problem 4 in Alan Pollak and Guillmin differential topology.
Clash Royale CLAN TAG#URR8PPP
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Let $a > 0$, and set $B_a = x in mathbbR^n : $. Let $phi : B_a to mathbbR^n$ be given by $phi(x) = fracaxsqrta^2 - $. Prove that $phi$ is a diffeomorphism of $B_a$ onto $mathbbR^n$.
I am stucked in proofing that the above function is 1-1 and onto because in the definition of the function the scalar part depends on the norm of x as a vector, so if I equate 2 functions say one of them of y and the other of x there will be a difference in the scalar value on both sides so what shall I do?
related links:
Show that the map is a diffeomorphism
general-topology functions differential-topology
add a comment |Â
up vote
0
down vote
favorite
Let $a > 0$, and set $B_a = x in mathbbR^n : $. Let $phi : B_a to mathbbR^n$ be given by $phi(x) = fracaxsqrta^2 - $. Prove that $phi$ is a diffeomorphism of $B_a$ onto $mathbbR^n$.
I am stucked in proofing that the above function is 1-1 and onto because in the definition of the function the scalar part depends on the norm of x as a vector, so if I equate 2 functions say one of them of y and the other of x there will be a difference in the scalar value on both sides so what shall I do?
related links:
Show that the map is a diffeomorphism
general-topology functions differential-topology
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $a > 0$, and set $B_a = x in mathbbR^n : $. Let $phi : B_a to mathbbR^n$ be given by $phi(x) = fracaxsqrta^2 - $. Prove that $phi$ is a diffeomorphism of $B_a$ onto $mathbbR^n$.
I am stucked in proofing that the above function is 1-1 and onto because in the definition of the function the scalar part depends on the norm of x as a vector, so if I equate 2 functions say one of them of y and the other of x there will be a difference in the scalar value on both sides so what shall I do?
related links:
Show that the map is a diffeomorphism
general-topology functions differential-topology
Let $a > 0$, and set $B_a = x in mathbbR^n : $. Let $phi : B_a to mathbbR^n$ be given by $phi(x) = fracaxsqrta^2 - $. Prove that $phi$ is a diffeomorphism of $B_a$ onto $mathbbR^n$.
I am stucked in proofing that the above function is 1-1 and onto because in the definition of the function the scalar part depends on the norm of x as a vector, so if I equate 2 functions say one of them of y and the other of x there will be a difference in the scalar value on both sides so what shall I do?
related links:
Show that the map is a diffeomorphism
general-topology functions differential-topology
general-topology functions differential-topology
asked Sep 10 at 19:46
user591766
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1 Answer
1
active
oldest
votes
up vote
0
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Hint: Show that the map $psi : Bbb R^n to B_a$ defined by the equation
$$psi(y) = fracaysqrta^2 + lvert yrvert^2$$ is the inverse of $phi$.
answered Sep 10 at 20:42
kobe
34.1k22146
Is that a prove for showing 1-1 & onto ..... how?
– user591766
Sep 10 at 20:53
Yes. Since $psi(phi(x)) = x$ for all $xin B_a$, $phi$ is one-to-one; since $phi(psi(y)) = y$ for all $yin Bbb R^n$, $phi$ is onto.
– kobe
Sep 10 at 20:55
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Hint: Show that the map $psi : Bbb R^n to B_a$ defined by the equation
$$psi(y) = fracaysqrta^2 + lvert yrvert^2$$ is the inverse of $phi$.
answered Sep 10 at 20:42
kobe
34.1k22146
Is that a prove for showing 1-1 & onto ..... how?
– user591766
Sep 10 at 20:53
Yes. Since $psi(phi(x)) = x$ for all $xin B_a$, $phi$ is one-to-one; since $phi(psi(y)) = y$ for all $yin Bbb R^n$, $phi$ is onto.
– kobe
Sep 10 at 20:55
add a comment |Â
up vote
0
down vote
accepted
Hint: Show that the map $psi : Bbb R^n to B_a$ defined by the equation
$$psi(y) = fracaysqrta^2 + lvert yrvert^2$$ is the inverse of $phi$.
answered Sep 10 at 20:42
kobe
34.1k22146
Is that a prove for showing 1-1 & onto ..... how?
– user591766
Sep 10 at 20:53
Yes. Since $psi(phi(x)) = x$ for all $xin B_a$, $phi$ is one-to-one; since $phi(psi(y)) = y$ for all $yin Bbb R^n$, $phi$ is onto.
– kobe
Sep 10 at 20:55
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Hint: Show that the map $psi : Bbb R^n to B_a$ defined by the equation
$$psi(y) = fracaysqrta^2 + lvert yrvert^2$$ is the inverse of $phi$.
answered Sep 10 at 20:42
kobe
34.1k22146
Hint: Show that the map $psi : Bbb R^n to B_a$ defined by the equation
$$psi(y) = fracaysqrta^2 + lvert yrvert^2$$ is the inverse of $phi$.
answered Sep 10 at 20:42
kobe
34.1k22146
answered Sep 10 at 20:42
kobe
34.1k22146
answered Sep 10 at 20:42
kobe
34.1k22146
answered Sep 10 at 20:42
kobe
34.1k22146
34.1k22146
Is that a prove for showing 1-1 & onto ..... how?
– user591766
Sep 10 at 20:53
Yes. Since $psi(phi(x)) = x$ for all $xin B_a$, $phi$ is one-to-one; since $phi(psi(y)) = y$ for all $yin Bbb R^n$, $phi$ is onto.
– kobe
Sep 10 at 20:55
add a comment |Â
Is that a prove for showing 1-1 & onto ..... how?
– user591766
Sep 10 at 20:53
Yes. Since $psi(phi(x)) = x$ for all $xin B_a$, $phi$ is one-to-one; since $phi(psi(y)) = y$ for all $yin Bbb R^n$, $phi$ is onto.
– kobe
Sep 10 at 20:55
Is that a prove for showing 1-1 & onto ..... how?
– user591766
Sep 10 at 20:53
Is that a prove for showing 1-1 & onto ..... how?
– user591766
Sep 10 at 20:53
Yes. Since $psi(phi(x)) = x$ for all $xin B_a$, $phi$ is one-to-one; since $phi(psi(y)) = y$ for all $yin Bbb R^n$, $phi$ is onto.
– kobe
Sep 10 at 20:55
Yes. Since $psi(phi(x)) = x$ for all $xin B_a$, $phi$ is one-to-one; since $phi(psi(y)) = y$ for all $yin Bbb R^n$, $phi$ is onto.
– kobe
Sep 10 at 20:55
add a comment |Â
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