some facts about injective and surjective linear maps

Let $k$ be a field and $V,W$ be vector spaces over $k$.

Proposition. Let $f:V\to W$ be an injective linear map. Then there exists a (surjective) linear map $g:W\to V$ such that $g\circ f={\mathrm{id}}_V$.

Proof. Of course $\mathrm{Im}(f)$ is a subspace of $W$ so $f:V\to \mathrm{Im}(f)$ is a linear isomorphism. Let ${(e_i)}_{i\in I}$ be a basis of $\mathrm{Im}(f)$ and ${(e_j)}_{j\in J}$ be its completion to the basis of $W$, i.e. ${(e_i)}_{i\in I\cup J}$ is a basis of $W$. Define $g:W\to V$ on the basis as follows:

$$g(e_i)=f^{-1}(e_i),\text{if }i\in I;$$

$$g(e_j)=0,\text{if }j\in J.$$

We will show that $g\circ f={\mathrm{id}}_V$.

Let $v\in V$. Then

$$f(v)=\sum _{i\in I}{\alpha }_i{e}_i,$$

where ${\alpha }_i\in k$ (note that the indexing set is $I$). Thus we have

$$(g\circ f)(v)=g(\sum _{i\in I}{\alpha }_i{e}_i)=\sum _{i\in I}{\alpha }_{i}g(e_i)=\sum _{i\in I}{\alpha }_i{f}^{-1}(e_i)=$$

$$=f^{-1}(\sum _{i\in I}{\alpha }_i{e}_i)=f^{-1}(f(v))=v.$$

It is clear that the equality $g\circ f={\mathrm{id}}_V$ implies that $g$ is surjective. $\mathrm{\square }$

Proposition. Let $g:W\to V$ be a surjective linear map. Then there exists a (injective) linear map $f:V\to W$ such that $g\circ f={\mathrm{id}}_V$.

Proof. Let ${(e_i)}_{i\in I}$ be a basis of $V$. Since $g$ is onto, then for any $i\in I$ there exist $w_i\in W$ such that $g(w_i)=e_i$.Now define $f:V\to W$ by the formula

$$f(e_i)=w_i.$$

It is clear that $g\circ f={\mathrm{id}}_V$, which implies that $f$ is injective. $\mathrm{\square }$

If we combine these two propositions, we have the following corollary:

Corollary. There exists an injective linear map $f:V\to W$ if and only if there exists a surjective linear map $g:W\to V$.