# Monomorphism and identity function

I want to show that if $$F: V \to W$$ is monomorphism, then there is linear $$G: W \to V$$ such that $$G \circ F = \mathrm{Id}_V$$. So my try is to define $$G$$ like this: $$G(w) = \begin{cases} F^{-1}(w) \quad &\text{if } w \in F[V] \\ \vec{0}_{\small V} \quad &\text{otherwise} \end{cases}$$ By $$F^{-1} : F[V] \to V$$ I mean inverse of $$F$$. So now $$(G \circ F)(v) = G(F(v)) = G(w) = v$$ since $$w \in F[V]$$. Is this a good solution? If not, could you give me any hints? Only thing I don't know how to do, is to prove $$G$$ is linear, but I think it should be.

Let $$\{v_i\}_{i\in I}$$ be a basis for $$V$$. Then, since $$F$$ is injective, $$V$$ will be isomorphic to $$F(V)$$ where $$\{F(v_i)\}_{i\in I}$$ is basis for $$F(V)$$.
Extend $$\{F(v_i)\}$$ to a basis on $$W$$, say $$\{F(v_i)\}_{i\in I}\cup\{w_j\}_{j\in J}$$.
Now you can define $$G(a_1F(v_1)+,...,+a_mF(v_m)+b_1w_1+,...,+b_nw_n)=a_1v_1+,...,+a_mv_m$$.
You should easily be able to show that this is linear and is the inverse of $$F$$ restricted to $$F(V)$$.