A commutative-aggregate map and any second map have a common hole (Erdős 624)

Let $U$ be a finite set with $n=2^k$ elements. For a map $g:\mathcal P(U)\to U$ and $Y\subseteq U$ write $I_g(Y)=\{g(A):A\subseteq Y\}$. For $B\subseteq U$ let $\mathcal K_B$ be the set of $k$-subsets $Y\supseteq B$.

Let $M$ be a finite commutative monoid, $v:U\to M$ and $\psi:M\to U$ arbitrary, and let $h(A)=\psi\bigl(\prod_{x\in A}v(x)\bigr)$ (a commutative-aggregate map). Write $b=2^{2t}$.

Theorem. Let $k\ge3$ and $t,d\ge0$ with $|M|<\binom{n-2t(d-1)}{t}$, $2td\le k-2$ and $64(b-1)^d\le b^d$. Then for every map $g:\mathcal P(U)\to U$ there are a $k$-set $Y\subseteq U$ and a point $u\in U$ with $u\notin I_h(Y)$ and $u\notin I_g(Y)$.

Context: Erdős Problem 624 asks whether $H(n)-\log_2 n\to\infty$, where $H(n)$ is the least $m$ such that some $f:\mathcal P(X)\to X$ ($|X|=n$) satisfies $f[\mathcal P(Y)]=X$ for all $|Y|\ge m$. By an earlier track-amplification result it would suffice to show that, for large $k$, any two maps $\mathcal P(U)\to U$ with $|U|=2^k$ have a common hole: a $k$-set $Y$ and a point $u$ missed by both maps on $\mathcal P(Y)$. These results do not settle Problem 624.

Formalization Note. Counting form with natural numbers (truncated subtraction is harmless since $|I_g(Y)|\le n$); $U$ is an arbitrary Fintype with Fintype.card U = 2 ^ k.