nyuuzyou/wikis huggingface dataset:
Quotient Group of Reals by Integers is Circle Group

== Theorem ==
Let $\struct {\Z, +}$ be the additive group of integers.
Let $\struct {\R, +}$ be the additive group of real numbers.
Let $K$ be the circle group.
Then the quotient group of $\struct {\R, +}$ by $\struct {\Z, +}$ is isomorphic to $K$.
== Proof ==
Define $\phi: \R / \Z \to K$ by:
:$\map \phi {x + \Z} = \map \exp {2 \pi i x}$
Then $\phi$ is well-defined.
For, if $x + \Z = y + \Z$, then $y = x + n$ for some $n \in \Z$, and:
:$\map \exp {2 \pi i \paren {x + n} } = \map \exp {2 \pi i x}$
by Complex Exponential Function has Imaginary Period.
Moreover, by Exponential of Sum:
:$\map \exp {x + y + \Z} = \map \exp {2 \pi i \paren {x + y} } = \map \exp {2 \pi i x} \, \map \exp {2 \pi i y}$
meaning $\phi$ is a group homomorphism.
By Euler's Formula:
:$\map \exp {2 \pi i x} = \map \cos {2 \pi i x} + i \, \map \sin {2 \pi i x}$
so that, by Sine and Cosine are Periodic on Reals:
:$\map \phi x = 1$ $x + \Z = 0 + \Z$
Hence, by Kernel is Trivial iff Group Monomorphism, $\phi$ is a monomorphism.
From polar form for complex numbers, it follows that all $z \in \C$ with $\cmod z = 1$ are of the form:
:$z = \map \exp {2 \pi i x}$
for some $x \in \R$.
Hence $\phi$ is also an epimorphism.
Thus $\phi: \R / \Z \to K$ is a group isomorphism.
== Sources ==
* : Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 47 \beta$
* : Chapter $1$: Introductory Concepts: $1$. Basic Building Blocks: $2$. GROUP: Example $3$
* : $\S 52$. The first isomorphism theorem
Category:Integers
Category:Real Numbers
Category:Complex Numbers
Category:Circle Group
Category:Examples of Quotient Groups
Quotient Mapping/Examples/Congruence Modulo 3

== Example of Quotient Mapping ==
Let $x \mathrel \RR y$ be the equivalence relation defined on the integers as congruence modulo $3$:
:$x \mathrel \RR y \iff x \equiv y \pmod 3$
defined as:
:$\forall x, y \in \Z: x \equiv y \pmod 3 \iff \exists k \in \Z: x - y = 3 k$
That is, if their difference $x - y$ is a multiple of $3$.
From Congruence Modulo $3$, the quotient set induced by $\RR$ is:
:$\Z / \RR = \set {\eqclass 0 3, \eqclass 1 3, \eqclass 2 3}$
Hence the quotient mapping $q_\RR: \Z \to \Z / \RR$ is defined as:
:$\forall x \in \Z: \map {q_\RR} x = \eqclass x 3 = \set {x + 3 k: k \in \Z}$
== Sources ==
* : Chapter $\text{I}$: Sets and Functions: Quotient Functions
Category:Examples of Quotient Mappings
Quotient Mapping/Examples/Modulo 2 pi as Angular Measurement

== Example of Quotient Mapping ==
Let $\RR$ denote the congruence relation modulo $2 \pi$ on the real numbers $\R$ defined as:
:$\forall x, y \in \R: \tuple {x, y} \in \RR \iff \text {$x$ and $y$}$ measure the same angle in radians
From Congruence Modulo $2 \pi$ as Angular Measurement, the quotient set induced by $\RR$ is:
:$\R / \RR = \set {\eqclass \theta {2 \pi}: 0 \le \theta < 2 \pi}$
where:
:$\eqclass \theta {2 \pi} = \set {\theta + 2 k \pi: k \in \Z}$
Hence the quotient mapping $q_\RR: \R \to \R / \RR$ is defined as:
:$\forall x \in \R: \map {q_\RR} x = \eqclass x {2 \pi} = \set {x + 2 k \pi: k \in \Z}$
== Sources ==
* : Chapter $\text{I}$: Sets and Functions: Quotient Functions
Category:Examples of Quotient Mappings
Quotient Mapping Induces Homeomorphism between Quotient Space and Image

