Mixing
Composition of an Affine Map and the Quotient Map is Injective
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In page 2 of the book Differential geometry: Cartan's generalization of Klein's Erlangen program by Sharpe, it stated the example of the projective plane $P(mathbbR^n+1)$.
The book said if $a:mathbbR^nrightarrow mathbbR^n+1$ is an affine map whose image does not contain the origin, then $p circ a$ is injective and continuous, where $p:mathbbR^n+1 setminus0 rightarrow P(mathbbR^n+1)$ is the quotient map.
I am confused with this part, since what I know about affine map is a map in the form $a(x)=Ax+b$, where $A: mathbbR^n rightarrowmathbbR^n+1$ is a linear transformation and $b in mathbbR^n+1$. In particular, if we take $A$ to be a zero transformation, then it is impossible for the map $p circ a$ to be injective.
Is that an error in the book or else?
geometry differential-geometry quotient-spaces affine-geometry
asked Jul 15 at 4:44
Jerry
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In page 2 of the book Differential geometry: Cartan's generalization of Klein's Erlangen program by Sharpe, it stated the example of the projective plane $P(mathbbR^n+1)$.
The book said if $a:mathbbR^nrightarrow mathbbR^n+1$ is an affine map whose image does not contain the origin, then $p circ a$ is injective and continuous, where $p:mathbbR^n+1 setminus0 rightarrow P(mathbbR^n+1)$ is the quotient map.
I am confused with this part, since what I know about affine map is a map in the form $a(x)=Ax+b$, where $A: mathbbR^n rightarrowmathbbR^n+1$ is a linear transformation and $b in mathbbR^n+1$. In particular, if we take $A$ to be a zero transformation, then it is impossible for the map $p circ a$ to be injective.
Is that an error in the book or else?
geometry differential-geometry quotient-spaces affine-geometry
asked Jul 15 at 4:44
Jerry
331213
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
In page 2 of the book Differential geometry: Cartan's generalization of Klein's Erlangen program by Sharpe, it stated the example of the projective plane $P(mathbbR^n+1)$.
The book said if $a:mathbbR^nrightarrow mathbbR^n+1$ is an affine map whose image does not contain the origin, then $p circ a$ is injective and continuous, where $p:mathbbR^n+1 setminus0 rightarrow P(mathbbR^n+1)$ is the quotient map.
I am confused with this part, since what I know about affine map is a map in the form $a(x)=Ax+b$, where $A: mathbbR^n rightarrowmathbbR^n+1$ is a linear transformation and $b in mathbbR^n+1$. In particular, if we take $A$ to be a zero transformation, then it is impossible for the map $p circ a$ to be injective.
Is that an error in the book or else?
geometry differential-geometry quotient-spaces affine-geometry
asked Jul 15 at 4:44
Jerry
331213
In page 2 of the book Differential geometry: Cartan's generalization of Klein's Erlangen program by Sharpe, it stated the example of the projective plane $P(mathbbR^n+1)$.
The book said if $a:mathbbR^nrightarrow mathbbR^n+1$ is an affine map whose image does not contain the origin, then $p circ a$ is injective and continuous, where $p:mathbbR^n+1 setminus0 rightarrow P(mathbbR^n+1)$ is the quotient map.
I am confused with this part, since what I know about affine map is a map in the form $a(x)=Ax+b$, where $A: mathbbR^n rightarrowmathbbR^n+1$ is a linear transformation and $b in mathbbR^n+1$. In particular, if we take $A$ to be a zero transformation, then it is impossible for the map $p circ a$ to be injective.
Is that an error in the book or else?
geometry differential-geometry quotient-spaces affine-geometry
asked Jul 15 at 4:44
Jerry
331213
edited Jul 15 at 5:48
asked Jul 15 at 4:44
Jerry
331213
asked Jul 15 at 4:44
Jerry
331213
asked Jul 15 at 4:44
Jerry
331213
331213
add a comment |Â
add a comment |Â
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You're right that to have any hope of $p circ a$ being injective, we need $a$ to be injective; in particular, if $a(x) = Ax + b$, then the matrix $A$ must have full column rank. (I don't have a copy of the book, so I can't say if this is an error or merely an assumption you've missed.)
