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understanding a MSE proof that collinearity preserving bijections from R^2 to R^2 are of the form Ax+b
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I am trying to understand the answer to this MSE question:
Bijection $f: mathbbR^2 rightarrow mathbbR^2$ preserves collinearity $iff f(x)=Ax+b$
which is the same question I am interested in.
The main issue I am having is that I can't understand why a line n and the line h(n) must be parallel as stated here: "What can we say about h(m)? Well, it's parallel to m" as well as here: "...lies on h(n), which must be parallel to n".
It was proven earlier in the answer that if lines n and m are parallel then h(n) and h(m) must be parallel however I am unable to find the exact point in which it is proven that n and h(n) must be parallel for any line n. It is possible that this is an obvious conclusion from the earlier work done in the answer however I am not very experienced in this field so it is not obvious to me. Also it is possible that I simply missed the proof in the answer. In either case, if this could be explained to me that would be great.
Thanks
PS: if you would rather prove this result using a completely different approach that doesn't use the lemma "h(n) is parallel to n for all lines n" that is quite alright, I will accept this as an answer as well.
Also I have not bothered defining h as it is defined in the question I linked to. I figure anyone reading this post aught to read that post as well. If anyone has a problem with this and thinks the terminology used in this post aught to be defined here please let me know and I'll change it, or if you'd like to change it yourself please go ahead.
linear-algebra affine-geometry
asked Jul 18 at 20:42
mathew
356114
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up vote
1
down vote
favorite
I am trying to understand the answer to this MSE question:
Bijection $f: mathbbR^2 rightarrow mathbbR^2$ preserves collinearity $iff f(x)=Ax+b$
which is the same question I am interested in.
The main issue I am having is that I can't understand why a line n and the line h(n) must be parallel as stated here: "What can we say about h(m)? Well, it's parallel to m" as well as here: "...lies on h(n), which must be parallel to n".
It was proven earlier in the answer that if lines n and m are parallel then h(n) and h(m) must be parallel however I am unable to find the exact point in which it is proven that n and h(n) must be parallel for any line n. It is possible that this is an obvious conclusion from the earlier work done in the answer however I am not very experienced in this field so it is not obvious to me. Also it is possible that I simply missed the proof in the answer. In either case, if this could be explained to me that would be great.
Thanks
PS: if you would rather prove this result using a completely different approach that doesn't use the lemma "h(n) is parallel to n for all lines n" that is quite alright, I will accept this as an answer as well.
Also I have not bothered defining h as it is defined in the question I linked to. I figure anyone reading this post aught to read that post as well. If anyone has a problem with this and thinks the terminology used in this post aught to be defined here please let me know and I'll change it, or if you'd like to change it yourself please go ahead.
linear-algebra affine-geometry
asked Jul 18 at 20:42
mathew
356114
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am trying to understand the answer to this MSE question:
Bijection $f: mathbbR^2 rightarrow mathbbR^2$ preserves collinearity $iff f(x)=Ax+b$
which is the same question I am interested in.
The main issue I am having is that I can't understand why a line n and the line h(n) must be parallel as stated here: "What can we say about h(m)? Well, it's parallel to m" as well as here: "...lies on h(n), which must be parallel to n".
It was proven earlier in the answer that if lines n and m are parallel then h(n) and h(m) must be parallel however I am unable to find the exact point in which it is proven that n and h(n) must be parallel for any line n. It is possible that this is an obvious conclusion from the earlier work done in the answer however I am not very experienced in this field so it is not obvious to me. Also it is possible that I simply missed the proof in the answer. In either case, if this could be explained to me that would be great.
Thanks
PS: if you would rather prove this result using a completely different approach that doesn't use the lemma "h(n) is parallel to n for all lines n" that is quite alright, I will accept this as an answer as well.
Also I have not bothered defining h as it is defined in the question I linked to. I figure anyone reading this post aught to read that post as well. If anyone has a problem with this and thinks the terminology used in this post aught to be defined here please let me know and I'll change it, or if you'd like to change it yourself please go ahead.
linear-algebra affine-geometry
asked Jul 18 at 20:42
mathew
356114
I am trying to understand the answer to this MSE question:
Bijection $f: mathbbR^2 rightarrow mathbbR^2$ preserves collinearity $iff f(x)=Ax+b$
which is the same question I am interested in.
