I was reading Axler's Linear Algebra Done Right, and the following question appears as exercise $12$ in chapter $3$, section $E$.
Suppose $U$ is a subspace of $V$ such that $V/U$ is finite-dimensional.
Prove that $V$ is isomorphic to $U \times (V/U)$.
I solve it as follows:
Since $V/U$ is finite-dimensional then we can find a basis, let it be:
$$(v_1 + U) + \cdots + (v_n + U) \quad;\enspace v_i \in V ,\text{for } i = 1,..,n$$
It is clear that $v_1, \cdots, v_n$ are independent.
Now for any list of vectors in $V$ that have $\text{length} \geq n + 1$, we know that this list is dependent because otherwise, we can construct an independent list in $V/U$ with a length greater than the length of its basis which is impossible.
So it must be that $V$ is also finite-dimensional and since $U$ is a subspace of $V$ it must be finite-dimensional.
Now we can proceed easily as follows:
$$\dim U \times (V/U) = \dim U + \dim V/U = \dim U + \dim V – \dim U = \dim V$$
And since we know that:
Two finite-dimensional vector spaces over $F$ are isomorphic if and only if they have the same dimension.
We can conclude that $V$ is isomorphic to $U \times (V/U)$.
Is my solution correct?
Also please note that I know the other solutions to this problem so I am not asking for other solutions.
Best Answer
Your proof is wrong, and here's an idea for you:
Let $\;U'\;$ be the complement of $\;U\;$ in V, meaning : if $\;B\;$ is a basis of $\;U\;$ , then we can complete it to a basis $\;B\cup S\;$ of all $\;V\;$ (we need AC for this here if the dimensions are not finite), so in fact $\;U'\,$= Span$\,S\;$ .
Clearly $\;U\cap U'=\{0\}\;$ and thus $\;V=U\oplus U' \;$ . This means there's a unique expression $\;v=u+u'\;,\;\;u\in U,\,\,u'\in U'\;$ , for all $\;v\in V\;$ .
Now define
$$\phi: V\to U\times V/U\;,\;\;\phi(v=u+u')=\left(u,\,u'+U\right)$$
By the first part this is well defined. Prove this is a linear map which is an isomorphism.
Question: where did you use that $\;\dim V/U<\infty\;$ ?