Note that several of these translates can be equal. Let's take a look at an example. Let $W=\{(x,0)\:|\: x\in \mathbb{R}\},v_1=(0,1),v_2=(1,1)$. Then $v_1+W=\{(x,1)\:|\:x\in \mathbb{R}\}=v_2+W$. Also notice that $v_2-v_1=(1,0)\in W$. Thus translating $W$ by $v_1$ or $v_2$ is the same. This motivates the following proposition.
Proposition (i): $v+W=v'+W \iff v-v'\in W$.
Proof: Suppose $v+W=v'+W$. Then $v=v'+w$ for some $w\in W$, implying $v-v'=w\in W$.
Conversely, suppose $v-v'\in W$. Note that if $z\in W$, then $z+W=W$ since $W$ is a subspace. Then $v+W=v'+(v-v')+W=v'+W.\:\Box$
According to the proposition, two elements $v$ and $v'$ of $V$ determine the same element of $V /W$ if and only if $v-v'\in W$. So we can think of $V/W$ as the set of all elements of $V$ under the extra condition that two elements of $V$ are declared to be the same if their difference is in $W$. Thus we can define $V/W$ to be the set of equivalence classes of the equivalence relation $∼$ on $V$ defined by $v \sim v'$ if $v-v'\in W$. Notice how similar this definition of relation is to congruence relation. So we can also say $v \equiv v' \:\text{mod}\:W$ if $v-v' \in W$. As an exercise, try to prove that the equivalence classes of $\equiv \text{mod}\:W$ are precisely the cosets of $W$.
Definition: A quotient set $V/W$, pronounced $V\:\text{mod}\:W$, is the set of all cosets of $W$, defined by $V/W=[v]=\{v+W\:|\:v\in V\}$.
[Informally, a quotient set is what we get when we "divide" a set $A$ by $B\subset A$, wherein we set all elements of $B$ to the identity in $A$. For example, if $A=\mathbb{Z}$ and $B=\{5n\:|\:n\in \mathbb{Z}\}$, then we're making all multiples of $5$ zero, so the quotient is $\{0,1,2,3,4\}$ (the set of all equivalence classes).]
We want to turn $V/W$ into a vector space (quotient space), so we have to define addition and scalar multiplication.
Definition: The sum of two elements $v+W$ and $v'+W$ of $V/W$ is defined by $(v+W)+(v'+W) := (v+v')+W$.
For $a\in \mathbb{F}$, the scalar multiplication of $a$ on $v+W$ is defined by $a\cdot (v+W) := av+W$.
It is necessary to check that these definitions are well defined, that is, these operations should not depend on which element of $v+W$ we choose to represent $v+W$.
Proposition (ii): Suppose $v+W=v'+W$. Then $(v+W)+(v''+W)=(v'+W)+(v''+W)$ for any $v''+W\in V/W$ and $a\cdot (v+W)=a\cdot (v'+W)$ for any $a\in \mathbb{F}$.
Proof:
WLOG it is enough to show that $(v+W)+(v''+W) \subset (v'+W)+(v''+W)$. Let $u\in(v+W)+(v''+W)=(v+v'')+W$. Then there exists $w\in W$ such that $u=(v+v'')+w$. Since $v+W=v'+W$, we have $v-v'\in W$. Denote this element by $w'$. Then $v=v'+w'$. Thus $u=(v+v'')+w=((v'+w')+v'')+w=(v'+v'')+(w'+w)$, showing that $u\in (v'+v'')+W$ since $w'+w\in W$.
We conclude that $(v+W)+(v''+W)=(v'+W)+(v''+W)$.
WLOG it is enough to show that $a\cdot (v+W)\subset a\cdot (v'+W)$. Let $u\in a\cdot (v+W)=av+W$. Then $u=av+w$ for some $w\in W$. Denote by $w'$ the element $v-v'\in W$. Then $u=av+w=a(v'+w')+w=av'+(aw'+w)\in av'+W$ since $aw'+w\in W$ as $W$ is closed under addition and scalar multiplication.
We have $a\cdot (v+W)=a\cdot (v'+W)$
Thus both addition and scalar multiplication on $V/W$ is well-defined. $\Box$
Note that the above depends crucially on the assumption that $W$ is a subspace of $V$; if $W$ were just an arbitrary subset of $V$, addition and scalar multiplication on $V/W$ would not be well-defined. With these operations then, $V/W$ becomes a vector space over $\mathbb{F}$.
