Let $V_{1,2}$ be two vector spaces over an arbitrary field and $U_{1,2}\subset V_{1,2}$ two lienar subspaces, respectively. Furthermore, let $f:V_{1}\to V_{2}$ be a linear map such that $f(U_{1})\subset U_{2}$. Then, it induces a well-defined map
$$[f]:\frac{V_{1}}{U_{1}}\to\frac{V_{2}}{U_{2}}$$
defined in the obvious way $[f]([v]):=[f(v)]$ for all $[v]\in V_{1}/U_{1}$. Now, in some lecture notes, I saw the following claim:
$[f]$ is injective if and only if $f^{-1}(U_{2})\subset U_{1}$.
where $f^{-1}(U_{2})=f^{-1}(U_{2}\cap f(V_{1}))$, since in general $U_{2}\not\subset f(V_{1})$.
I wanted to prove this statement, but wasn't quite able to conclude the proof. What I have so far is the following:
The map $[f]$ is injective if and only if $[f]([v])=0$ for some $[v]\in V_{1}/U_{1}$ implies that $[v]=0$. By definition of $[f]$ and the definition of quotient spaces, this is equivalent to say that the existence of a $u\in U_{2}$ such that $f(v)+u=0$ implies the existence of a $w\in U_{1}$ such that $v+w=0$. By linearity, the latter condition is equivalent to $0=f(v+w)=f(v)+f(w)$. To sum up, $[f]$ is injective, if and only if the existence of a $u\in U_{2}$ such that $f(v)+u=0$ implies the existence of a $w\in U_{1}$ such that $f(v)+f(w)=0$. Now, this can only be the case if $u=f(w)$, which shows that $[f]$ is injective if and only if the existence of a $u\in U_{2}$ such that $f(v)+u=0$ implies that $u\in f^{-1}(U_{2})$.
How to I go on from there?
First one direction is clear, if $[f]$ is injective $f^{-1}(U_2)$ must lie in $Ker([f]) = 0 = [U_1]$.
Now you are trying to prove the other direction and how you are doing it confuses me a bit so I will do it a bit different:
$[f] $ is injective if the only elements of $V_1$ that are mapped to $U_2$ by $f$ lie in $U_1$ (I think this is what you showed), so if $f(v) = u \in U_2$ implies that $v \in U_1$ and as $f^{-1}(U_2) = \{v\in V_1|f(v) = u \in U_2\}$, this means that $f^{-1}(U_2)\subset U_1$.