Proofing map $f:Bbb R^2 to Bbb R^3$ is injective ?
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Let $V subset Bbb R^2$ be open, $h:V to Bbb R$ differentialable and $f: V to Bbb R^3$ $(x,y) mapsto (x,y,h(x,y))$
I want to show that f is injective.
Just wanted to know what/how I can do this ? Jacobian ? Implicit function theorem ? Rank-Nullity theorem ?
functional-analysis
asked Nov 22 at 20:40
kratos88
496
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up vote
0
down vote
favorite
Let $V subset Bbb R^2$ be open, $h:V to Bbb R$ differentialable and $f: V to Bbb R^3$ $(x,y) mapsto (x,y,h(x,y))$
I want to show that f is injective.
Just wanted to know what/how I can do this ? Jacobian ? Implicit function theorem ? Rank-Nullity theorem ?
functional-analysis
asked Nov 22 at 20:40
kratos88
496
1
It's quite obvious from the first two coordinates, isn't it?
– Servaes
Nov 22 at 20:41
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $V subset Bbb R^2$ be open, $h:V to Bbb R$ differentialable and $f: V to Bbb R^3$ $(x,y) mapsto (x,y,h(x,y))$
I want to show that f is injective.
Just wanted to know what/how I can do this ? Jacobian ? Implicit function theorem ? Rank-Nullity theorem ?
functional-analysis
asked Nov 22 at 20:40
kratos88
496
Let $V subset Bbb R^2$ be open, $h:V to Bbb R$ differentialable and $f: V to Bbb R^3$ $(x,y) mapsto (x,y,h(x,y))$
I want to show that f is injective.
Just wanted to know what/how I can do this ? Jacobian ? Implicit function theorem ? Rank-Nullity theorem ?
functional-analysis
functional-analysis
asked Nov 22 at 20:40
kratos88
496
asked Nov 22 at 20:40
kratos88
496
asked Nov 22 at 20:40
kratos88
496
asked Nov 22 at 20:40
kratos88
496
asked Nov 22 at 20:40
kratos88
496
496
1
It's quite obvious from the first two coordinates, isn't it?
– Servaes
Nov 22 at 20:41
add a comment |
1
It's quite obvious from the first two coordinates, isn't it?
– Servaes
Nov 22 at 20:41
1
1
It's quite obvious from the first two coordinates, isn't it?
– Servaes
Nov 22 at 20:41
It's quite obvious from the first two coordinates, isn't it?
– Servaes
Nov 22 at 20:41
add a comment |
1 Answer
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It's much simpler than that. If $f(a, b) = f(c, d)$, then what can you say about the first two coordinates?
answered Nov 22 at 20:41
Patrick Stevens
28.2k52874
OMG - thanks ! I got it !
– kratos88
Nov 22 at 21:31
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
It's much simpler than that. If $f(a, b) = f(c, d)$, then what can you say about the first two coordinates?
answered Nov 22 at 20:41
Patrick Stevens
28.2k52874
OMG - thanks ! I got it !
– kratos88
Nov 22 at 21:31
add a comment |
up vote
2
down vote
accepted
It's much simpler than that. If $f(a, b) = f(c, d)$, then what can you say about the first two coordinates?
answered Nov 22 at 20:41
Patrick Stevens
28.2k52874
OMG - thanks ! I got it !
– kratos88
Nov 22 at 21:31
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
It's much simpler than that. If $f(a, b) = f(c, d)$, then what can you say about the first two coordinates?
answered Nov 22 at 20:41
Patrick Stevens
28.2k52874
It's much simpler than that. If $f(a, b) = f(c, d)$, then what can you say about the first two coordinates?
answered Nov 22 at 20:41
Patrick Stevens
28.2k52874
answered Nov 22 at 20:41
Patrick Stevens
28.2k52874
answered Nov 22 at 20:41
Patrick Stevens
28.2k52874
answered Nov 22 at 20:41
Patrick Stevens
28.2k52874
28.2k52874
OMG - thanks ! I got it !
– kratos88
Nov 22 at 21:31
add a comment |
OMG - thanks ! I got it !
– kratos88
Nov 22 at 21:31
OMG - thanks ! I got it !
– kratos88
Nov 22 at 21:31
OMG - thanks ! I got it !
– kratos88
Nov 22 at 21:31
add a comment |
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It's quite obvious from the first two coordinates, isn't it?
– Servaes
Nov 22 at 20:41