The main issue I am having is that I can't understand why a line n and the line h(n) must be parallel as stated here: "What can we say about h(m)? Well, it's parallel to m" as well as here: "...lies on h(n), which must be parallel to n".
It was proven earlier in the answer that if lines n and m are parallel then h(n) and h(m) must be parallel however I am unable to find the exact point in which it is proven that n and h(n) must be parallel for any line n. It is possible that this is an obvious conclusion from the earlier work done in the answer however I am not very experienced in this field so it is not obvious to me. Also it is possible that I simply missed the proof in the answer. In either case, if this could be explained to me that would be great.
Thanks
PS: if you would rather prove this result using a completely different approach that doesn't use the lemma "h(n) is parallel to n for all lines n" that is quite alright, I will accept this as an answer as well.
Also I have not bothered defining h as it is defined in the question I linked to. I figure anyone reading this post aught to read that post as well. If anyone has a problem with this and thinks the terminology used in this post aught to be defined here please let me know and I'll change it, or if you'd like to change it yourself please go ahead.
linear-algebra affine-geometry
asked Jul 18 at 20:42
mathew
356114
edited Jul 20 at 21:07
asked Jul 18 at 20:42
mathew
356114
asked Jul 18 at 20:42
mathew
356114
asked Jul 18 at 20:42
mathew
356114
356114
add a comment |Â
add a comment |Â
1 Answer
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I will use the proof by that link up until "h(n) is parallel to n" statement — everything before that seems to be entirely correct.
Now, for any $ainmathbb R, ane0$ we have $h(a,0)=(a',0)$ for some $a'$ (for the ease of notation I use $h(a,b)$ as a synonym for $h((a,b))$). Take a look at the line $x+y=a$. This line is parallel to the line $x+y=1$, which is preserved (setwise) by $h$, and therefore, turns into some other (or maybe the same) line parallel to it — so, to the line $x+y=a'$. Therefore, $h(0,a)=(0,a')$, since it's the point of intersection of lines $x=0$ (preserved) and $x+y=a'$.
Now, consider lines $x+ay=a$ and $x+ay=a^2$. They are, of course, parallel. First one contains points $(a,0)$ and $(0,1)$ — and, therefore, turns into the line, containing points $(a',0)$ and $(0,1)$ — that is, to the line $x+a'y=a'$. Therefore, the latter turns into $x+a'y=b$ for some $b$, since parallelism is preserved. On the other hand, since the original line $x+ay=a^2$ contained the point $(0,a)$, and we already know that $h(0,a)=(0,a')$, we see that $b=a^prime2$. Therefore, line $x+ay=a^2$ turns into $x+a'y=a^prime2$. Taking it's intersection with the line $y=0$, we see that $h(a^2,0)=(a^prime2,0)$.
This is VERY important. Because now we can conclude that if $a>0$ then $a'>0$ as well. Why? Because if $a>0$ then there is some $c>0$ such that $a=c^2$. Then $h(c,0)=(c',0)$ for some $c'$, and we conclude that $(a', 0) = h(a,0) = h(c^2,0) = (c^prime2,0)$. So, $a'=c^prime2>0$.
The rest is rather technical. We already know that $h$ preserves the origin and the points $(1,0)$, $(0,1)$, and $(1,1)$. Line $x+y=2$ is also preserved, since it's parallel to the line $x+y=1$ (which is preserved), and contains the point $(1,1)$. Therefore it's intersection with $y=0$ is preserved — which is $(2,0)$. The line $x=2$, which contains the last point and is parallel to $x=0$ is preserved as well, and so is it's intersection with $y=1$. So, $h(2,1)=(2,1)$. By induction you can easily prove that for $ainmathbb Z, binmathbb Z$ $h(a,b)=(a,b)$.
From integers it's easy to switch to rationals. Line $bx=a(y+1)$ (for integral nonzero $a$ and $b$) contains the points $(0,-1)$ and $(a,b-1)$ and is, therefore, preserved; so is it's intersection with the line $y=0$ — which is $(fracab,0)$. So, $h(fracab,0)=(fracab,0)$, which means that for any $rinmathbb Q$ $h(r,0)=(r,0)$. Similarly, $h(0,r)=(0,r)$.
