nyuuzyou/wikis huggingface dataset:
Identity Mapping is Automorphism

== Theorem ==
The identity mapping $I_S: \struct {S, \circ} \to \struct {S, \circ}$ on the algebraic structure $\struct {S, \circ}$ is an automorphism.
Its image is $S$.
=== Semigroup Automorphism ===
=== Ordered Semigroup Automorphism ===
=== Group Automorphism ===
=== Ring Automorphism ===
== Proof ==
By definition, an automorphism is an isomorphism from an algebraic structure onto itself.
An isomorphism, in turn, is a bijective homomorphism.
From Identity Mapping is Bijection, the identity mapping $I_S: S \to S$ on the set $S$ is a bijection from $S$ onto itself.
Now we need to show it is a homomorphism.
Let $x, y \in S$. Then:
Thus $\map {I_S} {x \circ y} = \map {I_S} x \circ \map {I_S} y$ and the morphism property holds/
This proves that $I_S: S \to S$ is a homomorphism.
As $I_S$ is a bijection, its image is $S$.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 6$: Isomorphisms of Algebraic Structures: Theorem $6.1: \ 1^\circ$
* : Chapter $\text I$: Algebraic Structures: $\S 6$: Isomorphisms of Algebraic Structures: Exercise $6.2$
Category:Automorphisms (Abstract Algebra)
Category:Identity Mappings
Identity Mapping is Automorphism/Groups

== Theorem ==
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Then $I_G: \struct {G, \circ} \to \struct {G, \circ}$ is a group automorphism.
Its kernel is $\set e$.
== Proof ==
The main result Identity Mapping is Automorphism holds directly.
As $I_G$ is a bijection, the only element that maps to $e$ is $e$ itself.
Thus the kernel is $\set e$.
== Sources ==
* : $\S 7.1$. Homomorphisms: Example $130$
* : Chapter $2$: Group Homomorphism and Isomorphism: $\S 64$
Category:Group Automorphisms
Category:Identity Mappings
Identity Mapping is Automorphism/Ordered Semigroups

== Theorem ==
Let $\struct {S, \circ, \preccurlyeq}$ be an ordered semigroup.
Then $I_S: \struct {S, \circ, \preccurlyeq} \to \struct {S, \circ, \preccurlyeq}$ is a ordered semigroup automorphism.
== Proof ==
From Identity Mapping is Semigroup Automorphism:
:$I_S: \struct {S, \circ} \to \struct {S, \circ}$ is a semigroup automorphism.
From Identity Mapping is Order Isomorphism:
:$I_S: \struct {S, \preccurlyeq} \to \struct {S, \preccurlyeq}$ is an order isomorphism.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 15$: Ordered Semigroups: Exercise $15.1$
Category:Semigroup Automorphisms
Category:Order Isomorphisms
Category:Ordered Semigroups
Category:Identity Mappings
Identity Mapping is Automorphism/Rings

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring whose zero is $0$.
Then $I_R: \struct {R, +, \circ} \to \struct {R, +, \circ}$ is a ring automorphism.
Its kernel is $\set 0$.
== Proof ==
The result Identity Mapping is Automorphism holds directly, for both $+$ and $\circ$.
As $I_R$ is a bijection, the only element that maps to $0$ is $0$ itself.
Thus the kernel is $\set 0$.
== Sources ==
* : Chapter $5$: Rings: $\S 24$. Homomorphisms: Example $43$
Category:Ring Automorphisms
Category:Identity Mappings
Identity Mapping is Automorphism/Semigroups

== Theorem ==
Let $\struct {S, \circ}$ be a semigroup.
Then $I_S: \struct {S, \circ} \to \struct {S, \circ}$ is a semigroup automorphism.
== Proof ==
The main result Identity Mapping is Automorphism holds directly.
Category:Semigroup Automorphisms
Category:Identity Mappings
Identity Mapping is Bijection

== Theorem ==
The identity mapping $I_S: S \to S$ on the set $S$ is a bijection.
== Proof ==
The identity mapping is:
:an injection, from Identity Mapping is Injection
:a surjection, from Identity Mapping is Surjection
and therefore a bijection.
== Sources ==
* : Exercise $1.3: \ 10$
* : $\S 3.5$. Identity mappings
* : Chapter $\text{I}$: Sets and Functions: Composition of Functions
* : $\S 5$. Induced mappings; composition; injections; surjections; bijections
* : Chapter $1$: Sets and mappings: $\S 1.3$: Mappings
* : Appendix $\text{A}.7$: Inverses: Proposition $\text{A}.7.5 \ (1)$
Category:Bijections
Category:Identity Mappings
Identity Mapping is Continuous

