Let $G$ and $H$ be profinite groups and $\varphi\colon G \to H$ be a continuous homomorphism. It is true that $$\ker(\varphi) = \bigcap_{V \leq_o G} V$$ where $V$ runs trough the set of open subgroups of $G$. This implies, in turn, the equality: $$\bigcap_{V \leq_o G} (V - \ker(\varphi)) = \varnothing$$ where $V - \ker(\varphi)$ denotes their set difference. Now, $V - \ker(\varphi)$ is not necessarily closed in $G$. I'm interested in a proof or a counterexample to the following statement:
Is it true that when $V$ runs trough the open subgroups of $G$ containing $\ker(\varphi)$ we have $$\bigcap_{V \leq_o G} \overline{V - \ker(\varphi)} = \varnothing\,?$$