Now, let's take some $uinmathbb R$ and $rinmathbb Q$. Assume $u>r$. Then $u-r>0$ and $h(0,u-r)=(0,w)$ for some $w>0$, as we learned above (in the VERY imporant paragraph). Also, $h(u,0)=(v,0)$ for some $v$. Then the line $y=u-r$ transforms to $y=w$, line $x=r$ is preserved, and so $h(r,u-r)=(r,w)$. Line $x+y=u$, parallel to (preserved) line $x+y=1$ and containing $(r,u-r)$, becomes therefore $x+y=r+w$, and it's intersection with $y=0$, which is $(u,0)$, becomes $(r+w,0)$. Therefore, $v=r+w>r$.
So, we proved that for any rational $r<u$, we also have $r<v$. In the same vein, if $r>u$, we'd have $r>v$. Which means $v=u$, and for any $uinmathbb R$ $h(u,0)=(u,0)$. Similarly, $h(0,u)=(0,u)$, and, given that any point is an intersection of a horizontal and vertical lines, $h$ is an identity.
answered Jul 24 at 14:40
MigMit
27113
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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
I will use the proof by that link up until "h(n) is parallel to n" statement — everything before that seems to be entirely correct.
Now, for any $ainmathbb R, ane0$ we have $h(a,0)=(a',0)$ for some $a'$ (for the ease of notation I use $h(a,b)$ as a synonym for $h((a,b))$). Take a look at the line $x+y=a$. This line is parallel to the line $x+y=1$, which is preserved (setwise) by $h$, and therefore, turns into some other (or maybe the same) line parallel to it — so, to the line $x+y=a'$. Therefore, $h(0,a)=(0,a')$, since it's the point of intersection of lines $x=0$ (preserved) and $x+y=a'$.
Now, consider lines $x+ay=a$ and $x+ay=a^2$. They are, of course, parallel. First one contains points $(a,0)$ and $(0,1)$ — and, therefore, turns into the line, containing points $(a',0)$ and $(0,1)$ — that is, to the line $x+a'y=a'$. Therefore, the latter turns into $x+a'y=b$ for some $b$, since parallelism is preserved. On the other hand, since the original line $x+ay=a^2$ contained the point $(0,a)$, and we already know that $h(0,a)=(0,a')$, we see that $b=a^prime2$. Therefore, line $x+ay=a^2$ turns into $x+a'y=a^prime2$. Taking it's intersection with the line $y=0$, we see that $h(a^2,0)=(a^prime2,0)$.
This is VERY important. Because now we can conclude that if $a>0$ then $a'>0$ as well. Why? Because if $a>0$ then there is some $c>0$ such that $a=c^2$. Then $h(c,0)=(c',0)$ for some $c'$, and we conclude that $(a', 0) = h(a,0) = h(c^2,0) = (c^prime2,0)$. So, $a'=c^prime2>0$.
The rest is rather technical. We already know that $h$ preserves the origin and the points $(1,0)$, $(0,1)$, and $(1,1)$. Line $x+y=2$ is also preserved, since it's parallel to the line $x+y=1$ (which is preserved), and contains the point $(1,1)$. Therefore it's intersection with $y=0$ is preserved — which is $(2,0)$. The line $x=2$, which contains the last point and is parallel to $x=0$ is preserved as well, and so is it's intersection with $y=1$. So, $h(2,1)=(2,1)$. By induction you can easily prove that for $ainmathbb Z, binmathbb Z$ $h(a,b)=(a,b)$.
From integers it's easy to switch to rationals. Line $bx=a(y+1)$ (for integral nonzero $a$ and $b$) contains the points $(0,-1)$ and $(a,b-1)$ and is, therefore, preserved; so is it's intersection with the line $y=0$ — which is $(fracab,0)$. So, $h(fracab,0)=(fracab,0)$, which means that for any $rinmathbb Q$ $h(r,0)=(r,0)$. Similarly, $h(0,r)=(0,r)$.
Now, let's take some $uinmathbb R$ and $rinmathbb Q$. Assume $u>r$. Then $u-r>0$ and $h(0,u-r)=(0,w)$ for some $w>0$, as we learned above (in the VERY imporant paragraph). Also, $h(u,0)=(v,0)$ for some $v$. Then the line $y=u-r$ transforms to $y=w$, line $x=r$ is preserved, and so $h(r,u-r)=(r,w)$. Line $x+y=u$, parallel to (preserved) line $x+y=1$ and containing $(r,u-r)$, becomes therefore $x+y=r+w$, and it's intersection with $y=0$, which is $(u,0)$, becomes $(r+w,0)$. Therefore, $v=r+w>r$.
