 Solution

This is König theorem for cardinal numbers. Let $D={⨆}_{\alpha <\beta }{\kappa }_{\alpha }$ and $P={\prod }_{\alpha <\beta }{\lambda }_{\alpha }$. For any proposed map $g:D\to P$, consider the values of the $\alpha$th coordinates on the $\alpha$th summand:This set has cardinality at most ${\kappa }_{\alpha }<{\lambda }_{\alpha }$. Choose $y_{\alpha }\in {\lambda }_{\alpha }\setminus E_{\alpha }$, using the axiom of choice, and assemble $y=(y_{\alpha })\in P$. For any $(\alpha ,\xi )\in D$, its image disagrees with $y$ at coordinate $\alpha$. Thus $y$ is outside the image of $g$, and no such $g$ is surjective.

Every ${\lambda }_{\alpha }$ is nonzero. The product is nonempty by choice; if $D$ is empty, strict inequality is immediate. Otherwise cardinal comparability principle in ZFC says that $|P|\le |D|$ would yield a surjection from $D$ onto the nonempty $P$, which the diagonal argument excludes. ThereforeThe diagonal coordinate is tied to the summand containing the argument; this is why a sum on the left pairs naturally with a product on the right.