Past exam of the geometry course of the University of Oxford / 2011 / 4

Choose representatives $v_1,\dots ,v_{n+1}$ of the first $n+1$ points. The general-position condition says that these vectors form a basis of ${\mathbb{C}}^{n+1}$. Write a representative of the last point as $v_{n+2}={\sum }_i{c}_i{v}_i$. Every $c_i$ is nonzero; otherwise the last point and the other $n$ basis points would lie in a hyperplane. Rescaling $v_i$ by $c_i$ gives representatives with $v_{n+2}={\sum }_i{v}_i$. Normalize representatives $w_i$ of the second frame in the same way.

There is a unique invertible linear map $L$ with $L(v_i)=w_i$ for $i\le n+1$. It also maps $v_{n+2}$ to $w_{n+2}$, so its projectivization has the required effect. For uniqueness, the quotient of two candidate projective transformations fixes the normalized frame. A representing matrix must be diagonal in the $v_i$ basis, with entries $a_i$. Fixing $[{\sum }_i{v}_i]$ forces all $a_i$ to be equal. Such a scalar represents the identity projective transformation. Thus the ordered projective frame determines the transformation uniquely.

Represent each point by a nonzero column $v_i$. The source's cross-ratio isRescaling representatives cancels between numerator and denominator. If the projective transformation is represented by $A\in {\mathrm{GL}}_2(\mathbb{C})$, then $\det (A{v}_i,A{v}_j)=\det (A)\det (v_i,v_j)$. Two factors of $\det (A)$ occur in both numerator and denominator and cancel. Hence the cross-ratio is invariant under every projective transformation.

Injectivity makes the images of any three distinct points distinct. By the ordered-frame result, there is a projective transformation $L$ agreeing with the given map on three chosen points. Composing by $L^{-1}$ reduces to an injective map fixing those points and preserving every cross-ratio. Change coordinates so that the three points are $\infty ,0,1$.

For any other point with affine coordinate $z$, the convention in the source givesPreservation therefore forces the image of $z$ to have the same coordinate $z$. The normalized map fixes every point and is the identity. Undoing the normalization proves the original map is exactly $L$, hence projective. No prior continuity or surjectivity assumption is needed.