hcju/proofwikiqa huggingface dataset:
We need to demonstrate the following properties: * 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: :$\forall x, y \in S: \phi \left({x}\right) *_k \phi \left({a}\right) = \phi \left({y}\right) *_k \phi \left({a}\right) \implies \phi \left({x}\right) = \phi \left({y}\right)$ * 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: :$\forall x, y \in S: \phi \left({a}\right) *_k \phi \left({x}\right) = \phi \left({a}\right) *_k \phi \left({y}\right) \implies \phi \left({x}\right) = \phi \left({y}\right)$ * To show the first of the properties above: {{begin-eqn}} {{eqn | o=       | r=\left({x \circ_k a = y \circ_k a \implies x = y}\right)       | c= }} {{eqn | o=\implies       | r=\left({\phi \left({x \circ_k a}\right) = \phi \left({y \circ_k a}\right) \implies \phi \left({x}\right) = \phi \left({y}\right)}\right)       | c=Mappings are many-to-one by definition }} {{eqn | o=\implies       | r=\left({\phi \left({x}\right) *_k \phi \left({a}\right) = \phi \left({y}\right) *_k \phi \left({a}\right) \implies \phi \left({x}\right) = \phi \left({y}\right)}\right)       | c=Morphism Property }} {{end-eqn}} and thus left cancellability is demonstrated. * And the second is like it, namely this: {{begin-eqn}} {{eqn | o=       | r=\left({a \circ_k x = a \circ_k y \implies x = y}\right)       | c= }} {{eqn | o=\implies       | r=\left({\phi \left({a \circ_k x}\right) = \phi \left({a \circ_k y}\right) \implies \phi \left({x}\right) = \phi \left({y}\right)}\right)       | c=Mappings are many-to-one by definition }} {{eqn | o=\implies       | r=\left({\phi \left({a}\right) *_k \phi \left({x}\right) = \phi \left({a}\right) *_k \phi \left({y}\right) \implies \phi \left({x}\right) = \phi \left({y}\right)}\right)       | c=Morphism Property }} {{end-eqn}} and thus right cancellability is demonstrated. {{qed}} Category:Homomorphisms renh9lnsqjymefimq6eqfxvw23fgjy7
The quotient mapping $q_\mathcal R: S \to S / \mathcal R$ is the canonical surjection from $S$ to $S / \mathcal R$. Next we show that this is a homomorphism: {{begin-eqn}} {{eqn | l = \map {q_\mathcal R} x       | r = \eqclass x {\mathcal R}       | c = {{Defof|Quotient Mapping}} }} {{eqn | l = \map {q_\mathcal R} y       | r = \eqclass y {\mathcal R}       | c = {{Defof|Quotient Mapping}} }} {{eqn | l = \map {q_\mathcal R} {x \circ y}       | r = \eqclass {x \circ y} {\mathcal R}       | c = {{Defof|Quotient Mapping}} }} {{eqn | l = \eqclass {x \circ y} {\mathcal R}       | r = \eqclass x {\mathcal R} \circ_\mathcal R \eqclass y {\mathcal R}       | c = {{Defof|Operation Induced on Quotient Set}} }} {{eqn | ll= \leadsto       | l = \map {q_\mathcal R} {x \circ y}       | r = \map {q_\mathcal R} x \circ_\mathcal R \map {q_\mathcal R} y       | c =  }} {{end-eqn}} Thus the morphism property is shown to hold. So the quotient mapping $q_\mathcal R: \struct {S, \circ} \to \struct {S / \mathcal R, \circ_\mathcal R}$ has been shown to be a homomorphism which is a surjection, and is thus an epimorphism. {{Qed}}
Let $\phi: S \to T$ be a homomorphism. Let $\operatorname{Im} \left({\phi}\right) = T'$ By Morphism Property Preserves Closure, $T'$ is closed. From Restriction of Mapping to Image is Surjection, $\phi \to \operatorname{Im} \left({\phi}\right)$ 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 stsi7puz9734bf4mk55ir5amqnele7l
