Left and Right inverses of linear operators Let $X$ and $U$ be vector spaces over a field $F$, and let $T : X \to U$.
(a) If there exists an operator $S : U \to X$ such that 
$S(T(x)) =x$ for all $x \in X$, then $S$ is called a left inverse of $T$.
(b) If there exists an operator $S : U \to X$ such that 
$T(S(u)) =u$ for all $u \in U$, then $S$ is called a right inverse of $T$.
I'm trying to prove the following theorem.
Let $X$ and $U$ be vector spaces over a field $F$, and let $T: X\to U$ be linear.
(a) There exists a left inverse $S$ of $T$ iff $T$ is injective.
(b) There exists a right inverse $S$ of $T$ iff $T$ is surjective.
So far my study on linear algebra has been largely restricted to finite dimensional vector spaces but this problem, I think applies to general vector spaces. How can I solve this without resorting to basis?
 A: Here is the proof for (a).
$(\Rightarrow)$ Suppose there exists a left inverse $S$ of $T$. To see that $T$ is injective, let $x,y\in X$ such that $T(x)=T(y)$. Then $$x=S(T(x))=S(T(y))=y$$ Hence $T$ is injective. (Note that this proof requires no linear algebra).
$(\Leftarrow)$ Suppose that $T$ is injective. Let $\{x_i:i\in I\}$ be a basis for $X$. Then $\gamma=\{T(x_i):i\in I\}$ is a linearly independent subset of $U$ (check this!) so there exists a basis $\beta$ of $U$ such that $\gamma\subset\beta$. Now, let $S:U\to X$ be the linear map defined on $\beta$ by 
$$
S(u)=
\begin{cases}
v & \text{if } u\in\gamma\text{ with }u=T(v) \\
0 & \text{if } u\notin\gamma
\end{cases}
$$
Then for $x=\sum\lambda_i x_i\in X$ we have
$$
ST(x)=ST\left(\sum\lambda_i x_i\right)=\sum\lambda_iST(x_i)=\sum\lambda_i x_i=x
$$
so that $S$ is a left inverse of $T$. $\Box$
The key result here is that if $\gamma$ is a linearly independent subset of a vector space $V$, then there exists a basis $\beta$ of $V$ such that $\gamma\subseteq\beta$. This is a standard result covered in most linear algebra courses.
The proof for (b) is very similar.
A: Hint: If $\exists$ left inverse:
$$x_1\ne x_2\implies S(T(x_1))\ne S(T(x_2))\implies T(x_1)\ne T(x_2)$$
(Why each step?)
A: Proving that if $T$ has a left inverse, then $f$ is injective actually has very little to do with linearity. It is an immediate set-theoretic consequence. Namely, if $T(x)=T(y)$, what happens when you apply $S$ on the left?
Similarly, if $T$ has a right inverse $S$, then $T$ is surjective. Indeed, if $y$ is an arbitrary element in the codomain of $T$, what can you say about $S(x)$?
The implications in the other directions are also valid set-theoretically and are not hard to prove. That is, for any function $f:A\to B$ if $f$ is injective, then there exists a left inverse $g:B\to A$. You will then need to prove that if there is also a linear structure present on the domain and codomain, and $f$ preserves it, then so does $g$. Similarly for surjectivity. 
