Let $V$ be a vector space and $U$ be a subspace of $V$. There is a canonical linear map $\pi : V → V/U$ which is defined by $\pi(v) = v + U$. Suppose that $W$ is another vector space and $T: V → W$ is a linear map such that $U ⊂ \ker(T)$. Prove that there exists a unique linear map $T′ : V/U → W$ such that $T′\circπ = T$.
I'm not sure where to get started with this proof. Specifically speaking, what does it mean by canonical and how to prove something is well-defined.
Any help would be greatly appreciated. Thanks in advance.
Here, “canonical” means that it is the standard map from $V$ onto $V/U$, the natural thing to think of as a map from $V$ onto $V/U$ in this context.
Now, you can define $T'\colon V/U\longrightarrow W$ by $T'(v+U)=T(v)$. Problem: does this even make sense? That is, suppose that $v+U=v'+U$; does it follow that $T(v)=T(v')$? Yes, it does, because\begin{align}v+U=v'+U&\iff v-v'\in U\\&\implies v-v'\in\ker T\\&\iff T(v-v')=0\\&\iff T(v)=T(v').\end{align}So, $T'$ is “well-defined”, in the sense that any element of $V/U$ has one and only one image.