Proof of equivalence theorem about left invertible matrices I am taking a course in Matrix Theory and we have a theorem that states (among other things) that:
The following conditions on the matrix $A$ of size $m \times n$ are equivalent:
(1) A has left inverse 
(2) The system $Ax=b$ has at most one solution for any column vector $b$. 
...
The proof that (1) $\implies$ (2) goes like this:
If $Ax=b$ and $V$ is a left inverse, then $VAx=Vb \implies x=Vb$, so we have at most one solution (if any).
The thing is, left inverses are not unique right? Take 
$A =
\left( 
\begin{matrix}
1 \\ 0
\end{matrix}
\right)$
That has left inverses
$V_1=
\left(
\begin{matrix}
1 & 0
\end{matrix}
\right)
$
and
$ V_2 =
\left( 
\begin{matrix}
1 & 1
\end{matrix}
\right)$
Does this mean that the proof is wrong or am I missing something?

Adding an intermediate step could help.
The matrix $A$ has a left inverse iff $A$ is injective iff the system $Ax=b$ has at most one solution for every $b$.