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
The identity mapping $I_S: S \to S$ defined as:
:$\forall x \in S: \map {I_S} x = x$
is a continuous mapping.
=== Metric Space ===
== Proof ==
Let $U \in \tau$.
We have Identity Mapping is Bijection.
So $I_S^{-1}$ is well-defined and:
:$\forall x \in U: \map {I_S^{-1} } x = x$
Thus $I_S^{-1} \sqbrk U = U \in \tau$.
Hence, by definition of continuous mapping, $I_S$ is continuous.
== Also see ==
* Definition:Finer Topology
* Definition:Coarser Topology
* Identity Mapping to Coarser Topology is Continuous
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.1$: Topological Spaces: Proposition $3.1.10 \ \text{(a)}$
* : $3$: Continuity generalized: topological spaces: Exercise $3.9: 5$
Category:Continuous Mappings
Category:Identity Mappings
Category:Identity Mapping is Continuous
Identity Mapping is Continuous/Metric Space

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
The identity mapping $I_A: A \to A$ defined as:
:$\forall x \in A: \map {I_A} x = x$
is a continuous mapping.
== Proof ==
Let $a \in A$.
Let $\epsilon \in \R_{>0}$.
Let $\delta = \epsilon$.
Then:
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 3$: Continuity: Theorem $3.4$
Category:Continuous Mappings on Metric Spaces
Category:Identity Mapping is Continuous
Identity Mapping is Frame Homomorphism

== Theorem ==
Let $L = \struct{S, \preceq}$ be a frame.
Let $\operatorname{id}_S$ denote the identity mapping on $S$.
Then:
:$\operatorname{id}_S$ is a frame homomorphism of $L$ to $L$
== Proof ==
=== $\operatorname{id}_S$ is Finite Meet Preserving ===
Let $F \subseteq S_1$ be a finite subset.
We have:
Since $F$ was arbitrary, it follows that $\operatorname{id}_S$ is finite meet preserving by definition.
=== $\operatorname{id}_S$ is Arbitrary Join Preserving ===
Let $A \subseteq S_1$ be any subset of $S$.
We have:
Since $A$ was arbitrary, it follows that $\operatorname{id}_S$ is arbitrary join preserving by definition.
By definition, $\operatorname{id}_S$ is a frame homomorphism
Category:Frame Homomorphisms
Identity Mapping is Group Endomorphism

== Theorem ==
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Then $I_G: \struct {G, \circ} \to \struct {G, \circ}$ is a group endomorphism.
== Proof $1$ ==
== Proof $2$ ==
== Sources ==
* : Chapter $2$: Group Homomorphism and Isomorphism: $\S 60$
Category:Identity Mapping is Group Endomorphism
Category:Identity Mappings
Category:Group Endomorphisms
Identity Mapping is Group Endomorphism/Proof 2

== Theorem ==
== Proof ==
From Identity Mapping is Group Automorphism, $G$ is a group automorphism.
By definition, a group endomorphism is a group homomorphism from $G$ to itself.
A group automorphism is a group isomorphism from $G$ to itself.
As a group isomorphism is a group homomorphism which is also a bijection, the result follows by definition.
Category:Identity Mapping is Group Endomorphism
Identity Mapping is Homeomorphism

== Theorem ==
Let $T$ be a topological space.
The identity mapping $I_T: T \to T$ defined as:
:$\forall x \in T: \map {I_T} x = x$
is a homeomorphism.
== Proof ==
We have Identity Mapping is Bijection.
We also have Identity Mapping is Continuous.
Hence, by definition, $I_T$ is a homeomorphism.
Category:Homeomorphisms (Topological Spaces)
Category:Identity Mappings
Identity Mapping is Idempotent

== Theorem ==
Let $S$ be a set.
Let $I_S: S \to S$ be the identity mapping on $S$.
Then $I_S$ is idempotent:
:$I_S \circ I_S = I_S$
== Proof ==
From Identity Mapping is Left Identity:
:$I_S \circ f = f$
for all mappings $f: S \to S$.
From Identity Mapping is Right Identity:
:$f \circ I_S = f$
for all mappings $f: S \to S$.
Substituting $I_S$ for $f$ in either one and the result follows.
== Sources ==
* : $\S 1.14$: Composition of Functions: Theorem $14.7$
Category:Identity Mappings
Identity Mapping is Injection

== Theorem ==
On any set $S$, the identity mapping $I_S: S \to S$ is an injection.
== Proof ==
From the definition of the identity mapping:
:$\forall x \in S: \map {I_S} x = x$
So:
:$\map {I_S} x = \map {I_S} y \implies x = y$
So by definition $I_S$ is an injection.
== Also see ==
* Identity Mapping is Surjection
== Sources ==
* : Chapter $\text{I}$: Sets and Functions: Composition of Functions
* : $\S 5$. Induced mappings; composition; injections; surjections; bijections: Example $5.3$
Category:Injections
Category:Identity Mappings
Identity Mapping is Lattice Homomorphism

== Theorem ==
Let $L = \struct{A, \vee, \wedge, \preceq}$ be a lattice.
Let $\operatorname{id}_A$ denote the identity mapping on $A$.
Then:
:$\operatorname{id}_A$ is a lattice homomorphism of $L$ to $L$
== Proof ==
Follows from Identity Mapping is Automorphism applied to $\struct{A, \vee}$ and $\struct{A, \wedge}$.
Category:Lattice Homomorphisms
Identity Mapping is Left Identity

