nyuuzyou/wikis huggingface dataset:
Quotient Ring of Ring with Unity is Ring with Unity

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring with unity whose zero is $0_R$ and whose unity is $1_R$.
Let $J$ be an ideal of $R$.
Let $\struct {R / J, +, \circ}$ be the quotient ring defined by $J$.
Then $\struct {R / J, +, \circ}$ is a ring with unity, and its unity is $1_R + J$.
== Proof ==
Let $\struct {R, +, \circ}$ be a ring with unity.
First, let $J \subsetneq R$.
By :
:$1_R \in J \implies J = R$
So $1_R \notin J$.
Thus $1_R + J \ne J$, so $1_R + J \ne 0_{R/J}$.
Now let $x \in R$.
Thus $R / J$ has a unity, and that unity is $1_R + J$.
Now suppose $J = R$.
Then $1_R + J = J$ and therefore $1_R = 0_R$.
The only ring to have $1_R = 0_R$ is the null ring.
This is appropriate, because:
:$R / J = R / R = \set {0_{R / R} }$
which is the null ring.
== Also see ==
* Ring Without Unity may have Quotient Ring with Unity
== Sources ==
* : $\S 60.2$ Factor rings: $\text{(iv)}$
* 
* 
Category:Quotient Rings
Category:Rings with Unity
Quotient Rule

* Quotient Rule for Derivatives
** Quotient Rule for Complex Derivatives
* Quotient Rule for Differentials
* Quotient Rule for Sequences
** Quotient Rule for Real Sequences
** Quotient Rule for Complex Sequences
* Quotient Rule for Continuous Functions
** Quotient Rule for Continuous Real Functions
** Quotient Rule for Continuous Complex Functions
* Quotient Rule for Limits of Functions
** Quotient Rule for Limits of Real Functions
** Quotient Rule for Limits of Complex Functions
Quotient Rule for Derivatives

== Theorem ==
Let $\map j x, \map k x$ be real functions defined on the open interval $I$.
Let $\xi \in I$ be a point in $I$ at which both $j$ and $k$ are differentiable.
Define the real function $f$ on $I$ by:
:$\ds \map f x = \begin {cases} \dfrac {\map j x} {\map k x} & : \map k x \ne 0 \\ 0 & : \text {otherwise} \end {cases}$
Then, if $\map k \xi \ne 0$, $f$ is differentiable at $\xi$, and furthermore:
:$\map {f'} \xi = \dfrac {\map {j'} \xi \map k \xi - \map j \xi \map {k'} \xi} {\paren {\map k \xi}^2}$
It follows from the definition of derivative that if $j$ and $k$ are both differentiable on the interval $I$, then:
:$\ds \forall x \in I: \map k x \ne 0 \implies \map {f'} x = \frac {\map {j'} x \map k x - \map j x \map {k'} x} {\paren {\map k x}^2}$
== Proof ==
== Examples ==
== Sources ==
* : $\text {II}$. Calculus: Differentiation
* : $3$: Elementary Analytic Methods: $3.3$ Rules for Differentiation and Integration: Derivatives: $3.3.4$
* : $\S 13$: General Rules of Differentiation: $13.9$
* : $5$. Differential Calculus: Appendix: Differentiation Rules: $4.$ Quotient rule
* 
* 
* : Appendix: Table $1$: Derivatives
* 
* : Appendix: Table $1$: Derivatives
* 
* 
Category:Quotient Rule for Derivatives
Category:Differential Calculus
Quotient Rule for Derivatives/Examples

== Examples of Use of Quotient Rule for Derivatives ==
=== Example: $\dfrac x {x + 1}$ ===
=== Example: $\dfrac {x + 1} {x - 1}$ ===
=== Example: $\dfrac {\sin x} x$ ===
=== Example: $\dfrac x {\cos x}$ ===
=== Example: $\dfrac {e^x} x$ ===
=== Example: $\dfrac {\paren {x - 1} \paren {2 x - 1} } {x - 2}$ ===
Category:Examples of Use of Quotient Rule for Derivatives
Quotient Rule for Derivatives/Examples/(x-1)(2x-1) over x-2

