= Solution
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 gives
$$
(AB:C\,\sigma(P))=(AB:CP).
$$
Let $T$ 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. Therefore
$$
\boxed{\tau=S\text{ is a projective transformation}.}
$$
No continuity or prior <surjectivity> assumption was needed. This is the rigidity assertion that <cross-ratio preserving maps are projective>.
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