hcju/proofwikips huggingface dataset:
Inverse Image Mapping of Relation is Mapping
Let $S$ and $T$ be sets. Let $\mathcal R \subseteq S \times T$ be a relation on $S \times T$. Let $\mathcal R^\gets$ be the inverse image mapping of $\mathcal R$: :$\mathcal R^\gets: \powerset T \to \powerset S: \map {\mathcal R^\gets} Y = \mathcal R^{-1} \sqbrk Y$ Then $\mathcal R^\gets$ is indeed a mapping.
Direct Image Mapping of Injection is Injection
Let $f: S \to T$ be an injection. Then the direct image mapping of $f$: :$f^\to: \powerset S \to \powerset T$ is an injection.
Direct Image Mapping of Surjection is Surjection
Let $f: S \to T$ be a surjection. Then the direct image mapping of $f$: :$f^\to: \powerset S \to \powerset T$ is a surjection.
Cantor's Theorem
There is no surjection from a set $S$ to its power set for any set $S$. That is, $S$ is strictly smaller than its power set.
Surjection Induced by Powerset is Induced by Surjection
Let $\mathcal R \subseteq S \times T$ be a relation. Let $\mathcal R^\to: \powerset S \to \powerset T$ be the direct image mapping $\mathcal R$. Let $\mathcal R^\to$ be a surjection. Let $X = \Preimg {\mathcal R}$, that is, the preimage of $\mathcal R$. Then $\mathcal R {\restriction_X} \subseteq X \times T$, that is, the restriction of $\mathcal R$ to $X$, is a surjection.
Mapping is Injection if its Direct Image Mapping is Injection
Let $f: S \to T$ be a mapping. Let $f^\to: \powerset S \to \powerset T$ be the direct image mapping of $f$. Let $f^\to$ be an injection. Then $f: S \to T$ is also an injection.
Direct Image Mapping is Bijection iff Mapping is Bijection
Let $\mathcal R \subseteq S \times T$ be a relation. Let $\mathcal R^\to: \powerset S \to \powerset T$ be the direct image mapping of $\mathcal R$. Then $\mathcal R \subseteq S \times T$ is a bijection  $\mathcal R^\to: \powerset S \to \powerset T$ is a bijection.
No Bijection from Set to its Power Set
Let $S$ be a set. Let $\powerset S$ denote the power set of $S$. There is no bijection $f: S \to \powerset S$.
Union of Power Sets
The union of the power sets of two sets $S$ and $T$ is a subset of the power set of their union: :$\powerset S \cup \powerset T \subseteq \powerset {S \cup T}$
Intersection of Power Sets
The intersection of the power sets of two sets $S$ and $T$ is equal to the power set of their intersection: :$\powerset S \cap \powerset T = \powerset {S \cap T}$
Cantor-Bernstein-Schröder Theorem
If a subset of one set is equivalent to the other, and a subset of the other is equivalent to the first, then the two sets are themselves equivalent: :$\forall S, T: T \sim S_1 \subseteq S \land S \sim T_1 \subseteq T \implies S \sim T$
Trivial Ordering is Universally Compatible
Let $S$ be a set. Let $\RR$ be the trivial ordering on $S$. Then $\RR$ is universally compatible.
Identity Mapping is Order Isomorphism
Let $\left({S, \preceq}\right)$ be an ordered set. The identity mapping $I_S$ is an order isomorphism from $\left({S, \preceq}\right)$ to itself.
Inverse of Order Isomorphism is Order Isomorphism
Let $\left({S, \preceq_1}\right)$ and $\left({T, \preceq_2}\right)$ be ordered sets. Let $\phi$ be a bijection from $\left({S, \preceq_1}\right)$ to $\left({T, \preceq_2}\right)$. Then: : $\phi: \left({S, \preceq_1}\right) \to \left({T, \preceq_2}\right)$ is an order isomorphism : : $\phi^{-1}: \left({T, \preceq_2}\right) \to \left({S, \preceq_1}\right)$ is also an order isomorphism.
