Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 1 / 16J / d / Solution

Choice well-orders every infinite $X$, so Hessenberg gives $X^2\cong X$. Conversely assume this square identity for every infinite set. For arbitrary $X$, form the disjoint union $Y=X\bigsqcup h(X)$. A pairing $Y^2\cong Y$, together with the well-order on $h(X)$, yields by the standard tracing/comparison argument an injection either $h(X)\to X$ or $X\to h(X)$. The first is forbidden by Hartogs theorem, so the second well-orders $X$. This is Tarski cardinal-square theorem, and hence choice follows.