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: $$g(e_i)=f^{-1}(e_i),\ \mbox{if } i\in I;$$ $$g(e_j)=0,\ \mbox{if } j\in J.$$ 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. $\square$

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. $\square$

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 $g:W\to V$ .