I am reading "Linear Algebra Done Right 3rd Edition" by Sheldon Axler.
I solved Exercise 3.E.12 on p.99 as follows.
I want to know if my solution is ok or not.
I also want to know a typical solution to this exercise.
3.E.12
Suppose $U$ is a subspace of $V$ such that $V/U$ is finite-dimensional. Prove that $V$ is isomorphic to $U\times(V/U)$.
Let $v_1+U,\dots,v_n+U$ be a basis of $V/U$.
Then, $v_1,\dots,v_n$ is linearly independent:
Proof:
If $\lambda_1v_1+\dots+\lambda_nv_n=0$, then $\lambda_1(v_1+U)+\dots+\lambda_n(v_n+U)=(\lambda_1v_1+\dots+\lambda_nv_n)+U=0+U$.
Since $v_1+U,\dots,v_n+U$ is linearly independent, $\lambda_1=\dots=\lambda_n=0$.So, $v_1,\dots,v_n$ is a basis of $\operatorname{span}(v_1,\dots,v_n)$.
Let $v$ be an arbitrary element of $V$.
Then, $v+U=\lambda_1(v_1+U)+\dots+\lambda_n(v_n+U)=(\lambda_1v_1+\dots+\lambda_nv_n)+U$ holds for some $\lambda_1,\dots,\lambda_n\in\mathbb{F}$.
So, $v-(\lambda_1v_1+\dots+\lambda_nv_n)\in U$.
Since $v_1+U,\dots,v_n+U$ is linearly independent, there exists a unique element $v'\in\operatorname{span}(v_1,\dots,v_n)$ such that $v-v'\in U$.
So, $V=U\oplus\operatorname{span}(v_1,\dots,v_n)$.
Since $\dim\operatorname{span}(v_1,\dots,v_n)=n$ and $\dim V/U=n$, $\operatorname{span}(v_1,\dots,v_n)$ and $V/U$ are isomorphic. (by 3.59 on p.82)
Since $U+\operatorname{span}(v_1,\dots,v_n)$ is a direct sum, $U\times\operatorname{span}(v_1,\dots,v_n)$ and $U\oplus\operatorname{span}(v_1,\dots,v_n)$ are isomorphic. (by 3.77 on p.93)
Therefore, $V=U\oplus\operatorname{span}(v_1,\dots,v_n)$, $U\times\operatorname{span}(v_1,\dots,v_n)$ and $U\times (V/U)$ are isomorphic.