Assume that $g\colon V\to W$ is linear. Show that if $f\colon W\to V$ is a left-inverse of $g$, then $f$ is linear.
I am not sure how to prove that. I need to show that $f(ax+by)=af(x)+bf(y)$ for scalars $a$ and $b$ and $x,y\in W$.
I think if the domain of $f$ is the range of $g$, i.e. each $x\in W$ can be expressed as $x=g(v)$ for some $v\in V$, then $$ f(ag(v_1)+bg(v_2))=f(g(av_1+bv_2))=av_1+b v_2=af(g(v_1))+bf(g(v_2)) $$
and $f$ is linear.
But in general, I do not think that the domain of $f$ needs to be the range of $g$. I rather think that it is possible that $$ \operatorname{dom}(f)\supsetneq\operatorname{range}(g). $$
In this case, I do not know how to prove that $f$ is linear or maybe it is even wrong.
Observe that $\;f\circ g=Id_V\;$ (and thus $\;g\;$ must be injective...) . But there is a problem here, for the composition $\;f\circ g\;$ is defined only on $\;Im\,g:=\{w\in W\;/\;w=g v\;,\;\;\text{for some}\;v\in V\}\;$ .
Assuming thus that $\;f: Im\,g\to V\;$ is s.t. $\;f\circ g=Id_V\;$ , then for $\;w_i,w_2\in \;Im\,g\;$, say $\;gv_i=w_i\;$ , we get:
$$f(w_1+w_2)=f(gv_1+gv_2)=f(g(v_1+v_2))=f\circ g(v_1+v_2)\stackrel{f\circ g=Id_V}=v_1+v_2$$
and OTOH:
$$fw_1+fw_2=f(gv_1)+f(gv_2)=f\circ g(v_1)+f\circ g(v_2)\stackrel{f\circ g=Id_V}=v_1+v_2$$
and likewise with scalar multiplication, so we get linearity of $\;f\;$.
But if $\;Im\,g\neq W\;$ (since otherwise $\;g\;$ is already an isomorphism and left and right inverses are thus the same). and we insist in defining $\;f\;$ on the whole $\;W\;$ , then it can easily be disproved that $\;f\;$ has to be linear. Read the comment under your question.