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I don't know how to prove formally this statements.

1) Let $V$, $W$ be two vector spaces (of finite dimension)

and let $f: V \to W$ be a linear surjective function.

Prove that there exist a function $g: W \to V$ such as $f(g(x))$ is the identity function of $W$

1) Let $f: V \to W$ be a linear injective function.

Prove that there exist a function $g: W \to V$ such as $g(f(x))$ is the identity function of $V$

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    Could you prove the same statements for finite sets? The extensions to f.d. vector spaces isn't much more difficult – 2017-02-19
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    well in theory I have a rough idea but I don't know how to write it properly – 2017-02-19
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    I did only 1), from there you should be able to do 2). – 2017-02-19

1 Answers 1

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Pick a base in $W$, say $\{u_1, \ldots, u_n\}$, since $f$ is surjective there are $n$ vectors in $V$, $\{v_1, \ldots, v_n\}$ such that $$f(v_j)=u_j,\text{ } j=1, \ldots, n$$

Let $g$ be the unique linear function from $W$ to $V$ such that $$g(u_j)=v_j$$

this obviously implies that $$f(g(u_j))=u_j,\text{ } j=1\ldots,n$$

That's the map we were looking for, indeed let's pick a $v\in W$, for some $\lambda_1, \ldots,\lambda_n\in\mathbb{R}$ it is $$v=\sum_{j=1}^n\lambda_j u_j$$ and, since both $f$ and $g$ are linear, then $f\circ g$ is linear too, $$f(g(v))=f(g(\sum_{j=1}^n\lambda_ju_j))=\sum_{j=1}^n\lambda_j f(g(u_j))=\sum_{j=1}^n\lambda_j u_j=v$$