nyuuzyou/wikis huggingface dataset:
Homeomorphic Topology of Initial Topology is Initial Topology

== Theorem ==
Let $\struct {X_\alpha, \tau_\alpha}, \struct {X_\beta, \tau_\beta}$ be topological spaces.
Let $\ds \family {\struct {Y_i, \tau_i} }_{i \mathop \in I}$ be an indexed family of topological spaces where $I$ is an arbitrary index set.
Let $\ds \family {f_i: X_\beta \to Y_i}_{i \mathop \in I}$ be an indexed family of mappings indexed by $I$.
Let $\tau_\beta$ be the initial topology on $X_\beta$ with respect to $\ds \family {f_i}_{i \mathop \in I}$.
Let $\phi : \struct {X_\alpha, \tau_\alpha} \to \struct {X_\beta, \tau_\beta}$ be a homeomorphism.
Then:
:$\tau_\alpha$ is the initial topology on $X_\alpha$ with respect to $\ds \family {f_i \circ \phi : X_\alpha \to Y_i}_{i \mathop \in I}$
== Proof ==
Let $\SS_\beta = \set {f_i^{-1} \sqbrk U: i \in I, U \in \tau_i}$.
By definition of initial topology:
:$\SS_\beta$ is a sub-basis for $\tau_\beta$
From Inverse of Homeomorphism is Homeomorphism:
:$\phi^{-1}$ is a homeomorphism
From Homeomorphic Image of Sub-Basis is Sub-Basis:
:$\SS_\alpha = \set {\phi^{-1} \sqbrk {f_i^{-1} \sqbrk U} : i \in I, U \in \tau_i}$ is a sub-basis for $\tau_\alpha$
From Preimage of Subset under Composite Mapping:
:$\forall i \in I, U \in \tau_i : \phi^{-1} \sqbrk {f_i^{-1} \sqbrk U} = \paren{f_i \circ \phi}^{-1} \sqbrk U$
Hence:
:$\SS_\alpha = \set {\paren {f_i \circ \phi}^{-1} \sqbrk U : i \in I, U \in \tau_i}$ is a sub-basis for $\tau_\alpha$
By definition, $\tau_\alpha$ is the initial topology on $X_\alpha$ with respect to $\ds \family {f_i \circ \phi}_{i \mathop \in I}$
Category:Homeomorphisms (Topological Spaces)
Category:Initial Topology
Homeomorphism Relation is Equivalence

== Theorem ==
Let $T_1$ and $T_2$ be topological spaces.
Let $T_1 \sim T_2$ denote that $T_1$ and $T_2$ are homeomorphic.
The relation $\sim$ is an equivalence relation.
== Proof ==
Checking in turn each of the criteria for equivalence:
=== Reflexivity ===
Let $T$ be a topological space.
From Identity Mapping is Homeomorphism, the identity mapping $I_T: T \to T$ is a homeomorphism.
So $T \sim T$, and $\sim$ has been shown to be reflexive.
=== Symmetry ===
Let $T_1$ and $T_2$ be topological spaces such that $T_1 \sim T_2$.
By definition, there exists a homeomorphism $f: T_1 \to T_2$.
From Inverse of Homeomorphism is Homeomorphism it follows that $f^{-1}: T_2 \to T_1$ is also a homeomorphism.
So $T_2 \sim T_1$, and $\sim$ has been shown to be symmetric.
=== Transitivity ===
Let $T_1, T_2, T_3$ be topological spaces such that $T_1 \sim T_2$ and $T_2 \sim T_3$.
By definition, there exist homeomorphisms $f: T_1 \to T_2$ and $g: T_2 \to T_3$.
From Composite of Homeomorphisms is Homeomorphism it follows that $g \circ f: T_1 \to T_3$ is also a homeomorphism.
So $T_1 \sim T_3$, and $\sim$ has been shown to be transitive.
$\sim$ has been shown to be reflexive, symmetric and transitive.
Hence by definition it is an equivalence relation.
== Sources ==
* : Chapter $\text {I}$: Topological Spaces: $1$. Open Sets and Closed Sets
* : $3$: Continuity generalized: topological spaces: $3.6$: Homeomorphisms: Definition $3.6.1$
* : $3$: Continuity generalized: topological spaces: Exercise $3.9: 16$
Category:Homeomorphisms (Topological Spaces)
Category:Examples of Equivalence Relations
Homeomorphism between Topological Spaces may not be Unique

