Collineation of affine plane extends uniquely to collineation of projective plane

Question

I'm trying to prove the following result, which is exercise 4.1 in Projective Planes by Hughes & Piper:

If $\alpha$ is a collineation of an affine plane $A = P^l,$ there is a unique collineation $\beta$ of the projective plane $P$ such that $\beta(l) = l$ and $\beta$ induces $\alpha.$

For a line $l, P^l$ is the plane with $l$ and all points on it removed. A collineation $\alpha$ is a 1-1 map of the points and lines such that if $X \in m, \alpha(X) \in \alpha(m).$ The textbook should say bijective map since all other sources do so and there's no way to derive onto from just 1-1, but that's small potatoes. The real trouble begins when I try to prove the result.

Suppose $\beta|_{P^l} = \alpha$ and $\beta$ is a collineation of $P.$ Then $\beta(l) = l$ as all other lines are accounted for. Fix $A \in P^l,$ then for any $B \in l, \beta(B)$ lies in $l$ and must also lie in $\beta(\overline{AB}) = \alpha(\overline{AB}),$ so $\beta(B) = \alpha(\overline{AB}) \cap l,$ which uniquely determines $\beta.$ We just need to check this is a collineation.

$\beta$ is bijective on lines since $\alpha$ is, and $\beta$ is 1-1 on points since $\alpha$ is and $\beta(B) = \beta(B') \Rightarrow \alpha(AB) = \alpha(AB') \Rightarrow AB = AB' \Rightarrow B = B'.$ Given $Y \in l,$ let $C$ be a 3rd point on $\alpha(A)Y,$ let $X = A\alpha^{-1}(C) \cap l,$ then $\beta(X) = \alpha(A)C \cap l = \alpha(A)Y \cap l = Y,$ so $\beta$ is onto on points. Thus, $\beta$ is bijective on points and lines.

Suppose $X \in m, X \not\in l.$ Then $\beta(X) = \alpha(X) \in \alpha(m) = \beta(m).$ If $X \in l = m,$ then $\beta(X) = \alpha(AX) \cap L \in L = \beta(L).$ Finally, suppose $X \in l, X \in m, m \ne l$ and write $m = XC.$ Then we want to show $\beta(X) = \alpha(AX) \cap l \in \beta(m) = \alpha(XC),$ which is equivalent to $\alpha(AX) \cap l = \alpha(CX) \cap l$ for all $C,$ i.e. the construction of $\beta$ didn't depend on which $A$ we chose at the start.

Suppose not and let $Y = \alpha(AX) \cap l, Z = \alpha(XC) \cap l, W = \alpha(A)\alpha(C) \cap YZ = \alpha(AC) \cap L.$ I've been trying to derive a contradiction from $Y \ne Z$ without any luck so far. How do I proceed, and how can the entire proof be simplified? There's no reason for the solution to be so long given this is only the first exercise.

Answer 1

Turns out you should consider $W = a(AX) \cap a(CX)$ instead. Since $Y \ne Z,$ we have $W \not\in l,$ so we can write $W = \alpha(D)$ for some $D \not\in l.$ In particular, $D \ne X.$

Now we show the inverse of a collineation is a collineation, that is $\alpha(X) \in \alpha(m) \Rightarrow X \in m.$ Pick $Y \in m,$ if $Y = X$ we're already done. Otherwise, $\alpha(Y), \alpha(X) \in \alpha(m)$ so $\alpha(m) = \alpha(X)\alpha(Y) = \alpha(XY) \Rightarrow m = XY \Rightarrow X \in m.$

This means $D \in AX \cap CX \Rightarrow D = X,$ contradiction.

Answer 2

Proof above is clean, very elegant; I’d like to try here:

Let $\beta$ be the unique collineation of $P$ extending $\alpha$, defined by $\beta(X) = \alpha(A X) \cap l$ for $X \in l$ and $\beta(X) = \alpha(X)$ for $X \notin l$. We must show $\beta$ is well-defined and a collineation.

$\beta$ is well-defined: If $X, Y \in l$ and $\alpha(A X) \cap l = \alpha(A Y) \cap l$, then $\alpha(A (X - Y)) = \alpha(0 _l) = l$, so $X = Y$ by injectivity of $\alpha$. $\beta$ is 1-1: If $\beta(X) = \beta(Y)$, then $\alpha(A X) \cap l = \alpha(A Y) \cap l$, so $X = Y$ by the above. $\beta$ is onto (points): Let $Y \in P$. If $Y \in l$ then $\beta(Y) = Y$. Otherwise, let $X = A ^{-1}(Y)$; then $\beta(X) = \alpha(A X) \cap l = Y$. $\beta$ is onto (lines): Clear since $\alpha$ is. $\beta$ preserves incidence: If $X \in m$ then $\beta(X) = \alpha(X) \in \alpha(m) = \beta(m)$. If $X \in l = m$ then $\beta(X) = \alpha(A X) \cap l \in l = \beta(l)$.

Uniqueness: Let $\beta'$ be another collineation extending $\alpha$. Then $\beta'(l) = l$, so $\beta'$ and $\beta$ agree on $l$. By injectivity of collineations, $\beta = \beta'$.