avewright/proofwiki-math huggingface dataset:
Theorem: Left and Right Inverses of Mapping are Inverse Mapping

Let $f: S \to T$ be a mapping such that:
:$(1): \quad \exists g_1: T \to S: g_1 \circ f = I_S$
:$(2): \quad \exists g_2: T \to S: f \circ g_2 = I_T$
Then:
: $g_1 = g_2 = f^{-1}$
where $f^{-1}$ is the inverse of $f$.

Proof:
From Left and Right Inverse Mappings Implies Bijection, we have that $f$ is a bijection.
Let $y \in T$ and let $x = \map {f^{-1} } y$.
Such an $x$ exists because $f$ is surjective, and unique within $S$ as $f$ is injective.
Then $y = \map f x$ and so:
{{begin-eqn}}
{{eqn | l = \map {f^{-1} } y
      | r = x
      | c = 
}}
{{eqn | r = \map {I_S} x
      | c = $I_S$ is the identity mapping on $S$
}}
{{eqn | r = \map {g_1 \circ f} x
      | c = by hypothesis
}}
{{eqn | r = \map {g_1} y
      | c = from above: $y = \map f x$
}}
{{end-eqn}}
So $f^{-1} = g_1$.
Also:
{{begin-eqn}}
{{eqn | l = \map f x
      | r = y
      | c = 
}}
{{eqn | r = \map {I_T} y
      | c = $I_T$ is the identity mapping on $T$
}}
{{eqn | r = \map {f \circ g_2} y
      | c = by hypothesis
}}
{{eqn | ll= \leadsto
      | l = \map {f^{-1} } y
      | r = x
      | c = as $f$ is injective
}}
{{eqn | r = \map {g_2} y
      | c = 
}}
{{end-eqn}}
So $f^{-1} = g_2$.
{{qed}}
Theorem: Left and Right Inverses of Mapping are Inverse Mapping

Let $f: S \to T$ be a mapping such that:
:$(1): \quad \exists g_1: T \to S: g_1 \circ f = I_S$
:$(2): \quad \exists g_2: T \to S: f \circ g_2 = I_T$
Then:
: $g_1 = g_2 = f^{-1}$
where $f^{-1}$ is the inverse of $f$.

Proof:
Because Composition of Mappings is Associative, brackets do not need to be used.
{{begin-eqn}}
{{eqn | n = 1
      | l = g_1 \circ f
      | r = I_S
      | c = 
}}
{{eqn | ll= \leadsto
      | l = g_1 \circ f \circ g_2
      | r = I_S \circ g_2
      | c = 
}}
{{eqn | r = g_2
      | c = {{Defof|Identity Mapping}}
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | n = 2
      | l = f \circ g_2
      | r = I_T
      | c = 
}}
{{eqn | ll= \leadsto
      | l = g_1 \circ f \circ g_2
      | r = g_1 \circ I_T
      | c = 
}}
{{eqn | r = g_1
      | c = {{Defof|Identity Mapping}}
}}
{{end-eqn}}
Thus $g_1 = g_2$.
Now suppose there exists $g_3: T \to S: g_3 \circ f = I_S$.
By the same argument as above, $g_3 = g_2$.
This means that $g_1 (= g_3)$ is the only left inverse of $f$.
Similarly, suppose there exists $g_4: T \to S: f \circ g_4 = I_T$.
By the same argument as above, $g_4 = g_1$.
This means that $g_2 (= g_4)$ is the only right inverse of $f$.
So $g_1 = g_2 = g_3 = g_4$ are all the same.
By Composite of Bijection with Inverse is Identity Mapping, it follows that this unique mapping is the inverse $f^{-1}$.
{{qed}}
Theorem: Composition of Mappings is Composition of Relations

Let $f_1: S_1 \to S_2$ and $f_2: S_2 \to S_3$ be mappings such that the domain of $f_2$ is the same as the codomain of $f_1$.
Let $f_2 \circ f_1$ be the composite of $f_1$ and $f_2$.
Let $f_1$ and $f_2$ be considered as relations on $S_1 \times S_2$ and $S_2 \times S_3$ respectively.
Then $f_2 \circ f_1$ defined as the composition of relations coincides with the definition of $f_2 \circ f_1$ as the composition of mappings.

