If a smooth map between manifolds is injective, is the induced map on the tangent spaces injective too?
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If $phi:Mlongrightarrow N$ is an injective smooth map between two manifolds, then is $dphi_m:M_mlongrightarrow N_{phi(m)}$, the induced map between the tangent spaces injective too?
I tried the following : If $vin M_m$ is such that $dphi_m(v)=0$, then for all $g$, $C^{infty}$ function in a neighbourhood of $phi(m)$, $dphi_m(v)(g)=0$, that is $v(gcircphi)=0$ for all such $g$. From this can we conclude that $v$ is the $0$ tangent vector.
I got this doubt when I was trying to understand the definition of an immersion. I was wondering if $phi$ being injective will automatically make it an immersion.
differential-geometry manifolds
edited Feb 21 '14 at 0:53
Brian Fitzpatrick
21.1k42958
asked Feb 11 '14 at 7:04
gradstudentgradstudent
1,293720
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add a comment |
$begingroup$
If $phi:Mlongrightarrow N$ is an injective smooth map between two manifolds, then is $dphi_m:M_mlongrightarrow N_{phi(m)}$, the induced map between the tangent spaces injective too?
I tried the following : If $vin M_m$ is such that $dphi_m(v)=0$, then for all $g$, $C^{infty}$ function in a neighbourhood of $phi(m)$, $dphi_m(v)(g)=0$, that is $v(gcircphi)=0$ for all such $g$. From this can we conclude that $v$ is the $0$ tangent vector.
I got this doubt when I was trying to understand the definition of an immersion. I was wondering if $phi$ being injective will automatically make it an immersion.
differential-geometry manifolds
edited Feb 21 '14 at 0:53
Brian Fitzpatrick
21.1k42958
asked Feb 11 '14 at 7:04
gradstudentgradstudent
1,293720
$endgroup$
add a comment |
$begingroup$
If $phi:Mlongrightarrow N$ is an injective smooth map between two manifolds, then is $dphi_m:M_mlongrightarrow N_{phi(m)}$, the induced map between the tangent spaces injective too?
I tried the following : If $vin M_m$ is such that $dphi_m(v)=0$, then for all $g$, $C^{infty}$ function in a neighbourhood of $phi(m)$, $dphi_m(v)(g)=0$, that is $v(gcircphi)=0$ for all such $g$. From this can we conclude that $v$ is the $0$ tangent vector.
I got this doubt when I was trying to understand the definition of an immersion. I was wondering if $phi$ being injective will automatically make it an immersion.
differential-geometry manifolds
edited Feb 21 '14 at 0:53
Brian Fitzpatrick
21.1k42958
asked Feb 11 '14 at 7:04
gradstudentgradstudent
1,293720
$endgroup$
If $phi:Mlongrightarrow N$ is an injective smooth map between two manifolds, then is $dphi_m:M_mlongrightarrow N_{phi(m)}$, the induced map between the tangent spaces injective too?
I tried the following : If $vin M_m$ is such that $dphi_m(v)=0$, then for all $g$, $C^{infty}$ function in a neighbourhood of $phi(m)$, $dphi_m(v)(g)=0$, that is $v(gcircphi)=0$ for all such $g$. From this can we conclude that $v$ is the $0$ tangent vector.
I got this doubt when I was trying to understand the definition of an immersion. I was wondering if $phi$ being injective will automatically make it an immersion.
differential-geometry manifolds
differential-geometry manifolds
edited Feb 21 '14 at 0:53
Brian Fitzpatrick
21.1k42958
asked Feb 11 '14 at 7:04
gradstudentgradstudent
1,293720
edited Feb 21 '14 at 0:53
Brian Fitzpatrick
21.1k42958
asked Feb 11 '14 at 7:04
gradstudentgradstudent
1,293720
edited Feb 21 '14 at 0:53
Brian Fitzpatrick
21.1k42958
edited Feb 21 '14 at 0:53
Brian Fitzpatrick
21.1k42958
edited Feb 21 '14 at 0:53
Brian Fitzpatrick
21.1k42958
21.1k42958
asked Feb 11 '14 at 7:04
gradstudentgradstudent
1,293720
asked Feb 11 '14 at 7:04
gradstudentgradstudent
1,293720
asked Feb 11 '14 at 7:04
gradstudentgradstudent
1,293720
1,293720
add a comment |
add a comment |
2 Answers
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The answer is no. Let $phi:mathbb Rrightarrowmathbb R^2$ be $phi(t)=(t^3,t^9)$. Then the derivative at $t=0$ is not injective.
answered Feb 11 '14 at 7:12
Brian FitzpatrickBrian Fitzpatrick
21.1k42958
$endgroup$
3
$begingroup$
You can also use the function $f(x)=x^3$ from the real line to itself.
