Exact sequences and direct sums of subspaces proof

Question

I've been reading Harvey Rose's book on Linear Algebra, and I am stuck trying to understand this proof.

The short exact sequence is $ \{0\} \rightarrow U \xrightarrow{f} V \xrightarrow{g} W \rightarrow \{0\}$

Since it splits, there exists $h:W \rightarrow V$ such that $g\circ h$ forms an identity function.

Here's my problem:

They consider a basis for W, i.e. $\{w_1,...,w_n\}$, and say that the set $B=\{h(w_1),...,h(w_n)\}$ forms a basis for the image of $W$ under $h$.

I understand that the set $B$ will span the $\operatorname{img}(h)$ but I do not understand why it will be linearly independent.

Any help will be really appreciated and thank you!

Answer 1

That's because $h$ is injective and injective maps map linearly independent sets into linearly independent sets.

And $h$ is injective because, if $w\in W$ and $h(w)=0$, you can take $v\in V$ such that $g(v)=w$ and then$$0=h(w)=h\bigl(g(v)\bigr)=v$$and therefore $w=g(v)=g(0)=0$.