Monomorphism and identity function

Question

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.

Answer 1

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 $F(V)$.