Why is this function well defined and $C^infty$?

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Let $MsubseteqmathbbR^k$ be an embedded submanifold of $mathbbR^k$, with dim$M=n$.



Let $(U,phi)$ be a smooth chart for $M$.



Then $phi^-1:phi(U)to U$ is a diffeomorphism and for each $xin phi(U)$ the map $d(phi^-1)_x:T_xphi(U)to T_pU$ is isomorphism of $mathbbR$-vector spaces, with $p=phi^-1(x)$.



My book defines $$G:phi(U) timesmathbbS^n-1 to mathbbS^k-1, quad (x,u)mapsto fracd(phi^-1)_x(u)lVert d(phi^-1)_x(u) rVert$$



and says that this function is $C^infty$.



I am not even able to see why this function is well defined.



For example, if $ uin mathbbS^n-1$, what is the meaning of $d(phi^-1)_x(u)?$



I know that $d(phi^-1)_x$ acts on elements of $T_xphi(U)$, form where it comes out $mathbbS^n-1$?




Please explain me with full (formal) details what it's going on here: what is the sense and the meaning of this functions and how can I prove that it is $C^infty$. Thank you in advance.





smooth-manifolds




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  • My guess: the tangent space is identifiable as $mathbbR^n$, so $u$ is just a unit vector in the tangent space.
    – Randall
    Aug 6 at 15:56














up vote
0
down vote

favorite












Let $MsubseteqmathbbR^k$ be an embedded submanifold of $mathbbR^k$, with dim$M=n$.



Let $(U,phi)$ be a smooth chart for $M$.



Then $phi^-1:phi(U)to U$ is a diffeomorphism and for each $xin phi(U)$ the map $d(phi^-1)_x:T_xphi(U)to T_pU$ is isomorphism of $mathbbR$-vector spaces, with $p=phi^-1(x)$.



My book defines $$G:phi(U) timesmathbbS^n-1 to mathbbS^k-1, quad (x,u)mapsto fracd(phi^-1)_x(u)lVert d(phi^-1)_x(u) rVert$$



and says that this function is $C^infty$.



I am not even able to see why this function is well defined.



For example, if $ uin mathbbS^n-1$, what is the meaning of $d(phi^-1)_x(u)?$



I know that $d(phi^-1)_x$ acts on elements of $T_xphi(U)$, form where it comes out $mathbbS^n-1$?




Please explain me with full (formal) details what it's going on here: what is the sense and the meaning of this functions and how can I prove that it is $C^infty$. Thank you in advance.





smooth-manifolds




share|cite|improve this question





















  • My guess: the tangent space is identifiable as $mathbbR^n$, so $u$ is just a unit vector in the tangent space.
    – Randall
    Aug 6 at 15:56












up vote
0
down vote

favorite









up vote
0
down vote

favorite











Let $MsubseteqmathbbR^k$ be an embedded submanifold of $mathbbR^k$, with dim$M=n$.



Let $(U,phi)$ be a smooth chart for $M$.



Then $phi^-1:phi(U)to U$ is a diffeomorphism and for each $xin phi(U)$ the map $d(phi^-1)_x:T_xphi(U)to T_pU$ is isomorphism of $mathbbR$-vector spaces, with $p=phi^-1(x)$.



My book defines $$G:phi(U) timesmathbbS^n-1 to mathbbS^k-1, quad (x,u)mapsto fracd(phi^-1)_x(u)lVert d(phi^-1)_x(u) rVert$$



and says that this function is $C^infty$.



I am not even able to see why this function is well defined.



For example, if $ uin mathbbS^n-1$, what is the meaning of $d(phi^-1)_x(u)?$



I know that $d(phi^-1)_x$ acts on elements of $T_xphi(U)$, form where it comes out $mathbbS^n-1$?




Please explain me with full (formal) details what it's going on here: what is the sense and the meaning of this functions and how can I prove that it is $C^infty$. Thank you in advance.





smooth-manifolds




share|cite|improve this question













Let $MsubseteqmathbbR^k$ be an embedded submanifold of $mathbbR^k$, with dim$M=n$.



Let $(U,phi)$ be a smooth chart for $M$.



Then $phi^-1:phi(U)to U$ is a diffeomorphism and for each $xin phi(U)$ the map $d(phi^-1)_x:T_xphi(U)to T_pU$ is isomorphism of $mathbbR$-vector spaces, with $p=phi^-1(x)$.



My book defines $$G:phi(U) timesmathbbS^n-1 to mathbbS^k-1, quad (x,u)mapsto fracd(phi^-1)_x(u)lVert d(phi^-1)_x(u) rVert$$



and says that this function is $C^infty$.



I am not even able to see why this function is well defined.



For example, if $ uin mathbbS^n-1$, what is the meaning of $d(phi^-1)_x(u)?$



I know that $d(phi^-1)_x$ acts on elements of $T_xphi(U)$, form where it comes out $mathbbS^n-1$?




Please explain me with full (formal) details what it's going on here: what is the sense and the meaning of this functions and how can I prove that it is $C^infty$. Thank you in advance.






smooth-manifolds





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edited Aug 7 at 17:01
























asked Aug 6 at 15:26









Minato

184111




184111











  • My guess: the tangent space is identifiable as $mathbbR^n$, so $u$ is just a unit vector in the tangent space.
    – Randall
    Aug 6 at 15:56
















  • My guess: the tangent space is identifiable as $mathbbR^n$, so $u$ is just a unit vector in the tangent space.
    – Randall
    Aug 6 at 15:56















My guess: the tangent space is identifiable as $mathbbR^n$, so $u$ is just a unit vector in the tangent space.
– Randall
Aug 6 at 15:56




My guess: the tangent space is identifiable as $mathbbR^n$, so $u$ is just a unit vector in the tangent space.
– Randall
Aug 6 at 15:56















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