Proof:
By definition of composition of mappings:
:$f_2 \circ f_1 = \set {\tuple {x, z} \in S_1 \times S_3: \exists y \in S_2: \tuple {x, y} \in f_1 \land \tuple {y, z} \in f_2}$
Let $\RR_1 \subseteq S_1 \times T_1$ and $\RR_2 \subseteq S_2 \times T_2$ be relations.
The composite of $\RR_1$ and $\RR_2$ is defined as:
:$\RR_2 \circ \RR_1 := \set {\tuple {x, z} \in S_1 \times T_2: \exists y \in S_2 \cap T_1: \tuple {x, y} \in \RR_1 \land \tuple {y, z} \in \RR_2}$
When $T_1 = S_2$ and $T_2 = S_3$, this becomes:
:$\RR_2 \circ \RR_1 := \set {\tuple {x, z} \in S_1 \times T_3: \exists y \in S_2: \tuple {x, y} \in \RR_1 \land \tuple {y, z} \in \RR_2}$
The definitions can be seen to be identical.
{{qed}}
Theorem: Null Sequence in Exponential Sequence

Let $\sequence {a_n}_{n \mathop \in \N} \in \C$ be a sequence of complex numbers such that:
:$\ds \lim_{n \mathop \to +\infty}a_n = 0$
Then:
:$\ds \lim_{n \mathop \to +\infty} \paren {1 + \dfrac {a_n} n}^n = 1$

Proof:
{{begin-eqn}}
{{eqn | l = \paren {1 + \frac {a_n} n}^n
      | r = \sum_{k \mathop = 0}^n {n \choose k} \paren {\frac {a_n} n}^k
      | c = Binomial Theorem
}}
{{eqn | r = {n \choose 0} \paren {\frac {a_n} n}^0 + \sum_{k \mathop = 1}^n {n \choose k} \paren {\frac {a_n} n}^k
}}
{{eqn | r = 1 + a_n \sum_{k \mathop = 1}^n \dbinom n k \frac { {a_n}^{k - 1} } {n^k}
}}
{{eqn | ll= \leadsto
      | l = \lim_{n \mathop \to +\infty} 1 + a_n \sum_{k \mathop = 1}^n \dbinom n k \frac { {a_n}^{k - 1} } {n^k}
      | r = 1 + \paren {\lim_{n \mathop \to +\infty} a_n} \paren {\lim_{n \mathop \to +\infty} \sum_{k \mathop = 1}^n \dbinom n k \frac { {a_n}^{k - 1} } {n^k} }
      | c = Combination Theorem for Sequences
}}
{{eqn | r = 1 + 0 \cdot \paren {\lim_{n \mathop \to +\infty} \sum_{k \mathop = 1}^n \dbinom n k \frac { {a_n}^{k - 1} } {n^k} }
}}
{{eqn | r = 1
}}
{{end-eqn}}
{{explain|A note on the finiteness of the right limit is necessary to be allowed to use the combination theorem}}
{{qed}}
Theorem: Null Sequence in Exponential Sequence

Let $\sequence {a_n}_{n \mathop \in \N} \in \C$ be a sequence of complex numbers such that:
:$\ds \lim_{n \mathop \to +\infty}a_n = 0$
Then:
:$\ds \lim_{n \mathop \to +\infty} \paren {1 + \dfrac {a_n} n}^n = 1$

Proof:
Let $\sequence {E_n}$ be the sequence of complex functions $E_n: \C \to \C$ defined by:
:$\map {E_n} z = \paren {1 + \dfrac z n}^n$
We have that:
:$\ds \lim_{n \mathop \to \infty} \map {E_n} z = \map \exp z$
where $\map \exp z$ is the complex exponential.
We also have that:
:$E_n \paren {a_n} = \paren {1 + \dfrac {a_n} n}^n$
By Convergent Sequence in Metric Space is Bounded, we have that $\sequence {a_n}$ is Bounded Complex Sequence.
Let this bound be $M$.
Let $K \subseteq \C$ be the closed disk of radius $M$.
By Closed Disk is Compact, $K$ is compact.
By Exponential Sequence is Uniformly Convergent on Compact Sets, $\sequence {E_n}$ is uniformly convergent on $K$.
Now the hypotheses of Uniformly Convergent Sequence Evaluated on Convergent Sequence are satisfied, so:
:$\ds \lim_{n \mathop \to \infty} \map {E_n} {a_n} = \map \exp 0 = 1$
Hence the result.
{{qed}}
Theorem: Identity Mapping is Right Identity

Let $S$ and $T$ be sets.
Let $f: S \to T$ be a mapping.
Then:
:$f \circ I_S = f$
where $I_S$ is the identity mapping on $S$, and $\circ$ signifies composition of mappings.