Let $\struct {S, \circ}$ be an algebraic structure in which $\circ$ is associative. Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an epimorphism. As an epimorphism 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}} {{qed}}
As $\left({S, \circ}\right)$ is a semigroup, then by definition it is closed. As $\phi$ is an epimorphism, it is by definition surjective. That is: :$T = \phi \left[{S}\right]$ where $\phi \left[{S}\right]$ denotes the image of $S$ under $\phi$. From Morphism Property Preserves Closure it follows that $\left({T, *}\right)$ is closed. As $\left({S, \circ}\right)$ is a semigroup, then by definition $\circ$ is associative. From Epimorphism Preserves Associativity, $*$ is therefore also associative. So: : $\left({T, *}\right)$ is closed and: : $*$ is associative. Therefore, by definition, $\left({T, *}\right)$ is a semigroup. {{qed}}
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 Category:Subsemigroups jw5hpjmdnguazj8xyel7nwh0pvngfyu
Let $\left({S, \circ}\right)$ be an algebraic structure in which $\circ$ is commutative. Let $\phi: \left({S, \circ}\right) \to \left({T, *}\right)$ be an epimorphism. As an epimorphism is surjective, it follows that: :$\forall u, v \in T: \exists x, y \in S: \phi \left({x}\right) = u, \phi \left({y}\right) = v$ So: {{begin-eqn}} {{eqn | l = u * v       | r = \phi \left({x}\right) * \phi \left({y}\right)       | c = $\phi$ is a surjection }} {{eqn | r = \phi \left({x \circ y}\right)       | c = Morphism Property }} {{eqn | r = \phi \left({y \circ x}\right)       | c = Commutativity of $\circ$ }} {{eqn | r = \phi \left({y}\right) * \phi {\left({x}\right)}       | c = Morphism Property }} {{eqn | r = v * u       | c = by definition as above }} {{end-eqn}} {{qed}}
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: {{begin-eqn}} {{eqn | l = \map \phi x       | r = \map \phi {x \circ e_S}       | c = $e_S$ is an identity for $\circ$ }} {{eqn | r = \map \phi x * \map \phi {e_S}       | c = Morphism property of $\circ$ under $\phi$ }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = \map \phi x       | r = \map \phi {e_S \circ x}       | c = $e_S$ is an identity for $\circ$ }} {{eqn | r = \map \phi {e_S} * \map \phi x       | c = Morphism property of $\circ$ under $\phi$ }} {{end-eqn}} The result follows because every element $y \in T$ is of the form $\map \phi x$ with $x \in S$. {{qed}}
Let $\left({S, \circ}\right)$ be an algebraic structure in which $\circ$ has an identity element $e_S$. From Epimorphism Preserves Identity, it follows that $\left({T, *}\right)$ also has an identity element, which is $\phi \left({e_S}\right)$. Let $y$ be an inverse of $x$ in $\left({S, \circ}\right)$.  Then: {{begin-eqn}} {{eqn | l = \phi \left({x}\right) * \phi \left({y}\right)       | r = \phi \left({x \circ y}\right)       | c =  }} {{eqn | r = \phi \left({e_S}\right)       | c =  }} {{eqn | r = \phi \left({y \circ x}\right)       | c =  }} {{eqn | r = \phi \left({y}\right) * \phi \left({x}\right)       | c =  }} {{end-eqn}} So $\phi \left({y}\right)$ is an inverse of $\phi \left({x}\right)$ in $\left({T, *}\right)$. {{qed}}