== Theorem ==
Let $T_1$ and $T_2$ be topological spaces.
Let $f$ be a homeomorphism from $T_1$ to $T_2$.
Then $f$ may not necessarily be unique.
== Proof ==
Let $\R$ be the real number line with the Euclidean topology.
Let $I_1 := \openint a b$ and $I_2 := \openint c d$ be non-empty open real intervals.
From Open Real Intervals are Homeomorphic, $I_1$ and $I_2$ are homeomorphic.
The example given of a homeomorphism was the real function $f: I_1 \to I_2$ defined as:
:$\forall x \in I_1: \map f x = c + \dfrac {\paren {d - c} \paren {x - a} } {b - a}$
However, note that the following real functions for all $n \in \Z_{>0}$:
are all homeomorphisms from $I_1$ to $I_2$.
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.6$: Homeomorphisms
Category:Homeomorphisms (Topological Spaces)
Homeomorphism iff Image of Closure equals Closure of Image

== Definition ==
Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.
Let $f: T_1 \to T_2$ be a bijection.
Then $f$ is a homeomorphism :
:$\forall H \subseteq T_1: f \sqbrk {H^-} = \paren {f \sqbrk H}^-$
where $H^-$ denotes the closure of $H$.
== Proof ==
From Bijection iff Inverse is Bijection, $f$ is a bijection $f^{-1}$ is a bijection.
From Continuity Defined by Closure we have that $f: T_1 \to T_2$ is continuous :
:$\forall H \subseteq S_1: f \sqbrk {H^-} \subseteq \paren {f \sqbrk H}^-$
Similarly, $f^{-1}: T_2 \to T_1$ is continuous :
:$\forall H \subseteq S_2: f^{-1} \sqbrk {H^-} \subseteq \paren {f^{-1} \sqbrk H}^-$
Thus:
:$\forall H \subseteq S_1: f \sqbrk {\paren {f^{-1} \sqbrk H}^-} \subseteq \paren {f \sqbrk {f^{-1} \sqbrk H} }^-$
That is:
:$\forall H \subseteq S_1: f \sqbrk {\paren {f^{-1} \sqbrk H}^-} \subseteq H^-$
Hence the result.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction: Functions
Category:Set Closures
Category:Homeomorphisms (Topological Spaces)
Homeomorphism may Exist between Non-Comparable Topologies

== Theorem ==
Let $S$ be a set.
Let $T_1 = \struct {S, \tau_1}$ and $T_2 = \struct {S, \tau_2}$ be topological spaces defined on the underlying set $S$.
Let $\tau_1$ and $\tau_2$ be non-comparable.
Then it may possibly be the case that $T_1$ and $T_2$ are homeomorphic.
== Proof ==
A counterexample is demonstrated in Homeomorphic Non-Comparable Particular Point Topologies.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction: Functions
Category:Homeomorphisms (Topological Spaces)
Homeomorphism of Metric Spaces is Equivalence Relation

