hcju/proofwikips huggingface dataset:
Image of Empty Set is Empty Set
Let $\RR \subseteq S \times T$ be a relation. The image of the empty set is the empty set: :$\RR \sqbrk \O = \O$
Domain of Composite Relation
Let $\RR_2 \circ \RR_1$ be a composite relation. Then the domain of $\RR_2 \circ \RR_1$ is the domain of $\RR_1$: :$\Dom {\RR_2 \circ \RR_1} = \Dom {\RR_1}$
Codomain of Composite Relation
Let $\RR_2 \circ \RR_1$ be a composite relation. Then the codomain of $\RR_2 \circ \RR_1$ is the codomain of $\RR_2$: :$\Cdm {\RR_2 \circ \RR_1} = \Cdm {\RR_2}$
Preimage of Mapping equals Domain
The preimage of a mapping is the same set as its domain: :$\Preimg f = \Dom f$
Preimage of Relation is Subset of Domain
Let $\mathcal R \subseteq S \times T$ be a relation. Then the preimage of $\mathcal R$ is a subset of its domain: :$\Preimg {\mathcal R} \subseteq S$
Preimage of Image under Left-Total Relation is Superset
Let $\RR \subseteq S \times T$ be a left-total relation. Then: :$A \subseteq S \implies A \subseteq \paren {\RR^{-1} \circ \RR} \sqbrk A$ where: :$\RR \sqbrk A$ denotes the image of $A$ under $\RR$ :$\RR^{-1} \sqbrk A$ denotes the preimage of $A$ under $\RR$ :$\RR^{-1} \circ \RR$ denotes composition of $\RR^{-1}$ and $\RR$. This can be expressed in the language and notation of direct image mappings and inverse image mappings as: :$\forall A \in \powerset S: A \subseteq \map {\paren {\RR^\gets \circ \RR^\to} } A$
Inverse of Composite Relation
Let $\RR_2 \circ \RR_1 \subseteq S_1 \times S_3$ be the composite of the two relations $\RR_1 \subseteq S_1 \times S_2$ and $\RR_2 \subseteq S_2 \times S_3$. Then: :$\paren {\RR_2 \circ \RR_1}^{-1} = \RR_1^{-1} \circ \RR_2^{-1}$
Image of Union under Relation
Let $S$ and $T$ be sets. Let $\RR \subseteq S \times T$ be a relation. Let $S_1$ and $S_2$ be subsets of $S$. Then: :$\RR \sqbrk {S_1 \cup S_2} = \RR \sqbrk {S_1} \cup \RR \sqbrk {S_2}$ That is, the image of the union of subsets of $S$ is equal to the union of their images.
Image of Intersection under Relation
Let $S$ and $T$ be sets. Let $\RR \subseteq S \times T$ be a relation. Let $S_1$ and $S_2$ be subsets of $S$. Then: :$\RR \sqbrk {S_1 \cap S_2} \subseteq \RR \sqbrk {S_1} \cap \RR \sqbrk {S_2}$ That is, the image of the intersection of subsets of $S$ is a subset of the intersection of their images.
Image of Set Difference under Relation
Let $\RR \subseteq S \times T$ be a relation. Let $A$ and $B$ be subsets of $S$. Then: :$\RR \sqbrk A \setminus \RR \sqbrk B \subseteq \RR \sqbrk {A \setminus B}$ where: :$\setminus$ denotes set difference :$\RR \sqbrk A$ denotes image of $A$ under $\RR$.
Preimage of Union under Relation
Let $S$ and $T$ be sets. Let $\RR \subseteq S \times T$ be a relation. Let $T_1$ and $T_2$ be subsets of $T$. Then: :$\RR^{-1} \sqbrk {T_1 \cup T_2} = \RR^{-1} \sqbrk {T_1} \cup \RR^{-1} \sqbrk {T_2}$
Preimage of Intersection under Relation
Let $S$ and $T$ be sets. Let $\mathcal R \subseteq S \times T$ be a relation. Let $C$ and $D$ be subsets of $T$. Then: :$\mathcal R^{-1} \left[{C \cap D}\right] \subseteq \mathcal R^{-1} \left[{C}\right] \cap \mathcal R^{-1} \left[{D}\right]$
Restriction is Subset of Relation
Let $\mathcal R \subseteq S \times T$ be a relation. Let $X \subseteq S$. Then the restriction of $\mathcal R$ to $X$ is a subset of $\mathcal R$.
