Null of Quotient Map is the Subspace Consider a subspace $U$ of $V$, where $V$ is a finite-dimensional vector space over a generic field, $F$.
Let the quotient map $\pi$ be the linear map $\pi:V\rightarrow V/U$ such that $\pi(v)=v+U$ for $v\in V$. Note $V/U$ is the quotient space such that $V/U=\{v+U:v\in V\}$.
$\textbf{My Question:}$ 


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*Why is $\operatorname{null}\pi=U$?
By the definition, $\operatorname{null}\pi=\{v\in V:\pi(v)=0\}$. So, then does this mean, for any $w\in U$, $v(w)=w+U=0$? How?

*Why is the representation of an affine subset parallel to $U$ not $\textbf{unique}$? In other words, why is it the case, we have for $v,v'\in V$, $v+U=v'+U$?
Reference:
Axler, Sheldon J. $\textit{Linear Algebra Done Right}$, New York: Springer, 2015.

Since $U$ is a subspace $w+U=U=0+U$ if $w \in U$. [Just verify that each side is contained in the other]. $0+U$ is the zero element of the quotient space. 
You  have already seen why $v+U=v'+U$ can be true with $v \neq v'$: just take any non-zero $v$ on $U$ and take $v'=0$. 