== Example of Use of Quotient Rule for Derivatives ==
:$\map {\dfrac \d {\d x} } {\dfrac {\paren {x - 1} \paren {2 x - 1} } {x - 2} } = \dfrac {2 x^2 - 8 x + 5} {\paren {x - 2}^2}$
== Proof ==
== Sources ==
* : $\text {II}$. Calculus: Differentiation: Exercises $\text {IX}$: $28$.
Category:Examples of Use of Quotient Rule for Derivatives
Quotient Rule for Derivatives/Examples/Exponential of x over x

== Example of Use of Quotient Rule for Derivatives ==
:$\map {\dfrac \d {\d x} } {\dfrac {e^x} x} = \dfrac {e^x \paren {x - 1} } {x^2}$
== Proof ==
== Sources ==
* : $\text {II}$. Calculus: Differentiation: Exercises $\text {IX}$: $26$.
Category:Examples of Use of Quotient Rule for Derivatives
Quotient Rule for Derivatives/Examples/Sine of x over x

== Example of Use of Quotient Rule for Derivatives ==
:$\map {\dfrac \d {\d x} } {\dfrac {\sin x} x} = \dfrac {x \cos x - \sin x} {x^2}$
== Proof ==
== Sources ==
* 
* 
Category:Examples of Use of Quotient Rule for Derivatives
Quotient Rule for Derivatives/Examples/x+1 over x-1

== Example of Use of Quotient Rule for Derivatives ==
:$\map {\dfrac \d {\d x} } {\dfrac {x + 1} {x - 1} } = -\dfrac 2 {\paren {x - 1}^2}$
== Proof ==
Category:Examples of Use of Quotient Rule for Derivatives
Quotient Rule for Derivatives/Examples/x over Cosine of x

== Example of Use of Quotient Rule for Derivatives ==
:$\map {\dfrac \d {\d x} } {\dfrac x {\cos x} } = \dfrac {\cos x + x \sin x} {\cos^2 x}$
== Proof ==
== Sources ==
* : $\text {II}$. Calculus: Differentiation: Exercises $\text {IX}$: $25$.
Category:Examples of Use of Quotient Rule for Derivatives
Quotient Rule for Derivatives/Examples/x over x+1

== Example of Use of Quotient Rule for Derivatives ==
:$\map {\dfrac \d {\d x} } {\dfrac x {x + 1} } = \dfrac 1 {\paren {x + 1}^2}$
== Proof ==
== Sources ==
* : $\text {II}$. Calculus: Differentiation: Exercises $\text {IX}$: $6$.
Category:Examples of Use of Quotient Rule for Derivatives
Quotient Rule for Derivatives/Proof

== Theorem ==
== Proof ==
Let $\xi$ be such that $\map k \xi \ne 0$.
From Differentiable Function is Continuous‎, $k$ is continuous at $\xi$.
It follows that there exists an $\epsilon > 0$, such that $\size h < \epsilon \implies \map k {\xi + h} \ne 0$.
So let $\size h < \epsilon$.
Then we have:
Hence:
Thus by:
:continuity of $k$ at $\xi$
:differentiability of $j$ and $k$ at $\xi$
:Combined Sum Rule for Limits of Real Functions:
it is concluded that:
:$\ds \lim_{h \mathop \to 0} \frac {\map f {\xi + h} - \map f \xi} h = \frac 1 {\map k \xi^2} \paren {\map {j'} \xi \map k \xi - \map j \xi \map {k'} \xi}$
From the definition of differentiability, $f$ is differentiable at $\xi$, with stated value.
== Sources ==
* : $\text {II}$. Calculus: Differentiation: Quotient
* : $\S 10.9 \ \text{(iii)}$
Category:Quotient Rule for Derivatives
Quotient Rule for Real Sequences/Corollary