It is important to keep in mind that the quotient space is in general not a subspace, nor naturally equivalent to one. But because of the presence of a natural inner product on $\mathbb{R}^n$, there is a natural way to choose a second subspace orthogonal to the given one, and that second subspace serves as a natural isomorphic model of the quotient space by the given subspace. Namely, if $V$ has an inner product, then $V/W$ is isomorphic to $W^\perp=\{v\:|\:\langle v,w \rangle =0\: \forall\: w\in W\}$.
Summary of concepts: sum of subspaces, coset (translate), subspaces, equivalence relation & equivalence class, congruence relation, quotient set
Examples:
The quotient space in $\mathbb{R}^3$ by a line is a two-dimensional vector space that is isomorphic to a plane. It does not necessarily equal a plane in $\mathbb{R}^3$. In particular, the quotient space doesn't need to be a subset of $\mathbb{R}^3$. The quotient space is something very abstract. It's an entirely new vector space.
The quotient space in $\mathbb{R}^3$ by a plane is the set of all planes parallel to the plane given by the equation $z=0$. So any plane with equation $z=k$ is an element of the quotient space, which is isomorphic to $\mathbb{R}$, but note that the quotient space does not necessarily equal a line in $\mathbb{R}^3$.
In the quotient $\mathbb{R}[x]/(x^2+1)$, the polynomial $x^2+1$ is equivalent to $0$. It follows from the fact that $0$ and $x^2+1$ are in the same equivalence class since $x^2+1=0+(x^2+1)$. So we have $x^2=-1$, which means $x=i$ or $x=-i$. Since an arbitrary element of $\mathbb{R}[x]$ looks like $a_0+a_1x+a_2x^2+\cdots+a^nx^n\; a_i\in \mathbb{R}$, in the quotient $\mathbb{R}[x]/(x^2+1)$ the polynomial becomes $f=a_0+a_1 i+a_2 i^2+\cdots+a^n i^n$. Since $i^2=-1$, we have $i^3=-i, i^4=1$. Collecting terms we can write $f$ as $a+bi$ for some $a,b\in \mathbb{R}$. Therefore, we can say $\mathbb{R}[x]/(x^2+1)\cong \mathbb{C}$.
Quotient space $\mathbb{P}_n/\mathbb{R}$
Let $V=\mathbb{P}_n=\{a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0\:|\:a_0,a_1,\cdots,a_n\in \mathbb{R}\}$ and $W=\{a_0\:|\:a_0\in \mathbb{R}\}=\mathbb{R}$.
Let $[f(x)], [g(x)]\in V/W$. Then $[f(x)]=[g(x)] \iff f(x)-g(x) \in W \iff f(x)-g(x)=c$ for $c\in \mathbb{R}$.
Let a differential operator be $D:V \to V$ where $D[f(x)]=\frac{df}{dx}$.
We have $D[c]=\frac{d}{dx}(c)=0$, which means $\text{ker}(D)=[c]=W$, the set of constant polynomials. This operator is not 1-1 because different functions after differentiation can give the same constant polynomial.
Now let's change the domain of the differential operator to $\overline{D}:V/W \to V$ where $\overline{D}[f(x)]=\frac{df}{dx}$. Now $\text{ker}(\overline{D})=[c]=W=[0]+W=0_{V/W}$. So the operator is 1-1 in the quotient space.
The key idea is this: if a mapping is not 1-1 in which different points map to the same point, we can take all those different points to be elements of a common coset. In other words, we are identifying points that are equivalent to one another in some equivalence relation. With the quotient space formed by squashing those points to the same coset, we have a 1-1 mapping. Similarly, we can modify the codomain and turn a mapping into a surjection, which leads to the following proposition.
Proposition (iii) Canonical epimorphism: The function $\pi:V\to V/W$ defined by $\pi(v)=[v]=v+W$ is a linear transformation with $\text{ker}(\pi)=W, \text{im}(\pi)=V/W$.
Proof:
$\pi(u+v)=(u+v)+W=(u+W)+(v+W)=\pi(u)+\pi(v)$
$\pi(av)=(av)+W=a(v+W)=a\pi(v)$
It follows that $\pi$ is a linear transformation.
The image of $\pi$ is $V/W$ by definition.
For the kernel of $\pi$, $v\in \text{ker}(\pi) \iff v+W=0+W \iff v\in W.\:\Box$
Corollory: Let $V$ be a vector space and $W$ be a subspace of $V$. Then $\text{dim}(W)+\text{dim}(V/W)=\text{dim}(V)$.
Proof: Let $\pi:V\to V/W$ be the canonical epimorphism. By the dimension theorem, we have $\text{dim}\:\text{ker}(\pi)+\text{dim}\:\text{im}(\pi)=\text{dim}(V)$. But $\text{ker}(\pi)=W$ and $\text{im}(\pi)=V/W$, so the claim follows. $\Box$
Theorem: Let $W$ be a subspace of $V$ with quotient mapping $T:V\to V/W$. A subspace $U$ is a complement to $W$ iff $T|_U:U\to V/W$ is an isomorphism.
