$T/ (\text{null } T)$ is injective iff $\text{null } T \cap \text{Range } T=\{0\}$ Let $T\in\mathcal{L} (V)$.
I want to prove that $T/ (\text{null } T)$ (a quotient operator) is injective iff  $\text{null } T \cap \text{Range } T=\{0\}$.
I thought of using the property that a linear map is injective iff its null space is $\{0\}$.
In this case , I get $\text{null } (T/\text{null } T) =\{0+\text{null } T\}$
$\iff \{v+\text{null } T:T(v)+\text{null } T=0+\text{null } T\}=\{0+\text{null } T \}$
$\iff \{v+\text{null } T:T(v) \in \text{null } T\}=\{0+\text{null } T \}$, from the definition of affine space.
I'm not sure how to proceed from here. 
Maybe we can get $\{v\in \text{null } T : T(v) \in \text{null } T\}=\{0\}$, from using again the definition of affine space, but I don't know how to justify the RHS being just$\{0\}$.
Any help would be appreciated.
 A: You are somehow not using your hypothesis. 
Assume first that $\text{null } T \cap \text{Range } T=\{0\}$. Then, if $v+\text{null }(T)\in\text{null }(T/\text{null } T)$ this means that 
$$
Tv+\text{null }T=0+\text{null }T.
$$
This implies that $Tv\in\text{null }T$, so $Tv\in\text{null } T \cap \text{Range } T=\{0\}$. That is $Tv=0$, and so $v\in\text{null }T$. In other words, $\text{null }(T/\text{null }T)=\text{null }T$, so $T/\text{null }T$ is injective. 
For the converse, assume that $T/\text{null }T$ is injective. Let $y\in\text{null } T \cap \text{Range } T$. Then $y=Tv$ for some $v$, and 
$$
(T/\text{null }T)(v+\text{null }T)=Tv+\text{null }T=y+\text{null }T=\text{null }T,
$$
since $y\in\text{null }T$. As $T/\text{null }T$ is injective, we get that $v+\text{null }T=\text{null }T$, i.e. $v\in\text{null }T$. But then $y=Tv=0$. 
A: My last intuition was wrong.
The deduction should be the following:
$ \{v+\text{null } T:T(v) \in \text{null } T\}=\{0+\text{null } T \} $
$\iff [ v+\text{null } T=0+\text{null } T \iff T(v) \in \text{null } T]$
$\iff [v\in \text{null } T \iff T(v) \in \text{null } T]$
$\iff [Tv=0\iff Tv \in \text{null } T]\iff [\text{null } T \cap \text{Range } T=\{0\}]$.
We've proved both directions in one go.