Composite of Order Isomorphisms is Order Isomorphism
Let $\left({S_1, \preceq_1}\right)$, $\left({S_2, \preceq_2}\right)$ and $\left({S_3, \preceq_3}\right)$ be ordered sets. Let: : $\phi: \left({S_1, \preceq_1}\right) \to \left({S_2, \preceq_2}\right)$ and: : $\psi: \left({S_2, \preceq_2}\right) \to \left({S_3, \preceq_3}\right)$ be order isomorphisms. Then $\psi \circ \phi: \left({S_1, \preceq_1}\right) \to \left({S_3, \preceq_3}\right)$ is also an order isomorphism.
Order Isomorphism is Equivalence Relation
Order isomorphism between ordered sets is an equivalence relation. So any given family of ordered sets can be partitioned into disjoint classes of isomorphic sets.
Dual Ordering is Ordering
Let $\left({S, \preceq}\right)$ be an ordered set. Let $\succeq$ denote the dual ordering of $\preceq$. Then $\succeq$ is an ordering on $S$.
Power Set is Complete Lattice
Let $S$ be a set. Let $\left({\mathcal P \left({S}\right), \subseteq}\right)$ be the relational structure defined on $\mathcal P \left({S}\right)$ by the relation $\subseteq$. Then $\left({\mathcal P \left({S}\right), \subseteq}\right)$ is a complete lattice.
Totally Ordered Set is Lattice
Every totally ordered set is a lattice.
Trichotomy Law (Ordering)
Let $\struct {S, \preceq}$ be an ordered set. Then $\preceq$ is a total ordering : :$\forall a, b \in S: \paren {a \prec b} \lor \paren {a = b} \lor \paren {a \succ b}$ That is, every element either strictly precedes, is the same as, or strictly succeeds, every other element. In other words,  $\prec$ is a trichotomy.
Complement of Reflexive Relation
Let $\RR \subseteq S \times S$ be a relation. Then $\RR$ is reflexive  its complement $\relcomp {S \times S} \RR \subseteq S \times S$ is antireflexive. Likewise, $\RR$ is antireflexive  its complement $\relcomp {S \times S} \RR \subseteq S \times S$ is reflexive.
Strictly Increasing Mapping is Increasing
A mapping that is strictly increasing is an increasing mapping.
Strictly Decreasing Mapping is Decreasing
A mapping that is strictly decreasing is a decreasing mapping.
Strictly Monotone Mapping with Totally Ordered Domain is Injective
Let $\struct {S, \preceq_1}$ be a totally ordered set. Let $\struct {T, \preceq_2}$ be an ordered set. Let $\phi: \struct {S, \preceq_1} \to \struct {T, \preceq_2}$ be a strictly monotone mapping. Then $\phi$ is injective.
Mapping from Totally Ordered Set is Order Embedding iff Strictly Increasing
Let $\struct {S, \preceq_1}$ be a totally ordered set. Let $\struct {T, \preceq_2}$ be an ordered set. Let $\phi: S \to T$ be a mapping. Then $\phi$ is an order embedding  $\phi$ is strictly increasing. That is: :$\forall x, y \in S: x \preceq_1 y \iff \map \phi x \preceq_2 \map \phi y$ : :$\forall x, y \in S: x \prec_1 y \implies \map \phi x \prec_2 \map \phi y$
Finite Totally Ordered Set is Well-Ordered
Every finite totally ordered set is well-ordered.
Subset of Well-Ordered Set is Well-Ordered
Every non-empty subset of a well-ordered set is itself well-ordered.
Element Commutes with Product of Commuting Elements
Let $(S, \circ)$ be a semigroup. Let $x, y, z \in S$. If $x$ commutes with both $y$ and $z$, then $x$ commutes with $y \circ z$.
