Quotient Space of Vector Spaces I'm trying to prove that for subspaces of $V,U$ of a vector space $W$ it holds that
$
V=(U\cap V) \oplus V'
$
where $V'$ is isomorphic to the quotient space $V/(U\cap V)$. I think I got the intuition right but I'm not really sure how to show it. Can anyone help me?
And why is
$U+V=U+((U\cap V)\oplus V')=U\oplus V'$
 A: The main fact is that every subspace $X$ of a vector space $V$ has a complement, that is, a subspace $X'$ such that $V=X\oplus X'$, which is an abbreviation for
$$
V=X+X',\qquad X\cap X'=\{0\}.
$$
This fact requires the axiom of choice when infinite dimensional vector spaces are concerned; on finite dimensional spaces it is just the fact that every linearly independent subset can be extended to a basis.
Now, consider $X=U\cap V$ as a subspace of $V$; then we have $V'=X'$ such that $V=(U\cap V)\oplus V'$ and we can consider the canonical map
$$
\pi\colon V\to V/(U\cap V)
$$
so that $\ker\pi=U\cap V$. The restriction $\pi'$ of $\pi$ to $V'$ is an isomorphism. Indeed, let $v+(U\cap V)\in V/(U\cap V)$; then $v=x+y$, with $x\in U\cap V$ and $y\in V'$ (because $V=(U\cap V)+V'$) and
$$
v+(U\cap V)=\pi(v)=\pi(x)+\pi(y)=0+\pi'(y)
$$
which shows $\pi'$ is surjective. If $y\in V'$ and $\pi'(y)=0$, then $\pi(y)=0$ and $y\in\ker\pi=U\cap V$. Since $(U\cap V)\cap V'=\{0\}$, we have $y=0$, so $\pi'$ is injective.
For the second part, let's first prove that $U\cap V'=\{0\}$. If $v\in U\cap V'$, then $v\in V$, so $v\in(U\cap V')\cap V=(U\cap V)\cap V'=\{0\}$. Therefore $U+V'=U\oplus V'$.
The inclusion $U\oplus V'\subseteq U+V$ is clear. Let $u\in U$, $v\in V$; then $v=x+y$, with $x\in U\cap V$ and $y\in V'$, so $u+v=(u+x)+y\in U\oplus V'$.
Since $(U\cap V)\oplus V'=V$, the equality $U+V=U+((U\cap V)\oplus V')$ is obvious.
A: It is true without any hypothesis on dimensions.
Consider the short exact sequence:
$$0\to U\cap V\xrightarrow{\ i\ } V\xrightarrow{\ p\ } V/U\cap V\to 0$$
Let $\mathcal B=(e_i)_{i\in I}$ be a basis of the quotient space $V/U\cap V$. Consider $\mathcal B'=(e'_i)_{i\in I}$, with $p(e'_i)=e_i$. This is a set of linearly independent vectors of $V$, and mapping each $e_i$ to $e'_i$ defines a linear section $s$ of $p$.
Now for any $v\in V$, one checks $v-s(p(v))\in\ker p=U\cap V$. Setting $V'=s(V/U\cap V)$, we have proved that $\;V=(U\cap V)+V'$.
This sum is direct: indeed, if $u\in U\cap V$ lies in $V'$, i.e. if $u=s(p(v))$ for some $v\in V$, then $p(u)=0$ by definition, so, as $s$ is a section of $p$, $\;p(s(p(v))=p(v)=0$ and thus  $\;u=s(p(v))=0$.
A: You can argue using dimensions (supposing that $W$ is finite dimensional). The we have
$$
\dim( V/(U\cap V)) = \dim(V)-\dim(U\cap V).
$$
By complementing $U\cap V$ with $V'$ such that $V =(U\cap V) \oplus V'$, then
$\dim(V') =\dim(V)-\dim(U\cap V)=\dim(V/(U\cap V))$. Hence $V'$ and $V/(U\cap V)$ are isomorphic.
This also shows $\dim(U+V) = \dim(U)+\dim(V')$. It remains to show $U\cap V'=\emptyset$ to answer your second question. Take $x\in U\cap V'$. Since $V'$ is a subspace of $V$ this implies $x\in V$, $x\in U\cap V$, and $x\in V'$. Since $V=(U\cap V)\oplus V'$ it follows $(U\cap V)\cap V'=\{0\}$, and $x=0$.
