Given a social choice function $F$, a subset $B\subset A$ of the candidates and a coalition $S\subset N$ of the voters, $\beta$-effectiveness of $S$ for $B$ is equivalent to $N\setminus S$ not being $\beta$-effective for $A\setminus B$.
However, for a social choice correspondence it is not true - the fact that $N\setminus S$ is not $\beta$-effective for $A\setminus B$ does not, in general, lead to $S$ being $\beta$-effective for $B$.
Could anyone give an example that shows why it is so? What is special about social choice functions that gives this symmetry?
The core difference is that a social choice function selects only one candidate - the fact that $N - S$ is not $\beta$-effective for $A - B$ with regards to a s.c.f F means that there exists $P^S \in L^S$ such that for every $Q^{N - S} \in L^{N - S}$, it does not hold that $F(P^S,Q^{N-S}) \in A - B$. Since F is a s.c.f, this means that $F(P^S, Q^{N-S}) \in B$ and therefore that $S$ is $\beta$-effective for $B$ with regards to F.
However, if we talk about $\beta$-effectiveness with regards to a s.c.c. $H$, the fact that it does not hold that $H(P^S, Q^{N-S}) \subset A-B$ does not necessarily mean that $H(P^S,Q^{N-S}) \subset B$ - we could have that H selects both members of $B$ and $A-B$. Is this clear enough for you to come up with a s.c.c that does not satisfy this condition?