Preimage of Singleton

Theorem
Let $\RR$ be a relation.

Let $\map {\RR^{-1} } t$ denote the preimage of $t$ under $\RR$.

Let $\RR^{-1} \sqbrk {\set t}$ denote the preimage of $\set t$ under $\RR$.

Then:
 * $\RR^{-1} \sqbrk {\set t} = \map {\RR^{-1} } t$