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

• $v+U$ is rather the image of $v$ by $\pi$, instead of the range of $\pi$. If you take $u\in U$, then $\pi(u)=u+U$, but $u+U=\{u+v:\ v\in U\}=U$, since $U$ is a vector subspace. Therefore $\pi(u)=U=0+U$. Likewise if $\pi(v)=U$, then $v+U=U$, which implies that $v=v+0\in U$. May 5, 2019 at 4:32

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}

• If I take $u \in U$, then $\pi(u) = u + U$. How can I see that $\pi(u) = 0$ from that?
– JOHN
May 5, 2019 at 4:56
• @JOHN, $u \in U$. It is closed under addition. Hence $u + U = U.$ May 5, 2019 at 5:48
• so $\pi(u) = U$. But I still cannot see how $\pi(u) = U$ becomes $\pi(u) = 0$
– JOHN
May 5, 2019 at 5:59
• @JOHN $0_{V/U} = U$ May 5, 2019 at 6:00
• After a month, I still a bit confused on why the zero vector of quotient space is U. the line $0_{V/U}=U$ seems like given which I don't understand
– JOHN
May 27, 2019 at 0:16

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 $$null \ \pi = U$$.

The null space of $$\pi$$ is the set of $$v\in V$$ such that $$\pi(v)=0\in V/U$$. Now using what you wrote, convince yourself that the zero element in the vector space $$V/U$$ is $$U$$.

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.

@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, $$\forall u \in U, \pi(u) = 0_{V/U}$$ $$\therefore U\subset \ker \pi$$ $$\therefore \ker \pi = U$$