[proofplan]
The only issue is that a coset has many representatives, so we first prove that changing representatives does not change the proposed sum or scalar multiple. This uses exactly the closure of the subspace $U$ under addition, additive inverses, and scalar multiplication. Once the operations are well-defined, every [vector space](/page/Vector%20Space) axiom follows by applying the corresponding axiom in $V$ and then passing to cosets.
[/proofplan]
[step:Relate equality of cosets to membership in the subspace]
For $v,v'\in V$, we claim that
\begin{align*}
v+U=v'+U \quad \iff \quad v-v'\in U.
\end{align*}
Indeed, if $v+U=v'+U$, then $v\in v+U=v'+U$, so there exists $u\in U$ such that $v=v'+u$. Hence $v-v'=u\in U$.
Conversely, suppose $v-v'\in U$. Let $x\in v+U$. Then there exists $u\in U$ such that $x=v+u$. Since $v-v'\in U$ and $u\in U$, closure of $U$ under addition gives $(v-v')+u\in U$, and therefore
\begin{align*}
x=v'+((v-v')+u)\in v'+U.
\end{align*}
Thus $v+U\subset v'+U$. Interchanging $v$ and $v'$ is valid because $v'-v=-(v-v')\in U$, so the same argument gives $v'+U\subset v+U$. Hence $v+U=v'+U$.
[/step]
[step:Show addition of cosets is independent of representatives]
Let $v,v',w,w'\in V$ satisfy
\begin{align*}
v+U=v'+U
\end{align*}
and
\begin{align*}
w+U=w'+U.
\end{align*}
By the coset criterion just proved, $v-v'\in U$ and $w-w'\in U$. Since $U$ is a linear subspace, it is closed under addition, so
\begin{align*}
(v+w)-(v'+w')=(v-v')+(w-w')\in U.
\end{align*}
Applying the coset criterion again gives
\begin{align*}
(v+w)+U=(v'+w')+U.
\end{align*}
Therefore the formula
\begin{align*}
(v+U)+(w+U)=(v+w)+U
\end{align*}
does not depend on the chosen representatives $v$ and $w$.
[guided]
The operation on cosets is supposed to use representatives: choose $v$ from $v+U$ and $w$ from $w+U$, add them in $V$, and then take the coset of $v+w$. To prove this is well-defined, we must show that choosing different representatives gives the same coset.
Assume $v,v',w,w'\in V$ satisfy
\begin{align*}
v+U=v'+U
\end{align*}
and
\begin{align*}
w+U=w'+U.
\end{align*}
The equality $v+U=v'+U$ means, by the coset criterion, that the difference of the representatives lies in the subspace:
\begin{align*}
v-v'\in U.
\end{align*}
Likewise, $w+U=w'+U$ gives
\begin{align*}
w-w'\in U.
\end{align*}
Now compare the two possible sums. Their difference is
\begin{align*}
(v+w)-(v'+w')=(v-v')+(w-w').
\end{align*}
Both summands on the right lie in $U$, and $U$ is a linear subspace, hence is closed under addition. Therefore
\begin{align*}
(v+w)-(v'+w')\in U.
\end{align*}
Applying the coset criterion once more gives
\begin{align*}
(v+w)+U=(v'+w')+U.
\end{align*}
Thus the coset produced by adding representatives is independent of the representatives chosen, so addition on $V/U$ is well-defined.
[/guided]
[/step]
[step:Show scalar multiplication of cosets is independent of representatives]
Let $a\in k$ and let $v,v'\in V$ satisfy $v+U=v'+U$. By the coset criterion, $v-v'\in U$. Since $U$ is a linear subspace, it is closed under scalar multiplication by elements of $k$, so
\begin{align*}
av-av'=a(v-v')\in U.
\end{align*}
Applying the coset criterion gives
\begin{align*}
av+U=av'+U.
\end{align*}
Therefore the formula
\begin{align*}
a(v+U)=av+U
\end{align*}
does not depend on the chosen representative $v$.
[/step]
[step:Verify the vector space axioms descend from $V$]
Let $0_V\in V$ denote the zero vector of $V$, and let $0_{V/U}:=0_V+U=U$ denote the zero coset. We verify the vector space axioms for arbitrary $a,b\in k$ and arbitrary cosets $v+U,w+U,z+U\in V/U$.
Associativity of addition follows from associativity in $V$:
\begin{align*}
((v+U)+(w+U))+(z+U)=((v+w)+z)+U=(v+(w+z))+U=(v+U)+((w+U)+(z+U)).
\end{align*}
Commutativity of addition follows from commutativity in $V$:
\begin{align*}
(v+U)+(w+U)=(v+w)+U=(w+v)+U=(w+U)+(v+U).
\end{align*}
The zero coset is an additive identity because
\begin{align*}
(v+U)+0_{V/U}=(v+0_V)+U=v+U.
\end{align*}
The additive inverse of $v+U$ is $(-v)+U$, since
\begin{align*}
(v+U)+((-v)+U)=(v-v)+U=0_V+U=0_{V/U}.
\end{align*}
The scalar multiplication axioms also follow from the corresponding axioms in $V$:
\begin{align*}
a((v+U)+(w+U))=a(v+w)+U=(av+aw)+U=a(v+U)+a(w+U).
\end{align*}
Also,
\begin{align*}
(a+b)(v+U)=((a+b)v)+U=(av+bv)+U=a(v+U)+b(v+U).
\end{align*}
Furthermore,
\begin{align*}
a(b(v+U))=a(bv+U)=a(bv)+U=(ab)v+U=(ab)(v+U).
\end{align*}
Finally, if $1_k\in k$ denotes the multiplicative identity of the field, then
\begin{align*}
1_k(v+U)=(1_kv)+U=v+U.
\end{align*}
All operations involved are well-defined by the preceding steps, and all vector space axioms hold. Hence $V/U$ is a vector space over $k$.
[/step]