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 = by the morphism property of $\circ$ under $\phi$ }} {{eqn | r = \map \phi {e_S}       | c = as $e_S$ is the identity of $\left({S, \circ}\right)$ }} {{end-eqn}} So $\map \phi {e_S}$ is idempotent in $\struct{T, *}$ and the result follows. {{qed}}
Let $\left({S, \circ}\right)$ be an algebraic structure in which $\circ$ has an identity $e_S$. Let $\phi: \left({S, \circ}\right) \to \left({T, *}\right)$ be a homomorphism. Let $\left({T, *}\right)$ be an algebraic structure in which $*$ has an identity $e_T = \phi \left({e_S}\right)$. {{begin-eqn}} {{eqn | l = \phi \left({x}\right) * \phi \left({x^{-1} }\right)       | r = \phi \left({x \circ x^{-1} }\right)       | c = Morphism property of $\phi$ }} {{eqn | r = \phi \left({e_S}\right) = e_T       | c =  }} {{eqn | r = \phi \left({x^{-1} \circ x}\right)       | c =  }} {{eqn | r = \phi \left({x^{-1} }\right) * \phi \left({x}\right)       | c = Morphism property of $\phi$ }} {{end-eqn}} {{qed}}
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}}
Let $\left({x_1, x_2}\right), \left({y_1, y_2}\right) \in S_1 \circ S_2$. Then: {{begin-eqn}} {{eqn | l = \left({\phi_1 \times \phi_2}\right) \left({\left({x_1, x_2}\right) \circ \left({y_1, y_2}\right)}\right)       | r = \left({\phi_1 \times \phi_2}\right) \left({\left({x_1 \circ_1 y_1, x_2 \circ_2 y_2}\right)}\right)       | c =  }} {{eqn | r = \left({\phi_1 \left({x_1 \circ_1 y_1}\right), \phi_2 \left({x_2 \circ_2 y_2}\right)}\right)       | c =  }} {{eqn | r = \left({\phi_1 \left({x_1}\right) \ast_1 \phi_1 \left({y_1}\right), \phi_2 \left({x_2}\right) \ast_2 \phi_2 \left({y_2}\right)}\right)       | c =  }} {{eqn | r = \left({\phi_1 \left({x_1}\right), \phi_2 \left({x_2}\right)}\right) \ast \left({\phi_1 \left({y_1}\right), \phi_2 \left({y_2}\right)}\right)       | c =  }} {{eqn | r = \left({\phi_1 \times \phi_2}\right) \left({\left({x_1, x_2}\right)}\right) \ast \left({\phi_1 \times \phi_2}\right) \left({\left({y_1, y_2}\right)}\right)       | c =  }} {{end-eqn}} thus demonstrating the morphism property. {{Qed}} Category:Homomorphisms Category:Cartesian Product sg0mmocq3mnm7li4x7qx66ymostexp0
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}}
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: :$\exists \phi^{-1}: \left({T, *}\right) \to \left({S, \circ}\right)$ It follows that: {{begin-eqn}} {{eqn | ll=\forall s \in S, t \in T:       | o=       | r=\phi \left({s}\right) = t \iff \phi^{-1} \left({t}\right) = s       | c=Inverse Element of Bijection }} {{eqn | o=\implies       | r=\phi \left({s_1 \circ s_2}\right) = t_1 * t_2       | c=Morphism Property }} {{eqn | o=\implies       | r=\phi^{-1} \left({t_1 * t_2}\right) = s_1 \circ s_2 = \phi^{-1} \left({t_1}\right) \circ \phi^{-1} \left({t_2}\right)       | c=Inverse Element of Bijection }} {{end-eqn}} So $\phi^{-1}: \left({T, *}\right) \to \left({S, \circ}\right)$ is a homomorphism. $\phi^{-1}$ is also (from above) a bijection. Thus, by definition, $\phi^{-1}$ is an isomorphism. Let $\phi^{-1}: \left({T, *}\right) \to \left({S, \circ}\right)$ be an isomorphism. Applying the same result as above in reverse, we have that $\left({\phi^{-1}}\right)^{-1}: \left({S, \circ}\right) \to \left({T, *}\right)$ is also an isomorphism. But by Inverse of Inverse of Bijection: : $\left({\phi^{-1}}\right)^{-1} = \phi$ and hence the result. {{qed}}
