Quotient space decomposition Let $V$ be a vector space(or module) with decomposition $V= V_1 \oplus V_2$. And let $W \subset V$ be a subspace with decomposition $W= W_1 \oplus W_2$ such that $W_1 \subset V_1$ and $W_2 \subset V_2$. show that
$$V/W = V_1/W_1 \oplus V_2/W_2$$
 A: Let $q_1\colon V\to V_1$ and $q_2\colon V\to V_2$ be the projections onto the direct summands; let $p_1\colon V_1\to V_1/W_1$ and $p_2\colon V_2\to V_2/W_2$ be the canonical maps. Consider the linear map
$$
f\colon V\to V_1/W_1\oplus V_2/W_2,
\qquad f(v)=(p_1(q_1(v)),p_2(q_2(v)))=(q_1(v)+W_1,q_2(v)+W_2)
$$
The kernel of $f$ is the set of vectors $v\in V$ such that $q_1(v)\in W_1$ and $q_2(v)\in W_2$. Since $v=q_1(v)+q_2(v)$ by definition, we see that $\ker f=W_1+W_2=W$.
A: Here we view $V_i/W_i$ as a subspace of $V/W$ through the monomorphism $V_i/W_i \to V/W$ that maps $v_i + W_i \mapsto v_i + W$. Note that this map is injective since the kernel of $V_i \to V/W$ is precisely $V_i \cap W = W_i$.
Any $v + W \in V/W$ decomposes as $v_1 + v_2 + W$, with $v_i \in V_i$. But, with the identification above, we have
$$\begin{align*}v + W &= v_1 + v_2 + W \\ &= (v_1 + W) + (v_2 + W) \\ &= (v_1 + W_1) + (v_2 + W_2). \end{align*}$$
Also, suppose we have some $v+W \in V_1/W_1 \cap V_2/W_2$. Then, as an element of $V/W$, this $v+W$ can be written as $v+W = v_1 + W = v_2 + W$, with $v_i \in V_i$. Thus $v_1 - v_2 \in W = W_1 \oplus W_2$, and so we must have $v_i \in W_i$. But this means that $v \in W$, i.e., $v+W$ is the trivial vector in $V/W$.