== Corollary to Quotient Rule for Real Sequences == 
Let $\sequence {x_n}$ be a sequence in $\R$.
Let $\sequence {x_n}$ be convergent to the following limit:
:$\ds \lim_{n \mathop \to \infty} x_n = l$
Then:
:$\ds \lim_{n \mathop \to \infty} \frac 1 {x_n} = \frac 1 l$
provided that $l \ne 0$.
== Proof ==
Follows directly from Quotient Rule for Real Sequences, setting
:$\sequence {y_n} := \sequence {x_n}$
and:
:$\sequence {x_n} := \tuple {1, 1, 1, \ldots}$
== Sources ==
* : $1$: Review of some real analysis: $\S 1.2$: Real Sequences: Proposition $1.2.11 \ \text {(c)}$
Category:Combination Theorems for Sequences
Quotient Set Determined by Relation Induced by Partition is That Partition

== Theorem ==
Let $S$ be a set.
Let $\PP$ be a partition of $S$.
Let $\RR$ be the equivalence relation induced by $\PP$.
Then the quotient set $S / \RR$ of $S$ is $\PP$ itself.
== Proof ==
Let $P \subseteq S$ such that $P \in \PP$.
Let $x \in P$.
Then:
Therefore:
:$P = \eqclass x \RR$
and so:
:$P \in S / \RR$
and so:
:$\PP \subseteq S / \RR$
Now let $x \in S$.
As $\PP$ is a partition:
:$\exists P \in \PP: x \in P$
Then by definition of $\RR$:
:$\tuple {x, y} \in \RR \iff y \in \eqclass x \RR$
Therefore:
:$\eqclass x \RR = P$
and so:
:$\eqclass x \RR \in \PP$
That is:
:$\SS / \RR \subseteq \PP$
It follows by definition of set equality that:
:$\SS / \RR = P$
Hence the result.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 10$: Equivalence Relations: Theorem $10.3$
Category:Quotient Sets
Category:Equivalence Relations
Quotient Space of Compact Space is Compact

== Theorem ==
Let $T = \struct {X, \tau}$ be a compact topological space.
Let $\RR \subseteq X \times X$ be an equivalence relation on $X$.
Then, the quotient space:
:$T / \RR$
is compact.
== Proof 1 ==
 
== Proof 2 ==
Category:Compact Spaces
Category:Quotient Spaces
Category:Quotient Space of Compact Space is Compact
Quotient Space of Compact Space is Compact/Proof 1

== Theorem ==
== Proof ==
Let $\UU$ be an open cover of $T / \RR$.
By definition of quotient topology, for every $U \in \UU$:
:$q_\RR^{-1} \sqbrk U \in \tau$
where $q_\RR$ is the quotient mapping induced by $\RR$.
Therefore:
:$\VV = \set {q_\RR^{-1} \sqbrk U : U \in \UU}$
is a set of open sets of $T$.
Let $x \in X$ be arbitrary.
By definition of cover, there is some $U \in \UU$ such that:
:$\eqclass x \RR \in U$
Therefore:
:$x \in q_\RR^{-1} \sqbrk U$
As $x \in X$ was arbitrary, it follows that $\VV$ is a cover of $X$.
As $\VV$ is a cover consisting of open sets, it is by definition an open cover of $T$.
Thus, by definition of compact space, there is some finite subcover $\set {V_i}_{1 \le i \le n} \subseteq \VV$.
But by definition of $\VV$, for every $1 \le i \le n$, there is some open set $U_i \in \UU$ such that:
:$V_i = q_\RR^{-1} \sqbrk U_i$
It remains to show that:
:$\set {U_i}_{1 \le i \le n}$
is a cover of $X / \RR$, the quotient set of $X$ induced by $\RR$.
Let $y \in X / \RR$ be arbitrary.
By definition of quotient set, there is some $x \in X$ for which:
:$y = \eqclass x \RR$
By definition of cover there is some $1 \le i \le n$ such that:
:$x \in V_i$
By definition of preimage:
:$\eqclass x \RR = \map {q_\RR} x \in U_i$
As $y \in X / \RR$ was arbitrary, it follows that:
:$\set {U_i}_{1 \le i \le n} \subseteq \UU$
is a finite subcover of $\UU$.
As $\UU$ was arbitrary, $T / \RR$ is compact by definition.
Quotient Space of Compact Space is Compact/Proof 2

