Let $T\in L(V, W)$ be a linear map between two vector spaces $V$ and $W$. Choose a subspace $S$ in the kernel of $T$.
Prove that there is a unique linear map $T': V/S\to W$ such that $T'\circ\pi=T$, where $\pi: V\to V/S$ is the natural quotient map.
And what is the kernel of $T'$?
Nobody has answered yet, maybe it's a quite basic exercise. I will follow the notations in "Wikipedia: Quotient space (linear algebra)".
For each $[v]\in V/S$, we know that there is at least one $w\in V$ such that $\pi(w)=[v]$. We define $T'([v])\equiv T(w)$.
There are many choices of $w$ in defining $T'([v])$. Assume we choose any other $w'\in V$ such that $\pi(w')=[v]$. Since $\pi(w')=\pi(w)=[v]$, we have $\pi(w'-w)=\pi(w')-\pi(w)=[0]$ so that $$ w'-w\in S\subset\ker T \Rightarrow T(w'-w)=T(w')-T(w)=0 \Rightarrow T(w')=T(w) $$ Now it makes sense to define $T'([v])=T(w)$ for any $w\in V$ satisfying $\pi(w)=[v]$.
From now on we will choose $v$ (instead of $w$) for a representative in $V$ whose image by $\pi$ is $[v]$. Then we always have $T(v)=T'\circ\pi(v)=T'([v])$.
From the definition of $T'$, it is trivial that $T'\circ\pi=T$.
It is easy to check that $T'$ is linear since it follows from the linearity of $T$. For $v,w\in V$, $$ T'([v]+[w])=T'([v+w])=T(v+w)=T(v)+T(w)=T'([v])+T'([w]) $$ Similarly, for a constant $c$ and $v\in V$, $$ T'(c[v])=T'([cv])=T(cv)=cT(v)=cT'([v]) $$ 3. $T'$ is unique
Assume that we have another $T''\colon V/S\to W$ satisfying $T''\circ\pi=T$. Since $T'\circ\pi=T$ for all $v\in V$ $$ T''([v])=T''\circ\pi(v)=T(v)=T'\circ\pi(v)=T'([v]) $$ Thus $T''=T'$.
$(\supseteq)$ Trivial. For any $v\in\ker T$, we have $T'([v])=T(v)=0$. Thus $[v]\in(\ker T)/S$ is also included in $\ker T'$.
$(\subseteq)$ For any $[v]\in\ker T'$, we have $T(v)=T'([v])=0$ and so $v\in\ker T \Rightarrow [v]\in(\ker T)/S$.