Suppose $V$ and $W$ are finite dimensional vector spaces, and that $f~:~V \to W$ is a linear map.
Suppose $\{e_1, \dots, e_n\} \subset V$ and that $\{f(e_1), \dots , f(e_n)\}$ is a basis of $W$.
Then which of the following are true?
- I. $\{e_1, \dots, e_n\}$ is a basis of $V$.
- II. There exists a linear map $g~:~W \to V$ such that $g \circ f = \text{Id}_V$
- III. There exists a linear map $g~:~W \to V$ such that $f \circ g = \text{Id}_W$
I know I is not right and thought that III would be the correct one because $f$ is surjective and it should have a right inverse. But it turns out that II is the only correct option and I have no clue how that could be possible. Any hints/explanations would be greatly appreciated.
The condition that $e_1,\dots, e_n$ is mapped to a basis $f(e_1,),\dots,f(e_n)$ means that the map is surjective as you figured out yourself. It is easy to write down a counterexample for II: Consider $f:\mathbb R^2\to \mathbb R$ given by the matrix $$A=\begin{pmatrix}1\\ 0\end{pmatrix}.$$ Let $e_1,e_2$ be the standard basis vectors. Then, $Ae_1=A(1,0)=1$ and $Ae_2=A(0,1)=0.$ Thus the kernel of $A$ is spanned by $e_2$ and $A$ cannot be injective.