Null Relation is Antireflexive, Symmetric and Transitive
Let $S$ be a set which is non-empty. Let $\RR \subseteq S \times S$ be the null relation. Then $\RR$ is antireflexive, symmetric and transitive. If $S = \O$ then Relation on Empty Set is Equivalence applies.
Symmetry of Relations is Symmetric
Let $\mathcal R$ be a relation on $S$ which is symmetric. Then: :$\left({x, y}\right) \in \mathcal R \iff \left({y, x}\right) \in \mathcal R$.
Relation is Symmetric and Antisymmetric iff Coreflexive
Let $S$ be a set. Let $\RR \subseteq S \times S$ be a relation in $S$. Then: :$\RR$ is both symmetric and antisymmetric : :$\RR$ is coreflexive.
Relation both Symmetric and Asymmetric is Null
Let $\RR$ be a relation in $S$ which is both symmetric and asymmetric. Then: :$\RR = \O$
Equivalent Characterizations of Abelian Group
Let $G$ be a group.  {{begin-eqn}} {{eqn | n = 1       | o =       | c = $G$ is abelian }} {{eqn | n = 2       | o =       | c = $\forall a, b \in G: \paren {a b}^{-1} = a^{-1} b^{-1}$ }} {{eqn | n = 3       | o =       | c = Cross cancellation property: $\forall a, b, c \in G: a b = c a \implies b = c$ }} {{eqn | n = 4       | o =       | c = Middle cancellation property: $\forall a, b, c, d, x \in G: a x b = c x d \implies a b = c d$  }} {{end-eqn}}
Asymmetric Relation is Antisymmetric
Let $\RR$ be an asymmetric relation. Then $\RR$ is also antisymmetric.
Asymmetric Relation is Antireflexive
Let $S$ be a set. Let $\RR \subseteq S \times S$ be a relation on $S$. Let $\RR$ be asymmetric. Then $\RR$ is also antireflexive.
Antireflexive and Transitive Relation is Asymmetric
Let $\mathcal R \subseteq S \times S$ be a relation which is not null. Let $\mathcal R$ be antireflexive and transitive. Then $\mathcal R$ is also asymmetric.
Antitransitive Relation is Antireflexive
Let $S$ be a set. Let $\mathcal R \subseteq S \times S$ be a relation on $S$. Let $\mathcal R$ be antitransitive. Then $\mathcal R$ is also antireflexive.
Symmetric Transitive and Serial Relation is Reflexive
Let $\RR$ be a relation which is: :symmetric :transitive :serial. Then $\RR$ is reflexive. Thus such a relation is an equivalence.
Inverse Relation Properties
Let $\mathcal R$ be a relation on a set $S$. If $\mathcal R$ has any of the properties: * Reflexive * Antireflexive * Non-reflexive * Symmetric * Asymmetric * Antisymmetric * Non-symmetric * Transitive * Antitransitive * Non-transitive ... then its inverse $\mathcal R^{-1}$ has the same properties.
Properties of Restriction of Relation
Let $S$ be a set. Let $\mathcal R \subseteq S \times S$ be a relation on $S$. Let $T \subseteq S$ be a subset of $S$. Let $\mathcal R \restriction_T \ \subseteq T \times T$ be the restriction of $\mathcal R$ to $T$. If $\mathcal R$ on $S$ has any of the properties: * Reflexive * Antireflexive * Symmetric * Antisymmetric * Asymmetric * Transitive * Antitransitive * Connected ... then $\mathcal R \restriction_T$ on $T$ has the same properties.
Relation is Antireflexive iff Disjoint from Diagonal Relation
Let $\mathcal R \subseteq S \times S$ be a relation on a set $S$. Then: : $\mathcal R$ is antireflexive iff : $\Delta_S \cap \mathcal R = \varnothing$ where $\Delta_S$ is the diagonal relation.
Inverse Relation Equal iff Subset
If a relation $\RR$ is a subset or superset of its inverse, then it equals its inverse. That is, the following are equivalent: {{begin-axiom}} {{axiom | n = 1         | m = \RR \subseteq \RR^{-1} }} {{axiom | n = 2         | m = \RR^{-1} \subseteq \RR }} {{axiom | n = 3         | m = \RR = \RR^{-1} }} {{end-axiom}}
Relation is Symmetric iff Inverse is Symmetric
A relation $\mathcal R$ is symmetric  its inverse $\mathcal R^{-1}$ is also symmetric.
