pull back of smooth covering space is injective
$begingroup$
I need to prove this but I don't really know where to start:
Let $p:Mto N$ be a smooth covering space between smooth manifolds. Show that $p^*:Omega(N)toOmega(M)$ is injective.
Where $Omega(M)$ is the space of the differential forms.
differential-geometry covering-spaces pullback
asked Jan 10 at 16:57
GianniGianni
387
$endgroup$
add a comment |
$begingroup$
I need to prove this but I don't really know where to start:
Let $p:Mto N$ be a smooth covering space between smooth manifolds. Show that $p^*:Omega(N)toOmega(M)$ is injective.
Where $Omega(M)$ is the space of the differential forms.
differential-geometry covering-spaces pullback
asked Jan 10 at 16:57
GianniGianni
387
$endgroup$
1
$begingroup$
$p^*$ is conventionally $Omega(N) rightarrow Omega(M)$.
$endgroup$
– Mindlack
Jan 10 at 17:00
$begingroup$
Oh sure, that was a typo. Edited, thanks
$endgroup$
– Gianni
Jan 10 at 17:09
add a comment |
$begingroup$
I need to prove this but I don't really know where to start:
Let $p:Mto N$ be a smooth covering space between smooth manifolds. Show that $p^*:Omega(N)toOmega(M)$ is injective.
Where $Omega(M)$ is the space of the differential forms.
differential-geometry covering-spaces pullback
asked Jan 10 at 16:57
GianniGianni
387
$endgroup$
I need to prove this but I don't really know where to start:
Let $p:Mto N$ be a smooth covering space between smooth manifolds. Show that $p^*:Omega(N)toOmega(M)$ is injective.
Where $Omega(M)$ is the space of the differential forms.
differential-geometry covering-spaces pullback
differential-geometry covering-spaces pullback
asked Jan 10 at 16:57
GianniGianni
387
asked Jan 10 at 16:57
GianniGianni
387
edited Jan 10 at 17:08
Gianni
asked Jan 10 at 16:57
GianniGianni
387
asked Jan 10 at 16:57
GianniGianni
387
asked Jan 10 at 16:57
GianniGianni
387
387
1
$begingroup$
$p^*$ is conventionally $Omega(N) rightarrow Omega(M)$.
$endgroup$
– Mindlack
Jan 10 at 17:00
$begingroup$
Oh sure, that was a typo. Edited, thanks
$endgroup$
– Gianni
Jan 10 at 17:09
add a comment |
1
$begingroup$
$p^*$ is conventionally $Omega(N) rightarrow Omega(M)$.
$endgroup$
– Mindlack
Jan 10 at 17:00
$begingroup$
Oh sure, that was a typo. Edited, thanks
$endgroup$
– Gianni
Jan 10 at 17:09
1
1
$begingroup$
$p^*$ is conventionally $Omega(N) rightarrow Omega(M)$.
$endgroup$
– Mindlack
Jan 10 at 17:00
$begingroup$
$p^*$ is conventionally $Omega(N) rightarrow Omega(M)$.
$endgroup$
– Mindlack
Jan 10 at 17:00
$begingroup$
Oh sure, that was a typo. Edited, thanks
$endgroup$
– Gianni
Jan 10 at 17:09
$begingroup$
Oh sure, that was a typo. Edited, thanks
$endgroup$
– Gianni
Jan 10 at 17:09
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I would proceed as follows:
1) Note that the differential $d_{q}p:T_{q}Mrightarrow T_{p(q)}N$ is a surjective linear map, at any point $qin M$.
2) Recall that the dual of a surjective linear map is injective.
3) Apply these facts pointwise to get the result.
answered Jan 10 at 17:15
studiosusstudiosus
2,124714
$endgroup$
$begingroup$
I think your differential is actually an isomorphism.
$endgroup$
– Mindlack
Jan 10 at 17:27
$begingroup$
@Mindlack Sure, since $p$ is a local diffeomorphism. But all we need is surjectivity.
$endgroup$
– studiosus
Jan 10 at 17:44
add a comment |
$begingroup$
Write your definitions properly and notice that $p$ induces an isomorphism on each fiber of the tangent bundle.
answered Jan 10 at 17:12
MindlackMindlack
4,815210
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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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$begingroup$
I would proceed as follows:
1) Note that the differential $d_{q}p:T_{q}Mrightarrow T_{p(q)}N$ is a surjective linear map, at any point $qin M$.
2) Recall that the dual of a surjective linear map is injective.
3) Apply these facts pointwise to get the result.
answered Jan 10 at 17:15
studiosusstudiosus
2,124714
$endgroup$
$begingroup$
I think your differential is actually an isomorphism.
