Proving isomorphism between between a subspace and a quotient space I've been thinking about this for a day or two now, and I think I've found a way to prove this, but am very unsure about how watertight this is:
To be proven:
Let $V$ be a vector space. If $V = U \oplus W$, then $V/W$ is isomorphic to $U$.
My thoughts on this so far:
Consider $\pi : V \to V/W$, defined naturally by $\pi(v) = [v]$. Then $N(\pi)$ = W, where $N$ denotes the null space.
By the dimension theorem, we have $dim W + dim V/W = dim V$
Now, since the sum $U \oplus W$ is direct, they have no elements in their respective bases in common, and so we have $dim U + dim W = dim V$. But this implies $dim U = dim V/W$
Now consider the sets of their respective bases: $A(U) = \{a_1, a_2, a_3 ...\}$ and $B(V/W) = \{b_1, b_2, b_3 ...\}$ and define $\zeta: A \to B$ by $\zeta(a_i) = b_i$. This mapping is obviously injective since $N(\zeta) = \{0\}$ and surjective by definition. 
Since the set of bases are isomorphic, the spaces must also be isomorphic. (how true is this? It certainly 'feels' very true). QED
Does this hold water? If not, why? 
I would also like to see alternative proofs if you got 'em, they are usually very illuminating.
Any appropriate theory is appreciated!
Thanks in advance!
 A: Your proof is on the right track: you've shown that $V/W$ and $U$ have the same dimensions, and any two vector spaces of the same dimension (over a given field) are isomorphic via a map that sends the $n$-th basis vector of one vector space to the $n$-th basis vector of the other. This is essentially what you've shown, but when you say that "the bases are isomorphic," it's not clear what you mean. To make this precise, you simply must note that the assignment $a_i \mapsto b_i$ defines a linear map, with inverse determined $b_i \mapsto a_i$. 
Another way to prove $U \simeq V/W$ would be to use the following fact: given a surjective linear map $\phi: A \rightarrow B$ of vector spaces, $B \simeq A/ \ker \phi$ (this is the first isomorphism theorem for vector spaces). With this in mind, consider the projection map $\pi: U \oplus W \rightarrow U$ given by $u+w \mapsto u$, where $u+w$ is the unique decomposition of an element as a sum of elements $u \in U$ and $w \in W$. It's clear that this map is surjective, and $\ker \pi = \{u+w \in V: u=0\} =W.$ Hence $V/W \simeq U$.
The (dabatable) advantage of this approach that it is intrinsic -- independent of a choice of basis for $U$ and $V/W$. Moreover, the map induced by $\pi$ is seemingly very natural, since $\pi$ restricts to the identity on $U$. 
A: Is the converse true? That is, for all subspaces $U\leq V$ is there a subspace $W\leq V$ so that 
$$W\cong V/U$$ 
and if so does it follow that 
$$V=U\oplus W$$?
