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.

Remember that addition in the quotient space is defined as
$$(a+U) + (b+U) := (a+b) + U$$
This also corresponds to addition of sets $a+U = \{a+u : u \in U\}$ and $b+U = \{b+u : u \in U\}$ obtaining $(a+b)+U = \{a+b+u : u \in U\}$.
Hence the zero element in $V/U$ is $0+U = \{0+u : u \in U\} =U$.
We have
$$v \in \ker \pi \iff \pi(v) = U \iff v+U = U \iff v \in U$$
For the last equivalence, if $v \in U$ then $v+U = \{v+u : u \in U\} = \{v+(u-v) : u \in U \} = U$. Conversely, if $v+U  = U$, then in particular $0 \in U = v+U$ so there exists $u \in U$ such that $0 =v+u$, which implies $v = -u \in U$. 
For the second question, consider:
$$v+U= v'+U \iff U= (v+U) - (v'+U) = (v-v')+U \iff v-v' \in U$$
so $v+U \ne v'+U$ as long as $v-v' \notin U$.
