Does $(V_1+V_2)/V_2=V_1/V_2$? Let $V_1,V_2\subseteq V$ be vector (sub)spaces. Is it true that $(V_1+V_2)/V_2=V_1/V_2$? I've tried showing this but I don't know if the steps are correct.
$$
\begin{align*}
(V_1+V_2)/V_2&=\{v_1+v_2+V_2:v_1\in V_1, v_2\in V_2\} \\
&=\{\{v_1+v_2+\tilde{v}_2:\tilde{v}_2\in V_2\}:v_1\in V_1,v_2\in V_2\} \\
&=\{\{v_1+\bar{v}_2:\bar{v}_2\in V_2\}:v_1\in V_1\} \\
&=\{v_1+V_2:v_1\in V_1\} \\
&=V_1/V_2
\end{align*}
$$
In the 3rd line I used that $v_2+\tilde{v}_2\in V_2$.
 A: It is false (and meaningless), unless $V_2\subset V_1$.
What is true is given by the Second isomorphism theorem in group theory:
\begin{align*}
V_1/V_1\cap V_2 &\xrightarrow{\;\sim\enspace}(V_1+V_2)/V_2\\
v_1+V_1\cap V_2& \xrightarrow{\quad\enspace} v_1+V_2
\end{align*}
A: If $V_2$ is not a subspace of $V_1$, then the quotient $V_1/V_2$ is usually not defined, so I would be careful with the notation $V_1/V_2$.
Instead, let $\pi \colon V \to V\!/V_2$, $v \mapsto v + V_2$ denote the canonical projection.
If $U \subseteq V$ is any subspace containing $V_2$, then we have the equality
$$
    U\!/V_2
  = \{ u + V_2 \mid u \in U \}
  = \{ \pi(u) \mid u \in U \}
  = \pi(U),
$$
so we can replace the quotient $U\!/V_2$ by the subspace $\pi(U)$.
This has the advantage that the image $\pi(U)$ is a well-defined subspace of $V\!/V_2$ for every subspace $U \subseteq V$, and not only those which contain $V_2$. (We may still think of $\pi(U)$ as the quotient $U\!/V_2$, even if this quotient is not defined.)
Your calculation then show that $\pi(V_1 + V_2) = \pi(V_1)$.