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: === Reflexive === From Identity Mapping is Automorphism, the identity mapping $I_S: S \to S$ is an automorphism, which is an isomorphism from a magma onto itself. So $S \cong S$, and isomorphism is seen to be reflexive. {{Qed|lemma}} === Symmetric === If $S \cong T$, then $\exists \phi: S \to T$ such that $\phi$ is an isomorphism. Thus, from Inverse of Algebraic Structure Isomorphism is Isomorphism, $\phi^{-1}: T \to S$ is also an isomorphism, and $T \cong S$. Thus we have shown that $S \cong T \implies T \cong S$, so $\cong$ is symmetric. {{Qed|lemma}} === Transitive === Let $S_1 \cong S_2$, and $S_2 \cong S_3$. We can define these two isomorphisms: : $\phi: S_1 \to S_2$ : $\psi: S_2 \to S_3$ From Composite of Isomorphisms in Algebraic Structure is Isomorphism, we have that $\psi \circ \phi$ is an isomorphism between $S_1$ and $S_3$. Thus we have shown that $\cong$ is transitive. {{Qed|lemma}} Thus isomorphism is reflexive, symmetric and transitive, and therefore an equivalence. {{Qed}}
From Projection is Surjection, $\pr_1$ and $\pr_2$ are surjections. We now need to show they are homomorphisms. Let $s, t \in \struct {S, \circ}$ 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}}
=== Proof of Congruence Relation === {{:Equivalence Induced by Epimorphism is Congruence Relation}}{{qed|lemma}} === Proof of Unique Isomorphism === {{:Unique Isomorphism from Quotient Mapping to Epimorphism Domain}}
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.
Let $\left({T, \oplus}\right)$ be an abelian group. Let $x, y \in S$. Then: {{begin-eqn}} {{eqn | l = f^* \left({x \circ y}\right)       | r = \left({f \left({x \circ y}\right)}\right)^{-1}       | c = Induced Structure Inverse }} {{eqn | r = \left({f \left({x}\right) \oplus f \left({y}\right)}\right)^{-1}       | c = $f$ is a homomorphism }} {{eqn | r = \left({f \left({y}\right) \oplus f \left({x}\right)}\right)^{-1}       | c = Commutativity of $\oplus$ }} {{eqn | r = \left({f \left({x}\right)}\right)^{-1} \oplus \left({f \left({y}\right)}\right)^{-1}       | c = Inverse of Group Product }} {{eqn | r = f^* \left({x}\right) \oplus f^* \left({y}\right)       | c = Induced Structure Inverse }} {{end-eqn}} {{qed}}
Let $H$ be the set of all homomorphisms from $\struct {S, \circ}$ into $\struct {T, \oplus}$. === {{GroupAxiom|0}} === From Homomorphism on Induced Structure, we have that $\forall f, g \in H: f \oplus g$ is a homomorphism from $\struct {S, \circ}$ into $\struct {T, \oplus}$. Hence $\struct {H, \oplus}$ is closed. === {{GroupAxiom|3}} === From Inverse Mapping in Induced Structure, if $g$ is a homomorphism then its Induced Structure Inverse $g^*$ is one also. Thus $g \in H \implies g^* \in H$. So by the Two-Step Subgroup Test, $\struct {H, \oplus}$ is a subgroup of $\struct {T^S, \oplus}$. {{qed}}