== Theorem ==
Let $\struct {S_1, \tau_1}$ and $\struct {S_2, \tau_2}$ be topological spaces.
Let $f: S_1 \to S_2$ be a quotient mapping.
Let $\RR_f \subseteq S_1 \times S_1$ be the equivalence on $S_1$ induced by $f$:
:$\tuple {s_1, s_2} \in \RR_f \iff \map f {s_1} = \map f {s_2}$
Let $q_{\RR_f}: S_1 \to S_1 / \RR_f$ be the quotient mapping induced by $\RR_f$.
Let $\tau_{\RR_f}$ be the quotient topology on $S_1 / \RR_f$ by $q_{\RR_f}$.
Then there exists a uniquely defined homeomorphism $\tilde f: S_1 / \RR_f \to S_2$ such that $f = \tilde f \circ q_{\RR_f}$.
The homeomorphism between the quotient space $\struct {S_1 / \RR_f, \tau_{\RR_f} }$ and $\struct {S_2, \tau_2}$ is defined by:
:$\map {\tilde f}{\eqclass s { \RR_f} } = \map f s$
== Proof ==
Quotient Mapping equals Surjective Identification Mapping shows that $\tau_2$ is the identification topology on $S_2$ with respect to $f$ and $\struct { S_1, \tau_1}$.
Identification Topology equals Quotient Topology on Induced Equivalence shows that $\struct {S_1 / \RR_f, \tau_{\RR_f} }$ and $\struct {S_2, \tau_2}$ are homeomorphic.
The proof of Identification Topology equals Quotient Topology on Induced Equivalence shows that a homeomorphism $\tilde f: S_1 / \RR_f \to S_2$ is defined by:
:$\map {\tilde f}{\eqclass s { \RR_f} } = \map f s$
As we have $\map { q_{\RR_f} } s = \eqclass s { \RR_f}$, it follows that $\tilde f \circ q_{\RR_f} = f$.
To show uniqueness, let $g: S_1 / \RR_f \to S_2$ be a mapping such that $g \circ q_{\RR_f} = f$.
For all $t \in S_2$, we can find $s \in S_1$ such that $\map f s = t$, as $f$ is surjective by definition of quotient mapping.
Then:
:$\map { g \circ q_{\RR_f} } s = \map f s = \map { \tilde f \circ q_{\RR_f} } s$
and as $\map {q_{\RR_f} } s$ is uniquely defined, we must have $\map g {\eqclass s { \RR_f} } = \map {\tilde f}{\eqclass s { \RR_f} }$.
As this is true for all $\map g {\eqclass s { \RR_f} } \in S_2$, it follows that $g = \tilde f$.
Category:Quotient Topology
Quotient Mapping Maps Unit Open Ball in Normed Vector Space to Unit Open Ball in Normed Quotient Vector Space

