If $T$ is one-to-one linear transformation, then can we prove that $T$ has a left inverse transformation without AC? Let $V,~W$ be two vector space of a field $F$. Let $T:V\to W$ be a linear transformation. Then does $T$ has a left inverse linear transformation? I'd tried very hard to think of it, but I still can't see the way. And if so, can we prove it without Axiom of Choice?
PS: The problem is that, the left inverse $S$, must be a function such that $S\big\vert_{\text{range}(T)}(y)$ be defined by the value $x\in V$, such that $T(x)=y$. However, how do we define the function value of $S$ outside $\text{range}(T)$, such that $S$ is linear?
 A: This statement is equivalent to the axiom of choice.  Indeed, assuming the axiom of choice, you can take any basis $B$ of $V$, and then extend the linearly independent set $T(B)\subset W$ to a basis $C$ of $W$.  Then define $S_0:C\to V$ by $S_0(c)=b$ if $c=T(b)$ for some $b\in B$ and $S_0(c)=0$ if $c\in C\setminus T(B)$.  Since $C$ is a basis for $W$, this $S_0$ extends uniquely to a linear map $S:W\to V$, and this extension is a left inverse to $T$ since $S(T(b))=b$ for any $b\in B$ and $B$ spans $V$.
Conversely, suppose any linear injection has a linear left inverse.  Then for any vector space $W$ and any subspace $V\subseteq W$, the inclusion map $T:V\to W$ has a left inverse $S:W\to V$.  The kernel of $S$ is then a linear complement of $V$ in $W$: that is, a subspace $U\subseteq W$ such that $U\cap V=0$ and $U+V=W$.  The existence of such a subspace in general is equivalent to the axiom of choice; see Existence of vector space complement and axiom of choice.
A: The space $T(V)$ has a complement $W^\star$ in $W$ (assuming the axiom of choice). So, if $T$ is one-to-one, you define $S:T(V)\longrightarrow V$ as its inverse, which is linear, and then you extend it to $W$ imposing, for instance, that $S|_{W^\star}\equiv0$. Then $S$ is a left-inverse of $T$ which is linear.
