Quotient space. Understanding and proving things Let $E$ be a vector space and $F$ a subspace of $E$. The quotient space $E/F$ is $\{x+F:x\in E\}.$ Prove that if $E/F$ has dimension $n$, then there exists a linearly independent set $\{l_1,\dots, l_n\}$ in the algebraic dual space $E^{\#}$ such that $F = \bigcap_{i=1}^{n} \ker(l_i)$.
The only thing I can think of is, if $a \in E/F$ then $a = x+f$ with $f \in F$ and $x \in E$. For example, if $a \in F$ then $x=0$.
Since $E/F$ has dimension $n$, there is a Hamel base $\{h_1,\dots,h_n\} \subset E/F$ such that $$a = c_1h_1+\dots+c_nh_n.$$ Also, $\ker(l_i) = \{x \in E: l_i(x) = 0\}$, so if $x \in \bigcap_{i=1}^{n} \ker(l_i)$ the $l_i(x) = 0$ for all $i=1,\dots,n$.
But I don't see how can I relate all of this to get the result. Any hints?
We don't know any big result on this topic. The only thing we have is the very definition of the quotient space.
 A: It's helpful to think of the elements in $\{x + F: x \in E\}$ as equivalence classes mod F, i.e. for $y,y' \in E / F$ we say $ y =_{mod ~ F} y'$ if $y - y' \in F$.
In this context, we can define the application $\pi: E \to E / F$ given by $\pi x = \bar{x}$ where $\bar{x}$ is the set of all $x' \in E$ such that $x - x' \in F$. Observe that $\pi$ is linear, surjective and $\ker(\pi) = F$. This last sentence means that all $y \in F$ are in the zero class: $y \in \bar{0} \in E / F$. On the other hand if $x \in \bar{0}$ then $x = 0  + f$ for some $f \in F$, so the zero class is precisely $F$.
If $E / F$ has dimension $n$, then we can represent any class $\bar{x}$ as a linear combination of $n$ independent classes:
\begin{equation}
\bar{x} = \lambda_1 \bar{e}_1 + ... + \lambda_n \bar{e}_n
\end{equation}
Since the components for each $\bar{x}$ in this basis are unique, we can define the linear functionals $\bar{l}_i : E / F \to \mathbb{K}$, $\bar{l}_i(\bar{x}) = \lambda_i$. It's clear that $\bar{l}_i(\bar{x}) = 0 $ if the i-th component of $\bar{x}$ is zero. Over this construction, we define $l_i : E \to \mathbb{K}$ by $l_i(x) = \bar{l}_i(\pi(x)) = \lambda_i$. In this case we have, $\cap_{i =1,..,n} \ker(l_i) = \{x \in E : l_i(x) = 0 ~~\forall i = 1,...,n\} = \{x \in E: \pi(x) = \bar{0} \} = \{ x \in E : x \in F\} = F$.
