avewright/proofwiki-math huggingface dataset:
Theorem: Morphism Property Preserves Cancellability

Let:
:$\phi: \struct {S, \circ_1, \circ_2, \ldots, \circ_n} \to \struct {T, *_1, *_2, \ldots, *_n}$
be a mapping from one algebraic structure:
:$\struct {S, \circ_1, \circ_2, \ldots, \circ_n}$
to another:
:$\struct {T, *_1, *_2, \ldots, *_n}$
Let $\circ_k$ have the morphism property under $\phi$ for some operation $\circ_k$ in $\struct {S, \circ_1, \circ_2, \ldots, \circ_n}$.
Then if an element $a \in S$ is either left cancellable or right cancellable under $\circ_k$, then $\map \phi a$ is correspondingly left cancellable or right cancellable under $*_k$.
Thus, the morphism property is seen to preserve cancellability.

Proof:
First let $S$ be the empty set.
It follows from the definition of an Definition:Morphism Property that $\circ_1, \circ_2, \ldots, \circ_n$ are all empty maps.
It also follows from the definition of an Definition:Morphism Property that $*_1, *_2, \ldots, *_n$ are all empty maps.
By Image of Empty Set is Empty Set, $T$ is also the empty set.
{{Finish|Prove that the morphism property preserves cancellability when $S$ is empty}}
Hence, the morphism property is seen to preserve cancellability when $S$ is the empty set.
{{qed|lemma}}
Now let $S \ne \O$.
We need to demonstrate the following properties:
$(1): \quad$ If $a \in S$ has the property that:
:$\forall x, y \in S: x \circ_k a = y \circ_k a \implies x = y$
then:
:$\forall x, y \in S: \map \phi x *_k \map \phi a = \map \phi y *_k \map \phi a \implies \map \phi x = \map \phi y$
Thus:
{{begin-eqn}}
{{eqn | l = x \circ_k a = y \circ_k a
      | o = \implies
      | r = x = y
      | c =
}}
{{eqn | ll= \leadsto
      | l = \map \phi {x \circ_k a} = \map \phi {y \circ_k a}
      | o = \implies
      | r = \map \phi x = \map \phi y
      | c = Mappings are many-to-one by definition
}}
{{eqn | ll= \leadsto
      | l = \map \phi x *_k \map \phi a = \map \phi y *_k \map \phi a
      | o = \implies
      | r = \map \phi x = \map \phi y
      | c = {{Defof|Morphism Property}}
}}
{{end-eqn}}
and thus left cancellability is demonstrated.
{{qed|lemma}}
$(2): \quad$ If $a \in S$ has the property that:
:$\forall x, y \in S: a \circ_k x = a \circ_k y \implies x = y$
then:
:$\forall x, y \in S: \map \phi a *_k \map \phi x = \map \phi a *_k \map \phi y \implies \map \phi x = \map \phi y$
Thus:
{{begin-eqn}}
{{eqn | l = a \circ_k x = a \circ_k y
      | o = \implies
      | r = x = y
      | c =
}}
{{eqn | ll= \leadsto
      | l = \map \phi {a \circ_k x} = \map \phi {a \circ_k y}
      | o = \implies
      | r = \map \phi x = \map \phi y
      | c = Mappings are many-to-one by definition
}}
{{eqn | ll= \leadsto
      | l = \map \phi a *_k \map \phi x = \map \phi a *_k \map \phi y
      | o = \implies
      | r = \map \phi x = \map \phi y
      | c = {{Defof|Morphism Property}}
}}
{{end-eqn}}
and thus right cancellability is demonstrated.
{{qed}}
Category:Morphism Property
cwaq5vx6xw9cbirb4xotx333ll20uol
Theorem: Quotient Mapping on Structure is Epimorphism