Proof:
=== Equality of Codomains ===
The codomains of $f$ and $f \circ I_S$ are both equal to $T$ from Codomain of Composite Relation.
{{qed|lemma}}
=== Equality of Domains ===
From Domain of Composite Relation:
:$\Dom {f \circ I_S} = \Dom {I_S}$
But from the definition of the identity mapping:
:$\Dom {I_S} = \Img {I_S} = S$
{{qed|lemma}}
=== Equality of Mappings ===
The composite of $I_S$ and $f$ is defined as:
:$f \circ I_S = \set {\tuple {x, z} \in S \times T: \exists y \in S: \tuple {x, y} \in I_S \land \tuple {y, z} \in f}$
But by definition of the identity mapping on $S$, we have that:
:$\tuple {x, y} \in I_S \implies x = y$
Hence:
:$f \circ I_S = \set {\tuple {y, z} \in S \times T: \exists y \in S: \tuple {y, y}\ \in I_S \land \tuple {y, z} \in f}$
But as $\forall y \in S: \tuple {y, y} \in I_S$, this means:
:$f \circ I_S = \set {\tuple {y, z} \in S \times T: \tuple {y, z} \in f}$
That is:
:$f \circ I_S = f$
{{qed|lemma}}
Hence the result, by Equality of Mappings.
{{qed}}
Theorem: Identity Mapping is Right Identity

Let $S$ and $T$ be sets.
Let $f: S \to T$ be a mapping.
Then:
:$f \circ I_S = f$
where $I_S$ is the identity mapping on $S$, and $\circ$ signifies composition of mappings.

Proof:
By definition, a mapping is also a relation.
Also by definition, the identity mapping is the same as the diagonal relation.
Thus Diagonal Relation is Right Identity can be applied directly.
{{qed}}
Theorem: Diagonal Relation is Left Identity

Let $\RR \subseteq S \times T$ be a relation on $S \times T$.
Then:
:$\Delta_T \circ \RR = \RR$
where $\Delta_T$ is the diagonal relation on $T$, and $\circ$ signifies composition of relations.

Proof:
We use the definition of relation equality, as follows:
Theorem: Isomorphism Preserves Associativity

Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an isomorphism.
Then $\circ$ is associative {{iff}} $*$ is associative.

Proof:
Let $\struct {S, \circ}$ be an algebraic structure in which $\circ$ is associative.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an isomorphism.
As an isomorphism is surjective, it follows that:
:$\forall u, v, w \in T: \exists x, y, z \in S: \map \phi x = u, \map \phi y = v, \map \phi z = w$
So:
{{begin-eqn}}
{{eqn | l = \paren {u * v} * w
      | r = \paren {\map \phi x * \map \phi y} * \map \phi z
      | c = as $\phi$ is a Surjection
}}
{{eqn | r = \map \phi {x \circ y} * \map \phi z
      | c = {{Defof|Morphism Property}}
}}
{{eqn | r = \map \phi {\paren {x \circ y} \circ z}
      | c = {{Defof|Morphism Property}}
}}
{{eqn | r = \map \phi {x \circ \paren {y \circ z} }
      | c = Associativity of $\circ$
}}
{{eqn | r = \map \phi x * \map \phi {y \circ z}
      | c = {{Defof|Morphism Property}}
}}
{{eqn | r = \map \phi x * \paren {\map \phi y * \map \phi z}
      | c = {{Defof|Morphism Property}}
}}
{{eqn | r = u * \paren {v * w}
      | c = by definition as above
}}
{{end-eqn}}
As $\phi$ is an isomorphism, it follows from Inverse of Algebraic Structure Isomorphism is Isomorphism that $\phi^{-1}$ is also an isomorphism.
Thus the result for $\phi$ can be applied to $\phi^{-1}$.
{{qed}}
Theorem: Isomorphism Preserves Associativity

Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an isomorphism.
Then $\circ$ is associative {{iff}} $*$ is associative.

Proof:
An isomorphism is {{afortiori}} an epimorphism.
The result follows from Epimorphism Preserves Associativity.
{{qed}}
Theorem: Isomorphism Preserves Commutativity

Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an isomorphism.
Then $\circ$ is commutative {{iff}} $*$ is commutative.