== Theorem ==
Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space.
Let $N$ be a closed linear subspace of $X$.
Let $\struct {X/N, \norm {\, \cdot \,} }$ be the normed quotient vector space associated with the quotient vector space $X/N$.
Let $B_X$ be the unit open ball in $\struct {X, \norm {\, \cdot \,} }$.
Let $B_{X/N}$ be the unit open ball in $\struct {X/N, \norm {\, \cdot \,} }$.
Let $\pi$ be the quotient mapping associated with $X/N$.
Then:
:$\map \pi {B_X} = B_{X/N}$
== Proof ==
From Quotient Mapping is Bounded in Normed Quotient Vector Space, we have: 
:$\norm {\map \pi x}_{X/N} \le \norm x$ 
So if $x \in B_X$, we have $\norm x < 1$ and hence: 
:$\norm {\map \pi x}_{X/N} < 1$
So $\map \pi x \in B_{X/N}$.
So we have: 
:$\map \pi {B_X} \subseteq B_{X/N}$
Conversely, let $\mathbf x \in B_{X/N}$ and pick $x \in X$ such that $\mathbf x = \map \pi x$.
Then we have:
:$\norm {\map \pi x}_{X/N} = \inf_{z \in N} \norm {x - z} < 1$ 
So there exists $z_\ast \in N$ such that: 
:$\norm {x - z_\ast} < 1$
so that: 
:$x - z_\ast \in B_X$
From Kernel of Quotient Mapping and Quotient Mapping is Linear Transformation, we have: 
:$\map \pi {x - z_\ast} = \map \pi x = \mathbf x$
So we have: 
:$\mathbf x \in \map \pi {B_X}$ 
So we have: 
:$B_{X/N} \subseteq \map \pi {B_X}$ 
Category:Quotient Mappings
Category:Normed Quotient Vector Spaces
Quotient Mapping and Continuous Mapping Induces Continuous Mapping

== Theorem ==
Let $T_1 = \struct {S_1, \tau_1}$, $T_2 = \struct {S_2, \tau_2}$, $T_3 = \struct {S_3, \tau_3}$ be topological spaces.
Let $p: S_1 \to S_2$ be a quotient mapping.
Let $g: S_2 \to S_3$ be a mapping such that for all $s_1, s_2 \in S_1$ with $\map p {s_1} = \map p {s_2}$, we have $\map g {s_1} = \map g {s_2}$.
Then $g$ induces a mapping $f: S_2 \to S_3$ such that $f \circ p = g$.
The induced mapping $f$ is a continuous mapping $g$ is a continuous mapping.
=== Corollary ===
::$\begin{xy} \xymatrix@L+2mu@+1em{
 S_1 \ar[r]^*{p}
 \ar[rd]_*{g}
&
 S_2 \ar@{-->}[d]^*{f}
\\ &
 S_3
}\end{xy}$
== Proof ==
For all $t \in S_2$ we can find $s \in S_1$ with $\map p s = t$, as $p$ is surjective.
Define $\map f t := \map g s$.
For all $s' \in S_1$ with $\map p {s'} = t$, we have $\map g s = \map g {s'}$ .
Hence $f$ is well-defined.
It follows that $\map g s = \map {f \circ p} s$.
Suppose $f$ is continuous.
As $p$ is continuous by definition of quotient mapping, Composite of Continuous Mappings is Continuous shows that $g = f \circ p$ is continuous.
Suppose $g$ is continuous.
Let $V \subseteq S_3$ be open in $T_3$.
As $g$ is continuous:
:$g^{-1} \sqbrk V$ is open in $T_1$.
As $g = f \circ p$:
:$g^{-1} \sqbrk V = p^{-1} \sqbrk {f^{-1} \sqbrk V}$
Hence $p^{-1} \sqbrk {f^{-1} \sqbrk V}$ is open in $T_1$. 
As $p$ is a quotient mapping:
:$f^{-1} \sqbrk V$ is open in $T_2$.
As $f^{-1} \sqbrk V$ is open in $T_2$:
:$f$ is continuous.
== Sources ==
* : $2$: Topological Spaces and Continuous Functions: $\S 22$: The Quotient Topology
Category:Quotient Topology
Quotient Mapping and Continuous Mapping Induces Continuous Mapping/Corollary