== Theorem ==
Let $M_1$ and $M_2$ be metric spaces.
Let $M_1 \sim M_2$ denote that $M_1$ and $M_2$ are homeomorphic.
The relation $\sim$ is an equivalence relation.
== Proof ==
Checking in turn each of the criteria for equivalence:
=== Reflexivity ===
Let $M$ be a metric space.
From Identity Mapping on Metric Space is Homeomorphism, the identity mapping $I_M: M \to M$ is a homeomorphism.
So $M \sim M$, and $\sim$ has been shown to be reflexive.
=== Symmetry ===
Let $M_1$ and $M_2$ be topological spaces such that $M_1 \sim M_2$.
By definition, there exists a homeomorphism $f: M_1 \to M_2$.
From Inverse of Homeomorphism between Metric Spaces is Homeomorphism it follows that $f^{-1}: M_2 \to M_1$ is also a homeomorphism.
So $M_2 \sim M_1$, and $\sim$ has been shown to be symmetric.
=== Transitivity ===
Let $M_1, M_2, M_3$ be metric spaces such that $M_1 \sim M_2$ and $M_2 \sim M_3$.
By definition, there exist homeomorphisms $f: M_1 \to M_2$ and $g: M_2 \to M_3$.
From Composite of Homeomorphisms between Metric Spaces is Homeomorphism it follows that $g \circ f: M_1 \to M_3$ is also a homeomorphism.
So $M_1 \sim M_3$, and $\sim$ has been shown to be transitive.
$\sim$ has been shown to be reflexive, symmetric and transitive.
Hence by definition it is an equivalence relation.
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 7$: Subspaces and Equivalence of Metric Spaces
Category:Homeomorphisms (Metric Spaces)
Category:Examples of Equivalence Relations
Homeomorphism yields Topologically Equivalent Metric

== Theorem ==
Let $M_1 = \struct {A, d_1}$ and $M_2 = \struct {A, d_2}$ be metric spaces on the same underlying set $A$.
Let $f: M_1 \to M_2$ be a homeomorphism.
Let $d_3: A^2 \to \R$ be a mapping defined as:
:$\map {d_3} {x, y} := \map {d_2} {\map f x, \map f y}$
Then $d_3$ yields a metric on $M_1$ which is topologically equivalent to $d_1$.
== Proof ==
== Sources ==
* : $2$: Continuity generalized: metric spaces: $2.4$: Equivalent metrics
Category:Topologically Equivalent Metrics
Homogeneity of Space implies Conservation of Linear Momentum

== Theorem ==
Because space is homogeneous in an inertial frame of reference, linear momentum is conserved if the system is closed
== Proof ==
In inertial frames of reference, space is homogeneous. Therefore, working exclusively with inertial frames of reference, in closed systems (those which do not interact in any ways with external agents and are therefore causally disconnected from anything external), we can make a translation of every particle in said system and nothing will change (broadly speaking, this means "no region of space is special or privileged"). In order to prove this, let us examine what effects it would have on the Lagrangian function describing a system of these characteristics if we were to make an inifinitesimally small translation of all the particles in it. For simplicity, the proof shall be carried out with a one-particle system, but it is trivial to extrapolate it to an arbitrary number of them.
Because the system is closed, the Lagrangian of the system (of the particle) does not depend on time, and it is a function exclusively of the particle's position and velocity: $\mathcal{L}=\mathcal{L}(\vec{r}_i, \dot{\vec{r}_i})$. Therefore, an infinitesimal change in the Lagrangian function can be written in cartesian coordinates as follows:
$\delta\mathcal{L}=\displaystyle\sum_i \frac{\partial \mathcal{L}}{\partial x_i}\delta x_i+\displaystyle\sum_i\frac{\partial \mathcal{L}}{\partial \dot x_i}\delta \dot x_i$
Now, taking into account we are just carrying out a (constant) translation of the position vector, this translation does not depend on time, and therefore: $\delta \dot x_i=\frac{d}{dt}\delta x_i=0$. Hence, we are only left with the first term. Since our hypothesis is the Lagrangian remains unchanged under translations in space, we must then impose $\delta \mathcal{L}=0$ and see what conditions must be met in order to guarantee this equality. Let us consider what we have so far:
$\delta \mathcal{L}=\displaystyle\sum_i \frac{\partial \mathcal{L}}{\partial x_i}\delta x_i=0$Since each infinitesimal displacement is independent from the rest, the only way in which this summation can be null is if each partial derivative is null:
$\therefore \displaystyle\sum_i \frac{\partial \mathcal{L}}{\partial x_i}=0$Now, remembering Euler-Lagrange's equations, we can state: $\frac{\partial \mathcal{L}}{\partial x_i}=-\frac{d}{dt}\frac{\partial \mathcal L}{\partial \dot x_i}$, we reach the conclusion that: $\frac{d}{dt}\frac{\partial \mathcal L}{\partial \dot x_i}=0$. Therefore, $\frac{\partial \mathcal L}{\partial \dot x_i}=\text{constant}$, but since $\frac{\partial \mathcal L}{\partial \dot x_i} = \frac{\partial (T+U)}{\partial \dot x_i}$ with $U=U(x_i)$, then we have: $\T=\displaystyle\sum_i \frac{1}{2}m\dot x_i²$, it is clear that: $\frac{\partial \mathcal L}{\partial \dot x_i}=p$, where $p$ is the particle's linear momentum. Therefore, we have:
$p=\text{constant}\blacksquare$
Homogeneous Cartesian Coordinates/Examples/Arbitrary Example 1

