Spivak Calculus on Manifolds, Theorem 5-2(2)
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In the proof of Theorem 5-2 of Spivak Calculus on Mannifolds, are the two sets ${f(a):(a,0)in V_1}$ and ${f(a):(a,0)in V_1'}$ the same?
calculus multivariable-calculus manifolds smooth-manifolds
asked Jul 5 '17 at 14:08
chenchen
655
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add a comment |
$begingroup$
In the proof of Theorem 5-2 of Spivak Calculus on Mannifolds, are the two sets ${f(a):(a,0)in V_1}$ and ${f(a):(a,0)in V_1'}$ the same?
calculus multivariable-calculus manifolds smooth-manifolds
asked Jul 5 '17 at 14:08
chenchen
655
$endgroup$
3
$begingroup$
I took the liberty in posting images of the proof. Next time you can either give a sketch or do the same. I don't think everyone has a copy of this book or for some, we aren't willing to torture ourselves again so providing some more details is always nice.
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– Faraad Armwood
Jul 5 '17 at 14:13
1
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@FaraadArmwood Thank you.
$endgroup$
– chen
Jul 5 '17 at 14:21
add a comment |
$begingroup$
In the proof of Theorem 5-2 of Spivak Calculus on Mannifolds, are the two sets ${f(a):(a,0)in V_1}$ and ${f(a):(a,0)in V_1'}$ the same?
calculus multivariable-calculus manifolds smooth-manifolds
asked Jul 5 '17 at 14:08
chenchen
655
$endgroup$
In the proof of Theorem 5-2 of Spivak Calculus on Mannifolds, are the two sets ${f(a):(a,0)in V_1}$ and ${f(a):(a,0)in V_1'}$ the same?
calculus multivariable-calculus manifolds smooth-manifolds
calculus multivariable-calculus manifolds smooth-manifolds
asked Jul 5 '17 at 14:08
chenchen
655
asked Jul 5 '17 at 14:08
chenchen
655
edited Jul 6 '17 at 3:33
chen
asked Jul 5 '17 at 14:08
chenchen
655
asked Jul 5 '17 at 14:08
chenchen
655
asked Jul 5 '17 at 14:08
chenchen
655
655
3
$begingroup$
I took the liberty in posting images of the proof. Next time you can either give a sketch or do the same. I don't think everyone has a copy of this book or for some, we aren't willing to torture ourselves again so providing some more details is always nice.
$endgroup$
– Faraad Armwood
Jul 5 '17 at 14:13
1
$begingroup$
@FaraadArmwood Thank you.
$endgroup$
– chen
Jul 5 '17 at 14:21
add a comment |
3
$begingroup$
I took the liberty in posting images of the proof. Next time you can either give a sketch or do the same. I don't think everyone has a copy of this book or for some, we aren't willing to torture ourselves again so providing some more details is always nice.
$endgroup$
– Faraad Armwood
Jul 5 '17 at 14:13
1
$begingroup$
@FaraadArmwood Thank you.
$endgroup$
– chen
Jul 5 '17 at 14:21
3
3
$begingroup$
I took the liberty in posting images of the proof. Next time you can either give a sketch or do the same. I don't think everyone has a copy of this book or for some, we aren't willing to torture ourselves again so providing some more details is always nice.
$endgroup$
– Faraad Armwood
Jul 5 '17 at 14:13
$begingroup$
I took the liberty in posting images of the proof. Next time you can either give a sketch or do the same. I don't think everyone has a copy of this book or for some, we aren't willing to torture ourselves again so providing some more details is always nice.
$endgroup$
– Faraad Armwood
Jul 5 '17 at 14:13
1
1
$begingroup$
@FaraadArmwood Thank you.
$endgroup$
– chen
Jul 5 '17 at 14:21
$begingroup$
@FaraadArmwood Thank you.
$endgroup$
– chen
Jul 5 '17 at 14:21
add a comment |
1 Answer
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I'm not sure whether my answer is right or not,if you have free time please check it. Any help will be very much appreciated !
$textbf{1.}$
$${g(a,0):(a,0)in V^{'}_1}={f(a):(a,0)in V^{'}_1}Rightarrow$$$$ V^{'}_{2}cap f(W)=Ucap f(W),$$ since $$V^{'}_{2}cap f(W)={g(a,0):(a,0)in V^{'}_1},Ucap f(W)={f(a):(a,0)in V^{'}_1}.$$
$textbf{2.}$
$$V^{'}_{2}cap f(W)=Ucap f(W)=V_{2}cap f(W) left ( V_2:= V^{'}_{2}cap Uright ),$$
owing to
$$Acap C=Bcap CRightarrow Acap B cap C=Acap C=Bcap C.$$
From above, we have $${f(a):(a,0)in V^{'}_1}=V_{2}cap f(W)={g(a,0):(a,0)in V_1}={f(a):(a,0)in V_1}.qquadqquadqquadqquadqquadqquadqquadqquadBox$$
answered Dec 4 '18 at 9:38
user553010user553010
688
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1 Answer
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1 Answer
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$begingroup$
I'm not sure whether my answer is right or not,if you have free time please check it. Any help will be very much appreciated !
