# canonical linear map from quotient space to another vector space

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.
• Does this argument holds for infinite dimensional spaces $V, U$ and $W$ ? there is little typo in the answer: $u \to v'$ in the line before the justification, thanks in advance for the clarification Commented Sep 25, 2020 at 23:05