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:
{{begin-eqn}}
{{eqn | l = \map {q_\RR} x
      | r = \eqclass x \RR
      | c = {{Defof|Quotient Mapping}}
}}
{{eqn | l = \map {q_\RR} y
      | r = \eqclass y \RR
      | c = {{Defof|Quotient Mapping}}
}}
{{eqn | l = \map {q_\RR} {x \circ y}
      | r = \eqclass {x \circ y} \RR
      | c = {{Defof|Quotient Mapping}}
}}
{{eqn | l = \eqclass {x \circ y} \RR
      | r = \eqclass x \RR \circ_\RR \eqclass y \RR
      | c = {{Defof|Operation Induced on Quotient Set}}
}}
{{eqn | ll= \leadsto
      | l = \map {q_\RR} {x \circ y}
      | r = \map {q_\RR} x \circ_\RR \map {q_\RR} y
      | c = {{Defof|Quotient Mapping}}
}}
{{end-eqn}}
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.
{{Qed}}
Theorem: Restriction of Homomorphism to Image is Epimorphism

Let $S$ and $T$ be algebraic structures.
Let $\phi: S \to T$ be a homomorphism.
Then a surjective restriction of $\phi$ can be produced by limiting the codomain of $\phi$ to its image $\Img \phi$.

Proof:
Let $\phi: S \to T$ be a homomorphism.
Let $\Img \phi = T'$
By Morphism Property Preserves Closure, $T'$ is closed.
From Restriction of Mapping to Image is Surjection, $\phi \to \Img \phi$ is a surjection.
Thus $\phi: S \to T$ is an epimorphism.
Therefore, by suitably restricting the codomain of a homomorphism, it is possible to regard it as an epimorphism.
{{qed}}
Category:Epimorphisms (Abstract Algebra)
mo23jvc86v447eeq61inuu1ckzjhcv8
Theorem: Epimorphism Preserves Semigroups

Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an epimorphism.
Let $\struct {S, \circ}$ be a semigroup.
Then $\struct {T, *}$ is also a semigroup.

Proof:
As $\struct {S, \circ}$ is a semigroup, then by definition it is closed.
As $\phi$ is an epimorphism, it is by definition surjective.
That is:
:$T = \phi \sqbrk S$
where $\phi \sqbrk S$ denotes the image of $S$ under $\phi$.
From Morphism Property Preserves Closure it follows that $\struct {T, *}$ is closed.
As $\struct {S, \circ}$ is a semigroup, then by definition $\circ$ is associative.
From Epimorphism Preserves Associativity, $*$ is therefore also associative.
So:
:$\struct {T, *}$ is closed
and:
:$*$ is associative.
Therefore, by definition, $\struct {T, *}$ is a semigroup.
{{qed}}
Theorem: Homomorphism Preserves Subsemigroups

Let $\struct {S, \circ}$ and $\struct {T, *}$ be semigroups.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be a homomorphism.
Let $S'$ be a subsemigroup of $S$.
Then $\phi \paren {S'}$ is a subsemigroup of $T$.

Proof:
By Restriction of Homomorphism to Image is Epimorphism, $\phi$ is an epimorphism onto its image .
Then by Epimorphism Preserves Semigroups, it follows that the image of $S'$ is a semigroup.
The result follows.
{{Qed}}
Category:Homomorphisms (Abstract Algebra)
Category:Subsemigroups
rcr8g5my8td1oh6ytpnzeo82lt644i6
Theorem: Epimorphism Preserves Commutativity

Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an epimorphism.
Let $\circ$ be a commutative operation.
Then $*$ is also a commutative operation.

