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The needed general-position theorem is sharp triple transitivity of the projective linear group on the projective line: for any two ordered triples of distinct projective points, exactly one projective transformation takes the first triple to the second.

In particular, let $T_P$ take $(P_0,P_1,P_2)$ to $(\infty ,0,1)$. In these coordinates,so cross-ratio invariance gives $T_P(P_3)=\lambda$, where $\lambda =(P_0{P}_1:P_2{P}_3)$. Distinctness implies $\lambda \ne 0,1$ and $\lambda$ is finite. The entire ordered quadruple is thus normalized toFor a second quadruple $(Q_0,Q_1,Q_2,Q_3)$ with the same cross-ratio, its normalizing map $T_Q$ yields the same quadruple. ConsequentlyThis proves the equivalence. Uniqueness on the first three points also shows that the projectivity carrying one ordered quadruple to the other is unique. The cross-ratio normalization of four projective points therefore gives a complete invariant for this ordered classification.

Choose three distinct projective points $A,B,C$. Their images under the injective map $\tau$ are distinct. Sharp triple transitivity gives a projective transformation $S$ agreeing with $\tau$ on these three points. The map $\sigma =S^{-1}\circ \tau$ fixes $A,B,C$ and preserves their cross-ratios.

For any other projective point $P$, invariance givesLet $T$ normalize $(A,B,C)$ to $(\infty ,0,1)$. By part (b), the left side is $T(\sigma (P))$ and the right side is $T(P)$. Since $T$ is a bijection, $\sigma (P)=P$. It fixes the original three points too, so $\sigma$ is the identity map everywhere. ThereforeNo continuity or prior surjectivity assumption was needed. This is the rigidity assertion that cross-ratio preserving maps are projective.

For representatives $p=(a,b,c)$ and $p^{\prime }=(a^{\prime },b^{\prime },c^{\prime })$, a point $[x:y:z]$ belongs to their projective line exactly when these three vectors are linearly dependent. The determinant criterion givesExpanding along the first row givesThe coefficient vector is the cross product $p\times p^{\prime }$. It is nonzero because the projective points are distinct. Rescaling either representative merely rescales all three coefficients, leaving the projective line unchanged. This is the join formula in projective joins and meets by cross products.

The real projective plane is the set of one-dimensional linear subspaces of ${\mathbb{R}}^3$. Equivalently,A projective point is written in homogeneous coordinates $[x:y:z]$. A projective line is the set of projective points whose representatives belong to one fixed two-dimensional linear subspace of ${\mathbb{R}}^3$. Such a linear subspace is the kernel of a linear map of a nonzero linear functional, so its equation has the form $\alpha x+\beta y+\gamma z=0$ with $(\alpha ,\beta ,\gamma )\ne 0$.

Two distinct projective points have linearly independent representatives $p,q$. Their linear span has dimension two and hence defines a projective line containing both. Conversely, any two-dimensional linear subspace containing both representatives contains their entire linear span and therefore equals it. Thus exactly one projective line passes through two distinct projective points.

The dual projective space is $\mathbb{P}(({\mathbb{R}}^3{)}^{\ast })$. A projective line with coefficients $[\alpha :\beta :\gamma ]$ becomes a projective point in this dual plane; a projective point $[a:b:c]$ becomes the dual projective line $a\alpha +b\beta +c\gamma =0$.

The point-line duality reverses incidence:Thus the dual statement is that the dual point $l^{\ast }$ lies on the dual line $P^{\ast }$. A join of projective points becomes an intersection of dual projective lines, and the converse holds too. In particular, collinear projective points correspond to concurrent dual projective lines.

The dual of Desargues's theorem is its converse: if the intersections of the three pairs of corresponding sides are collinear, then the three lines joining corresponding vertices are concurrent. In the same nondegenerate setting, perspectivity from a line therefore implies perspectivity from a point.

The corresponding side equations, computed using projective joins and meets by cross products, areSubtracting the two equations in each row and substituting back gives the three projective pointsFor example, the first pair forces $y=z$ and $x=3z$; the second pair forces $y=0$ and $x=-z$; the last pair forces $x=z=2y$. All three satisfysince $3-4+1=0$, $1-0-1=0$, and $2-4+2=0$. The coefficient vector is nonzero, so this is a projective line. Thus the corresponding-side intersections are collinear, explicitly confirming Desargues's theorem.

The diagram uses the affine chart $x+y+z\ne 0$ with coordinates $u=y/(x+y+z)$ and $v=z/(x+y+z)$. Here the center $O$ is finite, the axis is $u=1/5$, and $Y$ lies on the projective line at infinity. The apparent parallelism of $QR$ and $Q^{\prime }{R}^{\prime }$ records their common vertical point at infinity, which also belongs to the axis.

