Proposition 11

Let $R$ be an open equivalence relation on a topological space $X$ and let $\varphi$ be the canonical mapping $X \to X/R$. Then for each $x \in X$ and each filter base $\mathfrak{B}'$ on $X/R$ which converges to $\varphi(x)$, there is a filter base $\mathfrak{B}$ on $X$ which converges to $x$ and is such that $\varphi(\mathfrak{B})$ is equivalent to $\mathfrak{B}'$.

If $U$ is any neighbourhood of $x$ in $X$, then $\varphi(U)$ is a neighbourhood of $\varphi(x)$ in $X/R$ (§ 5, no. 3, Proposition 5), hence there is a set $M' \in \mathfrak{B}'$ such that $M' \subset \varphi(U)$; if we put $M = U \cap \varphi^{-1}(M')$, then $M' = \varphi(M)$. This shows that as $M'$ runs through $\mathfrak{B}'$ and $U$ runs through the neighbourhood filter of $x$, the sets $U \cap \varphi^{-1}(M')$ form a filter base $\mathfrak{B}$ on $X$; clearly $\mathfrak{B}$ converges to $x$ and $\varphi(\mathfrak{B})$ is equivalent to $\mathfrak{B}'$.