Let $I\subseteq R$ be an ideal of a ring $R$. For any $a\in R$, prove that $$l_{R/I}(a+I)=l_{R/I}(a),$$ where $l_{R/I}(a+I)=\{r+I\in R/I:(r+I)(a+I)=I\}$.
In trying to prove this I have the following: $l_{R/I}(a+I)=\{r+I\in R/I:(r+I)(a+I)=I\}=\{r+I\in R/I: ra+I=I\}=\{r+I\in R/I: ra\in I\}=l_{R/I}(a)$.
We notice that $l_{R/I}(a)=\{s+I\in R/I:(s+I)a=I\}$