it.
$endgroup$
– jgon
Mar 17 at 19:07
add a comment |
$begingroup$
Interesting! May I know how you intuitively know that boundary is a contractible manifold or in general how do you look at an arbitrary manifold and know if it's contractible? I know just know the definition: $X$ is contractible if $id_X $ is homotopic to a constant map i.e. $X$ can be smoothly shrunk to a point.
$endgroup$
– manifolded
Mar 17 at 19:03
$begingroup$
Also, may I know what you mean by purely topological that makes it homeomorphism invariant? Thanks.
$endgroup$
– manifolded
Mar 17 at 19:04
$begingroup$
@manifolded The boundary is not contractible, at least not generally, but any point in the boundary has a contractible punctured open neighborhood. As I mentioned, take a sufficiently small open ball around the boundary point, which is homeomorphic to an ordinary open ball in $Bbb{R}^n$ intersected with a half-space that passes through the center of the open ball. Deleting the center of the open ball leaves us with half an open ball minus the center point, and this remaining set is star shaped around any point not in the hyperplane that defined the half-space, and thus contractible.
$endgroup$
– jgon
Mar 17 at 19:06
$begingroup$
You may want to try proving from the definition that (nonempty) star shaped subsets of $Bbb{R}^n$ are contractible.
$endgroup$
– jgon
Mar 17 at 19:06
$begingroup$
By purely topological, I mean that the existence of a contractible punctured open neighborhood of a point is something that is preserved by homeomorphisms. If this is not obvious, I suggest trying to prove it.
$endgroup$
– jgon
Mar 17 at 19:07
$begingroup$
Interesting! May I know how you intuitively know that boundary is a contractible manifold or in general how do you look at an arbitrary manifold and know if it's contractible? I know just know the definition: $X$ is contractible if $id_X $ is homotopic to a constant map i.e. $X$ can be smoothly shrunk to a point.
$endgroup$
– manifolded
Mar 17 at 19:03
$begingroup$
Interesting! May I know how you intuitively know that boundary is a contractible manifold or in general how do you look at an arbitrary manifold and know if it's contractible? I know just know the definition: $X$ is contractible if $id_X $ is homotopic to a constant map i.e. $X$ can be smoothly shrunk to a point.
$endgroup$
– manifolded
Mar 17 at 19:03
$begingroup$
Also, may I know what you mean by purely topological that makes it homeomorphism invariant? Thanks.
$endgroup$
– manifolded
Mar 17 at 19:04
$begingroup$
Also, may I know what you mean by purely topological that makes it homeomorphism invariant? Thanks.
$endgroup$
– manifolded
Mar 17 at 19:04
$begingroup$
@manifolded The boundary is not contractible, at least not generally, but any point in the boundary has a contractible punctured open neighborhood. As I mentioned, take a sufficiently small open ball around the boundary point, which is homeomorphic to an ordinary open ball in $Bbb{R}^n$ intersected with a half-space that passes through the center of the open ball. Deleting the center of the open ball leaves us with half an open ball minus the center point, and this remaining set is star shaped around any point not in the hyperplane that defined the half-space, and thus contractible.
$endgroup$
– jgon
Mar 17 at 19:06
$begingroup$
@manifolded The boundary is not contractible, at least not generally, but any point in the boundary has a contractible punctured open neighborhood. As I mentioned, take a sufficiently small open ball around the boundary point, which is homeomorphic to an ordinary open ball in $Bbb{R}^n$ intersected with a half-space that passes through the center of the open ball. Deleting the center of the open ball leaves us with half an open ball minus the center point, and this remaining set is star shaped around any point not in the hyperplane that defined the half-space, and thus contractible.
$endgroup$
– jgon
Mar 17 at 19:06
$begingroup$
You may want to try proving from the definition that (nonempty) star shaped subsets of $Bbb{R}^n$ are contractible.
$endgroup$
– jgon
Mar 17 at 19:06
$begingroup$
You may want to try proving from the definition that (nonempty) star shaped subsets of $Bbb{R}^n$ are contractible.
$endgroup$
– jgon
Mar 17 at 19:06
$begingroup$
By purely topological, I mean that the existence of a contractible punctured open neighborhood of a point is something that is preserved by homeomorphisms. If this is not obvious, I suggest trying to prove it.
$endgroup$
– jgon
Mar 17 at 19:07
$begingroup$
By purely topological, I mean that the existence of a contractible punctured open neighborhood of a point is something that is preserved by homeomorphisms. If this is not obvious, I suggest trying to prove it.
$endgroup$
– jgon
Mar 17 at 19:07
add a comment |
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