Universal property of the quotient
Let $V$ be a vector space over $\mathbb{F}$ and $W$ be a subspace of $V$. Define $\pi:V\to V/W$ by $v \mapsto [v]$. If $W'$ is a vector space over $\mathbb{F}$ and $f:V\to W'$ is a linear map whose kernel contains $W$, then there exists a unique linear map $g:V/W\to W'$ such that $f=g\circ \pi$. This is summarized by the following commutative diagram:
$V \xrightarrow{\qquad f \qquad} W'\\ \; \pi \searrow \quad \qquad \nearrow g\\ \qquad\;\; V/W$
Proof:
Let $W'$ be a vector space over $\mathbb{F}$, $f:V\to W'$ be a linear map with $W \subset \text{ker}f$. Then define $g:V/W\to W'$ to be the map $[v]\mapsto f(v)$.
$g$ is well-defined because if $[v]\in V/W$ and $v_1,v_2\in V$ are both representatives of $[v]$, then there exists $w\in W$ such that $v_1=v_2+W$.
Hence $f(v_1)=f(v_2+w)=f(v_2)+f(w)=f(v_2)$ since f is linear.
The linearity of $g$ comes from linearity of $f$.
Also, by definition $f=g\circ \pi$.
It remains to prove $g$ is unique.
Suppose $h:V/W\to W'$ is such that $h\circ \pi=f=g\circ \pi$. Then $h([v])=g([v])$ for all $v\in V$. Since $\pi$ is surjective, $h=g.\:\Box$
First isomorphism theorem
Let $V,W$ be vector spaces over $\mathbb{F}$ and $T:V \to W$ a linear map. Then $V/\text{ker}(T) \cong \text{im}(T)$, namely $V/\text{ker}(T)$ is isomorphic to $\text{im}(T)$.
Proof: Consider the mapping $\overline{T}:V/\text{ker}(T) \to W \;\;\; \overline{T}:[v]=v+\text{ker}(T) \mapsto T(v)$.
A commutative diagram would express this as
$V \xrightarrow{\qquad T \qquad} W\\ \; \pi \searrow \quad \qquad \nearrow \overline{T}\\ \qquad V/ \text{ker}(T)$
We can use similar arguments as the previous example to prove that $\overline{f}$ is well-defined and linear.
Now suppose $\overline{T}([v])=0$. Then $T(v)=0$, which implies $v\in \text{ker}(T)$. This means $[v]=0_{V/\text{ker}(T)}$. Then $\overline{T}$ is injective. Hence if we confine the codomain of $\overline{T}$ to $\text{im}(T)$, then $\overline{T}$ becomes a bijection and hence $V/\text{ker}(T) \cong \text{im}(T).\:\Box$
Example:
$T:\mathbb{R}^{n\times n} \to \mathbb{R}^{n\times n}\quad T(A)=A-A^T$
$\text{ker}(T)=\{A\in \mathbb{R}^{n\times n}\:|\:A=A^T\}$ set of symmetric matrices
$\text{im}(T)=\{B=A-A^T\:|\:A\in \mathbb{R}^{n\times n}\}$
$B^T=A^T-A=-B$ set of antisymmetric matrices
$\mathbb{R}^{n\times n}/ \{n\times n\;\text{symmetric matrices}\} \cong \{n\times n\;\text{antisymmetric matrices}\}$
Strong form of the first isomorphism theorem
Let $T:V\to W'$ be a linear transformation of vector spaces over $\mathbb{F}$. Suppose $W \subset \text{ker}(T)$. Then there exists a unique linear transformation $\overline{T}:V/W\to \text{im}(T)$ such that $\overline{T}\circ \pi=T$, where $\pi:V\to V/W$, and $\overline{T}$ is surjective. Also, $\text{ker}(\overline{T})\cong \text{ker}(T)/W$ and $(V/W)/(\text{ker}(T)/W)\cong \text{im}(T)$.
Second isomorphism theorem
Let $U,W$ be subspaces $V$. Then $(U+W)/U \cong W/(U \cap W)$.
Third isomorphism theorem
Let $V$ be a vector space and $U\subset W\subset V$ be subspaces. Then $(V/U)/(W/U)\cong V/W$.
References:
An awesome video on quotient space
Definition of quotient set
Basis for $V/U$
Examples of quotient spaces
Intuition about quotient spaces
Constructing $\mathbb{C}$ from $\mathbb{R}$
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