== Theorem ==
Let $T_1 = \struct {S_1, \tau_1}$, $T_2 = \struct {S_2, \tau_2}$, $T_3 = \struct {S_3, \tau_3}$ be topological spaces.
Let $p: S_1 \to S_2$ be a quotient mapping.
Let $g: S_2 \to S_3$ be a mapping such that for all $s_1 , s_2 \in S_1$ with $\map p {s_1} = \map p {s_2}$, we have $\map g {s_1} = \map g {s_2}$.
Then $g$ induces a mapping $f: S_2 \to S_3$ such that $f \circ p = g$.
The induced mapping $f$ is a quotient mapping, $g$ is a quotient mapping.
::$\begin{xy} \xymatrix@L+2mu@+1em{
 S_1 \ar[r]^*{p}
 \ar[rd]_*{g}
&
 S_2 \ar@{-->}[d]^*{f}
\\ &
 S_3
}\end{xy}$
== Proof ==
Quotient Mapping and Continuous Mapping Induces Continuous Mapping shows that $f$ is well-defined.
Suppose $f$ is a quotient mapping.
Composite of Quotient Mappings in Topology is Quotient Mapping shows that $g = f \circ p$ is a quotient mapping.
Suppose $g$ is a quotient mapping.
It follows that $g$ is surjective.
That is, for all $t \in S_3$ there exists $s \in S_1$ such that $\map g s = t$.
As $t = \map g s = \map { f \circ p } s$, it follows that $f$ is surjective.
By definition of quotient mapping, $g$ is continuous.
Quotient Mapping and Continuous Mapping Induces Continuous Mapping shows that $f$ is continuous.
Let $U \subseteq S_3$ such that $f^{-1} \sqbrk U$ is open in $T_2$.
By definition of quotient mapping, $p$ is continuous.
By definition of continuity, it follows that $p^{-1} \sqbrk {f^{-1} \sqbrk U}$ is open in $T_1$.
As $f \circ p = g$, it follows that $p^{-1} \sqbrk {f^{-1} \sqbrk U} = g^{-1} \sqbrk U$.
By definition of quotient mapping, $U$ is open in $T_3$.
By definition of quotient mapping, $f$ is a quotient mapping.
== Sources ==
* : $2$: Topological Spaces and Continuous Functions: $\S 22$: The Quotient Topology
Category:Quotient Topology
Quotient Mapping equals Surjective Identification Mapping

== Theorem ==
Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.
Let $f: S_1 \to S_2$ be a mapping.
Then $f$ is a quotient mapping, :
:$f$ is surjective, and $\tau_2$ is the identification topology on $S_2$ with respect to $f$ and $T_1$.
== Proof ==
=== Sufficient condition ===
Let $f$ be surjective, and $\tau_2$ be the identification topology on $S_2$ with respect to $f$ and $T_1$.
Identification Mapping is Continuous shows that $f$ is continuous.
Let $U \subseteq S_2$ such that $f^{-1} \sqbrk U$ is open in $T_1$.
By definition of identification topology, it follows that $U$ is open in $T_2$.
Hence $f$ fulfills all the requirements to be a quotient mapping.
=== Necessary condition ===
Suppose $f$ is not surjective, or $\tau_2$ is not the identification topology on $S_2$ with respect to $f$ and $T_1$.
If $f$ is not surjective, then $f$ is by definition not a quotient mapping.
Existence and Uniqueness of Identification Topology shows that the identification topology is uniquely defined.
So if $\tau_2$ is not the identification topology, there exists a subset $U \subseteq S_2$ not open in $T_2$ such that $f^{-1} \sqbrk U$ is open in $T_1$.
Then $f$ does not fulfill the condition to be a quotient mapping.
== Sources ==
* : $2$: Topological Spaces and Continuous Functions: $\S 22$: The Quotient Topology
Category:Quotient Topology
Category:Identification Topology
Quotient Mapping is Bounded in Normed Quotient Vector Space