=== Existence === To show that $\oplus$ as defined above exists: Let $u, v \in S$, and let $x = f \left({u}\right), y = f \left({v}\right)$. Then as $f$ is a bijection, $u = f^{-1} \left({x}\right), v = f^{-1} \left({y}\right)$. Thus: {{begin-eqn}} {{eqn | l = f \left({u \circ v}\right)       | r = f \left({f^{-1} \left({x}\right) \circ f^{-1} \left({y}\right)}\right)       | c = Definition of $x$ and $y$ }} {{eqn | r = f \left({f^{-1} \left({x}\right)}\right) \oplus f \left({f^{-1} \left({y}\right)}\right)       | c = Morphism Property }} {{eqn | r = x \oplus y       | c = Inverse of Inverse of Bijection }} {{eqn | r = f \left({u}\right) \oplus  f \left({v}\right)       | c = Definition of $x$ and $y$ }} {{end-eqn}} It is seen that $f$ is an isomorphism as required. {{qed|lemma}} === Uniqueness === Let $f$ be the isomorphism whose existence has been proven above. Thus: {{begin-eqn}} {{eqn | l = f \left({f^{-1} \left({x}\right) \circ f^{-1} \left({y}\right)}\right)       | r = f \left({f^{-1} \left({x}\right)}\right) \oplus f \left({f^{-1} \left({y}\right)}\right)       | c = Morphism Property }} {{eqn | r = \left({f \circ f^{-1} \left({x}\right)}\right) \oplus \left({f \circ f^{-1} \left({y}\right)}\right)       | c = Definition of Composite Mapping }} {{eqn | r = \left({I_T \left({x}\right)}\right) \oplus \left({I_T \left({y}\right)}\right)       | c = Bijection has Left and Right Inverse }} {{eqn | r = x \oplus y       | c = Definition of Identity Mapping }} {{end-eqn}} So, if $\oplus$ is an operation on $T$ such that $f$ is an isomorphism from $\left({S, \circ}\right) \to \left({T, \oplus}\right)$, then $\oplus$ must be defined as by this theorem, and there can be no other such operations. {{qed}}
Consider the set: :$X = \left\{{y \in S: \left({\exists x \in T: \left({x, y}\right) \in S}\right)}\right\}$ That is, $X$ consists of all the elements of $S$ which are the second coordinate of some ordered pair which ''also'' happens to be in $S$ and whose first coordinate is in $T$. From Exists Element Not in Set, we have that $\exists z: z \notin X$. Now consider the cartesian product $T' = T \times \left\{{z}\right\}$. Suppose that $p \in T'$. Then $p = \left({c, z}\right)$ where $c \in T$. Suppose $\left({c, z}\right) \in S$. Then that would mean that $z \in X$. But we have specifically selected $z$ such that $z \notin X$. So $p = \left({c, z}\right) \notin S$. Thus we have that $T' \cap S = \varnothing$. There is an obvious bijection $g: T \to T'$: :$\forall t \in T: g \left({t}\right) = \left({t, z}\right)$ and hence the result. {{qed}}
$T$ is commutative, from Inverse Completion is Commutative Semigroup. The mapping $f \left({x, y}\right) = x \circ' y^{-1}$ is an epimorphism from the cartesian product of $\left({S, \circ}\right)$ and $\left({C, \circ \restriction_C}\right)$ onto $\left({T, \circ'}\right)$. By the Quotient Theorem for Epimorphisms, the proof follows. {{Explain|The mapping $f \left({x, y}\right) {{=}} x \circ' y^{-1}$ is an epimorphism from the cartesian product of $\left({S, \circ}\right)$ and $\left({C, \circ \restriction_C}\right)$ onto $\left({T, \circ'}\right)$.}} {{Proofread}} Category:Inverse Completions Category:Quotient Structures 5bw3dopvx32rcfb6a9lyp5eqyjb3m8v
