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A linear map $T:V\to W$ between finite-dimensional vector spaces admits a linear left inverse $S:W\to V$ if and only if its kernel of a linear map is $\{0\}$, equivalently if and only if it is injective.

Necessity follows from $Tv=0⟹v=STv=0$. For sufficiency, choose a basis $v_1,\dots ,v_n$ of $V$. Triviality of the kernel of a linear map implies that $T{v}_1,\dots ,T{v}_n$ are linearly independent. By basis extension, they extend to a basis $T{v}_1,\dots ,T{v}_n,w_{n+1},\dots ,w_m$ of $W$. Define $S(T{v}_i)=v_i$ and $S(w_j)=0$, and extend by linearity. Then $ST{v}_i=v_i$ on a basis, proving $ST=I_V$.

The freedom in $S$ lies in its values on a direct-sum complement of the image of a linear map. For example, for $T(x,y)=(x,y,0)$, every linear map $S(u,v,w)=(u+aw,v+bw)$ is a left inverse, with arbitrary scalars $a,b$. A right inverse instead requires $T$ to be a surjective linear map.

True. Choose a basis $u_1,\dots ,u_r$ of $U$ and extend it to a basis $u_1,\dots ,u_r,w_1,\dots ,w_s$ of the finite-dimensional vector space $V$. Put $W=\mathrm{span}(w_1,\dots ,w_s)$. Every basis vector belongs to $U+W$, so $U+W=V$. Linear independence of the complete basis makes a vector belonging to both spans necessarily zero. ThereforeThis constructs a direct-sum complement; it need not be unique.

Assume ${\varphi }^2=0$. ThenSet $W_1=\mathrm{im}\varphi$, choose a direct-sum complement $U$ such that $\ker \varphi =U\oplus W_1$, and choose a complement $W_2$ such that $V=\ker \varphi \oplus W_2$.

The restriction $\varphi {|}_{W_2}:W_2\to W_1$ is injective because $W_2\cap \ker \varphi =0$, and it is surjective because every image $\varphi (v)$ equals $\varphi (w_2)$ after decomposing $v=k+w_2$. It is therefore a isomorphism, so $\dim W_2=\dim W_1$ and $\varphi (W_2)=W_1$. Also $\varphi (U)=\varphi (W_1)=0$. Hencehas all the required properties.