== Theorem ==
Let $\Bbb F \in \set {\R, \C}$.
Let $X$ be a normed vector space over $\Bbb F$.
Let $N$ be a closed linear subspace of $X$. 
Let $\struct {X/N, \norm {\, \cdot \,}_{X/N} }$ be the normed quotient vector space associated with quotient vector space $X/N$.
Let $\pi : X \to X/N$ be the quotient mapping associated with $X/N$. 
Then $\pi$ is a bounded linear transformation.
== Proof ==
From Quotient Mapping is Linear Transformation:
:$\pi$ is a linear transformation.
Let $x \in X$. 
Note that from the definition of quotient norm, we have: 
:$\ds \norm {\map \pi x}_{X/N} = \inf_{z \in N} \norm {x - z}_X$
Note that since $0 \in N$, we have that: 
:$\norm x_X \in \set {\norm {x - z}_X : z \in N}$ 
so that: 
:$\ds \inf_{z \in N} \norm {x - z}_X \le \norm x$ 
from the definition of the infimum.
So we have: 
:$\norm {\map \pi x}_{X/N} \le \norm x$ 
for each $x \in X$, and so: 
:$\pi$ is a bounded linear transformation.
Category:Quotient Mappings
Category:Normed Quotient Vector Spaces
Category:Bounded Linear Transformations
Quotient Mapping is Coequalizer

== Theorem ==
Let $\mathbf{Set}$ be the category of sets.
Let $S$ be a Set.
Let $\RR \subseteq S \times S$ be an equivalence relation on $S$.
Let $r_1, r_2: \RR \to S$ be the projections corresponding to the inclusion mapping $\RR \hookrightarrow S \times S$.
Let $q: S \to S / \RR$ be the quotient mapping induced by $\RR$.
Then $q$ is a coequalizer of $r_1$ and $r_2$ in $\mathbf{Set}$.
== Proof ==
Let $f: S \to T$ be a mapping as in the following commutative diagram:
$\quad\quad \begin{xy}\xymatrix{
\RR 
\ar[r]<2pt>^*{r_1}
\ar[r]<-2pt>_*{r_2}
& 
S
\ar[r]^*{q}
\ar[rd]_*{f}
&
S / \RR
\ar@{.>}[d]^*{\bar f}
\\
& &
T
}\end{xy}$
This translates to, for $s_1, s_2 \in S$ with $s_1 \RR s_2$:
:$\map {f \circ r_1} {s_1, s_2} = \map {f \circ r_2} {s_1, s_2}$
that is:
:$\map f {s_1} = \map f {s_2}$
The commutativity of the diagram implies that we must define $\bar f: S / \RR \to T$ by:
:$\map {\bar f} {\eqclass {s_1} \RR} = \map f {s_1}$
since $\map q {s_1} = \eqclass {s_1} \RR$.
The above condition precisely states that $\bar f$ is well-defined.
In conclusion, for any $f$ with $f \circ r_1 = f \circ r_2$, there is a unique $\bar f$ making the diagram commute.
That is, $q$ is a coequalizer of $r_1$ and $r_2$.
== Sources ==
* : $\S 3.4$: Example $3.20$
Category:Category of Sets
Category:Quotient Mappings
Quotient Mapping is Injection iff Equality

== Theorem ==
Let $\RR$ be an equivalence relation on $S$.
Then the quotient mapping $q_\RR: S \to S / \RR$ is an injection $\RR$ is the equality relation.
== Proof ==
Let $\eqclass x \RR, \eqclass y \RR \in S / \RR$
=== Sufficient Condition ===
Let $q_\RR: S \to S / \RR$ be an injection.
Then:
That is:
:$\RR$ is the equality relation.
=== Necessary Condition ===
Let $q_\RR: S \to S / \RR$ be a mapping which is specifically not an injection.
Then:
That is:
:$a \ne b$
but:
:$a \mathrel \RR b$
and so $\RR$ is not the equality relation.
From Rule of Transposition it follows that:
:if $\RR$ is the equality relation then $q_\RR$ is an injection.
== Sources ==
* : Chapter $1$: Sets and mappings: $\S 1.4$: Equivalence relations
Category:Quotient Sets
Category:Injections
Quotient Mapping is Linear Transformation

