Is the following argument correct?
Suppose $W$ is finite-dimensional and $T\in\mathcal{L}(V,W)$. Prove that $T$ is injective if and only if there exists $S\in\mathcal{L}(W,V)$ such that $ST$ is the indentity map on $V$.
Proof. $(\Rightarrow)$. Assume that $T\in\mathcal{L}(V,W)$ is injective, from theorem $\textbf{2.34}$ there exists a subspace $U$ such that $W = \{Tv:v\in V\}\oplus U$. Now define the map $S$ such that $S(Tv+u)=v$.
Prior to continuing with the above argument we show that the above mapping is well defined. Assume that $S(Tv+u)=w_1$ and $S(Tv+u)=w_2$, then by definition of $S$ we have $Tv=w_1$ and $Tv=w_2$ then given the injectivity of $T$ we have $w_1=w_2$.
To prove linearity observe that $$S(c_1(Tv_1+u_1)+c_2(Tv_2+u_2)) = S(T(c_1v_1+c_2v_2)+c_1u_1+c_2u_2)$$ $$ = c_1v_1+c_2v_2$$ and $$S(c_1(Tv_1+u_1))+S(c_2(Tv_2+u_2)) = S(T(c_1v_1)+c_1u_1)+S(T(c_2v_2)+c_2u_2) $$ $$ = c_1v_1+c_2v_2$$
Thus $S\in\mathcal{L}(W,V)$ and given the definition of $S$ it is evident that $ST=I_V$.
$(\Leftarrow)$. Conversely assume that there exists a map $S\in\mathcal{L}(W,V)$ such that $ST=I_V$. Let $v\in V$ and assume that $v\in\operatorname{null}T$ that is $Tv=0$ then $STv=S(0) = 0 = v =I_Vv$ consequently $\operatorname{null}T = \{0\}$ then by theorem $\textbf{3.16}$ it follows that $T$ is injective.
$\blacksquare$
Note:
$\textbf{2.34}$ - Given a finite dimensional vector space $V$ and a subspace $W$ of $V$ there exists a susbspace $U$ of $V$ such that $V = W\oplus U$.
$\textbf{3.16}$ - A Linear map $T\in\mathcal{L}(V,W)$ is injective if and only if $\operatorname{null}T = \{0\}$.