== Theorem ==
== Proof ==
Let $q_\RR$ be the quotient mapping induced by $\RR$.
By the definition of the quotient topology, $q_\RR$ is continuous.
Further, $T / \RR = q_\RR \sqbrk T$. 
From Continuous Image of Compact Space is Compact, $q_\RR \sqbrk T$ is compact. 
Hence $T / \RR$ is compact.
Category:Quotient Space of Compact Space is Compact
Quotient Space of Hausdorff Space is not necessarily Hausdorff

== Theorem ==
Let $T = \struct {S, \tau}$ be a Hausdorff space.
Let $\RR \subseteq S \times S$ be an equivalence relation on $S$.
Let $T_\RR := \struct {S / \RR, \tau_\RR}$ be the quotient space of $S$ by $\RR$.
Then $T_\RR$ is not necessarily also a Hausdorff space.
== Proof ==
Consider the real number line with the Euclidean topology $\struct {\R, \tau}$.
By Real Number Line satisfies all Separation Axioms, $\struct {\R, \tau}$ is a Hausdorff space.
By Quotient Space of Real Line may not be Kolmogorov, there is a relation $\RR$ on $\R$ such that the quotient space $\struct {\R / \RR, \tau_\RR}$ is not a Kolmogorov space.
Thus the theorem holds by Sequence of Implications of Separation Axioms.
 
== Sources ==
* : $4$: The Hausdorff condition: $4.2$: Separation axioms
Category:Hausdorff Spaces
Category:Quotient Spaces
Quotient Space of Real Line may be Indiscrete

== Theorem ==
Let $\struct {\R, \tau}$ denote the real number line with the usual (Euclidean) topology.
Let $\Q$ be the rational numbers.
Let $\mathbb I$ be the irrational numbers.
Then $\set {\Q, \mathbb I}$ is a partition of $\R$.
Let $\sim$ be the equivalence relation induced on $\R$ by $\set {\Q, \mathbb I}$.
Let $T_\sim := \struct {\R / {\sim}, \tau_\sim}
$ be the quotient space of $\R$ by $\sim$.
Then $T_\sim$ is an indiscrete space.
== Proof ==
Let $\phi: \R \to \R/{\sim}$ be the quotient mapping.
Then:
:$\forall x \in \Q: \map \phi x = \Q$
:$\forall x \in \mathbb I: \map \phi x = \mathbb I$
 $\set {\mathbb I} \in \tau_\sim$.
Then by the definition of the quotient topology:
:$\O \subsetneqq \mathbb I = \phi^{-1} \sqbrk {\set {\mathbb I} } \in \tau$
Thus by Rationals are Everywhere Dense in Topological Space of Reals, $\mathbb I$ contains a rational number.
This is a contradiction.
 $\set \Q \in \tau_\sim$.
Then:
:$\O \subsetneqq \Q = \phi^{-1} \sqbrk {\set \Q} \in \tau$
Thus by Irrationals are Everywhere Dense in Topological Space of Reals, $\Q$ contains a irrational number.
This is a contradiction.
As $\R / {\sim}$ has exactly two elements, its only non-empty proper subsets are $\set \Q$ and $\set {\mathbb I}$.
As neither of these sets is $\tau_\sim$-open, $\struct {\R / {\sim}, \tau_\sim}$ is indiscrete.
Category:Quotient Spaces
Category:Real Number Line with Euclidean Topology
Quotient Space of Real Line may be Kolmogorov but not Fréchet