== Theorem ==
Let $K$ be a field.
Let $V$ be a vector space over $K$.
Let $M$ be a subspace of $V$.
Let $V / M$ be the quotient vector space.
Let $Q: V \to V / M$ be the quotient mapping.
Then $Q$ is a linear transformation.
== Proof ==
Let $\lambda, \mu \in K$ and $x, y \in V$.
From Quotient Vector Space is Vector Space, it follows that $V / M$ is a vector space.
Using the definition of vector addition and scalar multiplication for $V/M$, we have: 
== Sources ==
* : Appendix $\text{A}$ Preliminaries: $\S 1.$ Linear Algebra
Category:Linear Transformations
Category:Quotient Vector Spaces
Quotient Mapping is Surjection

== Theorem ==
Let $S$ be a set.
Let $\RR$ be an equivalence relation on $S$.
Then the quotient mapping $q_\RR: S \to S / \RR$ is a surjection.
== Proof ==
Suppose $S$ is empty. 
Then from Null Relation is Mapping iff Domain is Empty Set and Relation on Empty Set is Equivalence, it follows that $\RR$ is the null relation.
So it is vacuously true that the quotient mapping $q_\RR: S \to S / \RR$ is a surjection when $S$ is empty.
Suppose $S$ is non-empty.
From Equivalence Class is not Empty, we have that:
:$\forall \eqclass x \RR \in S / \RR: \exists x \in S: x \in \eqclass x \RR$
and the result follows.
== Sources ==
* : Introduction $\S 3$: Equivalence relations
* : $\S 6$. Indexed families; partitions; equivalence relations
* : Chapter $1$: Theory of Sets: $\S 7$: Relations
* : Chapter $1$: Sets and mappings: $\S 1.4$: Equivalence relations
* : Appendix $\text{A}$ Preliminaries: $\S 1.$ Linear Algebra
Category:Quotient Mappings
Category:Surjections
Quotient Mapping of Inverse Completion

== Theorem ==
Let $\struct {T, \circ'}$ be an inverse completion of a commutative semigroup $\struct {S, \circ}$, where $C$ is the set of cancellable elements of $S$.
Let $f: S \times C: T$ be the mapping defined as:
:$\forall x \in S, y \in C: \map f {x, y} = x \circ' y^{-1}$
Let $\RR_f$ be the equivalence relation induced by $f$.
Then:
:$\tuple {x_1, y_1} \mathrel {\RR_f} \tuple {x_2, y_2} \iff x_1 \circ y_2 = x_2 \circ y_1$
== Proof ==
By the definition of $\RR_f$:
:$\tuple {x_1, y_1} \mathrel {\RR_f} \tuple {x_2, y_2} \iff x_1 \circ' y_1^{-1} = x_2 \circ' y_2^{-1}$
Now:
which leads us to:
:$\tuple {x_1, y_1} \mathrel {\RR_f} \tuple {x_2, y_2} \iff x_1 \circ y_2 = x_2 \circ y_1$
Category:Inverse Completions
Quotient Mapping on Quotient Topological Vector Space is Open Mapping

== Theorem ==
Let $K$ be a topological field.
Let $\struct {X, \tau}$ be a topological vector space over $K$.
Let $N$ be a closed linear subspace of $X$.
Let $\struct {X/N, \tau_N}$ be the quotient topological vector space of $X$ modulo $N$.
Let $\pi : \struct {X, \tau} \to \struct {X/N, \tau_N}$ be the quotient mapping.
Then $\pi$ is an open mapping.
== Proof ==
Let $V \in \tau$.
From Quotient Mapping is Linear Transformation, $\pi$ is a linear transformation.
From Preimage of Image of Linear Transformation, we have:
:$\pi^{-1} \sqbrk {\pi \sqbrk V} = \ker \pi + V$
From Kernel of Quotient Mapping, we have $\ker \pi = N$ and so:
:$\pi^{-1} \sqbrk {\pi \sqbrk V} = N + V$
From Sum of Set and Open Set in Topological Vector Space is Open, $N + V \in \tau$.
So $\pi^{-1} \sqbrk {\pi \sqbrk V} = N + V$ is open in $\struct {X, \tau}$.
From the definition of the quotient topology, it follows that $\pi \sqbrk V$ is open in $\struct {X/N, \tau_N}$.
== Sources ==
* : $1.41$: Theorem
Category:Quotient Topological Vector Spaces
Category:Open Mappings
Quotient Mapping on Structure is Epimorphism