Proof:
{{Proofread| Check validity of proof for the case of $S$ being the empty set}}
Let $\struct {S, \circ}$ be an algebraic structure in which $\circ$ is commutative.
Let $\struct {S, \circ}$ be an algebraic structure in which $\circ$ is associative.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an epimorphism.
Suppose $S$ is the empty set.
It follows from the definition of an epimorphism that:
$\phi$ is a surjective homomorphism
By Empty Mapping to Empty Set is Bijective, the empty map is bijective
By definition of bijection, the empty map is an epimorphism.
Therefore, suppose $\phi$ is the empty map, which is indeed an epimorphism.
By definition of a homomorphism, $\phi$ can be defined as:
:$\forall \O \in S: \map \phi {\O \circ \O} = \map \phi \O * \map \phi \O$
By Image of Empty Set is Empty Set, $T$ is also the empty set.
It follows from the definition of the homomorphism that the binary operations $\circ$ and $*$ are also the empty map.
Hence, it is vacuously true that $\circ$ is commutative on $S$, when $S$ is empty, as required.
Let $\struct {S, \circ}$ be an algebraic structure in which $\circ$ is commutative.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an epimorphism.
Suppose $S$ is non-empty.
As an epimorphism is surjective, it follows that:
:$\forall u, v \in T: \exists x, y \in S: \map \phi x = u, \map \phi y = v$
So:
{{begin-eqn}}
{{eqn | l = u * v
      | r = \map \phi x * \map \phi y
      | c = {{Defof|Surjection}}
}}
{{eqn | r = \map \phi {x \circ y}
      | c = {{Defof|Morphism Property}}
}}
{{eqn | r = \map \phi {y \circ x}
      | c = {{Defof|Commutative Operation}}
}}
{{eqn | r = \map \phi y * \map \phi x
      | c = {{Defof|Morphism Property}}
}}
{{eqn | r = v * u
      | c = by definition as above
}}
{{end-eqn}}
{{qed}}
Theorem: Epimorphism Preserves Identity

Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an epimorphism.
Let $\struct {S, \circ}$ have an identity element $e_S$.
Then $\struct {T, *}$ has the identity element $\map \phi {e_S}$.

Proof:
Let $\struct {S, \circ}$ be an algebraic structure in which $\circ$ has an identity $e_S$.
Then:
:$\forall x \in S: x \circ e_S = x = e_S \circ x$
Thus:
:$\forall x \in S: x \circ e_S, e_S \circ x \in \Dom \phi$
Hence:
{{begin-eqn}}
{{eqn | l = \map \phi x
      | r = \map \phi {x \circ e_S}
      | c = {{Defof|Identity Element}}
}}
{{eqn | r = \map \phi x * \map \phi {e_S}
      | c = {{Defof|Morphism Property}}
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = \map \phi x
      | r = \map \phi {e_S \circ x}
      | c = {{Defof|Identity Element}}
}}
{{eqn | r = \map \phi {e_S} * \map \phi x
      | c = {{Defof|Morphism Property}}
}}
{{end-eqn}}
The result follows because every element $y \in T$ is of the form $\map \phi x$ with $x \in S$.
{{qed}}
Theorem: Epimorphism Preserves Inverses

Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an epimorphism.
Let $\struct {S, \circ}$ have an identity $e_S$.
Let $x^{-1}$ be an inverse element of $x$ for $\circ$.
Then $\map \phi {x^{-1} }$ is an inverse element of $\map \phi x$ for $*$.
That is:
:$\map \phi {x^{-1} } = \paren {\map \phi x}^{-1}$

Proof:
Let $\struct {S, \circ}$ be an algebraic structure in which $\circ$ has an identity element $e_S$.
From Epimorphism Preserves Identity, it follows that $\struct {T, *}$ also has an identity element, which is $\map \phi {e_S}$.
Let $y$ be an inverse of $x$ in $\struct {S, \circ}$. 
By definition of inverse element:
:$x \circ y = e_S = y \circ x$
Then:
{{begin-eqn}}
{{eqn | l = \map \phi x * \map \phi y
      | r = \map \phi {x \circ y}
      | c = {{Defof|Morphism Property}}
}}
{{eqn | r = \map \phi {e_S}
      | c = {{Defof|Inverse Element}}
}}
{{eqn | r = \map \phi {y \circ x}
      | c = {{Defof|Inverse Element}}
}}
{{eqn | r = \map \phi y * \map \phi x
      | c = {{Defof|Morphism Property}}
}}
{{end-eqn}}
So $\map \phi y$ is an inverse of $\map \phi x$ in $\struct {T, *}$.
{{qed}}
Theorem: Homomorphism with Cancellable Codomain Preserves Identity

Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be a homomorphism.
Let $\struct {S, \circ}$ have an identity $e_S$.
Let $\struct {T, *}$ have an identity $e_T$.
Let every element of $\struct {T, *}$ be cancellable.
Then $\map \phi {e_S}$ is the identity $e_T$.