== Theorem ==
Let $T = \struct {\R, \tau}$ denote the real number line with the usual (Euclidean) topology.
Let $\sim$ be an equivalence relation on $\R$.
Let $\struct {\R / {\sim}, \tau_\sim}$ be the quotient space of $\R$ by $\sim$.
Then it is possible for $\struct {\R / {\sim}, \tau_\sim}$ to be a Kolmogorov space but not a Fréchet space.
== Proof by Counterexample ==
Let $\struct {\R, \tau}$ denote the real number line with the usual (Euclidean) topology.
Define an equivalence relation $\sim$ by letting $x \sim y$ either:
:$x = y$
or:
:$x, y \in \Q$
Let $\struct {\R / {\sim}, \tau_\sim}$ be the quotient space of $\R$ by $\sim$.
Then $\struct {\R / {\sim}, \tau_\sim}$ is a Kolmogorov space but not a Fréchet space.
Let $Y = \R / {\sim}$.
Let $\phi: \R \to Y$ be the quotient mapping.
Note that:
:$\map \phi x = \set x$ if $x$ is irrational.
:$\map \phi x = \Q$ if $x$ is rational.
=== Kolmogorov ===
Let $x$ be irrational.
Then:
:$\phi^{-1} \sqbrk {Y \setminus \set x} = \R \setminus \set x$
Thus $Y \setminus \set x$ is open in $Y$.
Let $p, q \in Y$ such that $p \ne q$.
Then $\set p$ or $\set q$ must be a singleton containing an irrational number.
, suppose that $\set p$ is a singleton containing an irrational number.
Then as shown above, $Y \setminus P$ is open in $Y$.
Thus so $p$ and $q$ are distinguishable.
Since this holds for any two points in $Y$, the space is Kolmogorov.
=== Not Fréchet ===
 $\set \Q$ is closed in $Y$.
By Identification Mapping is Continuous, $\phi$ is continuous.
Thus $\phi^{-1} \sqbrk {\set \Q} = \Q$ is closed in $\R$.
But this contradicts the fact that $\Q \subsetneqq \R$ and Rationals are Everywhere Dense in Topological Space of Reals.
Thus the singleton $\set \Q$ is not closed in $Y$.
Hence $\struct {Y, \tau_\sim}$ is not a Fréchet space.
Category:Quotient Spaces
Category:Real Number Line with Euclidean Topology
Category:T0 Spaces
Category:T1 Spaces
Quotient Space of Real Line may not be Kolmogorov

== Theorem ==
Let $\struct {\R, \tau}$ denote the real number line with the usual (Euclidean) topology.
Then there exists an equivalence relation $\sim$ on $\R$ such that the quotient space $\struct {\R / {\sim}, \tau_\sim}$ is not Kolmogorov.
== Proof ==
By Quotient Space of Real Line may be Indiscrete, there is an equivalence relation $\sim$ on $\R$ such that the quotient space $\struct {\R / {\sim}, \tau_\sim}$ has two points and is indiscrete.
It follows directly from the definition of Kolmogorov space that $\struct {\R / {\sim}, \tau_\sim}$ is not a Kolmogorov space.
Category:Quotient Spaces
Category:Real Number Line with Euclidean Topology
Category:T0 Spaces
Quotient Structure is Similar to Structure

== Theorem ==
Let $\RR$ be a congruence relation on a algebraic structure $\struct {G, \circ}$.
Then the quotient structure $\struct {G / \RR, \circ_\RR}$ is a similar structure to $\struct {G, \circ}$.
== Proof ==
=== Quotient Structure of Semigroup is Semigroup ===
=== Quotient Structure of Monoid is Monoid ===
=== Quotient Structure of Group is Group ===
=== Quotient Structure of Abelian Group is Abelian Group ===
Category:Quotient Structures
Quotient Structure is Well-Defined

== Theorem ==
Let $\struct {S, \circ}$ be an algebraic structure.
Let $\RR$ be a congruence relation on $\struct {S, \circ}$.
Let $S / \RR$ be the quotient set of $S$ by $\RR$.
Let $\circ_\RR$ be the operation induced on $S / \RR$ by $\circ$.
Then $\circ_\RR$ is a well-defined operation in the quotient structure $\struct {S / \RR, \circ_\RR}$.
== Proof ==
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 11$: Quotient Structures
* : $\S 1.1$
Category:Quotient Structures
Quotient Structure of Abelian Group is Abelian Group