== Theorem ==
Let $\RR$ be a congruence relation on an algebraic structure $\struct {S, \circ}$.
Then the quotient mapping from $\struct {S, \circ}$ to the quotient structure $\struct {S / \RR, \circ_\RR}$ is an epimorphism:
:$q_\RR: \struct {S, \circ} \to \struct {S / \RR, \circ_\RR}: \forall x, y \in S: \map {q_\RR} {x \circ y} = \map {q_\RR} x \circ_\RR \map {q_\RR} y$
== Proof ==
The quotient mapping $q_\RR: S \to S / \RR$ is the canonical surjection from $S$ to $S / \RR$.
Next we show that this is a homomorphism:
Thus the morphism property is shown to hold.
So the quotient mapping $q_\RR: \struct {S, \circ} \to \struct {S / \RR, \circ_\RR}$ has been shown to be a homomorphism which is a surjection, and is thus an epimorphism.
== Also see ==
* Definition:Quotient Epimorphism
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms
* : Chapter $\text{II}$: Groups: A Little Number Theory
Category:Quotient Epimorphisms
Category:Quotient Mappings
Category:Quotient Structures
Category:Epimorphisms (Abstract Algebra)‎
Quotient Metric on Vector Space is Invariant Metric iff Vector Subspace is Closed

== Theorem ==
Let $K$ be a field.
Let $X$ be a vector space over $K$. 
Let $d$ be an invariant metric on $X$.
Let $N$ be a vector subspace of $X$.
Let $X/N$ be the quotient vector space of $X$ modulo $N$.
Let $\pi : X \to X/N$ be the quotient mapping.
Let $d_N$ be the quotient metric on $X/N$ induced by $d$.
Then $d_N$ is an invariant metric $N$ is closed.
== Proof ==
From Quotient Metric on Vector Space is Invariant Pseudometric, $d_N$ is an invariant pseudometric.
It remains to show that holds $N$ is closed.
Note that holds for $x, y \in X$:
:$\map {d_N} {\map \pi x, \map \pi y} = 0$
implies that $\map \pi x = \map \pi y$.
That is, for $x, y \in X$:
:$\ds \inf_{z \mathop \in N} \map d {x - y, z} = 0$
implies that $\map \pi x = \map \pi y$. 
That is, for $x, y \in X$:
:$\map d {x - y, N} = 0$
implies that $\map \pi x = \map \pi y$.
From Quotient Mapping is Linear Transformation and Kernel of Quotient Mapping, $\map \pi x = \map \pi y$ is equivalent to $x - y \in N$.
Setting $y = {\mathbf 0}_X$, we can see that:
:for $x, y \in X$:
::$\map d {x - y, N} = 0$
:implies that $x - y \in N$
implies that:
:for all $x \in X$:
::$\map d {x, N} = 0$
:implies that $x \in N$.
Clearly the converse implication also holds.
From Subset of Metric Space is Closed iff contains all Zero Distance Points, we have that:
:for all $x \in X$:
::$\map d {x, N} = 0$
:implies that $x \in N$
holds $N$ is closed.
Hence holds $N$ is closed.
Category:Quotient Metrics on Vector Spaces
Quotient Metric on Vector Space is Invariant Pseudometric