Inverse of Many-to-One Relation is One-to-Many
The inverse of a many-to-one relation is a one-to-many relation, and vice versa.
Many-to-One Relation Composite with Inverse is Transitive
Let $\mathcal R \subseteq S \times T$ be a relation which is many-to-one. Then the composites (both ways) of $\mathcal R$ and its inverse $\mathcal R^{-1}$, that is, both $\mathcal R^{-1} \circ \mathcal R$ and $\mathcal R \circ \mathcal R^{-1}$, are transitive.
One-to-Many Relation Composite with Inverse is Coreflexive
Let $\mathcal R \subseteq S \times S$ be a relation which is one-to-many. Then the composite of $\mathcal R$ with its inverse is a coreflexive relation: :$\mathcal R^{-1} \circ \mathcal R \subseteq \Delta_X$ {{refactor|This is a separate result.}} That is, by Relation is Symmetric and Antisymmetric iff Coreflexive, $\mathcal R^{-1} \circ \mathcal R$ is both symmetric and antisymmetric.
Element in its own Equivalence Class
Let $\mathcal R$ be an equivalence relation on a set $S$. Then every element of $S$ is in its own $\mathcal R$-class: :$\forall x \in S: x \in \eqclass x {\mathcal R}$
Equivalence Class of Element is Subset
Let $\mathcal R$ be an equivalence relation on a set $S$. The $\mathcal R$-class of every element of $S$ is a subset of the set the element is in: :$\forall x \in S: \eqclass x {\mathcal R} \subseteq S$
Equivalence Class is not Empty
Let $\mathcal R$ be an equivalence relation on a set $S$. Then no $\mathcal R$-class is empty.
Handshake Lemma
Let $G$ be a $\tuple {p, q}$-undirected graph, which may be a multigraph or a loop-graph, or both. Let $V = \set {v_1, v_2, \ldots, v_p}$ be the vertex set of $G$. Then: :$\displaystyle \sum_{i \mathop = 1}^p \map {\deg_G} {v_i} = 2 q$ where $\map {\deg_G} {v_i}$ is the degree of vertex $v_i$. That is, the sum of all the degrees of all the vertices of an graph is equal to twice its size. This result is known as the '''Handshake Lemma''' or '''Handshaking Lemma'''.
Equivalence Class holds Equivalent Elements
Let $\RR$ be an equivalence relation on a set $S$. Then: :$\tuple {x, y} \in \RR \iff \eqclass x \RR = \eqclass y \RR$
Equivalence Classes are Disjoint
Let $\RR$ be an equivalence relation on a set $S$. Then all $\RR$-classes are pairwise disjoint: :$\tuple {x, y} \notin \RR \iff \eqclass x \RR \cap \eqclass y \RR = \O$
Fundamental Theorem on Equivalence Relations
Let $\RR \subseteq S \times S$ be an equivalence on a set $S$. Then the quotient $S / \RR$ of $S$ by $\RR$ forms a partition of $S$.
Equivalence Class is Unique
Let $\mathcal R$ be an equivalence relation on $S$. For each $x \in S$, the one and only one $\mathcal R$-class to which $x$ belongs is $\eqclass x {\mathcal R}$.
Equivalence Class Equivalent Statements
Let $\mathcal R$ be an equivalence on $S$. {{refactor|Use TFAE template}} Then $\forall x, y \in S$, the following statements are all equivalent: :$(1): \quad x$ and $y$ are in the same $\mathcal R$-class :$(2): \quad \eqclass x {\mathcal R} = \eqclass y {\mathcal R}$ :$(3): \quad x \mathrel {\mathcal R} y$ :$(4): \quad x \in \eqclass y {\mathcal R}$ :$(5): \quad y \in \eqclass x {\mathcal R}$ :$(6): \quad \eqclass x {\mathcal R} \cap \eqclass y {\mathcal R} \ne \O$
Relation Induced by Partition is Equivalence
Let $\mathbb S$ be a partition of a set $S$. Let $\RR$ be the relation induced by $\mathbb S$. Then: :$(1): \quad \RR$ is unique :$(2): \quad \RR$ is an equivalence relation on $S$. Hence $\mathbb S$ is the quotient set of $S$ by $\RR$, that is: :$\mathbb S = S / \RR$
Relation Partitions Set iff Equivalence
Let $\RR$ be a relation on a set $S$. Then $S$ can be partitioned into subsets by $\RR$  $\RR$ is an equivalence relation on $S$. The partition of $S$ defined by $\RR$ is the quotient set $S / \RR$.