Proof:
Let $\struct {S, \circ}$ be an algebraic structure in which $\circ$ has an identity $e_S$.
Let $\struct {T, *}$ be an algebraic structure in which $*$ has an identity $e_T$.
Let $\struct {T, *}$ be such that every element is cancellable.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be a homomorphism.
Every element of $\struct {T, *}$ is cancellable.
Suppose there is an idempotent element in $\struct {T, *}$
So from Identity is only Idempotent Cancellable Element, it must be the identity $e_T$.
Thus:
{{begin-eqn}}
{{eqn | l = \map \phi {e_S} * \map \phi {e_S}
      | r = \map \phi {e_S \circ e_S}
      | c = {{Defof|Morphism Property}}
}}
{{eqn | r = \map \phi {e_S}
      | c = {{Defof|Identity Element}}
}}
{{end-eqn}}
So $\map \phi {e_S}$ is idempotent in $\struct {T, *}$ and the result follows.
{{qed}}
Theorem: Homomorphism with Identity Preserves Inverses

Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be a homomorphism.
Let $\struct {S, \circ}$ have an identity $e_S$.
Let $\struct {T, *}$ also have an identity $e_T = \map \phi {e_S}$.
If $x^{-1}$ is an inverse of $x$ for $\circ$, then $\map \phi {x^{-1} }$ is an inverse of $\map \phi x$ for $*$.
That is:
:$\map \phi {x^{-1} } = \paren {\map \phi x}^{-1}$

Proof:
Let $\struct {S, \circ}$ be an algebraic structure in which $\circ$ has an identity $e_S$.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be a homomorphism.
Let $\struct {T, *}$ be an algebraic structure in which $*$ has an identity $e_T = \map \phi {e_S}$.
Let $x^{-1}$ be an inverse of $x$ for $\circ$.
Then by existence of identity $e_S$:
{{tidy}}
:$\forall x \in S: x \circ x^{-1}, x^{-1} \circ x \in \Dom \phi$
Hence:
{{begin-eqn}}
{{eqn | l = \map \phi x * \map \phi {x^{-1} }
      | r = \map \phi {x \circ x^{-1} }
      | c = {{Defof|Morphism Property}}
}}
{{eqn | r = \map \phi {e_S} = e_T
      | c = 
}}
{{eqn | r = \map \phi {x^{-1} \circ x}
      | c = 
}}
{{eqn | r = \map \phi {x^{-1} } * \map \phi x
      | c = {{Defof|Morphism Property}}
}}
{{end-eqn}}
{{qed}}
Theorem: Homomorphism to Group Preserves Identity

Let $\struct {S, \circ}$ be an algebraic structure.
Let $\struct {T, *}$ be a group.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be a homomorphism.
Let $\struct {S, \circ}$ have an identity $e_S$.
Then:
:$\map \phi {e_S} = e_T$

Proof:
By hypothesis, $\struct {T, *}$ is a group.
By the Cancellation Laws, all elements of $T$ are cancellable.
Thus Homomorphism with Cancellable Codomain Preserves Identity can be applied.
{{Qed}}
Theorem: Homomorphism of External Direct Products

