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Diffeomorphism mapping interior point to boundary point of manifolds with boundary
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Why can a diffeomorphism $F: M to N$ between two smooth manifolds with boundary not take an interior point of $M$ to a boundary point of $N$?
Let $(U, varphi)$ be a smooth chart for $M$, $(V, psi)$ a smooth chart for $N$.
I believe it is because it would cause $(psi circ F circ varphi^-1): varphi(U) to psi(V)$ to be a diffeomorphism between an open set in $mathbbR^n$ and an open set in $mathbbH^n$ such that its intersection with $partial mathbbH^n$ is nonempty, and this cannot be. However, I don't know how to prove this. Any help would be appreciated
differential-geometry smooth-manifolds diffeomorphism
asked 1 hour ago
Emilio Minichiello
3057
add a comment |Â
up vote
2
down vote
favorite
Why can a diffeomorphism $F: M to N$ between two smooth manifolds with boundary not take an interior point of $M$ to a boundary point of $N$?
Let $(U, varphi)$ be a smooth chart for $M$, $(V, psi)$ a smooth chart for $N$.
I believe it is because it would cause $(psi circ F circ varphi^-1): varphi(U) to psi(V)$ to be a diffeomorphism between an open set in $mathbbR^n$ and an open set in $mathbbH^n$ such that its intersection with $partial mathbbH^n$ is nonempty, and this cannot be. However, I don't know how to prove this. Any help would be appreciated
differential-geometry smooth-manifolds diffeomorphism
asked 1 hour ago
Emilio Minichiello
3057
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Why can a diffeomorphism $F: M to N$ between two smooth manifolds with boundary not take an interior point of $M$ to a boundary point of $N$?
Let $(U, varphi)$ be a smooth chart for $M$, $(V, psi)$ a smooth chart for $N$.
I believe it is because it would cause $(psi circ F circ varphi^-1): varphi(U) to psi(V)$ to be a diffeomorphism between an open set in $mathbbR^n$ and an open set in $mathbbH^n$ such that its intersection with $partial mathbbH^n$ is nonempty, and this cannot be. However, I don't know how to prove this. Any help would be appreciated
differential-geometry smooth-manifolds diffeomorphism
asked 1 hour ago
Emilio Minichiello
3057
Why can a diffeomorphism $F: M to N$ between two smooth manifolds with boundary not take an interior point of $M$ to a boundary point of $N$?
Let $(U, varphi)$ be a smooth chart for $M$, $(V, psi)$ a smooth chart for $N$.
I believe it is because it would cause $(psi circ F circ varphi^-1): varphi(U) to psi(V)$ to be a diffeomorphism between an open set in $mathbbR^n$ and an open set in $mathbbH^n$ such that its intersection with $partial mathbbH^n$ is nonempty, and this cannot be. However, I don't know how to prove this. Any help would be appreciated
differential-geometry smooth-manifolds diffeomorphism
differential-geometry smooth-manifolds diffeomorphism
asked 1 hour ago
Emilio Minichiello
3057
asked 1 hour ago
Emilio Minichiello
3057
asked 1 hour ago
Emilio Minichiello
3057
asked 1 hour ago
Emilio Minichiello
3057
asked 1 hour ago
Emilio Minichiello
3057
3057
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
4
down vote
accepted
A diffeomorphism is in particular a homeomorphism, so a neighbourhood of a point must be mapped to a homeomorphic neighbourhood of the image of the point. In particular, it cannot take interior points to boundary points since they have non-homeomorphic neighbourhoods.
answered 1 hour ago
Javi
2,2631725
Ah, I think I see my confusion. Lee's Smooth Manifolds never explicitly says the nbds of the different kinds of points are not homeomorphic but he says that a point can never be both kinds of points, which is effectively saying the above. Thanks.
– Emilio Minichiello
1 hour ago
1
@EmilioMinichiello: Javi's answer is correct (and I've upvoted) - but do you know how to prove they have non-homeomorphic neighborhoods? (And I wouldn't say they are non-homeomorphic by definition. They are certainly presented differently, but that doesn't rule out some kind of weird homeomorphism. The usual trick is local homology.)