Associative Idempotent Anticommutative
Let $\circ$ be a binary operation on a set $S$. Let $\circ$ be associative. Then $\circ$ is anticommutative : : $(1): \quad \circ$ is idempotent and: : $(2): \quad \forall a, b \in S: a \circ b \circ a = a$.
Associative and Anticommutative
Let $\circ$ be a binary operation on a set $S$. Let $\circ$ be both associative and anticommutative. Then: :$\forall x, y, z \in S: x \circ y \circ z = x \circ z$
Constant Operation is Commutative
Let $S$ be a set. Let $x \sqbrk c y = c$ be a constant operation on $S$. Then $\sqbrk c$ is a commutative operation: :$\forall x, y \in S: x \sqbrk c y = y \sqbrk c x$
Constant Operation is Associative
Let $S$ be a set. Let $x \left[{c}\right] y = c$ be a constant operation on $S$. Then $\left[{c}\right]$ is an associative operation: :$\forall x, y, z \in S: \left({x \left[{c}\right] y}\right) \left[{c}\right] z = x \left[{c}\right] \left({y \left[{c}\right] z}\right)$
Left Operation is Idempotent
The left operation is idempotent: :$\forall x: x \leftarrow x = x$
Left Operation is Associative
The left operation is associative: :$\forall x, y, z: \paren {x \leftarrow y} \leftarrow z = x \leftarrow \paren {y \leftarrow z}$
Right Operation is Associative
The right operation is associative: :$\forall x, y, z: \paren {x \rightarrow y} \rightarrow z = x \rightarrow \paren {y \rightarrow z}$
Max and Min are Commutative
The Max and Min operations are commutative: : $\max \left({x, y}\right) = \max \left({y, x}\right)$ : $\min \left({x, y}\right) = \min \left({y, x}\right)$
Max and Min are Associative
The Max and Min operations are associative: : $\max \left({\max \left({x, y}\right), z}\right) = \max \left({x, \max \left({y, z}\right)}\right)$ : $\min \left({\min \left({x, y}\right), z}\right) = \min \left({x, \min \left({y, z}\right)}\right)$ Thus we are justified in writing $\max \left({x, y, z}\right)$ and $\min \left({x, y, z}\right)$.
Max and Min are Idempotent
The Max and Min operations are idempotent: : $\max \left({x, x}\right) = x$ : $\min \left({x, x}\right) = x$
Max and Min Operations are Distributive over Each Other
The Max and Min operations are distributive over each other: :$\max \set {x, \min \set {y, z} } = \min \set {\max \set {x, y}, \max \set {x, z} }$ :$\max \set {\min \set {x, y}, z} = \min \set {\max \set {x, z}, \max \set {y, z} }$ :$\min \set {x, \max \set {y, z} } = \max \set {\min \set {x, y}, \min \set {x, z} }$ :$\min \set {\max \set {x, y}, z} = \max \set {\min \set {x, z}, \min \set {y, z} }$
Equality of Algebraic Structures
Two algebraic structures $\struct {S, \circ}$ and $\struct {T, *}$ are equal : :$S = T$ :$\forall a, b \in S: a \circ b = a * b$
Restriction of Associative Operation is Associative
Let $\struct {S, \circ}$ be an semigroup. Let $T \subseteq S$. Let $T$ be closed under $\circ$. Then $\struct {T, \circ {\restriction_T} }$ is also a semigroup, where $\circ {\restriction_T}$ is the restriction of $\circ$ to $T$.
Restriction of Commutative Operation is Commutative
Let $\struct {S, \circ}$ be an algebraic structure. Let $T \subseteq S$. Let the operation $\circ$ be commutative on $\struct {S, \circ}$. Then the restriction $\circ {\restriction_T}$ of $\circ$ to $T$ is also commutative.
Idempotent Magma Element forms Singleton Submagma
Let $\struct {S, \circ}$ be a magma. Let $x \in S$ be an idempotent element of $\struct {S, \circ}$. Then  $\struct {\set x, \circ}$ is a submagma of $\struct {S, \circ}$.
