Why the null space of quotient map is $U$

linear algebralinear-transformations

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:
$$v-w \in U$$
$$v+U=w+U$$
$$(v+U) \cap (w+U) \neq \emptyset$$

Best Answer

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}

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