– Jason DeVito
1 hour ago
I'm on Chapter 2 of Smooth Manifolds, Lee doesn't give us the proof that the nbds are non-homeomorphic. He only says that a point cannot be a boundary point and an interior point simultaneously, and that every smooth chart will map a boundary point to the boundary of $mathbbH^n$. I'm trying to understand how this implies that $F(textIntM) = textIntN$.
– Emilio Minichiello
1 hour ago
add a comment |Â
up vote
3
down vote
The usual proof that neighborhoods that intersect $partialmathbbH$ are distinct from neighborhoods uses the local homology groups $H^ast(M,Msetminusp)$ for different choices of $p$. It has the advantage of showing that a homeomorphism must map boundary to boundary and interior to interior.
But since you're asking about diffeomorphisms, we should be able to work less hard. So how about this?
Suppose $f:Mrightarrow N$ is a diffeomorphism and that $pin operatornameint M$ and $f(p)in partial N$. Then $d_p f:T_p Mrightarrow T_f(p) N$ must be an isomorphism.
But I claim that is not. To see this, pick a $win T_f(p)$ which points outside of $N$. I claim there is no $vin T_p M$ with $d_p f(v) = w$.
To see this, suppose there is such a $v$, choose a smooth path $gamma_v:(-epsilon,epsilon)rightarrow M$ with $gamma_v(0) = p$ and $gamma_v'(0) = v$. Then $fcircgamma$ is a smooth curve in $N$. By picking a chart near $f(p)$, we can view all this as happening in $mathbbR^n$.
So, without loss of generality, we have a curve $gamma$ for which $gamma(0) = vec0in mathbbR^n$, $gammasubseteq H =(x_1,...,x_n)in mathbbR^n: x_nleq 0$, but $gamma'(0) notin H$. That is, the $n$th coordinate of $gamma'(0)$ is positive.
Since $gamma$ is smooth, $gamma'$ is continuous, so the $n$-th coordinate of $gamma'(t)$ is positive in a neighborhood of $0$. But now the mean value theorem applied to the $n$th coordinate of $gamma$ implies that $gamma(t)notin H$ for small $t$ with $t>0$.
answered 1 hour ago
Jason DeVito
29.8k474132
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
A diffeomorphism is in particular a homeomorphism, so a neighbourhood of a point must be mapped to a homeomorphic neighbourhood of the image of the point. In particular, it cannot take interior points to boundary points since they have non-homeomorphic neighbourhoods.
answered 1 hour ago
Javi
2,2631725
Ah, I think I see my confusion. Lee's Smooth Manifolds never explicitly says the nbds of the different kinds of points are not homeomorphic but he says that a point can never be both kinds of points, which is effectively saying the above. Thanks.
– Emilio Minichiello
1 hour ago
1
@EmilioMinichiello: Javi's answer is correct (and I've upvoted) - but do you know how to prove they have non-homeomorphic neighborhoods? (And I wouldn't say they are non-homeomorphic by definition. They are certainly presented differently, but that doesn't rule out some kind of weird homeomorphism. The usual trick is local homology.)
– Jason DeVito
1 hour ago
I'm on Chapter 2 of Smooth Manifolds, Lee doesn't give us the proof that the nbds are non-homeomorphic. He only says that a point cannot be a boundary point and an interior point simultaneously, and that every smooth chart will map a boundary point to the boundary of $mathbbH^n$. I'm trying to understand how this implies that $F(textIntM) = textIntN$.
– Emilio Minichiello
1 hour ago
add a comment |Â
up vote
4
down vote
accepted
A diffeomorphism is in particular a homeomorphism, so a neighbourhood of a point must be mapped to a homeomorphic neighbourhood of the image of the point. In particular, it cannot take interior points to boundary points since they have non-homeomorphic neighbourhoods.
answered 1 hour ago
Javi
2,2631725
Ah, I think I see my confusion. Lee's Smooth Manifolds never explicitly says the nbds of the different kinds of points are not homeomorphic but he says that a point can never be both kinds of points, which is effectively saying the above. Thanks.
– Emilio Minichiello
1 hour ago
1
@EmilioMinichiello: Javi's answer is correct (and I've upvoted) - but do you know how to prove they have non-homeomorphic neighborhoods? (And I wouldn't say they are non-homeomorphic by definition. They are certainly presented differently, but that doesn't rule out some kind of weird homeomorphism. The usual trick is local homology.)