$endgroup$
– Mindlack
Jan 10 at 17:27
$begingroup$
@Mindlack Sure, since $p$ is a local diffeomorphism. But all we need is surjectivity.
$endgroup$
– studiosus
Jan 10 at 17:44
add a comment |
$begingroup$
I would proceed as follows:
1) Note that the differential $d_{q}p:T_{q}Mrightarrow T_{p(q)}N$ is a surjective linear map, at any point $qin M$.
2) Recall that the dual of a surjective linear map is injective.
3) Apply these facts pointwise to get the result.
answered Jan 10 at 17:15
studiosusstudiosus
2,124714
$endgroup$
$begingroup$
I think your differential is actually an isomorphism.
$endgroup$
– Mindlack
Jan 10 at 17:27
$begingroup$
@Mindlack Sure, since $p$ is a local diffeomorphism. But all we need is surjectivity.
$endgroup$
– studiosus
Jan 10 at 17:44
add a comment |
$begingroup$
I would proceed as follows:
1) Note that the differential $d_{q}p:T_{q}Mrightarrow T_{p(q)}N$ is a surjective linear map, at any point $qin M$.
2) Recall that the dual of a surjective linear map is injective.
3) Apply these facts pointwise to get the result.
answered Jan 10 at 17:15
studiosusstudiosus
2,124714
$endgroup$
I would proceed as follows:
1) Note that the differential $d_{q}p:T_{q}Mrightarrow T_{p(q)}N$ is a surjective linear map, at any point $qin M$.
2) Recall that the dual of a surjective linear map is injective.
3) Apply these facts pointwise to get the result.
answered Jan 10 at 17:15
studiosusstudiosus
2,124714
answered Jan 10 at 17:15
studiosusstudiosus
2,124714
answered Jan 10 at 17:15
studiosusstudiosus
2,124714
answered Jan 10 at 17:15
studiosusstudiosus
2,124714
2,124714
$begingroup$
I think your differential is actually an isomorphism.
$endgroup$
– Mindlack
Jan 10 at 17:27
$begingroup$
@Mindlack Sure, since $p$ is a local diffeomorphism. But all we need is surjectivity.
$endgroup$
– studiosus
Jan 10 at 17:44
add a comment |
$begingroup$
I think your differential is actually an isomorphism.
$endgroup$
– Mindlack
Jan 10 at 17:27
$begingroup$
@Mindlack Sure, since $p$ is a local diffeomorphism. But all we need is surjectivity.
$endgroup$
– studiosus
Jan 10 at 17:44
$begingroup$
I think your differential is actually an isomorphism.
$endgroup$
– Mindlack
Jan 10 at 17:27
$begingroup$
I think your differential is actually an isomorphism.
$endgroup$
– Mindlack
Jan 10 at 17:27
$begingroup$
@Mindlack Sure, since $p$ is a local diffeomorphism. But all we need is surjectivity.
$endgroup$
– studiosus
Jan 10 at 17:44
$begingroup$
@Mindlack Sure, since $p$ is a local diffeomorphism. But all we need is surjectivity.
$endgroup$
– studiosus
Jan 10 at 17:44
add a comment |
$begingroup$
Write your definitions properly and notice that $p$ induces an isomorphism on each fiber of the tangent bundle.
answered Jan 10 at 17:12
MindlackMindlack
4,815210
$endgroup$
add a comment |
$begingroup$
Write your definitions properly and notice that $p$ induces an isomorphism on each fiber of the tangent bundle.
answered Jan 10 at 17:12
MindlackMindlack
4,815210
$endgroup$
add a comment |
$begingroup$
Write your definitions properly and notice that $p$ induces an isomorphism on each fiber of the tangent bundle.
answered Jan 10 at 17:12
MindlackMindlack
4,815210
$endgroup$
Write your definitions properly and notice that $p$ induces an isomorphism on each fiber of the tangent bundle.
answered Jan 10 at 17:12
MindlackMindlack
4,815210
answered Jan 10 at 17:12
MindlackMindlack
4,815210
answered Jan 10 at 17:12
MindlackMindlack
4,815210
answered Jan 10 at 17:12
MindlackMindlack
4,815210
4,815210
add a comment |
add a comment |
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1
$begingroup$
$p^*$ is conventionally $Omega(N) rightarrow Omega(M)$.
$endgroup$
– Mindlack
Jan 10 at 17:00
$begingroup$
Oh sure, that was a typo. Edited, thanks
$endgroup$
– Gianni
Jan 10 at 17:09