Intersection of Equivalences
The intersection of two equivalence relations is itself an equivalence relation.
Equivalence of Definitions of Equivalence Relation
Let $\RR$ be a relation on a set $S$. {{TFAE|def = Equivalence Relation}}
Equivalence iff Diagonal and Inverse Composite
Let $\RR$ be a relation on $S$. Then $\RR$ is an equivalence relation on $S$ : :$\Delta_S \subseteq \RR$ and: :$\RR = \RR \circ \RR^{-1}$
Equality of Elements in Range of Mapping
Let $f: S \to T$ be a mapping. Then: :$\exists y \in \Rng f: \tuple {x_1, y} \in f \land \tuple {x_2, y} \in f \iff \map f {x_1} = \map f {x_2}$
Inverse of Mapping is One-to-Many Relation
Let $f$ be a mapping. Then its inverse $f^{-1}$ is a one-to-many relation.
Equality of Mappings
Two mappings $f_1: S_1 \to T_1, f_2: S_2 \to T_2$ are equal : :$(1): \quad S_1 = S_2$ :$(2): \quad T_1 = T_2$ :$(3): \quad \forall x \in S_1: \map {f_1} x = \map {f_2} x$
Mapping is Constant iff Image is Singleton
A mapping is a constant mapping  its image is a singleton.
Diagonal Relation is Right Identity
Let $\RR \subseteq S \times T$ be a relation on $S \times T$. Then: :$\RR \circ \Delta_S = \RR$ where $\Delta_S$ is the diagonal relation on $S$, and $\circ$ signifies composition of relations.
Identity Mapping is Left Identity
Let $S$ and $T$ be sets. Let $f: S \to T$ be a mapping. Then: :$I_T \circ f = f$ where $I_T$ is the identity mapping on $T$, and $\circ$ signifies composition of mappings.
Image of Element under Inverse Mapping
Let $S$ and $T$ be sets. Let $f: S \to T$ be a mapping such that its inverse $f^{-1}: T \to S$ is also a mapping. Then: :$\forall x \in S, y \in T: \map f x = y \iff \map {f^{-1} } y = x$
Injection iff Left Cancellable
A mapping $f$ is an injection  $f$ is left cancellable.
Identity Mapping is Injection
On any set $S$, the identity mapping $I_S: S \to S$ is an injection.
Composite of Injections is Injection
A composite of injections is an injection. That is: :If $f$ and $g$ are injections, then so is $f \circ g$.
Injection if Composite is Injection
Let $f$ and $g$ be mappings such that their composite $g \circ f$ is an injection. Then $f$ is an injection.
Injection iff Left Inverse
A mapping $f: S \to T, S \ne \O$ is an injection : :$\exists g: T \to S: g \circ f = I_S$ where $g$ is a mapping. That is,  $f$ has a left inverse.
Inclusion Mapping is Injection
Let $S, T$ be sets such that $S$ is a subset of $T$. Then the inclusion mapping $i_S: S \to T$ defined as: :$\forall x \in S: \map {i_S} x = x$ is an injection. For this reason the inclusion mapping can be known as the '''canonical injection of $S$ to $T$'''.
Surjection iff Right Cancellable
Let $f$ be a mapping. Then $f$ is a surjection  $f$ is right cancellable.
Identity Mapping is Surjection
On any set $S$, the identity mapping $I_S: S \to S$ is a surjection.
Restriction of Mapping to Image is Surjection
Let $f: S \to T$ be a mapping. Let $g: S \to \Img f$ be the restriction of $f$ to $S \times \Img f$. Then $g$ is a surjective restriction of $f$.
Composite of Surjections is Surjection
A composite of surjections is a surjection. That is: :If $g$ and $f$ are surjections, then so is $g \circ f$.
Surjection if Composite is Surjection
Let $f: S_1 \to S_2$ and $g: S_2 \to S_3$ be mappings such that $g \circ f$ is a surjection. Then $g$ is a surjection.
Surjection iff Right Inverse
A mapping $f: S \to T, S \ne \O$ is a surjection : :$\exists g: T \to S: f \circ g = I_T$ where: :$g$ is a mapping :$I_T$ is the identity mapping on $T$. That is,  $f$ has a right inverse.