– Jason DeVito
1 hour ago
I'm on Chapter 2 of Smooth Manifolds, Lee doesn't give us the proof that the nbds are non-homeomorphic. He only says that a point cannot be a boundary point and an interior point simultaneously, and that every smooth chart will map a boundary point to the boundary of $mathbbH^n$. I'm trying to understand how this implies that $F(textIntM) = textIntN$.
– Emilio Minichiello
1 hour ago
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
A diffeomorphism is in particular a homeomorphism, so a neighbourhood of a point must be mapped to a homeomorphic neighbourhood of the image of the point. In particular, it cannot take interior points to boundary points since they have non-homeomorphic neighbourhoods.
answered 1 hour ago
Javi
2,2631725
A diffeomorphism is in particular a homeomorphism, so a neighbourhood of a point must be mapped to a homeomorphic neighbourhood of the image of the point. In particular, it cannot take interior points to boundary points since they have non-homeomorphic neighbourhoods.
answered 1 hour ago
Javi
2,2631725
edited 14 mins ago
answered 1 hour ago
Javi
2,2631725
answered 1 hour ago
Javi
2,2631725
answered 1 hour ago
Javi
2,2631725
2,2631725
Ah, I think I see my confusion. Lee's Smooth Manifolds never explicitly says the nbds of the different kinds of points are not homeomorphic but he says that a point can never be both kinds of points, which is effectively saying the above. Thanks.
– Emilio Minichiello
1 hour ago
1
@EmilioMinichiello: Javi's answer is correct (and I've upvoted) - but do you know how to prove they have non-homeomorphic neighborhoods? (And I wouldn't say they are non-homeomorphic by definition. They are certainly presented differently, but that doesn't rule out some kind of weird homeomorphism. The usual trick is local homology.)
– Jason DeVito
1 hour ago
I'm on Chapter 2 of Smooth Manifolds, Lee doesn't give us the proof that the nbds are non-homeomorphic. He only says that a point cannot be a boundary point and an interior point simultaneously, and that every smooth chart will map a boundary point to the boundary of $mathbbH^n$. I'm trying to understand how this implies that $F(textIntM) = textIntN$.
– Emilio Minichiello
1 hour ago
add a comment |Â
Ah, I think I see my confusion. Lee's Smooth Manifolds never explicitly says the nbds of the different kinds of points are not homeomorphic but he says that a point can never be both kinds of points, which is effectively saying the above. Thanks.
– Emilio Minichiello
1 hour ago
1
@EmilioMinichiello: Javi's answer is correct (and I've upvoted) - but do you know how to prove they have non-homeomorphic neighborhoods? (And I wouldn't say they are non-homeomorphic by definition. They are certainly presented differently, but that doesn't rule out some kind of weird homeomorphism. The usual trick is local homology.)
– Jason DeVito
1 hour ago
I'm on Chapter 2 of Smooth Manifolds, Lee doesn't give us the proof that the nbds are non-homeomorphic. He only says that a point cannot be a boundary point and an interior point simultaneously, and that every smooth chart will map a boundary point to the boundary of $mathbbH^n$. I'm trying to understand how this implies that $F(textIntM) = textIntN$.
– Emilio Minichiello
1 hour ago
Ah, I think I see my confusion. Lee's Smooth Manifolds never explicitly says the nbds of the different kinds of points are not homeomorphic but he says that a point can never be both kinds of points, which is effectively saying the above. Thanks.
– Emilio Minichiello
1 hour ago
Ah, I think I see my confusion. Lee's Smooth Manifolds never explicitly says the nbds of the different kinds of points are not homeomorphic but he says that a point can never be both kinds of points, which is effectively saying the above. Thanks.
– Emilio Minichiello
1 hour ago
1
1
@EmilioMinichiello: Javi's answer is correct (and I've upvoted) - but do you know how to prove they have non-homeomorphic neighborhoods? (And I wouldn't say they are non-homeomorphic by definition. They are certainly presented differently, but that doesn't rule out some kind of weird homeomorphism. The usual trick is local homology.)
