Let $X$ and $Y$ be topological compact abelian groups, and let $A$ and $B$ be dense subgroups of $X$ and $Y$ respectively. Let $\varphi\colon X \to Y$ be a continuous homomorphism such that $\varphi(A) \subseteq B$, i.e., we have the following commutative diagram:
$$\begin{matrix} A & \rightarrow & B\\ \downarrow & & \downarrow\\ X & \rightarrow & Y \end{matrix}$$
Question. If $\varphi_A\colon A \to B$ denotes the restriction of $\varphi\colon X \to Y$ to $A$, is $\ker(\varphi_A)$ a dense subgroup of $\ker(\varphi)$?
I'm specifically interested when $X$ and $Y$ are profinite (compact and totally disconnected). This is certainly true if $\ker(\varphi)$ is open in $X$, and I've tried to reduce the closed case to the open case by writing $\ker(\varphi) = \bigcap_{N \unlhd_o G} N$ as an intersection of the open normal subgroups containing it, but my argument breaks down in a crucial part.
This is not true even for profinite groups. Let $X=(\mathbb{Z}/2\mathbb{Z})^{\aleph_0}$, let $\varphi:X\to Y$ be the quotient by the subgroup of elements all of whose coordinates are the same (observe $\ker\varphi$ has only two elements), and let $A$ be the dense subgroup of $X$ of elements almost all of whose coordinates are zero. Then $A$ is dense, but its intersection with $\ker \varphi$ is trivial.