$textbf{1.}$
$${g(a,0):(a,0)in V^{'}_1}={f(a):(a,0)in V^{'}_1}Rightarrow$$$$ V^{'}_{2}cap f(W)=Ucap f(W),$$ since $$V^{'}_{2}cap f(W)={g(a,0):(a,0)in V^{'}_1},Ucap f(W)={f(a):(a,0)in V^{'}_1}.$$
$textbf{2.}$
$$V^{'}_{2}cap f(W)=Ucap f(W)=V_{2}cap f(W) left ( V_2:= V^{'}_{2}cap Uright ),$$
owing to
$$Acap C=Bcap CRightarrow Acap B cap C=Acap C=Bcap C.$$
From above, we have $${f(a):(a,0)in V^{'}_1}=V_{2}cap f(W)={g(a,0):(a,0)in V_1}={f(a):(a,0)in V_1}.qquadqquadqquadqquadqquadqquadqquadqquadBox$$
answered Dec 4 '18 at 9:38
user553010user553010
688
$endgroup$
add a comment |
$begingroup$
I'm not sure whether my answer is right or not,if you have free time please check it. Any help will be very much appreciated !
$textbf{1.}$
$${g(a,0):(a,0)in V^{'}_1}={f(a):(a,0)in V^{'}_1}Rightarrow$$$$ V^{'}_{2}cap f(W)=Ucap f(W),$$ since $$V^{'}_{2}cap f(W)={g(a,0):(a,0)in V^{'}_1},Ucap f(W)={f(a):(a,0)in V^{'}_1}.$$
$textbf{2.}$
$$V^{'}_{2}cap f(W)=Ucap f(W)=V_{2}cap f(W) left ( V_2:= V^{'}_{2}cap Uright ),$$
owing to
$$Acap C=Bcap CRightarrow Acap B cap C=Acap C=Bcap C.$$
From above, we have $${f(a):(a,0)in V^{'}_1}=V_{2}cap f(W)={g(a,0):(a,0)in V_1}={f(a):(a,0)in V_1}.qquadqquadqquadqquadqquadqquadqquadqquadBox$$
answered Dec 4 '18 at 9:38
user553010user553010
688
$endgroup$
add a comment |
$begingroup$
I'm not sure whether my answer is right or not,if you have free time please check it. Any help will be very much appreciated !
$textbf{1.}$
$${g(a,0):(a,0)in V^{'}_1}={f(a):(a,0)in V^{'}_1}Rightarrow$$$$ V^{'}_{2}cap f(W)=Ucap f(W),$$ since $$V^{'}_{2}cap f(W)={g(a,0):(a,0)in V^{'}_1},Ucap f(W)={f(a):(a,0)in V^{'}_1}.$$
$textbf{2.}$
$$V^{'}_{2}cap f(W)=Ucap f(W)=V_{2}cap f(W) left ( V_2:= V^{'}_{2}cap Uright ),$$
owing to
$$Acap C=Bcap CRightarrow Acap B cap C=Acap C=Bcap C.$$
From above, we have $${f(a):(a,0)in V^{'}_1}=V_{2}cap f(W)={g(a,0):(a,0)in V_1}={f(a):(a,0)in V_1}.qquadqquadqquadqquadqquadqquadqquadqquadBox$$
answered Dec 4 '18 at 9:38
user553010user553010
688
$endgroup$
I'm not sure whether my answer is right or not,if you have free time please check it. Any help will be very much appreciated !
$textbf{1.}$
$${g(a,0):(a,0)in V^{'}_1}={f(a):(a,0)in V^{'}_1}Rightarrow$$$$ V^{'}_{2}cap f(W)=Ucap f(W),$$ since $$V^{'}_{2}cap f(W)={g(a,0):(a,0)in V^{'}_1},Ucap f(W)={f(a):(a,0)in V^{'}_1}.$$
$textbf{2.}$
$$V^{'}_{2}cap f(W)=Ucap f(W)=V_{2}cap f(W) left ( V_2:= V^{'}_{2}cap Uright ),$$
owing to
$$Acap C=Bcap CRightarrow Acap B cap C=Acap C=Bcap C.$$
From above, we have $${f(a):(a,0)in V^{'}_1}=V_{2}cap f(W)={g(a,0):(a,0)in V_1}={f(a):(a,0)in V_1}.qquadqquadqquadqquadqquadqquadqquadqquadBox$$
answered Dec 4 '18 at 9:38
user553010user553010
688
edited Dec 5 '18 at 1:12
answered Dec 4 '18 at 9:38
user553010user553010
688
answered Dec 4 '18 at 9:38
user553010user553010
688
answered Dec 4 '18 at 9:38
user553010user553010
688
688
add a comment |
add a comment |
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I took the liberty in posting images of the proof. Next time you can either give a sketch or do the same. I don't think everyone has a copy of this book or for some, we aren't willing to torture ourselves again so providing some more details is always nice.
$endgroup$
– Faraad Armwood
Jul 5 '17 at 14:13
1
$begingroup$
@FaraadArmwood Thank you.
$endgroup$
– chen
Jul 5 '17 at 14:21