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?



enter image description here



enter image description here






calculus multivariable-calculus manifolds smooth-manifolds







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  • 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
















0












$begingroup$


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?



enter image description here



enter image description here






calculus multivariable-calculus manifolds smooth-manifolds







share|cite|improve this question











$endgroup$








  • 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.
    $endgroup$
    – Faraad Armwood
    Jul 5 '17 at 14:13








  • 1




    $begingroup$
    @FaraadArmwood Thank you.
    $endgroup$
    – chen
    Jul 5 '17 at 14:21














0












0








0





$begingroup$


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?



enter image description here



enter image description here






calculus multivariable-calculus manifolds smooth-manifolds







share|cite|improve this question











$endgroup$




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?



enter image description here



enter image description here






calculus multivariable-calculus manifolds smooth-manifolds




calculus multivariable-calculus manifolds smooth-manifolds






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jul 6 '17 at 3:33







chen

















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














  • 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










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 $${f(a):(a,0)in V^{'}_1}=V_{2}cap f(W)={g(a,0):(a,0)in V_1}={f(a):(a,0)in V_1}.qquadqquadqquadqquadqquadqquadqquadqquadBox$$






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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 $${f(a):(a,0)in V^{'}_1}=V_{2}cap f(W)={g(a,0):(a,0)in V_1}={f(a):(a,0)in V_1}.qquadqquadqquadqquadqquadqquadqquadqquadBox$$






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      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 $${f(a):(a,0)in V^{'}_1}=V_{2}cap f(W)={g(a,0):(a,0)in V_1}={f(a):(a,0)in V_1}.qquadqquadqquadqquadqquadqquadqquadqquadBox$$






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        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 $${f(a):(a,0)in V^{'}_1}=V_{2}cap f(W)={g(a,0):(a,0)in V_1}={f(a):(a,0)in V_1}.qquadqquadqquadqquadqquadqquadqquadqquadBox$$






        share|cite|improve this answer











        $endgroup$



        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 $${f(a):(a,0)in V^{'}_1}=V_{2}cap f(W)={g(a,0):(a,0)in V_1}={f(a):(a,0)in V_1}.qquadqquadqquadqquadqquadqquadqquadqquadBox$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 5 '18 at 1:12

























        answered Dec 4 '18 at 9:38









        user553010user553010

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