– Jason DeVito
1 hour ago
@EmilioMinichiello: Javi's answer is correct (and I've upvoted) - but do you know how to prove they have non-homeomorphic neighborhoods? (And I wouldn't say they are non-homeomorphic by definition. They are certainly presented differently, but that doesn't rule out some kind of weird homeomorphism. The usual trick is local homology.)
– Jason DeVito
1 hour ago
I'm on Chapter 2 of Smooth Manifolds, Lee doesn't give us the proof that the nbds are non-homeomorphic. He only says that a point cannot be a boundary point and an interior point simultaneously, and that every smooth chart will map a boundary point to the boundary of $mathbbH^n$. I'm trying to understand how this implies that $F(textIntM) = textIntN$.
– Emilio Minichiello
1 hour ago
I'm on Chapter 2 of Smooth Manifolds, Lee doesn't give us the proof that the nbds are non-homeomorphic. He only says that a point cannot be a boundary point and an interior point simultaneously, and that every smooth chart will map a boundary point to the boundary of $mathbbH^n$. I'm trying to understand how this implies that $F(textIntM) = textIntN$.
– Emilio Minichiello
1 hour ago
add a comment |Â
up vote
3
down vote
The usual proof that neighborhoods that intersect $partialmathbbH$ are distinct from neighborhoods uses the local homology groups $H^ast(M,Msetminusp)$ for different choices of $p$. It has the advantage of showing that a homeomorphism must map boundary to boundary and interior to interior.
But since you're asking about diffeomorphisms, we should be able to work less hard. So how about this?
Suppose $f:Mrightarrow N$ is a diffeomorphism and that $pin operatornameint M$ and $f(p)in partial N$. Then $d_p f:T_p Mrightarrow T_f(p) N$ must be an isomorphism.
But I claim that is not. To see this, pick a $win T_f(p)$ which points outside of $N$. I claim there is no $vin T_p M$ with $d_p f(v) = w$.
To see this, suppose there is such a $v$, choose a smooth path $gamma_v:(-epsilon,epsilon)rightarrow M$ with $gamma_v(0) = p$ and $gamma_v'(0) = v$. Then $fcircgamma$ is a smooth curve in $N$. By picking a chart near $f(p)$, we can view all this as happening in $mathbbR^n$.
So, without loss of generality, we have a curve $gamma$ for which $gamma(0) = vec0in mathbbR^n$, $gammasubseteq H =(x_1,...,x_n)in mathbbR^n: x_nleq 0$, but $gamma'(0) notin H$. That is, the $n$th coordinate of $gamma'(0)$ is positive.
Since $gamma$ is smooth, $gamma'$ is continuous, so the $n$-th coordinate of $gamma'(t)$ is positive in a neighborhood of $0$. But now the mean value theorem applied to the $n$th coordinate of $gamma$ implies that $gamma(t)notin H$ for small $t$ with $t>0$.
answered 1 hour ago
Jason DeVito
29.8k474132
add a comment |Â
up vote
3
down vote
The usual proof that neighborhoods that intersect $partialmathbbH$ are distinct from neighborhoods uses the local homology groups $H^ast(M,Msetminusp)$ for different choices of $p$. It has the advantage of showing that a homeomorphism must map boundary to boundary and interior to interior.
But since you're asking about diffeomorphisms, we should be able to work less hard. So how about this?
Suppose $f:Mrightarrow N$ is a diffeomorphism and that $pin operatornameint M$ and $f(p)in partial N$. Then $d_p f:T_p Mrightarrow T_f(p) N$ must be an isomorphism.
But I claim that is not. To see this, pick a $win T_f(p)$ which points outside of $N$. I claim there is no $vin T_p M$ with $d_p f(v) = w$.
To see this, suppose there is such a $v$, choose a smooth path $gamma_v:(-epsilon,epsilon)rightarrow M$ with $gamma_v(0) = p$ and $gamma_v'(0) = v$. Then $fcircgamma$ is a smooth curve in $N$. By picking a chart near $f(p)$, we can view all this as happening in $mathbbR^n$.
So, without loss of generality, we have a curve $gamma$ for which $gamma(0) = vec0in mathbbR^n$, $gammasubseteq H =(x_1,...,x_n)in mathbbR^n: x_nleq 0$, but $gamma'(0) notin H$. That is, the $n$th coordinate of $gamma'(0)$ is positive.
