Show that $V \simeq U \times (V/U)$ Let $V$ be a vector space and $U \subseteq V$ be a subspace such that $\dim{V/U} < +\infty$. I need to prove that $V$ is isomorphic to $U \times (V/U)$. The question had already been asked here, but no satisfying answer was given. Same as in the link, I am able to prove the required result in the case $\dim{V} < +\infty$, but not in the general case. 
Given a linear map $T: V \mapsto V$, we know $\newcommand{\im}{\operatorname{range }} V/\ker{T} \simeq \im{T}$. This is because $T$ induces a one-to-one map $T': V/\ker{T} \mapsto \im{T}$ such that $T'(v+\ker{T})=Tv$. 
If $V$ is finite-dimensional (so that also $U$ is) then we can easily construct a linear map $f: V \mapsto V$ such that $\ker{f}=U$ and we are done. I am not able to rigorously construct a similar map in the infinite-dimensional case though. How to use the hypothesis that $V/U$ is finite-dimensional then?
 A: Let $\mathcal B_0$ be a basis for $U$ and extend it to a basis $\mathcal B_0 \sqcup \mathcal B_1$ for $V$. Let $f: V \to V$ be the linear transformation which sends all elements of $\mathcal B_0$ to $0$ and all elements of $\mathcal B_1$ to themselves.
To avoid using the axiom of choice under the assumption that $V/U$ is finite dimensional: Let $\overline v_1, \ldots, \overline v_m$ be a basis for $V/U$, and let $v_1, \ldots, v_m \in V$ be lifts of $\overline v_i \in V/U$ to $V$. There are only finitely many choices here, so the axiom of choice is not required. Then you can define $f$ to be the linear transformation which is $0$ on $U$ and sends $v_i$ to $v_i$. It shouldn't be too hard to show that this linear transformation exists and is uniquely defined.
A: We must show that if $\dim V/U<\infty$ then $V\cong U\times V/U$.
Let $\dim V/U=m$ finite then choose some $\tilde b_j\in V\setminus U$ such that
$$
B:=\{\tilde b_j+U:j\in[m]\}
$$
is a basis of $V/U$. For each $x\in V$ define the $B$-representative of $x+U$ as
$$
\tilde x:=\sum_{j=1}^m \tilde b_j c_j:\; x+U=\tilde x+ U,\quad c_j\in\Bbb F
$$
Observe that such $\tilde x$ exists and is unique because $B$ is a basis of $V/U$, that is
$$
x+U=\sum_{j=1}^m c_j(\tilde b_j+U)=\left(\sum_{j=1}^m \tilde b_j c_j\right)+U=\tilde x+ U
$$
And also we knows that $x-\tilde x\in U$. Now define the map
$$
\varphi: V\to U\times V/U,\quad x\mapsto (x-\tilde x, \tilde x+U)
$$
Clearly the map is linear and injective, then it remains to show that $\varphi$ is also surjective.
Let $(z,\tilde y+U)$, then if we choose $z+\tilde y$ then
$$
\varphi(z+\tilde y)=(z+\tilde y-0-\tilde y,\tilde y+z+U)=(z,\tilde y+U)
$$
and the proof is complete.
A: You need the following facts:

Let $X, Y$ be vector spaces with bases $B$ and $B'$ respectively. Then
  $$B \times \{0\} \cup \{0\} \times B' = \{(b, 0) : b \in B\} \cup \{(0, b') : b' \in B'\}$$
is a basis for $X \times Y$.

and

Let $X$ be a vector space, and $M \le X$ its nontrivial subspace. Let $B$ be a basis for $M$, and $B' \supset B$ its extension to a basis to $X$. Then 
$$\pi(B'\setminus B) = \{b + M : b \in B' \setminus M\}$$
is a basis for $V/M$, where $\pi : V \to V/M$ is the canonical epimorphism.

Now, let $B$ be a basis for $U$ and $B'$ its extension to a basis for $V$.
We have that $\pi(B' \setminus B)$ is a basis for $V/U$, and then
$$B \times \{0 + U\} \cup \{0\}\times \pi(B'\setminus B)$$
is a basis for $U \times (V/U)$.
Now you can construct a linear map $\phi : V \to U \times (V / U)$ on the basis $B'$ for $V$:
$$\phi(b) = \begin{cases}
(b, 0 + U),  & \text{if $b \in B$} \\
(0, b + U), & \text{if $b \in B' \setminus B$}
\end{cases}$$
and extend it by linearity. It is an isomorphism since it maps a basis to a basis.

Notice that in the finite-dimensional case the dimensions of product and quotient obey the so-called logarithmic law:
$$\dim(X \times Y) = \dim X + \dim Y$$
$$\dim(X/M) = \dim X - \dim M$$
so in your case we have:
$$\dim U \times (V/U) = \dim U + \dim (V/U) =  \dim U  + \dim V - \dim U = \dim V$$
so $U \times (V/U) \cong V$ because their dimensions are equal.
A: Choose $v_1,\dots,v_n$ such that $B=\{\pi(v_1),\dots,\pi(v_n)\}$ is a basis for $V/U$ (this does not require the axiom of choice), where $\pi\colon V\to V/U$ is the canonical projection.
Denote by $\{f_1,\dots,f_n\}$ the dual basis for $B$, that is, $f_k\colon V/U\to F$ ($F$ is the base field) and
$$
f_k(\pi(v_l))=
\begin{cases}
1 & k=l \\
0 & k\ne l
\end{cases}
$$
Note that if $\pi(v)=\sum_{l=1}^n\alpha_l\pi(v_l)$, then
$$
f_k(\pi(v))=\sum_{l=1}^n\alpha_l f_k(\pi(v_l))=\alpha_k
$$
so, for every $v\in V$,
$$
\pi(v)=\sum_{k=1}^n f_k(\pi(v))\pi(v_k)\tag{*}
$$
Define $\varphi\colon V\to V$ by
$$
\varphi(v)=v-\sum_{k=1}^n f_k(\pi(v))v_k
$$
Then $\varphi$ is clearly linear and
$$
\pi(\varphi(v))=
\pi(v)-\sum_{k=1}^n f_k(\pi(v))\pi(v_k)=0
$$
by (*). Hence $\varphi(v)\in U$.
Now consider
$$
\psi\colon V\to U\times(V/U)
\qquad
\psi(v)=(\varphi(v),\pi(v))
$$
If $\psi(v)=(0,0)$, then $\pi(v)=0$ and
$$
0=\varphi(v)=v-\sum_{k=1}^n f_k(\pi(v))v_k=v
$$
Let now $(u,\pi(w))\in U\times(V/U)$. Then we wish to find $v$ such that $\psi(v)=(u,\pi(w))$. We need $\pi(v)=\pi(w)$ and
$$
u=v-\sum_{k=1}^n f_k(\pi(v))v_k=
v-\sum_{k=1}^n f_k(\pi(w))v_k
$$
Thus we have to try
$$
v=u+\sum_{k=1}^n f_k(\pi(w))v_k
$$
and verifying that indeed $\psi(v)=(u,\pi(w))$ is easy.