So, we proved that for any rational $r<u$, we also have $r<v$. In the same vein, if $r>u$, we'd have $r>v$. Which means $v=u$, and for any $uinmathbb R$ $h(u,0)=(u,0)$. Similarly, $h(0,u)=(0,u)$, and, given that any point is an intersection of a horizontal and vertical lines, $h$ is an identity.
answered Jul 24 at 14:40
MigMit
27113
add a comment |Â
up vote
2
down vote
accepted
I will use the proof by that link up until "h(n) is parallel to n" statement — everything before that seems to be entirely correct.
Now, for any $ainmathbb R, ane0$ we have $h(a,0)=(a',0)$ for some $a'$ (for the ease of notation I use $h(a,b)$ as a synonym for $h((a,b))$). Take a look at the line $x+y=a$. This line is parallel to the line $x+y=1$, which is preserved (setwise) by $h$, and therefore, turns into some other (or maybe the same) line parallel to it — so, to the line $x+y=a'$. Therefore, $h(0,a)=(0,a')$, since it's the point of intersection of lines $x=0$ (preserved) and $x+y=a'$.
Now, consider lines $x+ay=a$ and $x+ay=a^2$. They are, of course, parallel. First one contains points $(a,0)$ and $(0,1)$ — and, therefore, turns into the line, containing points $(a',0)$ and $(0,1)$ — that is, to the line $x+a'y=a'$. Therefore, the latter turns into $x+a'y=b$ for some $b$, since parallelism is preserved. On the other hand, since the original line $x+ay=a^2$ contained the point $(0,a)$, and we already know that $h(0,a)=(0,a')$, we see that $b=a^prime2$. Therefore, line $x+ay=a^2$ turns into $x+a'y=a^prime2$. Taking it's intersection with the line $y=0$, we see that $h(a^2,0)=(a^prime2,0)$.
This is VERY important. Because now we can conclude that if $a>0$ then $a'>0$ as well. Why? Because if $a>0$ then there is some $c>0$ such that $a=c^2$. Then $h(c,0)=(c',0)$ for some $c'$, and we conclude that $(a', 0) = h(a,0) = h(c^2,0) = (c^prime2,0)$. So, $a'=c^prime2>0$.
The rest is rather technical. We already know that $h$ preserves the origin and the points $(1,0)$, $(0,1)$, and $(1,1)$. Line $x+y=2$ is also preserved, since it's parallel to the line $x+y=1$ (which is preserved), and contains the point $(1,1)$. Therefore it's intersection with $y=0$ is preserved — which is $(2,0)$. The line $x=2$, which contains the last point and is parallel to $x=0$ is preserved as well, and so is it's intersection with $y=1$. So, $h(2,1)=(2,1)$. By induction you can easily prove that for $ainmathbb Z, binmathbb Z$ $h(a,b)=(a,b)$.
From integers it's easy to switch to rationals. Line $bx=a(y+1)$ (for integral nonzero $a$ and $b$) contains the points $(0,-1)$ and $(a,b-1)$ and is, therefore, preserved; so is it's intersection with the line $y=0$ — which is $(fracab,0)$. So, $h(fracab,0)=(fracab,0)$, which means that for any $rinmathbb Q$ $h(r,0)=(r,0)$. Similarly, $h(0,r)=(0,r)$.
Now, let's take some $uinmathbb R$ and $rinmathbb Q$. Assume $u>r$. Then $u-r>0$ and $h(0,u-r)=(0,w)$ for some $w>0$, as we learned above (in the VERY imporant paragraph). Also, $h(u,0)=(v,0)$ for some $v$. Then the line $y=u-r$ transforms to $y=w$, line $x=r$ is preserved, and so $h(r,u-r)=(r,w)$. Line $x+y=u$, parallel to (preserved) line $x+y=1$ and containing $(r,u-r)$, becomes therefore $x+y=r+w$, and it's intersection with $y=0$, which is $(u,0)$, becomes $(r+w,0)$. Therefore, $v=r+w>r$.