Let:
:$\struct {S_1 \times S_2, \circ}$ be the external direct product of two algebraic structures $\struct {S_1, \circ_1}$ and $\struct {S_2, \circ_2}$
:$\struct {T_1 \times T_2, *}$ be the external direct product of two algebraic structures $\struct {T_1, *_1}$ and $\struct {T_2, *_2}$
:$\phi_1$ be a homomorphism from $\struct {S_1, \circ_1}$ onto $\struct {T_1, *_1}$
:$\phi_2$ be a homomorphism from $\struct {S_2, \circ_2}$ onto $\struct {T_2, *_2}$.
Then the mapping $\phi_1 \times \phi_2: \struct {S_1 \times S_2, \circ} \to \struct {T_1 \times T_2, *}$ defined as:
:$\map {\paren {\phi_1 \times \phi_2} } {x, y} = \tuple {\map {\phi_1} x, \map {\phi_2} y}$
is a homomorphism from $\struct {S_1 \times S_2, \circ}$ to $\struct {T_1 \times T_2, *}$.

Proof:
Let $\tuple {x_1, x_2}, \tuple {y_1, y_2} \in S_1 \circ S_2$.
Then:
{{begin-eqn}}
{{eqn | l = \map {\paren {\phi_1 \times \phi_2} } {\tuple {x_1, x_2} \circ \tuple {y_1, y_2} }
      | r = \map {\paren {\phi_1 \times \phi_2} } {x_1 \circ_1 y_1, x_2 \circ_2 y_2}
      | c = 
}}
{{eqn | r = \tuple {\map {\phi_1} {x_1 \circ_1 y_1}, \map {\phi_2} {x_2 \circ_2 y_2} }
      | c = 
}}
{{eqn | r = \tuple {\map {\phi_1} {x_1} \ast_1 \map {\phi_1} {y_1}, \map {\phi_2} {x_2} \ast_2 \map {\phi_2} {y_2} }
      | c = 
}}
{{eqn | r = \tuple {\map {\phi_1} {x_1}, \map {\phi_2} {x_2} } \ast \tuple {\map {\phi_1} {y_1}, \map {\phi_2} {y_2} }
      | c = 
}}
{{eqn | r = \map {\paren {\phi_1 \times \phi_2} } {x_1, x_2} \ast \map {\paren {\phi_1 \times \phi_2} } {y_1, y_2}
      | c = 
}}
{{end-eqn}}
thus demonstrating the morphism property.
{{Qed}}
Category:Homomorphism of External Direct Products
Category:Homomorphisms (Abstract Algebra)
Category:External Direct Products
98gb4lhktpkdgjhsia2k94wxntu47ah
Theorem: Monomorphism Image is Isomorphic to Domain

The image of a monomorphism is isomorphic to its domain.
That is, if $\phi: S_1 \to S_2$ is a monomorphism, then:
:$\phi: S_1 \to \Img \phi$
is an isomorphism.

Proof:
Let $\struct {S_1, \circ_1}$ and $\struct {S_2, \circ_2}$ be closed algebraic structures.
Let $\phi$ be a monomorphism from $\struct {S_1, \circ_1}$ to $\struct {S_2, \circ_2}$.
Let $T = \Img \phi$ be the image of $\phi$.
By Morphism Property Preserves Closure, $\struct {T, \circ_2}$ is closed.
As $\phi$ is a monomorphism, it is by definition an injective homomorphism.
From Restriction of Injection is Injection, $\phi: S_1 \to \Img \phi$ is an injection.
From Restriction of Mapping to Image is Surjection, $\phi: S_1 \to \Img \phi$ is a surjection.
Thus $\phi \to \Img \phi$ is by definition a bijection.
Thus $\phi: S_1 \to \Img \phi$ is a bijective homomorphism.
Hence, by definition, $\phi: S_1 \to \Img \phi$ is an isomorphism.
{{qed}}
Theorem: Inverse of Algebraic Structure Isomorphism is Isomorphism

Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be a mapping.
Then $\phi$ is an isomorphism {{iff}} $\phi^{-1}: \struct {T, *} \to \struct {S, \circ}$ is also an isomorphism.

