I'm new to projective geometry and I just read a few pages of Hartshone's Foundation of Projective Geometry. On page 5 he defines two projective planes to be isomorphic if there exists, a bijection from one to another that takes collinear points into collinear points. However, is ''isomorphic'' really an equivalence relation? i.e. does a bijection like that take collinear points back to collinear points? I didn't find out a way to prove that..
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1Identity is an isomorphism in this sense. If $f$ takes collinear point to collinear points and so does $g$ then also $f\circ g$ does. If . If $f$ takes collinear point to collinear points then so does $f^{-1}$. – Maja Blumenstein Dec 19 '19 at 14:42
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2@MajaBlumenstein Why”so does $f^{-1}$”? – Zichen Gao Dec 19 '19 at 15:56
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1@MajaBlumenstein I don't think it's obvious.. – Zichen Gao Dec 19 '19 at 15:58
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You're right: if 'collinear' were an arbitrary relation, it wouldn't necessarily follow. But I guess, here we can use some properties of projective spaces. – Berci Dec 19 '19 at 17:18
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@Berci yes you're right, I just wrote a proof, check it if you're interested :-) – Zichen Gao Dec 20 '19 at 01:17
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I think I solved it.
Suppose $f$ is a bijective mapping that takes collinear points in $S$ to collinear points in $S'$, we wish to prove that $f^{-1}$ also has this property, i.e. if $f$ takes $A, B C$ to $A',B',C'$ which are collinear, then $A,B,C$ must be collinear:
If not, then the lines $AB$ and $BC$ intersect in only the point $B$. For any point $D$ on $AB$, since $A,B,D$ are collinear, the imageof $D$ must lie on the line $A'B'C'$. For the same reason, so is the image of each point lying on the line $BC$. Since we know that the image of every point on $AB$ and $BC$ is on$A'B'C'$, we can use the property of $f$ again to see that the image of every line that intersects $AB$ and $BC$ at different points must also be on$A'B'C'$. However, through any point in $S$ except $A,B,C$,such a line exists. Hence we know that $f$ takes all the points in $S$ into the line $A'B'C'$, which is a contradiction to the fact that $f$ is a bijection, since every projective plane (here we indicatd $S'$) must have points which are not collinear.
Zichen Gao
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