If $\mathcal R$ is a not empty collections whose elements are binary relations then it is not complicated to show that even $\bigcap\mathcal R$ is a binary relation so that I tried to determine its domain, range and image.
So I would like to prove or disprove the following equialities. $$ \tag{1}\label{eq: domain intersection}\operatorname{dom}\left(\bigcap_{R\in\mathcal R}R\right)=\bigcap_{R\in\mathcal R}R^{-1}\left[\bigcap_{S\in\mathcal R}\operatorname{ran}S\right] $$
$$ \tag{2}\label{eq: inverse intersection}\left(\bigcap_{R\in\mathcal R}R\right)^{-1}=\bigcap_{R\in\mathcal R}R^{-1} $$
$$ \tag{3}\label{eq: range intersection}\operatorname{ran}\left(\bigcap_{R\in\mathcal R}R\right)=\bigcap_{R\in\mathcal R}R\left[\bigcap_{S\in\mathcal R}\operatorname{dom}S\right] $$
$$ \tag{4}\label{eq: image intersection}\left(\bigcap_{R\in\mathcal R}R\right)[X]=\bigcap_{R\in\mathcal R}R\left[X\cap\left(\bigcap_{S\in\mathcal R}\operatorname{dom}S\right)\right] $$
$$ \tag{5}\label{eq: preimage intersection}\left(\bigcap_{R\in\mathcal R}R\right)^{-1}[X]=\bigcap_{R\in\mathcal R}R^{-1}\left[X\cap\left(\bigcap_{S\in\mathcal R}\operatorname{ran}S\right)\right] $$
So I was apparently able to prove eq. \eqref{eq: inverse intersection} and to follow I show a possibile proof.
So if $(y,x)$ is in $\left(\bigcap_{R\in\mathcal R}R\right)^{-1}$ then $(x,y)$ is in $\mathcal R$ for any $R$ in $\mathcal R$ so that $(y,x)$ is in $\mathcal R^{-1}$ for any $R$ in $\mathcal R$ and so in $\bigcap_{R\in\mathcal R}R^{-1}$: we conclude the inclusion $$ \tag{2.1}\label{incl: inverse intersection 1}\left(\bigcap_{R\in\mathcal R}R\right)^{-1}\subseteq\bigcap_{R\in\mathcal R}R^{-1} $$ Conversely if $(y,x)$ is in $\bigcap_{R\in\mathcal R}R^{-1}$ then $(x,y)$ is in $R$ for any $R$ in $\mathcal R$ and so in $\bigcap_{R\in\mathcal R}$: we conclude that $(y,x)$ is in $\left(\bigcap_{R\in\mathcal R}R\right)^{-1}$ so that the inclusion $$ \tag{2.2}\label{incl: inverse intersection 2}\bigcap_{R\in\mathcal R}R^{-1}\subseteq\left(\bigcap_{R\in\mathcal R}R\right)^{-1} $$ hods. Finally, by incl. \eqref{incl: inverse intersection 1} and by incl. \eqref{incl: inverse intersection 2} we conclude that eq. \eqref{eq: inverse intersection} holds.
So I ask if actually equalities 1-5 holds and so if I well proved eq. \eqref{eq: inverse intersection}: could someone help me, please?
Only $(2)$ holds.
Here is a counterexample to $(1)$ and $(3):$
$$\mathcal R=\{R_1,R_2\},\quad R_1=\{(0,0),(1,1)\}=R_1^{-1},\quad R_2=\{(0,1),(1,0)\}=R_2^{-1}.$$ We have $\operatorname{dom}\left(R_1\cap R_2\right)=\operatorname{dom}(\varnothing)=\varnothing,$ whereas $\bigcap_{R\in\mathcal R}R^{-1}\left[\operatorname{ran}R_1\cap\operatorname{ran}R_2\right]=\bigcap_{R\in\mathcal R}R^{-1}\left(\{0,1\}\right)=\{0,1\}.$
It may also serve as a counterexample to $(4)$ and $(5),$ with $X=\{0,1\}.$