Why the null space of quotient map is $U$? I am reading the textbook Linear Algebra Done Right Chapter 3 section E on Products and Quotients of Vectors Spaces.
It tried to prove the dimension of a quotient space is equal to $\text{dim }V/U = \text{dim }V -\text{dim }U$.
Before that it defines the quotient map $\pi$ as follow:

Suppose $U$ is a subspace of $V$. The quotient map $\pi$ is the linear map $\pi:V \to V/U$ defined by
  $$\pi(v) = v+U$$
  for $v \in V$.

I can understand that the range of $\pi$ is $v+U$ which is $V/U$ according to the definition of $V/U$.
But I don't understand why the null space of $\pi$ is $U$. The book said it is due to the proof like this following:

Suppose $U$ is a subspace of $V$ and $v,w \in V$. Then the following are equivalent:
  $$v-w \in U$$
$$v+U=w+U$$
$$(v+U) \cap (w+U) \neq \emptyset$$

 A: The definition of $\ker\pi$ is
\begin{align}
\ker \pi &= \{ x: \pi(x) = 0_{V/U} \}\\
&= \{x : x + U = 0_{V/U}\}\\
&= \{x :x + U = U \}\\
&= U
\end{align}
A: Here's another way of thinking about it. Suppose that $\nu' \in$ $null\ \pi$. Then, we can see that:
$\pi(\nu') = \nu' + U = 0 + U$
*Because $0 + U$ is the additive identity of $V/U$.
And according to the additional proof that you provided, we see that $\nu'- 0 = \nu' \in U$.
So we see that whenever we map $\nu' \in U$ through $\pi$, we will always get $0 + U = U$ because the affine set of $\{\nu' + u : \nu',u \in U\} = \{0 + (\nu' + u)\}$. Since taking any vector from $U$ will map to the null space of $\pi$, we can see now that $null \ \pi = U$.
A: Not the answer but intuition to understand why $U_{0} (= U)$  should be the Nullspace:
The $0$ of any vector space is an element that gives you back what you added it to. In the case of a quotient space, $U$ is only such 'element' as:
\begin{align}
v + U + U_{0} = v + U
\end{align} as proved in Question 15 of Exercise 1.C
Note: Both $U$ and $U_{0}$ are one and the same.
A: @IAmNoOne has already given a good explanation. I want to add some details:
First, note that the additive identity of $V/U$ is $0+U$, which equals U. In other words, $0_{V/U} = U$.
Second, let $\pi(v) = 0_{V/U}$, solve $v$:
$$\because \pi(v) = v + U, v\in V$$
$$\therefore v+U=0_{V/U}=U$$
which can be written as:$$v+U = 0_V+U$$
From 3.85:
$$v-0_V=v\in U$$
$$\therefore \ker \pi\subset U$$
Apparently, $\forall u \in U, \pi(u) = 0_{V/U}$
$$\therefore U\subset \ker \pi$$
$$\therefore \ker \pi = U$$
