I want to see if it is possible to force the existence of a function $F:\aleph_2 \times \aleph_2\rightarrow \aleph_1$ such that:

a) $F(a,b)=F(b,a)$, for all $a,b\in \aleph_2$ and

b) for all distinct $a,b$, the set $\{x|F(a,x)=F(b,x)\}$ is finite.

What is known:

1) Under CH, there is no such function.

2) If such a function exists, let A be a subset of $\aleph_2$ of size $\aleph_1$. Consider the functions $F(a,\cdot)$ restricted to A. We have a family of $\aleph_2$ many such functions from $\aleph_1$ to $\aleph_1$ and any two of them agree only on a finite set. That's a strongly almost disjoint family of size $\aleph_2$. Baumgartner call this $A(\aleph_1,\aleph_2,\aleph_1,\aleph_0)$. It is consistent together with the negation of CH that either $A(\aleph_1,\aleph_2,\aleph_1,\aleph_0)$ or its negation hold. In particular, $\neg CH$ + "there is no such function" is consistent.

I want to see if anyone knows any result on the positive side.