# some facts about injective and surjective linear maps

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

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 $g:W\to V$.

Title some facts about injective and surjective linear maps SomeFactsAboutInjectiveAndSurjectiveLinearMaps 2013-03-22 18:32:22 2013-03-22 18:32:22 joking (16130) joking (16130) 6 joking (16130) Derivation  msc 15A04