Let $K'\subset K$ be fields, $V$ be a right $K$-vector space and $V'$ a $K'$-subspace of $V$ such that the $K$-linear mapping $\lambda:V'\otimes_{K'}K\rightarrow V$ such that $\lambda(x'\otimes \xi)=x'\xi$, for all $x'\in V'$ and $\xi\in K$, is bijective. Every $K'$-subspace of $V$ satisfying this condition is referred to as a $K'$-structure on $V$.
A $K$-subspace $W$ of $V$ is said to be rational over $K'$ if it is generated over $K$ by a subset of $V'$.
Let $V_1,V_2$ be two right vector spaces over $K$ with $K'$-structures $V'_1$ and $V'_2$ respectively. Suppose $f:V_1\rightarrow V_2$ is a $K$-linear mapping such that $f(V'_1)\subset V'_2$. For every $K$-subspace $W_2$ of $V_2$ rational over $K'$, the $K$-subspace $f^{-1}(W_2)$ is rational over $K'$.
Attempt:
Let $W_2$ be a $K$-subspace of $V_2$ rational over $K'$. There exists a $W'_2\subset V'_2$ such that $W_2=W'_2.K$, where $W'_2.K$ denotes the $K$-subspace of $V_2$ consisting of the linear combinations of elements of $W'_2$ with coefficients in $K$. Take $x\in f^{-1}(W_2)$. Then $f(x)=\sum_{y'\in W'_2}y'\xi_{y'}$ for some $\xi\in K^{(W'_2)}$. I have to find a subset of $W'_1$ of $V'_1$ such that $f^{-1}(W_2)=W'_1.K$ but I have no idea how to proceed. Any suggestions?
Edit: Does this problem make sense as stated? i.e. is there a hidden identification being employed somewhere?