== Theorem ==
Let $K$ be a field.
Let $X$ be a vector space over $K$. 
Let $d$ be an invariant metric on $X$.
Let $N$ be a vector subspace of $X$.
Let $X/N$ be the quotient vector space of $X$ modulo $N$.
Let $\pi : X \to X/N$ be the quotient mapping.
Let $d_N$ be the quotient metric on $X/N$ induced by $d$.
Then $d_N$ is an invariant pseudometric.
== Proof ==
=== Proof of ===
Let $x, y \in X$.
Then, we have:
:$\ds \map {d_N} {\map \pi x, \map \pi x} = \inf_{z \mathop \in N} \map d {x - x, z} = \inf_{z \mathop \in N} \map d { {\mathbf 0}_X, z}$
Since $N$ is a vector subspace, we have ${\mathbf 0}_X \in N$.
From , we have $\map d { {\mathbf 0}_X, {\mathbf 0}_X} = 0$, and so:
:$\ds \inf_{z \mathop \in N} \map d { {\mathbf 0}_X, z} \le 0$
Since we also have:
:$\ds \inf_{z \mathop \in N} \map d { {\mathbf 0}_X, z} \ge 0$
we obtain:
:$\ds \inf_{z \mathop \in N} \map d { {\mathbf 0}_X, z} = 0$
Hence we have proved for $d_N$.
=== Proof of ===
Let $x, y, z \in X$. 
Applying to $d$, we have:
:$\map d {x, z + n} \le \map d {x, y + n'} + \map d {y + n', z + n}$
for each $n, n' \in N$.
That is, using the translation invariance of $d$:
:$\map d {x - z, n} \le \map d {x - y, n'} + \map d {y - z, n - n'}$
for each $n, n' \in N$.
Taking the infimum over $n \in N$, we have: 
Taking the infimum over $n' \in N$, using Infimum preserves Inequalities and Infimum Plus Constant, we have:
:$\map {d_N} {\map \pi x, \map \pi z} \le \map {d_N} {\map \pi x, \map \pi y} + \map {d_N} {\map \pi y, \map \pi z}$
Hence we have proved for $d_N$.
=== Proof of ===
Let $x, y \in X$. 
Then we have: 
Hence we have proved for $d_N$. 
=== Proof of translation invariance ===
Let $\map \pi x, \map \pi y, \map \pi z \in X/N$ for $x, y, z \in X$.
We have:
So $d_N$ is an invariant pseudometric.
Category:Invariant Pseudometrics on Vector Spaces
Category:Quotient Metrics on Vector Spaces
Quotient Metric on Vector Space is Well-Defined

== Theorem ==
Let $K$ be a field.
Let $X$ be a vector space over $K$. 
Let $d$ be an invariant metric on $X$.
Let $N$ be a vector subspace of $X$.
Let $X/N$ be the quotient vector space of $X$ modulo $N$.
Let $\pi : X \to X/N$ be the quotient mapping. 
Then the mapping $d_N : X/N \times X/N \to \hointr 0 \infty$ defined by:
:$\ds \map {d_N} {\map \pi x, \map \pi y} = \inf_{z \mathop \in N} \map d {x - y, z}$
for each $\map \pi x, \map \pi y \in X/N$, is well-defined. 
== Proof ==
Let $x, y \in X$.
Then $\map d {x - y, z} \ge 0$ for all $z \in N$, and so:
:$\ds \inf_{z \mathop \in N} \map d {x - y, z}$ exists as a real number.
Let $x', y' \in X$ be such that $\map \pi x = \map \pi {x'}$ and $\map \pi y = \map \pi {y'}$.
We now need to show that if $x', y' \in X$ are such that:
:$\map \pi x = \map \pi {x'}$
and:
:$\map \pi y = \map \pi {y'}$
then:
:$\ds \inf_{z \mathop \in N} \map d {x - y, z} = \inf_{z \mathop \in N} \map d {x' - y', z}$
From Quotient Mapping is Linear Transformation, we have $\map \pi {x' - x} = {\mathbf 0}_{X/N}$ and $\map \pi {y' - y} = {\mathbf 0}_{X/N}$. 
From Kernel of Quotient Mapping, we obtain $x' \in x + N$ and $y' \in y + N$.
Then there exists $z_1, z_2 \in N$ such that $x' = x + z_1$ and $y' = y + z_2$. 
Then we have:
as required.
Category:Quotient Metrics on Vector Spaces