Proof:
Let $\struct {S, \circ}$ be an algebraic structure in which $\circ$ is commutative.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an isomorphism.
As an isomorphism 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 = as $\phi$ is a surjection
}}
{{eqn | r = \map \phi {x \circ y}
      | c = Morphism Property
}}
{{eqn | r = \map \phi {y \circ x}
      | c = Commutativity of $\circ$
}}
{{eqn | r = \map \phi y * \map \phi x
      | c = Morphism Property
}}
{{eqn | r = v * u
      | c = by definition as above
}}
{{end-eqn}}
As $\phi$ is an isomorphism, it follows from Inverse of Algebraic Structure Isomorphism is Isomorphism that $\phi^{-1}$ is also a isomorphism.
Thus the result for $\phi$ can be applied to $\phi^{-1}$.
{{qed}}
Theorem: Isomorphism Preserves Commutativity

Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an isomorphism.
Then $\circ$ is commutative {{iff}} $*$ is commutative.

Proof:
We have that an isomorphism is a homomorphism which is also a bijection.
By definition, an epimorphism is a homomorphism which is also a surjection.
That is, an isomorphism is an epimorphism which is also an injection.
Thus Epimorphism Preserves Commutativity can be applied.
{{qed}}
Theorem: Isomorphism Preserves Inverses

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

Proof:
An group isomorphism is by definition a group epimorphism.
The result follows from Epimorphism Preserves Inverses.
{{Qed}}
Theorem: Isomorphism Preserves Inverses

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

Proof:
Let $g \in G$.
{{begin-eqn}}
{{eqn | l = \map \phi g * \map \phi {g^{-1} }
      | r = \map \phi {g \circ g^{-1} }
      | c = {{Defof|Group Isomorphism}}
}}
{{eqn | r = \map \phi {e_G}
      | c = {{Defof|Inverse Element}}
}}
{{eqn | r = e_H
      | c = Group Isomorphism Preserves Identity
}}
{{end-eqn}}
It follows from Inverse in Group is Unique that $\map \phi {g^{-1} }$ is the unique inverse element of $\map \phi g$ in $\struct {H, *}$.
That is:
:$\forall g \in G: \map \phi {g^{-1} } = \paren {\map \phi g}^{-1}$
{{qed}}
Theorem: Isomorphism Preserves Inverses

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

Proof:
Let $\struct {S, \circ}$ be an algebraic structure in which $\circ$ has an identity $e_S$.
From Epimorphism Preserves Identity, it follows that $\struct {T, *}$ also has an identity, which is $\map \phi {e_S}$.
Let $y$ be an inverse of $x$ in $\struct {S, \circ}$. 
Then:
{{begin-eqn}}
{{eqn | l = \map \phi x * \map \phi y
      | r = \map \phi {x \circ y}
      | c = 
}}
{{eqn | r = \map \phi {e_S}
      | c = 
}}
{{eqn | r = \map \phi {y \circ x}
      | c = 
}}
{{eqn | r = \map \phi y * \map \phi x
      | c = 
}}
{{end-eqn}}
So $\map \phi y$ is an inverse of $\map \phi x$ in $\struct {T, *}$.
As $\phi$ is an isomorphism, it follows from Inverse of Algebraic Structure Isomorphism is Isomorphism that $\phi^{-1}$ is also a isomorphism.
Thus the result for $\map \phi x$ can be applied to $\map {\phi^{-1} } {\map \phi x}$.
{{qed}}
Theorem: Isomorphism Preserves Inverses

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

Proof:
We have that an isomorphism is {{afortiori}} an epimorphism.
The result follows from Epimorphism Preserves Inverses.
{{qed}}
Theorem: Isomorphism Preserves Identity

Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an isomorphism.
Then $\circ$ has an identity $e_S$ {{iff}} $\map \phi {e_S}$ is the identity for $*$.

Proof:
An group isomorphism is by definition a group epimorphism.
The result follows from Epimorphism Preserves Identity.
{{Qed}}
Theorem: Isomorphism Preserves Identity

Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an isomorphism.
Then $\circ$ has an identity $e_S$ {{iff}} $\map \phi {e_S}$ is the identity for $*$.

