For $V$ vector space, $W$ subspace, $V \simeq W \oplus V/W$? I thought this seemed intuitively true. So I searched for it but never found it. I thought that if it were true, surely it would appear somewhere in the Rank-Nullity Theorem page on Wikipedia. But no luck. So I tried to prove it and I think I have done. But I'm obviously very sceptical about my own work here. Is this correct ?
Let $V$ be a vector space over some field $K$ and let $W$ be a subspace of $V$. 
Then $V/W$ is a vector space over $K$, so  it has a basis $\mathcal B$. Define a linear map $ \phi : V/W \rightarrow V$ by choosing $\phi(x)$ for any $x \in \mathcal B$. More precisely, do it such that $\phi(x) \in x \quad \forall x \in \mathcal B$, in which case this property extends to all $x \in V/W$
For variables in $V$ I shall use $v$ instead of $x$ to avoid confusion. Let $p$ be the projection onto $V/W$. 
My claim is that $f : \begin{array}{cccc} V & \rightarrow & W \oplus V/W \\ v & \mapsto & (v-\phi \circ p (v),\, p(v)) \end{array}$ is an isomorphism.
As an ordered pair of sums and compositions of linear maps, $f$ is linear. 
Let's prove that $f$ is injective. $f(v) = 0 \iff (\,p(v)=0 \,\text{and}\, v- \phi \circ p(v)=0\,) \implies v-\phi(0)=0 \implies v=0$ 
$f$ is onto since for any $(\alpha,\, y) \in  W \oplus V/W$ we have $f(\alpha + \phi(y))=(\alpha+\phi(y)-\phi \circ p (\alpha + \phi(y)),\,  p (\alpha + \phi(y))\,)$ 
On the right hand side of the ordered pair, $p(\alpha)=0$ since $\alpha \in W$, and also $p(\phi(y))=y$ as $\phi(y)\in y \in V/W$ so in total we're left with $y$. 
As for the left hand side, we $\alpha + \phi(y) - \phi(\text{the right hand side})$ which adds up to $\alpha$, QED.
 A: This is true, irrespective of $V$ being finitely generated or not, but it relies on the axiom of choice, that's used to prove every subspace has a complement.
If you accept the existence of a complement (which usually people do), then you have a projection along it and you get the result. Namely, if $W'$ is a subspace so that $W\cap W'=\{0\}$ and $W+W'=V$, you can define $p\colon V\to W$ by $p(v)\in W$ and $v-p(v)\in W'$. Observe $p$ is linear.
Then you can define $g\colon V\to W'$ with $g(v)=v-p(v)$; this map is surjective and has kernel $W$, so it induces an isomorphism $q\colon V/W\to W'$.
Then define $f\colon V\to W\oplus V/W$ by
$$
f(v)=\bigl(p(v),q^{-1}(v-p(v))\bigr)
$$
and prove it is an isomorphism.
A: Let $V$ be a vector space over field $K$ and $W$ be a $K$-sub space of $V$. 
Now let us consider the natural surjective map $\eta : V \rightarrow V/W$, where  $\eta (x) = x + W$.
Note that,  ker $\eta = W$.
Let $\{ x_{i} + W \}_{i \in I}$ be a $K$-basis of $V/W$.
Now let $y_{i} \in V$ such that $\eta(y_{i}) = x_{i} +W$ for all $i \in I$.
Note that $y_{i}$'s are linearly independent and let $W'$ be the subspace spanned by $y_{i}$'s.
Extend the set of linearly independent vectors $A = \{ y_{i} \}_{i \in I}$ to a basis set $B$ of $V$.
Note that $B \setminus A$ is a basis set of $W$.
So, $V = W \oplus W'$.
And $W' \cong V/W$ implies that $V \cong W \oplus V/W$.  
A: Vector spaces are isomorphic iff they have the same dimension.
