I am currently working on a combinatorial problem that reduces to the following counting problem:
Let $i$ be a positive fixed integer. Given a set $S$, Is there a family of subsets $\mathcal{B}^i_S$ of bounded size (called the set of representatives), such that for any set of $i$ subsets $Y_1 \dots Y_i$ of $S$ the following holds: $$|S\cap Y_j|\equiv_2 1 \text{ for all }j\in[k]\iff |\{X\in \mathcal{B}^i_S\colon \forall j, |X\cap Y_j|\equiv_2 1\}|\text{ is odd}.$$
That means for any choice of $i$ sets: all have odd intersection with $S$, iff odd many representatives have odd intersection with all these sets.
Of course we can take $\mathcal{B}^i_S = \{S\}$. However, for a ground set $U$, I am looking for a fixed family $\mathcal{T}$ of bounded size such that for each $S\subseteq U$ it holds that $\mathcal{B}^i_S \subseteq \mathcal{T}$.
For odd values of $i$ and $k := |S|$, I have conjectured that taking all subsets of odd size at most $i$ suffices $$\mathcal{B}^i_S = \bigcup\limits_{r\leq i\\r\equiv_21}\binom{S}{r}.$$I have tried plenty of examples and it seems to work, but I cannot prove it.
I appreciate any help towards how to prove this.