By the definition of $\RR_f$: :$\tuple {x_1, y_1} \mathrel {\RR_f} \tuple {x_2, y_2} \iff x_1 \circ' y_1^{-1} = x_2 \circ' y_2^{-1}$ Now: {{begin-eqn}} {{eqn | l = x_1 \circ' y_1^{-1}       | r = x_2 \circ' y_2^{-1}       | c =  }} {{eqn | ll= \leadsto       | l = x_1 \circ' y_1^{-1} \circ' y_1 \circ' y_2       | r = x_2 \circ' y_2^{-1} \circ' y_1 \circ' y_2       | c = Elements $y_1$ and $y_2$ are cancellable }} {{eqn | ll= \leadsto       | l = x_1 \circ' y_2       | r = x_2 \circ' y_1       | c = {{Defof|Inverse Element}}, and commutativity of $\circ'$ }} {{eqn | ll= \leadsto       | l = x_1 \circ y_2       | r = x_2 \circ y_1       | c = $\circ'$ extends $\circ$ }} {{end-eqn}} which leads us to: :$\tuple {x_1, y_1} \mathrel {\RR_f} \tuple {x_2, y_2} \iff x_1 \circ y_2 = x_2 \circ y_1$ {{qed}} Category:Inverse Completions 2om0id39oam1sk3ljj5keesgt7lf0w0
Note that by definition of inverse completion, $\left({T, \circ'}\right)$ is a semigroup. Thus $\circ'$ is associative. First it is demonstrated that $S \circ' C^{-1}$ is a semigroup. Let $x, z \in S$. Let $y, w \in C$. Then: {{begin-eqn}} {{eqn | l = \left({x \circ' y^{-1} }\right) \circ' \left({z \circ' w^{-1} }\right)       | r = x \circ' \left({y^{-1} \circ' z}\right) \circ' w^{-1}       | c = Associativity of $\circ'$ }} {{eqn | r = x \circ' \left({z \circ' y^{-1} }\right) \circ' w^{-1}       | c = Commutation with Inverse in Monoid }} {{eqn | r = \left({x \circ' z}\right) \circ' \left({y^{-1} \circ' w^{-1} }\right)       | c = Associativity of $\circ'$ }} {{eqn | r = \left({x \circ' z}\right) \circ' \left({w \circ' y}\right)^{-1}       | c = Inverse of Product in Associative Structure }} {{eqn | r = \left({x \circ z}\right) \circ' \left({w \circ y}\right)^{-1}       | c = $\circ'$ extends $\circ$ }} {{end-eqn}} Thus: :$\left({x \circ z}\right)\circ' \left({w \circ y}\right)^{-1} \in S \circ' C^{-1}$ proving that $S \circ' C^{-1}$ is closed. Therefore by Subsemigroup Closure Test: :$S \circ' C^{-1}$ is a subsemigroup of $\left({T, \circ'}\right)$ and thus a semigroup. {{qed|lemma}} It remains to be shown that $\circ'$ is a commutative operation. Let $\left({x \circ' y^{-1} }\right)$ and $\left({z \circ' w^{-1} }\right)$ be two arbitrary elements of $S \circ' C^{-1}$. By Element Commutes with Product of Commuting Elements, $x, y, z, w$ all commute with each other under $\circ$. As $\circ'$ is an extension of $\circ$, it follows that $x, y, z, w$ also all commute with each other under $\circ'$. Then: {{begin-eqn}} {{eqn | l = \left({x \circ' y^{-1} }\right) \circ' \left({z \circ' w^{-1} }\right)       | r = x \circ' \left({y^{-1} \circ' z}\right) \circ' w^{-1}       | c = Associativity of $\circ'$ }} {{eqn | r = x \circ' \left({z \circ' y^{-1} }\right) \circ' w^{-1}       | c = Commutation with Inverse in Monoid }} {{eqn | r = \left({x \circ' z}\right) \circ' \left({y^{-1} \circ' w^{-1} }\right)       | c = Associativity of $\circ'$ }} {{eqn | r = \left({z \circ' x}\right) \circ' \left({w^{-1} \circ' y^{-1} }\right)       | c = Commutation of Inverses in Monoid }} {{eqn | r = z \circ' \left({x \circ' w^{-1} }\right) \circ' y^{-1}       | c = Associativity of $\circ'$ }} {{eqn | r = z \circ' \left({w^{-1} \circ' x}\right) \circ' y^{-1}       | c = Commutation with Inverse in Monoid }} {{eqn | r = \left({z \circ' w^{-1} }\right) \circ' \left({x \circ' y^{-1} }\right)       | c = Associativity of $\circ'$ }} {{end-eqn}} So $x \circ' y^{-1}$ commutes with $z \circ' w^{-1}$. It follows by definition that $S \circ' C^{-1}$ is a commutative semigroup. {{qed}}