Proof:
{{begin-eqn}}
{{eqn | l = \map \phi {e_G}
      | r = \map \phi {e_G \circ e_G}
      | c = {{Defof|Identity Element}}
}}
{{eqn | r = \map \phi {e_G} * \map \phi {e_G}
      | c = {{Defof|Group Isomorphism}}
}}
{{eqn | r = \map \phi {e_G} * \map \phi {e_G}
      | c = {{Defof|Group Isomorphism}}
}}
{{end-eqn}}
It follows from Identity is only Idempotent Element in Group that $\map \phi {e_G}$ is the identity of $H$.
That is:
:$\map \phi {e_G} = e_H$
{{qed}}
Theorem: Isomorphism Preserves Identity

Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an isomorphism.
Then $\circ$ has an identity $e_S$ {{iff}} $\map \phi {e_S}$ is the identity for $*$.

Proof:
Let $\struct {S, \circ}$ be an algebraic structure in which $\circ$ has an identity $e_S$.
Then $\forall x \in S: x \circ e_S = x = e_S \circ x$.
The result follows directly from the morphism property of $\circ$ under $\phi$:
{{begin-eqn}}
{{eqn | l = \map \phi {x \circ e_S}
      | r = \map \phi x
      | c =
}}
{{eqn | r = \map \phi {e_S \circ x}
      | c =
}}
{{eqn | l = \map \phi x * \map \phi {e_S}
      | r = \map \phi x
      | c = 
}}
{{eqn | r = \map \phi {e_S} * \map \phi x
      | c = 
}}
{{end-eqn}}
As $\phi$ is an isomorphism, it follows from Inverse of Algebraic Structure Isomorphism is Isomorphism that $\phi^{-1}$ is also a isomorphism.
Thus the result for $\map \phi {e_S}$ can be applied to $\map {\phi^{-1} } {\map \phi {e_S} }$.
{{qed}}
Theorem: Isomorphism Preserves Identity

Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an isomorphism.
Then $\circ$ has an identity $e_S$ {{iff}} $\map \phi {e_S}$ is the identity for $*$.

Proof:
We have that an isomorphism is a homomorphism which is also a bijection.
By definition, an epimorphism is a homomorphism which is also a surjection.
That is, an isomorphism is an epimorphism which is also an injection.
Thus Epimorphism Preserves Identity can be applied.
{{qed}}
Theorem: Non-Zero Natural Numbers under Multiplication form Commutative Semigroup

Let $\N_{>0}$ be the set of natural numbers without zero, that is, $\N_{>0} = \N \setminus \set 0$.
The structure $\struct {\N_{>0}, \times}$ forms an infinite commutative semigroup.

Proof:
=== {{Semigroup-axiom|0|nolink}} ===
From Natural Numbers have No Proper Zero Divisors:
:$\forall m, n \in \N: m \times n = 0 \iff m = 0 \lor n = 0$
It follows that:
:$\forall m, n \in \N_{>0}: m \times n \ne 0$
and so:
:$\forall m, n \in \N_{>0}: m \times n \in \N_{>0}$
That is, $\struct {\N_{>0}, \times}$ is closed.
{{qed|lemma}}
Theorem: Isomorphism Preserves Groups

Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an isomorphism.
If $\struct {S, \circ}$ is a group, then so is $\struct {T, *}$.

Proof:
From Isomorphism Preserves Semigroups, if $\struct {S, \circ}$ is a semigroup then so is $\struct {T, *}$.
From Isomorphism Preserves Identity, if $\struct {S, \circ}$ has an identity $e_S$, then $\map \phi {e_S}$ is the identity for $*$.
From Isomorphism Preserves Inverses, if $x^{-1}$ is an inverse of $x$ for $\circ$, then $\map \phi {x^{-1} }$ is an inverse of $\map \phi x$ for $*$.
The result follows from the definition of group.
{{qed}}
Theorem: Isomorphism Preserves Groups

Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an isomorphism.
If $\struct {S, \circ}$ is a group, then so is $\struct {T, *}$.

Proof:
An isomorphism is an epimorphism.
The result follows as a direct corollary of Epimorphism Preserves Groups.
{{qed}}
Theorem: Isomorphism Preserves Semigroups

Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: S \to T$ be an isomorphism.
If $\struct {S, \circ}$ is a semigroup, then so is $\struct {T, *}$.

Proof:
If $\struct {S, \circ}$ is a semigroup, then by definition it is closed.
From Morphism Property Preserves Closure, $\struct {T, *}$ is therefore also closed.
If $\struct {S, \circ}$ is a semigroup, then by definition $\circ$ is associative.
From Isomorphism Preserves Associativity, $*$ is therefore also associative.
So $\struct {T, *}$ is closed, and $*$ is associative, and therefore by definition, $\struct {T, *}$ is a semigroup.
{{qed}}
Theorem: Isomorphism Preserves Semigroups

Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: S \to T$ be an isomorphism.
If $\struct {S, \circ}$ is a semigroup, then so is $\struct {T, *}$.