$endgroup$
– Moishe Cohen
Feb 11 '14 at 10:33
add a comment |
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If the map $phi:Mlongrightarrow N$ is an diffeomorphism then $dphi_m:M_mlongrightarrow N_{phi(m)}$ is an isomorphism.
edited Dec 8 '18 at 17:19
dantopa
6,45942242
answered Feb 20 '14 at 9:43
ramya msuramya msu
162
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add a comment |
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2 Answers
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active
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2 Answers
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active
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$begingroup$
The answer is no. Let $phi:mathbb Rrightarrowmathbb R^2$ be $phi(t)=(t^3,t^9)$. Then the derivative at $t=0$ is not injective.
answered Feb 11 '14 at 7:12
Brian FitzpatrickBrian Fitzpatrick
21.1k42958
$endgroup$
3
$begingroup$
You can also use the function $f(x)=x^3$ from the real line to itself.
$endgroup$
– Moishe Cohen
Feb 11 '14 at 10:33
add a comment |
$begingroup$
The answer is no. Let $phi:mathbb Rrightarrowmathbb R^2$ be $phi(t)=(t^3,t^9)$. Then the derivative at $t=0$ is not injective.
answered Feb 11 '14 at 7:12
Brian FitzpatrickBrian Fitzpatrick
21.1k42958
$endgroup$
3
$begingroup$
You can also use the function $f(x)=x^3$ from the real line to itself.
$endgroup$
– Moishe Cohen
Feb 11 '14 at 10:33
add a comment |
$begingroup$
The answer is no. Let $phi:mathbb Rrightarrowmathbb R^2$ be $phi(t)=(t^3,t^9)$. Then the derivative at $t=0$ is not injective.
answered Feb 11 '14 at 7:12
Brian FitzpatrickBrian Fitzpatrick
21.1k42958
$endgroup$
The answer is no. Let $phi:mathbb Rrightarrowmathbb R^2$ be $phi(t)=(t^3,t^9)$. Then the derivative at $t=0$ is not injective.
answered Feb 11 '14 at 7:12
Brian FitzpatrickBrian Fitzpatrick
21.1k42958
answered Feb 11 '14 at 7:12
Brian FitzpatrickBrian Fitzpatrick
21.1k42958
answered Feb 11 '14 at 7:12
Brian FitzpatrickBrian Fitzpatrick
21.1k42958
answered Feb 11 '14 at 7:12
Brian FitzpatrickBrian Fitzpatrick
21.1k42958
21.1k42958
3
$begingroup$
You can also use the function $f(x)=x^3$ from the real line to itself.
$endgroup$
– Moishe Cohen
Feb 11 '14 at 10:33
add a comment |
3
$begingroup$
You can also use the function $f(x)=x^3$ from the real line to itself.
$endgroup$
– Moishe Cohen
Feb 11 '14 at 10:33
3
3
$begingroup$
You can also use the function $f(x)=x^3$ from the real line to itself.
$endgroup$
– Moishe Cohen
Feb 11 '14 at 10:33
$begingroup$
You can also use the function $f(x)=x^3$ from the real line to itself.
$endgroup$
– Moishe Cohen
Feb 11 '14 at 10:33
add a comment |
$begingroup$
If the map $phi:Mlongrightarrow N$ is an diffeomorphism then $dphi_m:M_mlongrightarrow N_{phi(m)}$ is an isomorphism.
edited Dec 8 '18 at 17:19
dantopa
6,45942242
answered Feb 20 '14 at 9:43
ramya msuramya msu
162
$endgroup$
add a comment |
$begingroup$
If the map $phi:Mlongrightarrow N$ is an diffeomorphism then $dphi_m:M_mlongrightarrow N_{phi(m)}$ is an isomorphism.
edited Dec 8 '18 at 17:19
dantopa
6,45942242
answered Feb 20 '14 at 9:43
ramya msuramya msu
162
$endgroup$
add a comment |
$begingroup$
If the map $phi:Mlongrightarrow N$ is an diffeomorphism then $dphi_m:M_mlongrightarrow N_{phi(m)}$ is an isomorphism.
edited Dec 8 '18 at 17:19
dantopa
6,45942242
answered Feb 20 '14 at 9:43
ramya msuramya msu
162
$endgroup$
If the map $phi:Mlongrightarrow N$ is an diffeomorphism then $dphi_m:M_mlongrightarrow N_{phi(m)}$ is an isomorphism.
edited Dec 8 '18 at 17:19
dantopa
6,45942242
answered Feb 20 '14 at 9:43
ramya msuramya msu
162
edited Dec 8 '18 at 17:19
dantopa
6,45942242
edited Dec 8 '18 at 17:19
dantopa
6,45942242
edited Dec 8 '18 at 17:19
dantopa
6,45942242
6,45942242
answered Feb 20 '14 at 9:43
ramya msuramya msu
162
answered Feb 20 '14 at 9:43
ramya msuramya msu
162
answered Feb 20 '14 at 9:43
ramya msuramya msu
162
162
add a comment |
add a comment |
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