Give three finite multisets $A,B,C$ I am trying to devise a sufficient condition that implies equality of both $A$ and $B$ with $C$. More formally, if $A,B,C$ are defined over a space $\Omega$, I am looking for a function $f : \Omega \times \Omega \to \Omega$ such that $$f(A,B) = C, \text{ implies that } A = C \text{ and } B = C.$$ Naturally, equality is as multisets.
The motivation for this question was noticing the basic observation that if $A\cup B \subseteq C$ as multisets, then $A \subseteq C$ and $B \subseteq C$ as multisets. I was wondering if something similar happens with the equality "$=$" operator instead of the subset "$\subseteq$" operator. In the subset example, the function $f$ is the union of multisets $\cup$ defined as $$f(A,B) = A \cup B := \{[x,\max\{m_A(x),m_B(x)\}] : x \in A \text{ or } x \in B\},$$ where $m_Z(x)$ counts the number of times that $x$ appears in some multiset $Z$.
To summarize the discussion in the comments:
No such function can exist (at least, assuming $\Omega$ is non-empty).
To see this, suppose we had such a function and let $\omega \in \Omega$ be some element, with $A=\{\omega\}$. Then we consider $f(A,\emptyset)$ as some set (or multiset). By definition we must have $A=\emptyset=f(A,\emptyset)$ so, in particular, $A=\emptyset$, counter to the definition of $A$, and we have a contradiction.
Technical point: Writing $f:\Omega\times \Omega\to \Omega$ can't be what was intended. That would mean a function that took two elements from $\Omega$ and returned a third. I assume that what was meant was a function, or more likely a functor, from $\mathbf {MSets}_{\Omega}\times \mathbf {MSets}_{\Omega}\to \mathbf {MSets}_{\Omega}$. That is, something that takes two multisets (of elements taken from $\Omega$) and returns a third. That's what the contradiction I described assumes.