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 $F(V)$.

  • $\begingroup$ Thank you! Could you also recommend me a good book for linear algebra where I could learn how to do such proofs? $\endgroup$ – chandx Apr 3 at 15:00
  • 1
    $\begingroup$ @chandx I have been out of school for a while but if you search this site you will find a few books that are highly recommended. I know I have seen "Linear Algebra Done Right" recommended but I have never even browsed it. $\endgroup$ – John Douma Apr 3 at 15:26
  • 1
    $\begingroup$ @chandx This book was used in my junior-level Linear Algebra class. $\endgroup$ – John Douma Apr 3 at 15:35
  • $\begingroup$ Thanks for recommendation! $\endgroup$ – chandx Apr 3 at 22:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.