Identity Mapping is Bijection
The identity mapping $I_S: S \to S$ on the set $S$ is a bijection.
Identity Mapping is Permutation
The identity mapping $I_S: S \to S$ on the set $S$ is a permutation.
Bijection iff Left and Right Inverse
Let $f: S \to T$ be a mapping. $f$ is a bijection : :$(1): \quad \exists g_1: T \to S: g_1 \circ f = I_S$ :$(2): \quad \exists g_2: T \to S: f \circ g_2 = I_T$ where both $g_1$ and $g_2$ are mappings. It also follows that it is necessarily the case that $g_1 = g_2$ for such to be possible.
Inverse of Bijection is Bijection
Let $f: S \to T$ be a bijection in the sense that: :$(1): \quad f$ is an injection :$(2): \quad f$ is a surjection. Then the inverse $f^{-1}$ of $f$ is itself a bijection by the same definition.
Inverse Element of Bijection
Let $S$ and $T$ be sets. Let $f: S \to T$ be a bijection. Then: :$\map {f^{-1} } y = x \iff \map f x = y$ where $f^{-1}$ is the inverse mapping of $f$.
Composite of Bijection with Inverse is Identity Mapping
Let $f: S \to T$ be a bijection. Then: : $f^{-1} \circ f = I_S$ : $f \circ f^{-1} = I_T$ where $I_S$ and $I_T$ are the identity mappings on $S$ and $T$ respectively.
Inverse of Inverse of Bijection
Let $f: S \to T$ be a bijection. Then: :$\paren {f^{-1} }^{-1} = f$ where $f^{-1}$ is the inverse of $f$.
Composite of Bijections is Bijection
Let $f$ and $g$ be mappings such that $\Dom f = \Cdm g$. Then: :If $f$ and $g$ are both bijections, then so is $f \circ g$ where $f \circ g$ is the composite mapping of $f$ with $g$.
Inverse of Composite Bijection
Let $f$ and $g$ be bijections such that $\Dom g = \Cdm f$. Then: :$\paren {g \circ f}^{-1} = f^{-1} \circ g^{-1}$ and $f^{-1} \circ g^{-1}$ is itself a bijection.
Composite of Permutations is Permutation
Let $f, g$ are permutations of a set $S$. Then their composite $g \circ f$ is also a permutation of $S$.
Inverse of Permutation is Permutation
If $f$ is a permutation of $S$, then so is its inverse $f^{-1}$.
Left Inverse Mapping is Surjection
Let $f: S \to T$ be an injection. Let $g: T \to S$ be a left inverse of $f$. Then $g$ is a surjection.
Right Inverse Mapping is Injection
Let $f: S \to T$ be a mapping. Let $g: T \to S$ be a right inverse of $f$. Then $g$ is an injection.
Set Equivalence is Equivalence Relation
Set equivalence is an equivalence relation.
Set Equivalence Less One Element
Let $S$ and $T$ be sets which are equivalent: :$S \sim T$ Let $a \in S$ and $b \in T$. Then: : $S \setminus \left\{{a}\right\} \sim T \setminus \left\{{b}\right\}$ where $\setminus$ denotes set difference.
Image of Union under Mapping
Let $S$ and $T$ be sets. Let $f: S \to T$ be a mapping. Let $A$ and $B$ be subsets of $S$. Then: :$f \sqbrk {A \cup B} = f \sqbrk A \cup f \sqbrk B$ This can be expressed in the language and notation of direct image mappings as: :$\forall A, B \in \powerset S: \map {f^\to} {A \cup B} = \map {f^\to} A \cup \map {f^\to} B$
Image of Intersection under Mapping
Let $S$ and $T$ be sets. Let $f: S \to T$ be a mapping. Let $S_1$ and $S_2$ be subsets of $S$. Then: :$f \sqbrk {S_1 \cap S_2} \subseteq f \sqbrk {S_1} \cap f \sqbrk {S_2}$ This can be expressed in the language and notation of direct image mappings as: :$\forall S_1, S_2 \in \powerset S: \map {f^\to} {S_1 \cap S_2} \subseteq \map {f^\to} {S_1} \cap \map {f^\to} {S_2}$
Image of Intersection under Injection
Let $S$ and $T$ be sets. Let $f: S \to T$ be a mapping. Then: :$\forall A, B \subseteq S: f \sqbrk {A \cap B} = f \sqbrk A \cap f \sqbrk B$  $f$ is an injection.