Since $gamma$ is smooth, $gamma'$ is continuous, so the $n$-th coordinate of $gamma'(t)$ is positive in a neighborhood of $0$. But now the mean value theorem applied to the $n$th coordinate of $gamma$ implies that $gamma(t)notin H$ for small $t$ with $t>0$.
answered 1 hour ago
Jason DeVito
29.8k474132
add a comment |Â
up vote
3
down vote
up vote
3
down vote
The usual proof that neighborhoods that intersect $partialmathbbH$ are distinct from neighborhoods uses the local homology groups $H^ast(M,Msetminusp)$ for different choices of $p$. It has the advantage of showing that a homeomorphism must map boundary to boundary and interior to interior.
But since you're asking about diffeomorphisms, we should be able to work less hard. So how about this?
Suppose $f:Mrightarrow N$ is a diffeomorphism and that $pin operatornameint M$ and $f(p)in partial N$. Then $d_p f:T_p Mrightarrow T_f(p) N$ must be an isomorphism.
But I claim that is not. To see this, pick a $win T_f(p)$ which points outside of $N$. I claim there is no $vin T_p M$ with $d_p f(v) = w$.
To see this, suppose there is such a $v$, choose a smooth path $gamma_v:(-epsilon,epsilon)rightarrow M$ with $gamma_v(0) = p$ and $gamma_v'(0) = v$. Then $fcircgamma$ is a smooth curve in $N$. By picking a chart near $f(p)$, we can view all this as happening in $mathbbR^n$.
So, without loss of generality, we have a curve $gamma$ for which $gamma(0) = vec0in mathbbR^n$, $gammasubseteq H =(x_1,...,x_n)in mathbbR^n: x_nleq 0$, but $gamma'(0) notin H$. That is, the $n$th coordinate of $gamma'(0)$ is positive.
Since $gamma$ is smooth, $gamma'$ is continuous, so the $n$-th coordinate of $gamma'(t)$ is positive in a neighborhood of $0$. But now the mean value theorem applied to the $n$th coordinate of $gamma$ implies that $gamma(t)notin H$ for small $t$ with $t>0$.
answered 1 hour ago
Jason DeVito
29.8k474132
The usual proof that neighborhoods that intersect $partialmathbbH$ are distinct from neighborhoods uses the local homology groups $H^ast(M,Msetminusp)$ for different choices of $p$. It has the advantage of showing that a homeomorphism must map boundary to boundary and interior to interior.
But since you're asking about diffeomorphisms, we should be able to work less hard. So how about this?
Suppose $f:Mrightarrow N$ is a diffeomorphism and that $pin operatornameint M$ and $f(p)in partial N$. Then $d_p f:T_p Mrightarrow T_f(p) N$ must be an isomorphism.
But I claim that is not. To see this, pick a $win T_f(p)$ which points outside of $N$. I claim there is no $vin T_p M$ with $d_p f(v) = w$.
To see this, suppose there is such a $v$, choose a smooth path $gamma_v:(-epsilon,epsilon)rightarrow M$ with $gamma_v(0) = p$ and $gamma_v'(0) = v$. Then $fcircgamma$ is a smooth curve in $N$. By picking a chart near $f(p)$, we can view all this as happening in $mathbbR^n$.
So, without loss of generality, we have a curve $gamma$ for which $gamma(0) = vec0in mathbbR^n$, $gammasubseteq H =(x_1,...,x_n)in mathbbR^n: x_nleq 0$, but $gamma'(0) notin H$. That is, the $n$th coordinate of $gamma'(0)$ is positive.
Since $gamma$ is smooth, $gamma'$ is continuous, so the $n$-th coordinate of $gamma'(t)$ is positive in a neighborhood of $0$. But now the mean value theorem applied to the $n$th coordinate of $gamma$ implies that $gamma(t)notin H$ for small $t$ with $t>0$.
answered 1 hour ago
Jason DeVito
29.8k474132
answered 1 hour ago
Jason DeVito
29.8k474132
answered 1 hour ago
Jason DeVito
29.8k474132
answered 1 hour ago
Jason DeVito
29.8k474132
29.8k474132
add a comment |Â
add a comment |Â
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