Proof:
Let $\phi$ be an isomorphism.
Then by definition $\phi$ is a bijection.
Thus $\exists \phi^{-1}$ such that $\phi^{-1}$ is also a bijection from Bijection iff Inverse is Bijection.
That is:
:$\exists \phi^{-1}: \struct {T, *} \to \struct {S, \circ}$
It follows that:
{{begin-eqn}}
{{eqn | q = \forall s \in S, t \in T
      | l = \map \phi s
      | r = t \iff \map {\phi^{-1} } t = s
      | c = Inverse Element of Bijection
}}
{{eqn | ll= \leadsto
      | l = \map \phi {s_1 \circ s_2}
      | r = t_1 * t_2
      | c = {{Defof|Morphism Property}}
}}
{{eqn | ll= \leadsto
      | l = \map {\phi^{-1} } {t_1 * t_2}
      | r = s_1 \circ s_2 = \map {\phi^{-1} } {t_1} \circ \map {\phi^{-1} } {t_2}
      | c = Inverse Element of Bijection
}}
{{end-eqn}}
So $\phi^{-1}: \struct {T, *} \to \struct {S, \circ}$ is a homomorphism.
$\phi^{-1}$ is also (from above) a bijection.
Thus, by definition, $\phi^{-1}$ is an isomorphism.
Let $\phi^{-1}: \struct {T, *} \to \struct {S, \circ}$ be an isomorphism.
Applying the same result as above in reverse, we have that $\paren {\phi^{-1} }^{-1}: \struct {S, \circ} \to \struct {T, *}$ is also an isomorphism.
But by Inverse of Inverse of Bijection:
:$\paren {\phi^{-1} }^{-1} = \phi$
and hence the result.
{{qed}}
Theorem: Isomorphism is Equivalence Relation

Isomorphism is an equivalence on a set of magmas.
This result applies to all magmas: rings, groups, R-algebraic structures etc.

Proof:
From the formal definitions:
=== Simple Graph ===
A (simple) graph $G$ is a non-empty set $V$ together with an antireflexive, symmetric relation $\RR$ on $V$.
=== Digraph ===
A digraph $D$ is a non-empty set $V$ together with an antireflexive relation $\RR$ on $V$.
=== Loop-graph ===
A loop-graph $P$ is a non-empty set $V$ together with a symmetric relation $\RR$ on $V$.
=== Loop-Digraph ===
A loop-digraph $D$ is a non-empty set $V$ together with a relation $\RR$ on $V$.
It can be seen from these definitions that all the above are relational structures with more or less restriction on the relation.
Hence the result from Relation Isomorphism is Equivalence Relation.
{{qed}}
Theorem: Isomorphism is Equivalence Relation

Isomorphism is an equivalence on a set of magmas.
This result applies to all magmas: rings, groups, R-algebraic structures etc.

Proof:
To prove a relation is an equivalence, we need to prove it is reflexive, symmetric and transitive.
So, checking in turn each of the criteria for equivalence:
Theorem: Isomorphism is Equivalence Relation

Isomorphism is an equivalence on a set of magmas.
This result applies to all magmas: rings, groups, R-algebraic structures etc.

Proof:
Let $\struct {S_1, \preccurlyeq_1} \cong \struct {S_2, \preccurlyeq_2}$ denote that $\struct {S_1, \preccurlyeq_1}$ is isomorphic to $\struct {S_2, \preccurlyeq_2}$.
Checking in turn each of the criteria for equivalence:
=== Reflexivity ===
{{:Order Isomorphism is Reflexive}}
Thus $\cong$ is seen to be reflexive.
{{qed|lemma}}
=== Symmetric ===
{{:Order Isomorphism is Symmetric}}
Thus $\cong$ is seen to be symmetric.
{{qed|lemma}}
=== Transitive ===
{{:Order Isomorphism is Transitive}}
Thus $\cong$ is seen to be transitive.
{{qed|lemma}}
$\cong$ has been shown to be reflexive, symmetric and transitive.
Hence the result.
{{qed}}
Theorem: Isomorphism is Equivalence Relation

Isomorphism is an equivalence on a set of magmas.
This result applies to all magmas: rings, groups, R-algebraic structures etc.