Apply projective joins and meets by cross products to each corresponding pair. The cross products of their representatives are proportional to $(0,1,0)$, $(0,0,1)$, and $(0,1,-1)$, respectively. HenceTheir common point isIt is a valid projective point, because its representative is nonzero. The three equations force $y=z=0$, so this common projective point is unique. Thus the two projective triangles are in perspectivity from a point $O$.

Choose nonzero vectors $p_i\in {\mathbb{F}}^2$ representing $P_i$, and write $[u,v]=\det (u,v)$. Use the cross-ratio conventionDistinctness makes the denominator nonzero. Every rescaling of a representative occurs once in numerator and denominator, so the value depends only on the projective points. Swapping both pairs simply exchanges the numerator's factors and the denominator's factors:Under an invertible linear map $A$, $[Au,Av]=\det (A)[u,v]$. Two factors of $\det (A)$ occur above and below the fraction, proving projective transformation invariance.

Finally choose homogeneous coordinates making $P_0=[1:0]$, $P_1=[0:1]$, $P_2=[1:1]$. This normalization exists by the allowed theorem about a projective frame. A point distinct from $P_0$ is uniquely $[t:1]$, and the displayed cross-ratio is $t$. Thus equality of the two cross-ratios forces equal coordinates andThis also shows that for fixed first three distinct points the fourth point is determined by its cross-ratio.

Four projective points in general linear position in the complex projective plane impose four independent linear conditions on the six coefficients of a quadratic polynomial. To see independence, put them at the three coordinate points and $[1:1:1]$ using a projective frame. The remaining equations are $Axy+Bxz+Cyz=0$ with $A+B+C=0$, a two-dimensional vector space. Projectivizing its nonzero elements gives a pencil. It has nonsingular bases, for example the forms with coefficient triples $(1,1,-2)$ and $(1,-2,1)$, because the determinant of the symmetric coefficient matrix is $ABC/4$. Its three singular members are exactly the three pairs of lines obtained by partitioning the four base points into two pairs.

Two projective triangles are in perspectivity from a point when the three projective lines joining corresponding vertices meet in one projective point, called their center of perspectivity.

In the affine coordinate $z\mapsto [z:1]$, the complex projective line consists of the complex plane together with the single extra projective point $[1:0]$, called its point at infinity. It is the only point whose second homogeneous coordinate vanishes. This coordinate choice distinguishes the point; it is not intrinsically special under projective transformations. The map $[u:v]\mapsto [v:u]$ exchanges zero and infinity. The resulting one-point compactification is the Riemann sphere.

An ordered projective frame consists of $n+2$ projective points such that any $n+1$ representative vectors are linearly independent. After mapping the first $n+1$ points to the coordinate axes, the last point has every coordinate nonzero; rescaling those axes sends it to $[1:\cdots :1]$. This proves existence and uniqueness of the projective transformation carrying one ordered frame to another. In the projective plane a frame consists of four points, no three collinear.

In the real projective plane, represent a projective point by a nonzero vector and a projective line by its nonzero coefficient covector, identified with a vector in chosen coordinates. The join of distinct projective points represented by $p,q$ has coefficient vector $p\times q$, using the cross product. It is nonzero because the representatives are linearly independent, and its dot product with each representative vanishes. Dually, two distinct projective lines with coefficients $\ell ,m$ meet in the projective point represented by $\ell \times m$. Scaling an input scales the output and leaves the projective object unchanged. This makes joins, intersections, concurrency, and collinearity exact calculations with homogeneous coordinates.

After choosing the affine chart $z\ne 0$ in the projective plane, its complement $z=0$ is the projective line at infinity. Affine parallel lines meet at the same projective point on this line, corresponding to their common direction. The choice of this line is part of the affine chart, not an intrinsic distinction preserved by every projective transformation.

A projective triangle is an ordered triple of noncollinear projective points in a projective plane, together with their three joining projective lines. In the real projective plane this means that three chosen representative vectors are linearly independent. Pairing the vertices of two such triangles determines corresponding sides and the three corresponding-vertex joins.

For a projective plane curve $C=V(F)$ and a nonsingular projective point $p$, the tangent is the line $F_x(p)x+F_y(p)y+F_z(p)z=0$. The coefficients are not all zero, and the Euler theorem for homogeneous functions makes this line pass through $p$. On an affine chart this is precisely the first-order part of a local defining equation, so the definition agrees with the usual tangent line and is invariant under projective transformations.

The degree-$d$ Veronese embedding sends a projective point to all monomials of degree $d$ in its homogeneous coordinates. For example,is a closed immersion with image the conic $uw-v^2=0$.