The point of the theorem is that the quotient map $p$ takes lines through the origin to points in projective space, and the image of any affine map usually intersects such a line in at most one point. The exception is when the image contains the origin; in that case, except in the very degenerate case when the image is $0$-dimensional, there will be some line(s) through the origin entirely contained in the image of $a$.
As a result, when the image of $a$ doesn't contain $0$, there can never be any "further loss of injectivity" by applying $p$.
answered Jul 15 at 6:06
Misha Lavrov
35.7k54691
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
You're right that to have any hope of $p circ a$ being injective, we need $a$ to be injective; in particular, if $a(x) = Ax + b$, then the matrix $A$ must have full column rank. (I don't have a copy of the book, so I can't say if this is an error or merely an assumption you've missed.)
The point of the theorem is that the quotient map $p$ takes lines through the origin to points in projective space, and the image of any affine map usually intersects such a line in at most one point. The exception is when the image contains the origin; in that case, except in the very degenerate case when the image is $0$-dimensional, there will be some line(s) through the origin entirely contained in the image of $a$.
As a result, when the image of $a$ doesn't contain $0$, there can never be any "further loss of injectivity" by applying $p$.
answered Jul 15 at 6:06
Misha Lavrov
35.7k54691
add a comment |Â
up vote
1
down vote
You're right that to have any hope of $p circ a$ being injective, we need $a$ to be injective; in particular, if $a(x) = Ax + b$, then the matrix $A$ must have full column rank. (I don't have a copy of the book, so I can't say if this is an error or merely an assumption you've missed.)
The point of the theorem is that the quotient map $p$ takes lines through the origin to points in projective space, and the image of any affine map usually intersects such a line in at most one point. The exception is when the image contains the origin; in that case, except in the very degenerate case when the image is $0$-dimensional, there will be some line(s) through the origin entirely contained in the image of $a$.
As a result, when the image of $a$ doesn't contain $0$, there can never be any "further loss of injectivity" by applying $p$.
answered Jul 15 at 6:06
Misha Lavrov
35.7k54691
add a comment |Â
up vote
1
down vote
up vote
1
down vote
You're right that to have any hope of $p circ a$ being injective, we need $a$ to be injective; in particular, if $a(x) = Ax + b$, then the matrix $A$ must have full column rank. (I don't have a copy of the book, so I can't say if this is an error or merely an assumption you've missed.)
The point of the theorem is that the quotient map $p$ takes lines through the origin to points in projective space, and the image of any affine map usually intersects such a line in at most one point. The exception is when the image contains the origin; in that case, except in the very degenerate case when the image is $0$-dimensional, there will be some line(s) through the origin entirely contained in the image of $a$.
As a result, when the image of $a$ doesn't contain $0$, there can never be any "further loss of injectivity" by applying $p$.
answered Jul 15 at 6:06
Misha Lavrov
35.7k54691
You're right that to have any hope of $p circ a$ being injective, we need $a$ to be injective; in particular, if $a(x) = Ax + b$, then the matrix $A$ must have full column rank. (I don't have a copy of the book, so I can't say if this is an error or merely an assumption you've missed.)
The point of the theorem is that the quotient map $p$ takes lines through the origin to points in projective space, and the image of any affine map usually intersects such a line in at most one point. The exception is when the image contains the origin; in that case, except in the very degenerate case when the image is $0$-dimensional, there will be some line(s) through the origin entirely contained in the image of $a$.
As a result, when the image of $a$ doesn't contain $0$, there can never be any "further loss of injectivity" by applying $p$.
answered Jul 15 at 6:06
Misha Lavrov
35.7k54691
answered Jul 15 at 6:06
Misha Lavrov
35.7k54691
answered Jul 15 at 6:06
Misha Lavrov
35.7k54691
answered Jul 15 at 6:06
Misha Lavrov
35.7k54691
35.7k54691
add a comment |Â
add a comment |Â
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