Direct sum of two subspaces is isomorphic to the whole space 
Let $W$ be a subspace of the finite-dimensional vector space $V$. Show that there is a subspace $U$ of $V$ such that $V \cong U \oplus W$

I know of a couple approaches to this problem. One way is to extend the basis of $W$ to a basis for $V$ and let the extension be the basis for $U$. Another way is to take the orthogonal complement. But my textbook has a hint to consider the exact sequence
$$0 \to W \to V \to V/W \to 0.$$
I believe it is suggesting $V \cong V/W \oplus W$, but this doesn't make sense because $V/W$ is not a subspace of $V$. What is this hint suggesting at?
 A: They're really the "same" approach, although this may not be self-evident, at first.
Suppose $\{w_i\}$ is your basis for $W$, and that $\{w_1,\dots,w_k,u_1,\dots,u_{n-k}\}$ (where $k = \dim (W)$ and $n = \dim(V)$) is your basis for $V$.
Show that $u_j \mapsto u_j + W$ is (induces) an isomorphism of $U$ with $V/W$.
Yes, $V/W$ is not a subspace of $V$, but there is such a thing as "external direct product" of vector spaces, where you take the (formal) direct sum of two (almost) disjoint spaces (I say "almost" because you must identify the zero vectors of each).
Working "the other way", if you find $n-k$ linearly independent vectors $u_j$ in $V$ such that $\{u_j + W\}$ form a basis for $V/W$, their span forms a basis for $U$.
A: To answer the abstract nonsense part at the heart of your question:
It is true that a short exact sequence $0 \to A \to B \to C \to 0$ of Abelian groups (or more generally, modules) need not split, i.e., there is generally no natural way of embedding the quotient $C \cong B/A$ back into $B$ in a way that plays well with the quotient map. However, such an embedding, called a section, always exists for vector spaces, because one can just lift a basis for the quotient space back up into $B$. More generally, all the short exact sequences terminating at $C$ split iff $C$ is a projective module.
