Let $U \subset V \subset W$ be vector spaces over a field $F$, show that $V/U \leq W/U$ and $W/V \cong \frac{W/U}{V/U}$.
My progress is like this.
Let $U \subset V \subset W$ be vector spaces over $F$. Define the equivalence relation ~ on $V$ such that for some $x$, $y \in V$, if $x - y \in U$, then $x$ ~ $y$.
$\rightarrow$ $V/U = \{ [v] \mid [v] = v+u \text{ for all }u \in U, v \in V \}$
$\rightarrow$ $W/U = \{ [w] \mid [w] = w+u \text{ for all }u \in U,w \in W \}$.
Obviously, $[0] = \{ 0 + u \mid u \in U \} \in V/U \subset W/U$
Let $[v_1], [v_2]$ be equivalence classes in $V/U$ and $\alpha$, $\beta \in F$. $$[\alpha v_1] + [\beta v_2] = \{ (\alpha v_1 + u) + (\beta v_2 + u) \mid u \in U \} = \{ (\alpha v_1 + \beta v_2) + u \mid u \in U \} \in V/U$$
Therefore, $$V/U \leq W/U.$$
Then I’m stuck. At first, I suppose $T \colon W \rightarrow \frac{W/U}{V/U}$ to be a linear transformation, but I cannot figure out what would $T$ look like and how to find its kernel and its image.
Does anyone have any clue or roadmap for me to prove it?
Hint: Consider the map $$S: W/U \to W/V: w+ U \mapsto w+ V.$$
Then prove the following:
(1) $S$ is a well-defined linear map.
(2) $\ker(S) = V/U$.
(3) $S$ is surjective.
(4) Use the first isomorphism theorem to find the desired isomorphism.