== Theorem ==
Let $\RR$ be a congruence relation on an abelian group $\struct {G, \circ}$.
Then the quotient structure $\struct {G / \RR, \circ_\RR}$ is an abelian group.
== Proof ==
From Quotient Structure of Group is Group we have that $\struct {G / \RR, \circ_\RR}$ is a group.
Let $\eqclass x \RR, \eqclass y \RR \in S / \RR$.
Hence $\circ_\RR$ is commutative.
Hence $\struct {G / \RR, \circ_\RR}$ is an abelian group.
== Sources ==
* : $\text{I}$: Groups: $\S 1$: Semigroups, Monoids and Groups
Category:Quotient Structures
Category:Abelian Groups
Quotient Structure of Group is Group

== Theorem ==
Let $\RR$ be a congruence relation on a group $\struct {G, \circ}$.
Then the quotient structure $\struct {G / \RR, \circ_\RR}$ is a group.
== Proof ==
From Quotient Structure of Monoid is Monoid $\struct {G / \RR, \circ_\RR}$ is a monoid with $\eqclass e \RR$ as its identity.
Let $\eqclass x \RR \in G / \RR$.
Let $x \in G$ be arbitrary.
Let $-x$ denote the inverse of $x \in G$ under $\circ$.
Consider the equivalence class $\eqclass {-x} \RR \in G / \RR$.
We need to show that:
We have:
Furthermore:
Hence $\eqclass {-x} \RR$ is the inverse of $\eqclass x \RR$.
Hence $\struct {G / \RR, \circ_\RR}$ is a group.
== Sources ==
* : $\text{I}$: Groups: $\S 1$: Semigroups, Monoids and Groups: Theorem $1.5$
Category:Quotient Structures
Category:Group Theory
Quotient Structure 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:
:$\forall x \in S, y \in C: \map f {x, y} = x \circ' y^{-1}$
Then the mapping $g: \paren {S \times C} / \RR_f \to T$ defined by $\map g {\eqclass {x, y} {\RR_f} } = x \circ' y^{-1}$,
where $\paren {S \times C} / \RR_f$ is a quotient structure, is an isomorphism.
== Proof ==
$T$ is commutative, from Inverse Completion is Commutative Semigroup.
The mapping $\map f {x, y} = x \circ' y^{-1}$ is an epimorphism from the cartesian product of $\struct {S, \circ}$ and $\struct {C, \circ \restriction_C}$ onto $\struct {T, \circ'}$.
By Quotient Theorem for Epimorphisms, the proof follows.
Category:Inverse Completions
Category:Quotient Structures
Quotient Structure of Monoid is Monoid

== Theorem ==
Let $\RR$ be a congruence relation on a monoid $\struct {S, \circ}$ with an identity $e$.
Then the quotient structure $\struct {S / \RR, \circ_\RR}$ is a monoid.
== Proof ==
From Quotient Structure of Semigroup is Semigroup $\struct {S / \RR, \circ_\RR}$ is a semigroup.
Let $\eqclass x {\RR} \in S / \RR$.
Consider $\eqclass e \RR$:
Furthermore:
Hence $\eqclass e \RR$ is an identity.
Hence $\struct {S / \RR, \circ_\RR}$ is a monoid.
== Also see==
*Quotient Structure of Group is Group
== Sources ==
* : $\text{I}$: Groups: $\S 1$: Semigroups, Monoids and Groups: Theorem $1.5$
Category:Quotient Structures
Category:Monoids
Quotient Structure of Semigroup is Semigroup

== Theorem ==
Let $\RR$ be a congruence relation on a semigroup $\struct {S, \circ}$.
Then the quotient structure $\struct {S / \RR, \circ_\RR}$ is a semigroup.
== Proof ==
From Quotient Structure is Well-Defined we have that $\circ_\RR$ is closed on $S / \RR$.
Let $\eqclass x \RR, \eqclass y \RR, \eqclass z \RR \in S / \RR$ under $\circ_\RR$ be arbitrary.
We shall prove that $\circ_\RR$ is associative:
Hence $\struct {S / \RR, \circ_\RR}$ is a semigroup.
== Also see ==
* Quotient Structure of Monoid is Monoid
* Quotient Structure of Group is Group
Category:Quotient Structures
Category:Semigroups
Quotient Structure on Group defined by Congruence equals Quotient Group