== Examples of Homogeneous Cartesian Coordinates ==
Consider the polynomial equation $\map P {x, y}$:
:$2 x^2 + x + 7 = y$
This can be expressed in homogeneous Cartesian coordinates $\map P {X, Y, Z}$ as:
:$2 X^2 + X Z + 7 Z^2 = Y Z$
== Proof ==
We confirm this is a homogeneous equation:
and homogeneity has been confirmed.
== Sources ==
* 
* 
Category:Examples of Homogeneous Cartesian Coordinates
Homogeneous Cartesian Coordinates represent Unique Finite Point

== Theorem ==
Let $\CC$ denote the Cartesian plane.
Let $P$ be an arbitrary point in $\CC$ which is not the point at infinity.
Let $Z \in \R_{\ne 0}$ be fixed.
Then $P$ can be expressed uniquely in homogeneous Cartesian coordinates as:
:$P = \tuple {X, Y, Z}$
== Proof ==
Let $P$ be expressed in (conventional) Cartesian coordinates as $\tuple {x, y}$.
As $Z$ is fixed and non-zero, there exists a unique $X \in \R$ and a unique $Y \in \R$ such that:
Then by definition of homogeneous Cartesian coordinates, $\tuple {X, Y, Z}$ uniquely represents $P$.
== Sources ==
* : Chapter $\text {II}$. The Straight Line: $9$. Parallel lines. Points at infinity
Category:Homogeneous Cartesian Coordinates
Homogeneous Cyclotomic Polynomial is Symmetric

== Theorem ==
Let $n>1$ be a natural number.
Let $\Phi_n(x,y)$ be the $n$th homogeneous cyclotomic polynomial.
Then $\Phi_n(x,y) = \Phi_n(y,x)$, that is, $\Phi_n(x,y)$ is symmetric.
== Proof ==
Category:Cyclotomic Polynomials
Homogeneous Differential Equation/Examples

== Definition ==
This page gathers examples of homogeneous differential equations of the first order:
:$\map M {x, y} + \map N {x, y} \dfrac {\d y} {\d x} = 0$
where both $M$ and $N$ are homogeneous functions of the same degree.
=== $\paren {x + y} \d x = \paren {x - y} \d y$ ===
=== $\paren {x^2 - 2 y^2} \d x + x y \rd y = 0$ ===
=== $x^2 y' - 3 x y - 2 y^2 = 0$ ===
=== $x^2 y' = 3 \paren {x^2 + y^2} \arctan \dfrac y x + x y$ ===
=== $x \sin \dfrac y x \dfrac {\d y} {\d x} = y \sin \dfrac y x + x$ ===
=== $x y' = y + 2 x e^{-y/x}$ ===
Category:Homogeneous Differential Equations
Homogeneous Equation/Examples/Arbitrary Example 2