So, we proved that for any rational $r<u$, we also have $r<v$. In the same vein, if $r>u$, we'd have $r>v$. Which means $v=u$, and for any $uinmathbb R$ $h(u,0)=(u,0)$. Similarly, $h(0,u)=(0,u)$, and, given that any point is an intersection of a horizontal and vertical lines, $h$ is an identity.
answered Jul 24 at 14:40
MigMit
27113
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
I will use the proof by that link up until "h(n) is parallel to n" statement — everything before that seems to be entirely correct.
Now, for any $ainmathbb R, ane0$ we have $h(a,0)=(a',0)$ for some $a'$ (for the ease of notation I use $h(a,b)$ as a synonym for $h((a,b))$). Take a look at the line $x+y=a$. This line is parallel to the line $x+y=1$, which is preserved (setwise) by $h$, and therefore, turns into some other (or maybe the same) line parallel to it — so, to the line $x+y=a'$. Therefore, $h(0,a)=(0,a')$, since it's the point of intersection of lines $x=0$ (preserved) and $x+y=a'$.
Now, consider lines $x+ay=a$ and $x+ay=a^2$. They are, of course, parallel. First one contains points $(a,0)$ and $(0,1)$ — and, therefore, turns into the line, containing points $(a',0)$ and $(0,1)$ — that is, to the line $x+a'y=a'$. Therefore, the latter turns into $x+a'y=b$ for some $b$, since parallelism is preserved. On the other hand, since the original line $x+ay=a^2$ contained the point $(0,a)$, and we already know that $h(0,a)=(0,a')$, we see that $b=a^prime2$. Therefore, line $x+ay=a^2$ turns into $x+a'y=a^prime2$. Taking it's intersection with the line $y=0$, we see that $h(a^2,0)=(a^prime2,0)$.
This is VERY important. Because now we can conclude that if $a>0$ then $a'>0$ as well. Why? Because if $a>0$ then there is some $c>0$ such that $a=c^2$. Then $h(c,0)=(c',0)$ for some $c'$, and we conclude that $(a', 0) = h(a,0) = h(c^2,0) = (c^prime2,0)$. So, $a'=c^prime2>0$.
The rest is rather technical. We already know that $h$ preserves the origin and the points $(1,0)$, $(0,1)$, and $(1,1)$. Line $x+y=2$ is also preserved, since it's parallel to the line $x+y=1$ (which is preserved), and contains the point $(1,1)$. Therefore it's intersection with $y=0$ is preserved — which is $(2,0)$. The line $x=2$, which contains the last point and is parallel to $x=0$ is preserved as well, and so is it's intersection with $y=1$. So, $h(2,1)=(2,1)$. By induction you can easily prove that for $ainmathbb Z, binmathbb Z$ $h(a,b)=(a,b)$.
From integers it's easy to switch to rationals. Line $bx=a(y+1)$ (for integral nonzero $a$ and $b$) contains the points $(0,-1)$ and $(a,b-1)$ and is, therefore, preserved; so is it's intersection with the line $y=0$ — which is $(fracab,0)$. So, $h(fracab,0)=(fracab,0)$, which means that for any $rinmathbb Q$ $h(r,0)=(r,0)$. Similarly, $h(0,r)=(0,r)$.
Now, let's take some $uinmathbb R$ and $rinmathbb Q$. Assume $u>r$. Then $u-r>0$ and $h(0,u-r)=(0,w)$ for some $w>0$, as we learned above (in the VERY imporant paragraph). Also, $h(u,0)=(v,0)$ for some $v$. Then the line $y=u-r$ transforms to $y=w$, line $x=r$ is preserved, and so $h(r,u-r)=(r,w)$. Line $x+y=u$, parallel to (preserved) line $x+y=1$ and containing $(r,u-r)$, becomes therefore $x+y=r+w$, and it's intersection with $y=0$, which is $(u,0)$, becomes $(r+w,0)$. Therefore, $v=r+w>r$.
So, we proved that for any rational $r<u$, we also have $r<v$. In the same vein, if $r>u$, we'd have $r>v$. Which means $v=u$, and for any $uinmathbb R$ $h(u,0)=(u,0)$. Similarly, $h(0,u)=(0,u)$, and, given that any point is an intersection of a horizontal and vertical lines, $h$ is an identity.
answered Jul 24 at 14:40
MigMit
27113
I will use the proof by that link up until "h(n) is parallel to n" statement — everything before that seems to be entirely correct.