== Theorem ==
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $\RR$ be a congruence relation for $\circ$.
Let $\struct {G / \RR, \circ_\RR}$ be the quotient structure defined by $\RR$.
Let $N = \eqclass e \RR$ be the normal subgroup induced by $\RR$.
Let $\struct {G / N, \circ_N}$ be the quotient group of $G$ by $N$.
Then $\struct {G / \RR, \circ_\RR}$ is the subgroup $\struct {G / N, \circ_N}$ of the semigroup $\struct {\powerset G, \circ_\PP}$.
== Proof ==
Let $\eqclass x \RR \in G / \RR$.
By Congruence Relation on Group induces Normal Subgroup:
:$\eqclass x \RR = x N$
where $x N$ is the (left) coset of $N$ in $G$.
Similarly, let:
:$y N \in G / N$
Then from Normal Subgroup induced by Congruence Relation defines that Congruence:
:$y N = \eqclass x \RR$
where:
:$\eqclass x \RR$ is the equivalence class of $y$ under $\RR$
:$\RR$ is the equivalence relation defined by $N$.
Hence the result.
== Also see ==
* Congruence Relation induces Normal Subgroup
* Congruence Relation on Group induces Normal Subgroup
* Normal Subgroup induced by Congruence Relation defines that Congruence
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 11$: Quotient Structures: Theorem $11.5$
Category:Normal Subgroups
Category:Congruence Relations
Quotient Structure on Subset Product

== Theorem ==
Let $\struct {S, \circ}$ be an algebraic structure.
Let $\RR$ be a congruence for $\circ$ on $S$.
Then:
:$\forall X, Y \in S / \RR: X \circ_\PP Y \subseteq X \circ_\RR Y$
where:
:$S / \RR$ is the quotient of $S$ by $\RR$
:$\circ_\PP$ is the operation induced on $\powerset S$ by $\circ$
:$\circ_\RR$ is the operation induced on $S / \RR$ by $\circ$
== Proof ==
By definition of subset product:
:$X \circ_\PP Y = \set {x \circ y: x \in X, y \in Y}$
Thus:
:$X \circ_\RR Y = \set {x \circ y: x \in X, y \in Y} \cup \set {x \circ y: x \in \eqclass X \RR, y \in \eqclass Y \RR}$
The result follows from Subset of Union.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 11$: Quotient Structures
Category:Quotient Structures
Category:Subset Products
Quotient Theorem

* Polynomial Factor Theorem
* Quotient-Remainder Theorem, also known as the Division Theorem
* A series of theorems in abstract algebra:
** Quotient Theorem for Epimorphisms
** Quotient Theorem for Group Epimorphisms
** Quotient Theorem for Monomorphisms
** Quotient Theorem for Sets
** Quotient Theorem for Surjections
The terms factor and quotient are frequently interchangeable.
Category:Abstract Algebra
Quotient Theorem for Epimorphisms

== Theorem ==
Let $\struct {S, \oplus}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \oplus} \to \struct {T, *}$ be an epimorphism.
Let $\RR_\phi$ be the equivalence induced by $\phi$.
Let $S / \RR_\phi$ be the quotient of $S$ by $\RR_\phi$.
Let $q_{\RR_\phi}: S \to S / \RR_\phi$ be the quotient mapping induced by $\RR_\phi$.
Let $\struct {S / \RR_\phi, \oplus_{\RR_\phi} }$ be the quotient structure defined by $\RR_\phi$.
Then:
:The induced equivalence $\RR_\phi$ is a congruence relation for $\oplus$
:There exists a unique isomorphism $\psi: \struct {S / \RR_\phi, \oplus_{\RR_\phi} } \to \struct {T, *}$ which satisfies $\psi \circ q_{\RR_\phi} = \phi$
where $\circ$ denotes composition of mappings.
== Proof ==
=== Proof of Congruence Relation ===
=== Proof of Unique Isomorphism ===
== Also known as ==
Some authors call this the Factor Theorem for Epimorphisms.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Theorem $12.5$
Category:Isomorphisms (Abstract Algebra)
Category:Quotient Epimorphisms
Category:Named Theorems
Category:Quotient Theorems