Smooth bijection and tangent spaces
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Let $f:Mto N$ be a smooth bijection between manifolds with same dimension. Do we necessarily have
$$df_p(T_pM)=T_{f(p)}N.$$
I think it is probably not true. But I can't give a counterexample...
differential-geometry smooth-manifolds
asked Dec 11 '18 at 16:34
No OneNo One
2,0341519
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add a comment |
$begingroup$
Let $f:Mto N$ be a smooth bijection between manifolds with same dimension. Do we necessarily have
$$df_p(T_pM)=T_{f(p)}N.$$
I think it is probably not true. But I can't give a counterexample...
differential-geometry smooth-manifolds
asked Dec 11 '18 at 16:34
No OneNo One
2,0341519
$endgroup$
add a comment |
$begingroup$
Let $f:Mto N$ be a smooth bijection between manifolds with same dimension. Do we necessarily have
$$df_p(T_pM)=T_{f(p)}N.$$
I think it is probably not true. But I can't give a counterexample...
differential-geometry smooth-manifolds
asked Dec 11 '18 at 16:34
No OneNo One
2,0341519
$endgroup$
Let $f:Mto N$ be a smooth bijection between manifolds with same dimension. Do we necessarily have
$$df_p(T_pM)=T_{f(p)}N.$$
I think it is probably not true. But I can't give a counterexample...
differential-geometry smooth-manifolds
differential-geometry smooth-manifolds
asked Dec 11 '18 at 16:34
No OneNo One
2,0341519
asked Dec 11 '18 at 16:34
No OneNo One
2,0341519
asked Dec 11 '18 at 16:34
No OneNo One
2,0341519
asked Dec 11 '18 at 16:34
No OneNo One
2,0341519
asked Dec 11 '18 at 16:34
No OneNo One
2,0341519
2,0341519
add a comment |
add a comment |
2 Answers
2
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The map $f:mathbb{R}tomathbb{R}$ given by $f:xmapsto x^3$ is a smooth bijection but $df_0(T_0mathbb{R}) = 0neq T_{f(0)}mathbb{R}$.
answered Dec 11 '18 at 16:41
NealNeal
23.6k23886
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How about $f: mathbb{R} to mathbb{R}$ by $f(x)=x^3$? Certainly smooth, certainly bijective. However, $df_0$ will map the $1$-dimensional tangent space to a point. The root issue is that a smooth bijection can still have critical points.
answered Dec 11 '18 at 16:41
RandallRandall
9,71111230
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2 Answers
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2 Answers
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$begingroup$
The map $f:mathbb{R}tomathbb{R}$ given by $f:xmapsto x^3$ is a smooth bijection but $df_0(T_0mathbb{R}) = 0neq T_{f(0)}mathbb{R}$.
answered Dec 11 '18 at 16:41
NealNeal
23.6k23886
$endgroup$
add a comment |
$begingroup$
The map $f:mathbb{R}tomathbb{R}$ given by $f:xmapsto x^3$ is a smooth bijection but $df_0(T_0mathbb{R}) = 0neq T_{f(0)}mathbb{R}$.
answered Dec 11 '18 at 16:41
NealNeal
23.6k23886
$endgroup$
add a comment |
$begingroup$
The map $f:mathbb{R}tomathbb{R}$ given by $f:xmapsto x^3$ is a smooth bijection but $df_0(T_0mathbb{R}) = 0neq T_{f(0)}mathbb{R}$.
answered Dec 11 '18 at 16:41
NealNeal
23.6k23886
$endgroup$
The map $f:mathbb{R}tomathbb{R}$ given by $f:xmapsto x^3$ is a smooth bijection but $df_0(T_0mathbb{R}) = 0neq T_{f(0)}mathbb{R}$.
answered Dec 11 '18 at 16:41
NealNeal
23.6k23886
answered Dec 11 '18 at 16:41
NealNeal
23.6k23886
answered Dec 11 '18 at 16:41
NealNeal
23.6k23886
answered Dec 11 '18 at 16:41
NealNeal
23.6k23886
23.6k23886
add a comment |
add a comment |
$begingroup$
How about $f: mathbb{R} to mathbb{R}$ by $f(x)=x^3$? Certainly smooth, certainly bijective. However, $df_0$ will map the $1$-dimensional tangent space to a point. The root issue is that a smooth bijection can still have critical points.
answered Dec 11 '18 at 16:41
RandallRandall
9,71111230
$endgroup$
add a comment |
$begingroup$
How about $f: mathbb{R} to mathbb{R}$ by $f(x)=x^3$? Certainly smooth, certainly bijective. However, $df_0$ will map the $1$-dimensional tangent space to a point. The root issue is that a smooth bijection can still have critical points.
answered Dec 11 '18 at 16:41
RandallRandall
9,71111230
$endgroup$
add a comment |
$begingroup$
How about $f: mathbb{R} to mathbb{R}$ by $f(x)=x^3$? Certainly smooth, certainly bijective. However, $df_0$ will map the $1$-dimensional tangent space to a point. The root issue is that a smooth bijection can still have critical points.
answered Dec 11 '18 at 16:41
RandallRandall
9,71111230
$endgroup$
How about $f: mathbb{R} to mathbb{R}$ by $f(x)=x^3$? Certainly smooth, certainly bijective. However, $df_0$ will map the $1$-dimensional tangent space to a point. The root issue is that a smooth bijection can still have critical points.
answered Dec 11 '18 at 16:41
RandallRandall
9,71111230
answered Dec 11 '18 at 16:41
RandallRandall
9,71111230
answered Dec 11 '18 at 16:41
RandallRandall
9,71111230
answered Dec 11 '18 at 16:41
RandallRandall
9,71111230
9,71111230
add a comment |
add a comment |
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