Image of Intersection under One-to-Many Relation
Let $S$ and $T$ be sets. Let $\RR \subseteq S \times T$ be a relation. Then: :$\forall S_1, S_2 \subseteq S: \RR \sqbrk {S_1 \cap S_2} = \RR \sqbrk {S_1} \cap \RR \sqbrk {S_2}$   $\RR$ is one-to-many.
One-to-Many Image of Set Difference
Let $\mathcal R \subseteq S \times T$ be a relation. Let $A$ and $B$ be subsets of $S$. Then: :$(1): \quad \mathcal R \left[{A}\right] \setminus \mathcal R \left[{B}\right] = \mathcal R \left[{A \setminus B}\right]$  $\mathcal R$ is one-to-many.
Preimage of Union under Mapping
Let $S$ and $T$ be sets. Let $f: S \to T$ be a mapping. Let $T_1$ and $T_2$ be subsets of $T$. Then: :$f^{-1} \sqbrk {T_1 \cup T_2} = f^{-1} \sqbrk {T_1} \cup f^{-1} \sqbrk {T_2}$ This can be expressed in the language and notation of inverse image mappings as: :$\forall T_1, T_2 \in \powerset T: \map {f^\gets} {T_1 \cup T_2} = \map {f^\gets} {T_1} \cup \map {f^\gets} {T_2}$
Preimage of Intersection under Mapping
Let $f: S \to T$ be a mapping. Let $T_1$ and $T_2$ be subsets of $T$. Then: :$f^{-1} \sqbrk {T_1 \cap T_2} = f^{-1} \sqbrk {T_1} \cap f^{-1} \sqbrk {T_2}$ This can be expressed in the language and notation of inverse image mappings as: :$\forall T_1, T_2 \in \powerset T: \map {f^\gets} {T_1 \cap T_2} = \map {f^\gets} {T_1} \cap \map {f^\gets} {T_2}$
Preimage of Set Difference under Mapping
Let $f: S \to T$ be a mapping. Let $T_1$ and $T_2$ be subsets of $T$. Then: :$f^{-1} \sqbrk {T_1 \setminus T_2} = f^{-1} \sqbrk {T_1} \setminus f^{-1} \sqbrk {T_2}$ where: :$\setminus$ denotes set difference :$f^{-1} \sqbrk {T_1}$ denotes preimage.
Projection is Surjection
Let $S$ and $T$ be non-empty sets. Let $S \times T$ be the Cartesian product of $S$ and $T$. Let $\pr_1: S \times T \to T$ and $\pr_2: S \times T \to T$ be the first projection and second projection respectively on $S \times T$. Then $\pr_1$ and $\pr_2$ are both surjections.
Sum Rule for Derivatives
Let $\map f x, \map j x, \map k x$ be real functions defined on the open interval $I$. Let $\xi \in I$ be a point in $I$ at which both $j$ and $k$ are differentiable. Let $\map f x = \map j x + \map k x$. Then $f$ is differentiable at $\xi$ and: :$\map {f'} \xi = \map {j'} \xi + \map {k'} \xi$ It follows from the definition of derivative that if $j$ and $k$ are both differentiable on the interval $I$, then: :$\forall x \in I: \map {f'} x = \map {j'} x + \map {k'} x$
Quotient Mapping is Surjection
Let $\RR$ be an equivalence relation on $S$. Then the quotient mapping $q_\RR: S \to S / \RR$ is a surjection.
Trivial Quotient is a Bijection
Let $\Delta_S$ be the diagonal relation on a set $S$. Let $q_{\Delta_S}: S \to S / \Delta_S$ be the trivial quotient of $S$. Then $q_{\Delta_S}: S \to S / \Delta_S$ is a bijection.
Renaming Mapping is Well-Defined
Let $f: S \to T$ be a mapping. Let $r: S / \RR_f \to \Img f$ be the renaming mapping,  defined as: :$r: S / \RR_f \to \Img f: \map r {\eqclass x {\RR_f} } = \map f x$ where: :$\RR_f$ is the equivalence induced by the mapping $f$ :$S / \RR_f$ is the quotient set of $S$ determined by $\RR_f$ :$\eqclass x {\RR_f}$ is the equivalence class of $x$ under $\RR_f$. The renaming mapping is always well-defined.