== Example of Homogeneous Equation ==
The equation:
:$x^2 + y^2 = 3$
is not a homogeneous equation.
We note that $x^2 + y^2$ is a homogeneous expression:
However, by definition of homogeneous equation, it has to be equated to $0$.
We further observe that $k^2 \paren {x^2 + y^2} = 3 k^2$ not $3$, showing that the complete equation as an expression is not homogeneous.
== Sources ==
* 
* 
Category:Examples of Homogeneous Equations
Homogeneous Expression/Examples/Arbitrary Example 1

== Example of Homogeneous Expression ==
The polynomial:
:$\map E {x, y} = x^2 \map \sin {x / y} + y^2 \map \cos {x / y}$
is a homogeneous expression.
This is demonstrated by replacing $x$ by $k x$ and $y$ by $k y$:
== Sources ==
* 
* 
Category:Examples of Homogeneous Expressions
Homogeneous Linear Equations with More Unknowns than Equations

== Theorem ==
Let $\alpha_{ij}$ be elements of a field $F$, where $1 \le i \le m, 1 \le j \le n$.
Let $n > m$.
Then there exist $x_1, x_2, \ldots, x_n \in F$ not all zero, such that:
:$\ds \forall i: 1 \le i \le m: \sum_{j \mathop = 1}^n \alpha_{ij} x_j = 0$
Alternatively, this can be expressed as:
If $n > m$, the following system of homogeneous linear equations:
has at least one solution such that not all of $x_1, \ldots, x_n$ is zero.
== Proof ==
Consider these vectors for $1 \le k \le n$:
:$\mathbf a_k = \tuple {\alpha_{1k}, \alpha_{2k}, \dots, \alpha_{mk}} \in F^m$
Since $n > m$, by Cardinality of Linearly Independent Set is No Greater than Dimension, $\set {\mathbf a_1, \mathbf a_2, \dots, \mathbf a_n}$ is linearly dependent.
By definition of linearly dependent:
:$\ds \exists \set {\lambda_k: 1 \le k \le n} \subseteq F: \sum_{k \mathop = 1}^n \lambda_k \mathbf a_k = \mathbf 0$
where at least one of $\lambda_k$ is not equal to $0$.
The system of homogeneous linear equations above can be written as:
:$\ds \sum_{k \mathop = 1}^n x_k \mathbf a_k = \mathbf 0$
The result follows from taking $x_k = \lambda_k$.
== Sources ==
* : Chapter $\text {I}$: Elementary Definitions: $\S 4$. Vector Spaces
Category:Algebra
Category:Linear Algebra
Homogeneous Polynomial/Examples/Arbitrary Example 1

== Example of Homogeneous Polynomial ==
The polynomial:
:$\map P {x, y} = x^2 + 3 x y + y^2$
is a homogeneous polynomial in which the degree of each term is $2$.
That this is also a homogeneous expression is demonstrated by replacing $x$ by $k x$ and $y$ by $k y$:
== Sources ==
* 
* 
Category:Examples of Homogeneous Polynomials
Homogeneous Quadratic Equation for Straight Lines Parallel to those Passing through Origin

== Theorem ==
Let $\LL_1$ and $\LL_2$ represent $2$ straight lines in the Cartesian plane which are represented by a quadratic equation $E$ in two variables:
:$a x^2 + b y^2 + 2 h x y + 2 g x + 2 f y + c = 0$
Then $\LL_1$ and $\LL_2$ are parallel respectively to the $2$ straight lines through the origin represented by the homogeneous quadratic equation:
:$a x^2 + 2 h x y + b y^2$
== Proof ==
From Characteristic of Quadratic Equation that Represents Two Straight Lines we have the conditions in which $E$ does indeed represent $2$ straight lines.
Let $E$ be written as:
Comparing coefficients of equivalent terms:
Hence:
:$a x^2 + 2 h x y + b y^2 = b \paren {y - \mu_1 x} \paren {y - \mu_2 x}$
From Homogeneous Quadratic Equation represents Two Straight Lines through Origin, it follows that:
:$y = \mu_1 x$
:$y = \mu_2 x$
are two straight lines through the origin represented by the homogeneous quadratic equation:
:$a x^2 + 2 h x y + b y^2$
By definition of slope of a straight line, these have the same slope as the straight lines $\LL_1$ and $\LL_2$:
:$y = \mu_1 x + b_1$
:$y = \mu_2 x + b_2$
which are described by $E$.
== Sources ==
* : Chapter $\text {II}$. The Straight Line: $17$.
Category:Straight Lines
Homogeneous Quadratic Equation representing Coincident Straight Lines