Proof:
An isomorphism is an epimorphism.
The result follows as a direct corollary of Epimorphism Preserves Semigroups.
{{qed}}
Theorem: Left Cancellable iff Left Regular Representation Injective

Let $\struct {S, \circ}$ be an algebraic structure.
Then $a \in S$ is left cancellable {{iff}} the left regular representation $\map {\lambda_a} x$ is injective.

Proof:
Suppose $a \in S$ is left cancellable.
Then:
:$\forall x, y \in S: a \circ x = a \circ y \implies x = y$
From the definition of the left regular representation:
:$\map {\lambda_a} x = a \circ x$
Thus:
:$\forall x, y \in S: \map {\lambda_a} x = \map {\lambda_a} y \implies x = y$
and so the left regular representation is injective.
{{qed|lemma}}
Suppose $\map {\lambda_a} x$ is injective.
Then:
:$\forall x, y \in S: \map {\lambda_a} x = \map {\lambda_a} y \implies x = y$
From the definition of the left regular representation:
:$\map {\lambda_a} x = a \circ x$
Thus:
:$\forall x, y \in S: a \circ x = a \circ y \implies x = y$
and so $a$ is left cancellable.
{{Qed}}
Theorem: Right Cancellable iff Right Regular Representation Injective

Let $\struct {S, \circ}$ be an algebraic structure.
Then $a \in S$ is right cancellable {{iff}} the right regular representation $\map {\rho_a} x$ is injective.

Proof:
Suppose $a \in S$ is right cancellable.
Then:
:$\forall x, y \in S: x \circ a = y \circ a \implies x = y$
From the definition of the right regular representation:
:$\map {\rho_a} x = x \circ a$
Thus:
:$\forall x, y \in S: \map {\rho_a} x = \map {\rho_a} y \implies x = y$
and so the right regular representation is injective.
{{qed|lemma}}
Suppose $\map {\rho_a} x$ is injective.
Then:
:$\forall x, y \in S: \map {\rho_a} x = \map {\rho_a} y \implies x = y$
From the definition of the right regular representation:
:$\map {\rho_a} x = x \circ a$
Thus:
:$\forall x, y \in S: x \circ a = y \circ a \implies x = y$
and so $a$ is right cancellable.
{{Qed}}
Theorem: Isomorphism Preserves Cancellability

Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an isomorphism.
Then:
:$a \in S$ is cancellable in $\struct {S, \circ}$ {{iff}} $\map \phi a \in T$ is cancellable in $\struct {T, *}$.

Proof:
Let $\struct {S, \circ}$ be an algebraic structure in which $a$ is cancellable.
From Isomorphism Preserves Left Cancellability and Isomorphism Preserves Right Cancellability:
:$a \in S$ is left cancellable in $\struct {S, \circ}$ {{iff}} $\map \phi a \in T$ is left cancellable in $\struct {T, *}$
and
:$a \in S$ is right cancellable in $\struct {S, \circ}$ {{iff}} $\map \phi a \in T$ is right cancellable in $\struct {T, *}$.
It follows that:
:$a \in S$ is cancellable in $\struct {S, \circ}$
{{iff}}:
:$\map \phi a \in T$ is cancellable in $\struct {T, *}$.
{{qed}}
Theorem: Isomorphism Preserves Left Cancellability

Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an isomorphism.
Then:
:$a \in S$ is left cancellable in $\struct {S, \circ}$
{{iff}}:
:$\map \phi a \in T$ is left cancellable in $\struct {T, *}$.

Proof:
Let $a$ be left cancellable in $\struct {S, \circ}$.
Let $x \in S$ and $y \in S$ be arbitrary.
Then:
{{begin-eqn}}
{{eqn | l = \map \phi a * \map \phi x
      | r = \map \phi a * \map \phi y
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \map \phi {a \circ x}
      | r = \map \phi {a \circ y}
      | c = Morphism Property
}}
{{eqn | ll= \leadsto
      | l = \map \phi x
      | r = \map \phi y
      | c = as $a$ is left cancellable
}}
{{end-eqn}}
That is, $\map \phi a$ is left cancellable in $\struct {T, *}$.
As $\phi$ is an isomorphism, then so is $\phi^{-1}$.
So the same proof works in reverse in exactly the same way.
{{qed}}