Proof:
An ordered set is a relational structure where order isomorphism is a special case of relation isomorphism.
The result follows directly from Relation Isomorphism is Equivalence Relation.
{{qed}}
Theorem: Projection is Epimorphism

Let $\struct {\SS, \circ}$ be the external direct product of the algebraic structures $\struct {S_1, \circ_1}$ and $\struct {S_2, \circ_2}$.
Then:
:$\pr_1$ is an epimorphism from $\struct {\SS, \circ}$ to $\struct {S_1, \circ_1}$
:$\pr_2$ is an epimorphism from $\struct {\SS, \circ}$ to $\struct {S_2, \circ_2}$
where $\pr_1$ and $\pr_2$ are the first and second projection respectively of $\struct {\SS, \circ}$.

Proof:
From Projection is Surjection, $\pr_1$ and $\pr_2$ are surjections.
We now need to show they are homomorphisms.
Let $s, t \in \struct {\SS, \circ}$ where $s = \tuple {s_1, s_2}$ and $t = \tuple {t_1, t_2$.
Then:
{{begin-eqn}}
{{eqn | l = \map {\pr_1} {s \circ t}
      | r = \map {\pr_1} {\tuple {s_1, s_2} \circ \tuple {t_1, t_2} }
      | c = 
}}
{{eqn | r = \map {\pr_1} {\tuple {s_1 \circ_1 t_1, s_2 \circ_2 t_2} }
      | c = 
}}
{{eqn | r = s_1 \circ_j t_1
      | c = 
}}
{{eqn | r = \map {\pr_1} s \circ_1 \map {\pr_1} t
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map {\pr_2} {s \circ t}
      | r = \map {\pr_2} {\tuple {s_1, s_2} \circ \tuple {t_1, t_2} }
      | c = 
}}
{{eqn | r = \map {\pr_2} {\tuple {s_1 \circ_1 t_1, s_2 \circ_2 t_2} }
      | c = 
}}
{{eqn | r = s_2 \circ_j t_2
      | c = 
}}
{{eqn | r = \map {\pr_2} s \circ_2 \map {\pr_2} t
      | c = 
}}
{{end-eqn}}
Thus the morphism property is demonstrated for both $\pr_1$ and $\pr_2$.
{{Qed}}
Theorem: Quotient Theorem for Epimorphisms

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 ===
{{:Equivalence Induced by Epimorphism is Congruence Relation}}{{qed|lemma}}
Theorem: Homomorphism on Induced Structure to Commutative Semigroup

Let $\struct {S, \circ}$ be an algebraic structure.
Let $\struct {T, \oplus}$ be a commutative semigroup.
Let $T^S$ be the set of all mappings from $S$ to $T$.
Let $f$ and $g$ be homomorphisms from $S$ into $T$.
Let $f \oplus' g$ be the pointwise operation on $T^S$ induced by $\oplus$.
Then $f \oplus' g$ is a homomorphism from $\struct {S, \circ}$ into $\struct {T, \oplus}$.

Proof:
Let $\struct {T, \oplus}$ be a commutative semigroup.
Let $x, y \in S$.
Then:
{{begin-eqn}}
{{eqn | l = \map {\paren {f \oplus' g} } {x \circ y}
      | r = \map f {x \circ y} \oplus \map g {x \circ y}
      | c = {{Defof|Pointwise Operation}}
}}
{{eqn | r = \paren {\map f x \oplus \map f y} \oplus \paren {\map g x \oplus \map g y}
      | c = $f$ and $g$ are homomorphisms
}}
{{eqn | r = \paren {\map f x \oplus \map g x} \oplus \paren {\map f y \oplus \map g y}
      | c = $\oplus$ is Commutative and Associative
}}
{{eqn | r = \map {\paren {f \oplus' g} } x \oplus \map {\paren {f \oplus' g} } Y
      | c = {{Defof|Pointwise Operation}}
}}
{{end-eqn}}
{{qed}}
Notice that for this to work, $\oplus$ needs to be both associative ''and'' commutative.