== Theorem ==
Let $E$ be a homogeneous quadratic equation in two variables:
:$E: \quad a x^2 + 2 h x y + b y^2$
Let $h^2 - a b = 0$.
Then 
Then $E$ represents $2$ straight lines in the Cartesian plane which completely coincide:
== Proof ==
From Homogeneous Quadratic Equation represents Two Straight Lines through Origin, $E$ represents $2$ straight lines in the Cartesian plane:
:$y = \dfrac {h \pm \sqrt {h^2 - a b} } b x$
But when $h^2 - a b = 0$, we get:
:$y = \dfrac h b x$
== Sources ==
* : Chapter $\text {II}$. The Straight Line: $15$. The homogeneous equation of second degree
Category:Quadratic Equations
Category:Straight Lines
Homogeneous Quadratic Equation representing Imaginary Straight Lines

== Theorem ==
Let $E$ be a homogeneous quadratic equation in two variables:
:$E: \quad a x^2 + 2 h x y + b y^2$
Let $h^2 - a b < 0$.
Then 
Then $E$ represents a conjugate pair of imaginary straight lines in the Cartesian plane:
== Proof ==
From Homogeneous Quadratic Equation represents Two Straight Lines through Origin, $E$ represents $2$ straight lines in the Cartesian plane:
:$y = \dfrac {h \pm \sqrt {h^2 - a b} } b x$
But when $h^2 - a b$, $\sqrt {h^2 - a b}$ is not defined as a real number.
Instead we have:
:$y = \dfrac {h \pm i \sqrt {h^2 - a b} } b x$
== Sources ==
* : Chapter $\text {II}$. The Straight Line: $15$. The homogeneous equation of second degree
Category:Quadratic Equations
Category:Straight Lines
Homogeneous Quadratic Equation represents Two Straight Lines through Origin

== Theorem ==
Let $E$ be a homogeneous quadratic equation in two variables:
:$E: \quad a x^2 + 2 h x y + b y^2 = 0$
Then $E$ represents $2$ straight lines in the Cartesian plane:
:$y = \dfrac {h \pm \sqrt {h^2 - a b} } b x$
== Proof ==
From Characteristic of Quadratic Equation that Represents Two Straight Lines, $E$ represents $2$ straight lines in the Cartesian plane 
:$a b c + 2 f g h - a f^2 - b g^2 - c h^2 = 0$
where in this case $c = f = g = 0$, giving:
:$a b \times 0 + 2 \times 0 \times 0 \times h - a \times 0^2 - b \times 0^2 - 0 \times h^2 = 0$
The result follows from using the Quadratic Formula on $E$.
Setting $x = 0$ gives $y = 0$, confirming that $\tuple {0, 0}$ is a point on both straight lines.
== Sources ==
* : Chapter $\text {II}$. The Straight Line: $15$. The homogeneous equation of second degree
Category:Quadratic Equations
Category:Straight Lines
Homogeneous System has Zero Vector as Solution/Corollary

== Corollary to Homogeneous System has Zero Vector as Solution ==
Every homogeneous system of linear equations is consistent.
== Proof ==
From Homogeneous System has Zero Vector as Solution, every homogeneous system of linear equations has the zero vector as a solution.
The result follows from the definition of consistency.
Category:Linear Algebra
Category:Null Spaces
Homomorphic Image of Cyclic Group is Cyclic Group

