Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 4 / 16H / c / Solution

Letbe the injection witnessing $\kappa \lambda \le \kappa +\lambda$, and fix $(k_{\ast },l_{\ast })\in K\times L$. Extend the inverse of $i$ to a surjectionby settingThis construction uses only the fixed pair, not the axiom of choice.

Restrict $p$ to the $K$-summand and take its second coordinate:If $r$ is onto, it is the required surjective function from $K$ to $L$.

Otherwise choose one $l_0\in L\setminus r[K]$. For each $k\in K$, surjectivity of $p$ gives a preimage of $(k,l_0)$. No such preimage lies in the $K$-summand, by the choice of $l_0$, so it lies in the $L$-summand. Unless $(k,l_0)=(k_{\ast },l_{\ast })$, this preimage is exactly $i(k,l_0)$ and is unique: all points outside the range of $i$ map only to the default pair. Distinct $k$ give distinct preimages because $i$ is injective.

If $l_0\ne l_{\ast }$, these unique preimages therefore define an injective function $K\to L$. The only exceptional case is $l_0=l_{\ast }$, where they initially define an injectionIf $j$ is onto, extend it arbitrarily at $k_{\ast }$ to obtain a surjection $K\to L$. If it is not onto, choose one $l^{\prime }\in L\setminus j[K\setminus \{k_{\ast }\}]$ and set $j(k_{\ast })=l^{\prime }$; this gives an injection $K\to L$. These cases prove the product-sum comparison lemma entirely in ZF.

Solved by gpt-5.6-sol high.