Magma Subset Product with Self
Let $\left({S, \circ}\right)$ be a magma. Let $T \subseteq S$. Then $\left({T, \circ}\right)$ is a magma  $T \circ T \subseteq T$, where $T \circ T$ is the subset product of $T$ with itself.
Subset Product within Semigroup is Associative
Let $\struct {S, \circ}$ be a semigroup. Then the operation $\circ_\PP$ induced on the power set of $S$ is also associative.
Subset Product within Commutative Structure is Commutative
Let $\struct {S, \circ}$ be a magma. If $\circ$ is commutative, then the operation $\circ_\mathcal P$ induced on the power set of $S$ is also commutative.
Subset of Subset Product
Let $\struct {S, \circ}$ be a magma. Let $\powerset S$ be the power set of $S$. Let $X, Y, Z \in \powerset S$. Then: :$X \subseteq Y \implies \paren {X \circ Z} \subseteq \paren {Y \circ Z}$ :$X \subseteq Y \implies \paren {Z \circ X} \subseteq \paren {Z \circ Y}$ where $X \circ Z$ etc. denotes subset product.
Cancellable Element is Cancellable in Subset
Let $\left ({S, \circ}\right)$ be an algebraic structure. Let $\left ({T, \circ}\right) \subseteq \left ({S, \circ}\right)$. Let $x \in T$ be cancellable in $S$. Then $x$ is also cancellable in $T$.
Subsemigroup Closure Test
To show that an algebraic structure $\left({T, \circ}\right)$ is a subsemigroup of a semigroup $\struct {S, \circ}$, we need to show only that: : $(1): \quad T \subseteq S$ : $(2): \quad \circ$ is a closed operation in $T$.
Cancellable iff Regular Representations Injective
Let $\struct {S, \circ}$ be an algebraic structure. Then $a \in S$ is cancellable : ::the left regular representation $\map {\lambda_a} x$ is injective and ::the right regular representation $\map {\rho_a} x$ is injective.
More than one Left Identity then no Right Identity
Let $\struct {S, \circ}$ be an algebraic structure. If $\struct {S, \circ}$ has more than one left identity, then it has no right identity.
Element under Left Operation is Right Identity
Let $\struct {S, \leftarrow}$ be an algebraic structure in which the operation $\leftarrow$ is the left operation. Then no matter what $S$ is, $\struct {S, \leftarrow}$ is a semigroup all of whose elements are right identities. Thus it can be seen that any right identity in a semigroup is not necessarily unique.
Element under Right Operation is Left Identity
Let $\struct {S, \rightarrow}$ be an algebraic structure in which the operation $\rightarrow$ is the right operation. Then $\struct {S, \rightarrow}$ is a semigroup all of whose elements are left identities. Thus it can be seen that any left identity in a semigroup is not necessarily unique.
Left and Right Identity are the Same
Let $\struct {S, \circ}$ be an algebraic structure. Let $e_L \in S$ be a left identity, and $e_R \in S$ be a right identity. Then: :$e_L = e_R$ that is, both the left identity and right identity are the same, and are therefore an identity $e$. Furthermore, $e$ is the only left identity and right identity for $\circ$.
Identity Property in Semigroup
Let $\struct {S, \circ}$ be a semigroup. Let $s \in S$ be such that: :$\forall a \in S: \exists x, y \in S: s \circ x = a = y \circ s$ Then $\struct {S, \circ}$ has an identity.
Identity of Monoid is Cancellable
The identity of a monoid is cancellable.
Identity is only Idempotent Cancellable Element
Let $e_S$ is the identity of an algebraic structure $\left({S, \circ}\right)$. Then $e_S$ is the only cancellable element of $\left({S, \circ}\right)$ that is idempotent.
