Finite join of binary Cartesian products and identity relation

Question

Let $X_0$, ..., $X_n$, $Y_0$, ..., $Y_n$ ($n\in\mathbb{N}$) be subsets of a set $U$.

Does the following statement necessarily hold?

Statement $(X_0\times Y_0)\cup\dots\cup(X_n\times Y_n)\supseteq \operatorname{id}_U$ iff $(X_0\cap Y_0)\cup\dots\cup(X_n\cap Y_n)=U$ (here $\operatorname{id}_U$ is the identity relation on $U$).

I'm sure I can solve myself, but why not to share this fun problem (which appeared during my research)?

Answer 1

$(X_0\times Y_0)\cup\dots\cup(X_n\times Y_n)\supseteq \operatorname{id}_U \Leftrightarrow\\ ((X_0\times Y_0)\cap \operatorname{id}_U)\cup\dots\cup((X_n\times Y_n)\cap \operatorname{id}_U) = \operatorname{id}_U \Leftrightarrow\\ \operatorname{id}_{X_0\cap Y_0} \cup\dots\cup \operatorname{id}_{X_n\cap Y_n} = \operatorname{id}_U \Leftrightarrow\\ \operatorname{id}_{(X_0\cap Y_0)\cup\dots\cup(X_n\cap Y_n)} = \operatorname{id}_U \Leftrightarrow (X_0\cap Y_0)\cup\dots\cup(X_n\cap Y_n)=U$