Direct sum decomposition of a quotient of vector spaces I just ran into a statement which I can't manage to prove.
Let $A,B,C$ be (edit: finite-dimensional) vector spaces, such that $B \leq A$. Then I believe (and am trying to prove) that $\dfrac{A}{B} \cong \dfrac{A + C}{B + C} \oplus \dfrac{A\cap C}{B\cap C}$. I can't really see how to go about proving that? It sounds a bit confusing.
(I'm not entirely sure the general statement is correct, so if not, here is the context:
I am reading this homology paper, which (third page, first column) introduces a filtration $C_0 \subset C_1 \subset C_2 = C$ of some vector space $C$, and if we denote by $Z$ and $B$ the respective cycles and boundaries in $C$, we also get filtrations $Z_0 \subset Z_1 \subset Z_2 = Z$, $B_0 \subset B_1 \subset B_2 = B$. They just go on to state (right before (7)) that $\dfrac{Z}{B} \cong \dfrac{Z+C_1}{B + C_1} \oplus \dfrac{Z\cap C_1}{B \cap C_1}$, which is the statement I don't know how to prove).
 A: The sequence
$$0\to\frac{A\cap C}{B\cap C}\to\frac{A}{B}\to\frac{A+C}{B+C}\to0,$$
where the maps are induced by the inclusions
$$A\cap C\hookrightarrow A\hookrightarrow A+C,$$
is a short exact sequence of vector spaces, and so the middle term is isomorphic to the direct sum of the other terms.
If you do this with abelian groups, you still have the short exact sequence, but it might not split, so the middle term is not necessarily isomorphic to the direct sum of the other terms. For example, if $A=\mathbb{Z}$, $B=4\mathbb{Z}$, and $C=2\mathbb{Z}$, then the middle term of the short exact sequence is cyclic of order $4$, and the other two terms are both cyclic of order $2$.
A: Based on Rob Arthan's answer, here is an attempt at a proof:
We are assuming $A, B, C$ to be finite-dimensional vector spaces, with $B \leq A$. We want to prove that $\dfrac{A}{B} \cong \dfrac{A + C}{B+C} \oplus \dfrac{A\cap C}{B\cap C}$.
This means that $\dim \dfrac{A}{B}$ = $\dim A - \dim B = \dim\left(\dfrac{A+C}{B+C}\oplus\dfrac{A\cap C}{B\cap C}\right)$, by calculation.
Then, left-hand side and right-hand side have the same dimension, to it suffices to produce an injective map from one to the other.
Let $v$ be a vector in $A$, and $[v]_B, [v]_{B+C}, [v]_{B\cap C}$ denote its respective equivalence classes in $\dfrac{A}{B}, \dfrac{A+C}{B+C}, \dfrac{A\cap C}{B \cap C}$. For an equivalence class $[w]$, let $\{[w]\}$ denote its tuple of coordinates (in the appropriate vector space). Moreover, let $P$ denote the natural projection operator from $A$ to $A\cap C$ (which exists since we're talking vector spaces).
Now, consider $\phi : \dfrac{A}{B} \to \dfrac{A + C}{B+C}\oplus\dfrac{A\cap C}{B\cap C} : [v]_B \mapsto \left(\{[v]_{B+C}\}, \{[Pv]_{B\cap C}\} \right)$
I understand the notation is a bit messy lol. Let's assume $v \notin B$ else everything is trivial.

*

*If $v$ was originally in $A \cap C$, then $[v]_{B+C} = 0$ (because $v \in C \implies v \in B + C$). So the first part of the coordinates is $0$. However, the second part isn't, since the projection acts as the identity.

*Vice versa, if $v \notin A \cap C$, then $v \notin C$. Then, the second part of the coordinates will be $0$, but the first will not. Because, since $v \in A \subset A + C$, the only way $[v] = 0$ in $\dfrac{A + C}{B + C}$ is if $[v] = 0$ in $\dfrac{A}{B}$.

So this ensures injectivity of $\phi$, which, by construction, is linear, so it should be an isomorphism?
Does that sound correct? :)
If so, that still only applies to vector spaces, while the original statements is for groups. In some cases, homology/cohomology groups do turn out to be vector spaces, but I guess there exists a more general proof in the case $A,B,C$ are groups?