Set of all Self-Maps is Monoid
Let $S$ be a set. Let $S^S$ be the set of all mappings from $S$ to itself. Let the operation $\circ$ represent composition of mappings. Then the algebraic structure $\struct {S^S, \circ}$ is a monoid whose identity element is the identity mapping on $S$.
Test for Submonoid
To show that $\left({T, \circ}\right)$ is a submonoid of a monoid $\left({S, \circ}\right)$, we need to show that: :$(1): \quad T \subseteq S$ :$(2): \quad \left({T, \circ}\right)$ is a magma (i.e. that it is closed) :$(3): \quad \left({T, \circ}\right)$ has an identity.
Identity of Cancellable Monoid is Identity of Submonoid
Let $\struct {S, \circ}$ be a monoid, all of whose elements are cancellable. Let $\struct {T, \circ}$ be a submonoid of $\struct {S, \circ}$. Then the identity of $\struct {T, \circ}$ is the same element as the identity of $\struct {S, \circ}$.
Cancellable Elements of Monoid form Submonoid
The cancellable elements of a monoid $\struct {S, \circ}$ form a submonoid of $\struct {S, \circ}$.
Product of Semigroup Element with Left Inverse is Idempotent
Let $\struct {S, \circ}$ be a semigroup with a left identity $e_L$. Let $x \in S$ such that $\exists x_L: x_L \circ x = e_L$, that is $x$ has a left inverse with respect to the left identity. Then: :$\paren {x \circ x_L} \circ \paren {x \circ x_L} = x \circ x_L$ That is, $x \circ x_L$ is idempotent.
Product of Semigroup Element with Right Inverse is Idempotent
Let $\struct {S, \circ}$ be a semigroup with a right identity $e_R$. Let $x \in S$ such that $\exists x_R: x \circ x_R = e_R$, i.e. $x$ has a right inverse with respect to the right identity. Then: :$\paren {x_R \circ x} \circ \paren {x_R \circ x} = x_R \circ x$ That is, $x_R \circ x$ is idempotent.
Left Inverse for All is Right Inverse
Let $\struct {S, \circ}$ be a semigroup with a left identity $e_L$ such that: :$\forall x \in S: \exists x_L: x_L \circ x = e_L$ That is, every element of $S$ has a left inverse with respect to the left identity. Then $x \circ x_L = e_L$, that is, $x_L$ is also a right inverse with respect to the left identity.
Left Inverse and Right Inverse is Inverse
Let $\struct {S, \circ}$ be a monoid with identity element $e_S$. Let $x \in S$ such that $x$ has both a left inverse and a right inverse. That is: :$\exists x_L \in S: x_L \circ x = e_S$ :$\exists x_R \in S: x \circ x_R = e_S$ Then $x_L = x_R$, that is, $x$ has an inverse. Furthermore, that element is the ''only'' inverse (both right and left) for $x$
Left and Right Inverses of Product
Let $\struct {S, \circ}$ be a monoid whose identity is $e_S$. Let $x, y \in S$. Let: :$(1): \quad x \circ y$ have a left inverse for $\circ$ :$(2): \quad y \circ x$ have a right inverse for $\circ$. Then both $x$ and $y$ are invertible for $\circ$.
Equivalence of Definitions of Self-Inverse
Let $\left({S, \circ}\right)$ be a monoid whose identity is $e_S$. Let $x \in S$. {{TFAE|def = Self-Inverse Element|context = Abstract Algebra}}
Inverse of Identity Element is Itself
Let $\struct {S, \circ}$ be an algebraic structure with an identity element $e$. Let the inverse of $e$ be $e^{-1}$. Then: : $e^{-1} = e$ That is, $e$ is self-inverse.
Invertible Element of Associative Structure is Cancellable
Let $\left({S, \circ}\right)$ be an algebraic structure where $\circ$ is associative. Let $\left({S, \circ}\right)$ have an identity element $e_S$. An element of $\left({S, \circ}\right)$ which is invertible is also cancellable.
