Let $k$ be a field of characteristic zero. By definition, a $k[[t_1, ...., t_n]]$-module $M$ is called topologically free if there is an isomorphism of $k[[t_1, ...., t_n]]$-modules $M\cong V[[t_1, ..., t_n]]$, where $V$ is a (not necessarily finite dimensional ) k-vector space.
Let $A$ be an associative $k$-algebra. Let $A[[t_1, ...,t_n]]$ be the natural topologically free $k[[t_1, ..., t_n]]$-algebra. Assume that $B$ is a $k[[t_1, ..., t_n]]$-algebra, which is topologically free as a module, and such that there is an embedding of $k[[t_1, ..., t_n]]$-algebras $$\phi: B\hookrightarrow A[[t_1, ...,t_n]].$$ Then, the image $\textrm{Im}(\phi)$ is a topologically free $k[[t_1, ..., t_n]]$-subalgebra of $A[[t_1, ...,t_n]]$.
Can we conclude immediately from the above that $\textrm{Im}(\phi)=A'[[t_1, ..., t_n]]$ for some $k$-subalgebra $A'$ of $A$?