Suppose $T \in \mathcal{L}(V)$. Prove that $T / \text{range} \ T = 0$. Suppose $T \in \mathcal{L}(V)$. Prove that $T / \text{range} \ T = 0$. 
I encountered this exercise in Sheldon Axler's Linear Algebra Done Right in the chapter on eigenvalues and invariant subspaces. I don't understand what  $T / \text{range} \ T$ is... I thought that notation meant the quotient space, but $T$ is a linear operator, but that doesn't make much sense to me. 
My attempt is to suppose that $x \in T / \text{range} \ T$ and show that $x = 0$, but I am stuck after that. 

Let $U$ be a $T$-invariant subspace of $V$, i.e. $T(U) \subset U$. The linear map
$$T/U: V/U \to V/U, v+U \mapsto Tv+U$$ is well defined.
$\operatorname{range}T$ is $T$-invariant, hence we can look at
$$T/\operatorname{range}T: V/\operatorname{range}T \to V/\operatorname{range}T, v+\operatorname{range}T \mapsto Tv+\operatorname{range}T$$
and easily show that is is the zero-map, since $Tv \in \operatorname{range}T$ for all $v \in V$. Thus we have $Tv+\operatorname{range}T=0 \in V/\operatorname{range}T$.