Regular Representation of Invertible Element is Permutation
Let $\struct {S, \circ}$ be a monoid. Let $a \in S$ be invertible. Then the left regular representation $\lambda_a$ and the right regular representation $\rho_a$ are permutations of $S$.
Group is Inverse Semigroup with Identity
A group is an inverse semigroup with an identity.
Invertible Elements of Monoid form Subgroup of Cancellable Elements
Let $\struct {S, \circ}$ be an monoid whose identity is $e_S$. Let $C$ be the set of all cancellable elements of $S$. Let $T$ be the set of all invertible elements of $S$. Then $\struct {T, \circ}$ is a subgroup of $\struct {C, \circ}$.
Structure Induced by Associative Operation is Associative
Let $\struct {T, \circ}$ be an algebraic structure, and let $S$ be a set. Let $\struct {T^S, \oplus}$ be the structure on $T^S$ induced by $\circ$. Let $\circ$ be associative. Then the pointwise operation $\oplus$ induced on $T^S$ by $\circ$ is also associative.
Structure Induced by Commutative Operation is Commutative
Let $\struct {T, \circ}$ be an algebraic structure, and let $S$ be a set. Let $\struct {T^S, \oplus}$ be the structure on $T^S$ induced by $\circ$. Let $\circ$ be a commutative operation. Then the pointwise operation $\oplus$ induced on $T^S$ by $\circ$ is also commutative.
Induced Structure Identity
Let $\struct {T, \circ}$ be an algebraic structure, and let $S$ be a set. Let $\struct {T^S, \oplus}$ be the structure on $T^S$ induced by $\circ$. Let $e$ be an identity for $\circ$. Then the constant mapping $f_e: S \to T$ defined as: :$\forall x \in S: \map {f_e} x = e$ is the identity for the pointwise operation $\oplus$ induced on $T^S$ by $\circ$.
Structure Induced by Abelian Group Operation is Abelian Group
Let $\struct {G, \circ}$ be an abelian group whose identity is $e$. Let $S$ be a set. Let $\struct {G^S, \oplus}$ be the structure on $G^S$ induced by $\circ$. Then $\struct {G^S, \oplus}$ is an abelian group.
Power Set with Union is Commutative Monoid
Let $S$ be a set and let $\powerset S$ be its power set. Then $\struct {\powerset S, \cup}$ is a commutative monoid whose identity is $\O$. The only invertible element of this structure is $\O$. Thus (except in the degenerate case $S = \O$) $\struct {\powerset S, \cup}$ cannot be a group.
Power Set with Intersection is Commutative Monoid
Let $S$ be a set and let $\powerset S$ be its power set. Then $\struct {\powerset S, \cap}$ is a commutative monoid whose identity is $S$. The only invertible element of this structure is $S$. Thus (except in the degenerate case $S = \O$) $\struct {\powerset S, \cap}$ cannot be a group.
Diagonal Relation is Universally Compatible
The diagonal relation $\Delta_S$ on a set $S$ is universally compatible with every operation on $S$.
Trivial Relation is Universally Congruent
The trivial relation $\mathcal R = S \times S$ on a set $S$ is universally congruent with every closed operation on $S$.
Quotient Structure is Well-Defined
Let $\struct {S, \circ}$ be an algebraic structure. Let $\RR$ be a congruence relation on $\struct {S, \circ}$. Let $S / \RR$ be the quotient set of $S$ by $\RR$. Let $\circ_\RR$ be the operation induced on $S / \RR$ by $\circ$. Then $\circ_\RR$ is a well-defined operation in the quotient structure $\struct {S / \RR, \circ_\RR}$.
External Direct Product Closure
Let $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$ be algebraic structures. Let $\struct {S \times T, \circ}$ be the external direct product of $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$. Let $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$ be closed. Then $\struct {S \times T, \circ}$ is also closed.
External Direct Product Associativity
Let $\struct {S \times T, \circ}$ be the external direct product of the two algebraic structures $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$. Let $\circ_1$ and $\circ_2$ be associative. Then $\circ$ is also associative.