Now, for any $ainmathbb R, ane0$ we have $h(a,0)=(a',0)$ for some $a'$ (for the ease of notation I use $h(a,b)$ as a synonym for $h((a,b))$). Take a look at the line $x+y=a$. This line is parallel to the line $x+y=1$, which is preserved (setwise) by $h$, and therefore, turns into some other (or maybe the same) line parallel to it — so, to the line $x+y=a'$. Therefore, $h(0,a)=(0,a')$, since it's the point of intersection of lines $x=0$ (preserved) and $x+y=a'$.
Now, consider lines $x+ay=a$ and $x+ay=a^2$. They are, of course, parallel. First one contains points $(a,0)$ and $(0,1)$ — and, therefore, turns into the line, containing points $(a',0)$ and $(0,1)$ — that is, to the line $x+a'y=a'$. Therefore, the latter turns into $x+a'y=b$ for some $b$, since parallelism is preserved. On the other hand, since the original line $x+ay=a^2$ contained the point $(0,a)$, and we already know that $h(0,a)=(0,a')$, we see that $b=a^prime2$. Therefore, line $x+ay=a^2$ turns into $x+a'y=a^prime2$. Taking it's intersection with the line $y=0$, we see that $h(a^2,0)=(a^prime2,0)$.
This is VERY important. Because now we can conclude that if $a>0$ then $a'>0$ as well. Why? Because if $a>0$ then there is some $c>0$ such that $a=c^2$. Then $h(c,0)=(c',0)$ for some $c'$, and we conclude that $(a', 0) = h(a,0) = h(c^2,0) = (c^prime2,0)$. So, $a'=c^prime2>0$.
The rest is rather technical. We already know that $h$ preserves the origin and the points $(1,0)$, $(0,1)$, and $(1,1)$. Line $x+y=2$ is also preserved, since it's parallel to the line $x+y=1$ (which is preserved), and contains the point $(1,1)$. Therefore it's intersection with $y=0$ is preserved — which is $(2,0)$. The line $x=2$, which contains the last point and is parallel to $x=0$ is preserved as well, and so is it's intersection with $y=1$. So, $h(2,1)=(2,1)$. By induction you can easily prove that for $ainmathbb Z, binmathbb Z$ $h(a,b)=(a,b)$.
From integers it's easy to switch to rationals. Line $bx=a(y+1)$ (for integral nonzero $a$ and $b$) contains the points $(0,-1)$ and $(a,b-1)$ and is, therefore, preserved; so is it's intersection with the line $y=0$ — which is $(fracab,0)$. So, $h(fracab,0)=(fracab,0)$, which means that for any $rinmathbb Q$ $h(r,0)=(r,0)$. Similarly, $h(0,r)=(0,r)$.
Now, let's take some $uinmathbb R$ and $rinmathbb Q$. Assume $u>r$. Then $u-r>0$ and $h(0,u-r)=(0,w)$ for some $w>0$, as we learned above (in the VERY imporant paragraph). Also, $h(u,0)=(v,0)$ for some $v$. Then the line $y=u-r$ transforms to $y=w$, line $x=r$ is preserved, and so $h(r,u-r)=(r,w)$. Line $x+y=u$, parallel to (preserved) line $x+y=1$ and containing $(r,u-r)$, becomes therefore $x+y=r+w$, and it's intersection with $y=0$, which is $(u,0)$, becomes $(r+w,0)$. Therefore, $v=r+w>r$.
So, we proved that for any rational $r<u$, we also have $r<v$. In the same vein, if $r>u$, we'd have $r>v$. Which means $v=u$, and for any $uinmathbb R$ $h(u,0)=(u,0)$. Similarly, $h(0,u)=(0,u)$, and, given that any point is an intersection of a horizontal and vertical lines, $h$ is an identity.
answered Jul 24 at 14:40
MigMit
27113
edited Jul 24 at 14:59
answered Jul 24 at 14:40
MigMit
27113
answered Jul 24 at 14:40
MigMit
27113
answered Jul 24 at 14:40
MigMit
27113
27113
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StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password