[proofplan] We use only the definitions of the [quotient space](/page/Quotient%20Space) and quotient map. Surjectivity follows because every element of $V/U$ is, by definition, a coset $v+U$ for some $v\in V$, and this coset is $q(v)$. The kernel computation reduces to identifying which vectors map to the zero coset $U$, and the equality $v+U=U$ is equivalent to $v\in U$. [/proofplan] [step:Show every coset is the image of a vector] Let $C\in V/U$. By the definition of the quotient space, there exists $v\in V$ such that \begin{align*} C=v+U. \end{align*} Since the quotient map $q:V\to V/U$ is defined by $q(v)=v+U$, we have \begin{align*} C=q(v). \end{align*} Thus every element of $V/U$ lies in the image of $q$, so $q$ is surjective. [guided] We must prove that $q:V\to V/U$ is surjective, meaning that every element of the codomain $V/U$ is hit by $q$. Let $C\in V/U$ be arbitrary. The quotient space $V/U$ is the set of cosets of $U$ in $V$, so there is some vector $v\in V$ with \begin{align*} C=v+U. \end{align*} The quotient map is precisely the map sending a vector to its coset: \begin{align*} q(v)=v+U. \end{align*} Therefore \begin{align*} C=q(v). \end{align*} Since the arbitrary element $C\in V/U$ has been written as the image of an element $v\in V$, the map $q$ is surjective. [/guided] [/step] [step:Identify the vectors mapped to the zero coset] The zero element of the quotient [vector space](/page/Vector%20Space) $V/U$ is the coset \begin{align*} 0_{V/U}=0_V+U=U. \end{align*} For $v\in V$, the definition of the kernel gives \begin{align*} v\in\ker(q)\iff q(v)=0_{V/U}. \end{align*} Using $q(v)=v+U$ and $0_{V/U}=U$, this becomes \begin{align*} v\in\ker(q)\iff v+U=U. \end{align*} We now show that $v+U=U$ is equivalent to $v\in U$. If $v+U=U$, then $v=v+0_V\in v+U=U$, since $0_V\in U$. Conversely, if $v\in U$, then for every $u\in U$ we have $v+u\in U$ because $U$ is closed under addition, so $v+U\subset U$. Also, for every $u\in U$, we have $u=v+(u-v)$, and $u-v\in U$ because $U$ is a linear subspace, so $U\subset v+U$. Hence $v+U=U$. [/step] [step:Conclude the kernel is exactly the subspace] From the previous step, for every $v\in V$, \begin{align*} v\in\ker(q)\iff v\in U. \end{align*} Therefore the two subsets of $V$ have the same elements, and \begin{align*} \ker(q)=U. \end{align*} Together with the surjectivity proved above, this proves the theorem. [/step]