External Direct Product of Semigroups
The external direct product of two semigroups is itself a semigroup.
External Direct Product Commutativity
Let $\left({S \times T, \circ}\right)$ be the external direct product of the two algebraic structures $\left({S, \circ_1}\right)$ and $\left({T, \circ_2}\right)$. Let $\circ_1$ and $\circ_2$ be commutative operations. Then $\circ$ is also a commutative operation.
External Direct Product Identity
Let $\struct {S \times T, \circ}$ be the external direct product of the two monoids $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$. Let: : $e_S$ be the identity for $\struct {S, \circ_1}$ and: : $e_T$ be the identity for $\struct {T, \circ_2}$. Then $\tuple {e_S, e_T}$ is the identity for $\struct {S \times T, \circ}$.
External Direct Product Inverses
Let $\struct {S \times T, \circ}$ be the external direct product of the two monoids $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$. Let: :$s^{-1}$ be an inverse of $s \in \struct {S, \circ_1}$ and: :$t^{-1}$ be an inverse of $t \in \struct {T, \circ_2}$. Then $\tuple {s^{-1}, t^{-1} }$ is an inverse of $\tuple {s, t} \in \struct {S \times T, \circ}$.
Morphism Property Preserves Closure
Let $\phi: \left({S, \circ_1, \circ_2, \ldots, \circ_n}\right) \to \left({T, *_1, *_2, \ldots, *_n}\right)$ be a mapping from one algebraic structure $\left({S, \circ_1, \circ_2, \ldots, \circ_n}\right)$ to another $\left({T, *_1, *_2, \ldots, *_n}\right)$. Let $\circ_k$ have the morphism property under $\phi$ for some operation $\circ_k$ in $\left({S, \circ_1, \circ_2, \ldots, \circ_n}\right)$. Then the following properties hold: : If $S' \subseteq S$ is closed under $\circ_k$, then $\phi \left[{S'}\right]$ is closed under $*_k$ : If $T' \subseteq T$ is closed under $*_k$, then $\phi^{-1} \left[{T'}\right]$ is closed under $\circ_k$ where $\phi \left[{S'}\right]$ denotes the image of $S'$.
Morphism Property Preserves Cancellability
Let: : $\phi: \left({S, \circ_1, \circ_2, \ldots, \circ_n}\right) \to \left({T, *_1, *_2, \ldots, *_n}\right)$ be a mapping from one algebraic structure: : $\left({S, \circ_1, \circ_2, \ldots, \circ_n}\right)$ to another: : $\left({T, *_1, *_2, \ldots, *_n}\right)$ Let $\circ_k$ have the morphism property under $\phi$ for some operation $\circ_k$ in $\left({S, \circ_1, \circ_2, \ldots, \circ_n}\right)$. Then if an element $a \in S$ is either left cancellable or right cancellable under $\circ_k$, then $\phi \left({a}\right)$ is correspondingly left cancellable or right cancellable under $*_k$. Thus, the morphism property is seen to preserve cancellability.
Quotient Mapping on Structure is Canonical Epimorphism
Let $\mathcal R$ be a congruence relation on an algebraic structure $\struct {S, \circ}$. Then the quotient mapping from $\struct {S, \circ}$ to the quotient structure $\struct {S / \mathcal R, \circ_\mathcal R}$ is an epimorphism: :$q_\mathcal R: \struct {S, \circ} \to \struct {S / \mathcal R, \circ_\mathcal R}: \forall x, y \in S: \map {q_\mathcal R} {x \circ y} = \map {q_\mathcal R} x \circ_\mathcal R \map {q_\mathcal R} y$ This is sometimes called the '''canonical epimorphism''' from $\struct {S, \circ}$ to $\struct {S / \mathcal R, \circ_\mathcal R}$.
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 $\operatorname{Im} \left({\phi}\right)$.