== Theorem ==
Let $G$ be a cyclic group with generator $g$.
Let $H$ be a group.
Let $\phi: G \to H$ be a (group) homomorphism.
Let $\Img G$ denote the homomorphic image of $G$ under $\phi$.
Then $\Img G$ is a cyclic group with generator $\map \phi g$.
That is:
:$\phi \sqbrk {\gen g} = \gen {\map \phi g}$
== Proof ==
Let $y \in \Img G$.
Then $\exists x \in G: y = \map \phi x$.
As $G$ be a cyclic group with generator $g$, $x = g^n$ for some $n \in \Z$.
Thus by Homomorphism of Power of Group Element:
:$y = \paren {\map \phi g}^n$
and so is a power of $\map \phi g$.
As $y$ is arbitrary, it follows that all elements of $\Img G$ are powers of $\map \phi g$.
Hence the result, by definition of cyclic group.
== Sources ==
* : Chapter $7$: Homomorphisms: Exercise $7$
* : Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $4 \ \text{(i)}$
Category:Cyclic Groups
Category:Group Homomorphisms
Homomorphic Image of Group Element is Coset

== Theorem ==
Let $\phi: \struct {G, \circ} \to \struct {H, *}$ be a group homomorphism.
Let $\map \ker \phi$ be the kernel of $\phi$.
Let $h \in H$.
Then $\Preimg h$ is either the empty set or a coset of $\map \ker \phi$.
== Proof ==
There are two possibilities for any $h \in H$.
:$(1): \quad \Preimg h = \O$
:$(2): \quad \Preimg h \ne \O$
If $(1)$, then the first disjunct of the result is satisfied.
Now suppose $(2)$ holds.
Let $e_G$ and $e_H$ be the identity elements of $G$ and $H$ respectively.
Let $K = \map \ker \phi$.
Let $x, y \in G$ such that $\map \phi x = \map \phi y$.
Then:
Thus:
:$\set {x: \map \phi x = \map \phi y}$ is a subset of $y K$.
From Kernel is Normal Subgroup of Domain we have that $y K = K y$.
Now suppose $x \in K y$.
Then, by definition, $x = k y$ for some $k \in K$.
Thus:
A similar process gives that $x \in y K \implies \map \phi x = \map \phi y$.
Hence the result.
== Sources ==
* : $\S 1.10$: Theorem $22 \ (2)$
Category:Group Homomorphisms
Category:Cosets
Homomorphic Image of Quotient Group under Epimorphism

== Theorem ==
Let $G_1$ and $G_2$ be groups whose identities are $e_{G_1}$ and $e_{G_2}$ respectively.
Let $\phi: G_1 \to G_2$ be a group epimorphism.
Let $K := \map \ker \phi$ be the kernel of $\phi$.
Let $N$ be a normal subgroup of $G_1$ such that $K \subseteq N$.
Then:
:$\dfrac {G_1} N \cong \dfrac {G_2} {\map \phi N}$
where $\dfrac {G_1} N$ denotes the quotient group of $G_1$ by $N$.
== Proof ==
From Group Epimorphism Preserves Normal Subgroups, $\map \phi N$ is normal in $G_2$.
Let $N' := \map \phi N$.
From Quotient Group Epimorphism is Epimorphism, we construct the (group) epimorphism $q: G_2 \to \dfrac {G_2} {N'}$.
Now consider the composite mapping $q \circ \phi: G \to \dfrac {G_2} {N'}$ defined as:
:$\forall q \in G: \map {q \circ \phi} g = \map q {\map \phi g}$
From Composite of Group Epimorphisms is Epimorphism, $q \circ \phi$ is an epimorphism.
Let $g \in G$ such that:
:$\map {q \circ \phi} g = e_{G_2 / N'}$
where $e_{G_2 / N'}$ is the identity of $\dfrac {G_2} {N'}$.
Then $\map \phi g \in \map \ker q$.
But then by Group Epimorphism Induces Bijection between Subgroups it must be that $g \in N$.
Hence from the First Isomorphism Theorem, $\dfrac {G_2} {N'} \cong \dfrac {G_1} N$.
== Sources ==
* : $\S 1.10$: Theorem $29$: Corollary
Category:Group Epimorphisms
Category:Quotient Groups