Powers of Group Elements
Let $\left({G, \circ}\right)$ be a group whose identity is $e$. Let $a \in G$. Then the following results hold:
Renaming Mapping is Bijection
Let $f: S \to T$ be a mapping. Let $r: S / \RR_f \to \Img f$ be the renaming mapping, defined as: :$r: S / \RR_f \to \Img f: \map r {\eqclass x {\RR_f} } = \map f x$ where: :$\RR_f$ is the equivalence induced by the mapping $f$ :$S / \RR_f$ is the quotient set of $S$ determined by $\RR_f$ :$\eqclass x {\RR_f}$ is the equivalence class of $x$ under $\RR_f$. The renaming mapping is a bijection.
Composite of Mapping with Inverse
Let $f: S \to T$ be a mapping. Then: :$\forall x \in S: \map {f^{-1} \circ f} x = \eqclass x {\RR_f}$ where: :$\RR_f$ is the equivalence induced by $f$ :$\eqclass x {\RR_f}$ is the $\RR_f$-equivalence class of $x$.
Factoring Mapping into Surjection and Inclusion
Every mapping $f:S \to T$ can be uniquely '''factored into''' a surjection $g$ followed by the inclusion mapping $i_T$. That is, $f = i_T \circ g$ where: :$g: S \to \Img f: \map g x = \map f x$ :$i_T: \Img f \to T: \map {i_T} x = x$ This can be illustrated using a commutative diagram as follows: ::$\begin{xy}\xymatrix@L+2mu@+1em {  S \ar@{-->}[r]^*{g}      \ar[rd]_*{f = i_T \circ g} &  \Img f \ar@{-->}[d]^*{i_T} \\ &  T }\end{xy}$
Quotient Theorem for Surjections
Let $f: S \to T$ be a surjection. Then there is one and only one bijection $r: S / \RR_f \to T$ such that: :$r \circ q_{\RR_f} = f$ where: :$\RR_f$ is the equivalence induced by $f$ :$r: S / \RR_f \to T$ is the renaming mapping :$q_{\RR_f}: S \to S / \RR_f$ is the quotient mapping induced by $\RR_f$. This can be illustrated using a commutative diagram as follows: ::$\begin {xy} \xymatrix@L + 2mu@ + 1em {  S \ar@{-->}[rr]^*{f = r \circ q_{\RR_f} }    \ar[dd]_*{q_{\RR_f} } && T \\ \\ S / \RR_f \ar[urur]_*{r}  } \end {xy}$
Quotient Theorem for Sets
A mapping $f: S \to T$ can be uniquely '''factored into''' a surjection, followed by a bijection, followed by an injection. Thus: :$f = i \circ r \circ q_{\RR_f}$ where: {{begin-eqn}} {{eqn | ll= q_{\RR_f}:       | lo= S \to S / \RR_f:       | l = \map {q_{\RR_f} } s       | r = \eqclass s {\RR_f}       | c = Quotient Mapping }} {{eqn | ll= r:       | lo= S / \RR_f \to \Img f:       | l = \map r {\eqclass s {\RR_f} }       | r = \map f s       | c = Renaming Mapping }} {{eqn | ll= i:       | lo= \Img f \to T:       | l = \map i t       | r = t       | c = Inclusion Mapping }} {{end-eqn}} where: :$\RR_f$ is the equivalence induced by $f$ :$S / \RR_f$ is the quotient set of $S$ induced by $\RR_f$ This can be illustrated using a commutative diagram as follows: ::$\begin {xy} \xymatrix@L + 2mu@ + 1em {  S \ar@{-->}[rrr]^*{f = i_T \circ r \circ q_{\RR_f} }    \ar[d]_*{q_{\RR_f} } & & & T \\ S / \RR_f \ar[rrr]_*{r} & & & \Img f \ar[u]_*{i_T} } \end {xy}$
Empty Set is Element of Power Set
The empty set is an element of all power sets: :$\forall S: \O \in \powerset S$
Direct 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 $\mathcal R^\to: \powerset S \to \powerset T$ be the direct image mapping of $\mathcal R$: :$\forall X \in \powerset S: \map {\mathcal R^\to} X = \set {t \in T: \exists s \in X: \tuple {s, t} \in \mathcal R}$ Then $\mathcal R^\to$ is indeed a mapping.