== Theorem ==
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.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Identity Mapping is Right Identity
* Diagonal Relation is Left Identity
* Diagonal Relation is Right Identity
== Sources ==
* : Introduction $\S 2$: Product sets, mappings
* : Chapter $\text I$: Algebraic Structures: $\S 5$: Composites and Inverses of Functions
* : Chapter $\text{I}$: Sets and Functions: Composition of Functions
* : Chapter $1$: Mappings: $\S 16$
* : $\S 1.14$: Composition of Functions: Theorem $14.8$
* : Chapter $1$: Sets and mappings: $\S 1.3$: Mappings
* : Chapter $2$: Maps and relations on sets: Definition $2.11$
* : Appendix $\text{A}.5$: Identity, One-one, and Onto Functions
* : $\S 1.2$
Category:Identity Mappings
Category:Identity Mapping is Left Identity
Identity Mapping is Left Identity/Proof 1

== Theorem ==
== Proof ==
=== Equality of Domains ===
The domains of $f$ and $I_T \circ f$ are equal from Domain of Composite Relation:
:$\Dom {I_T \circ f} = \Dom f$
=== Equality of Codomains ===
The codomains of $f$ and $f \circ I_S$ are also easily shown to be equal.
From Codomain of Composite Relation, the codomains of $I_T \circ f$ and $I_T$ are both equal to $T$.
But from the definition of the identity mapping, the codomain of $I_T$ is $\Dom {I_T} = T$
=== Equality of Relations ===
The composite of $f$ and $I_T$ is defined as:
:$I_T \circ f = \set {\tuple {x, z} \in S \times T: \exists y \in T: \tuple {x, y} \in f \land \tuple {y, z} \in I_T}$
But by definition of the identity mapping on $T$, we have that:
:$\tuple {y, z} \in I_T \implies y = z$
Hence:
:$I_T \circ f = \set {\tuple {x, y} \in S \times T: \exists y \in T: \tuple {x, y} \in f \land \tuple {y, y} \in I_T}$
But as $\forall y \in T: \tuple {y, y} \in I_T$, this means:
:$I_T \circ f = \set {\tuple {x, y} \in S \times T: \tuple {x, y} \in f}$
That is:
:$I_T \circ f = f$
Hence the result, by Equality of Mappings.
== Sources ==
* : $\S 3.5$. Identity mappings: Example $52$
* : Chapter $\text{I}$: Sets and Functions: Problem $\text{BB}$: Categorical Matters
* : $\S 24.3 \ \text{(ii)}$: Composition of Mappings
Category:Identity Mapping is Left Identity
Identity Mapping is Localic Mapping

== Theorem ==
Let $L = \struct {S, \preceq}$ be a locale.
Let $\operatorname{id}_S : S \to S$ be the identity mapping on $S$.
Then:
:$\operatorname{id}_S : L \to L$ is a localic mapping.
== Proof ==
By definition of localic mapping, we need to show that $\operatorname{id}_S$ is the upper adjoint of a Galois connection where the lower adjoint is a frame homomorphism of $L$ to $L$.
From Identity Mapping forms Galois Connection:
:$\tuple{\operatorname{id}_S, \operatorname{id}_S}$ is a Galois connection
From Identity Mapping is Frame Homomorphism
:$\operatorname{id}_S$ is a frame homomorphism of $L$ to $L$
The result follows.
== Sources ==
* : Chapter II: Frames and Locales. Spectra, $\S 2.2$
Category:Localic Mappings
Identity Mapping is Order Isomorphism

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
The identity mapping $I_S$ is an order isomorphism from $\struct {S, \preceq}$ to itself.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Theorem $14.1: \ 1^\circ$
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.6$
* : $\S 6.30$
Category:Examples of Order Isomorphisms
Category:Identity Mappings
Category:Identity Mapping is Order Isomorphism
Identity Mapping is Order Isomorphism/Proof 1

== Theorem ==
== Proof ==
By definition:
:$\forall x \in S: \map {I_S} x = x$
So:
:$x \preceq y \implies \map {I_S} x \preceq \map {I_S} y$
As $I_S$ is a bijection, we also have:
:$\map {I_S^{-1} } x = x$
So:
:$x \preceq y \implies \map {I_S^{-1} } x \preceq \map {I_S^{-1} } y$
Category:Identity Mapping is Order Isomorphism
Identity Mapping is Ordered Ring Automorphism

== Theorem ==
Let $\struct {S, +, \circ, \preceq}$ be an ordered ring.
Then the identity mapping $I_S: S \to S$ is an ordered ring automorphism.
== Proof ==
We have that:
:an identity mapping is an order isomorphism
:an identity mapping is a group automorphism
:an identity mapping is a semigroup automorphism
Hence the result by definition of ordered ring automorphism.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $23$. The Field of Rational Numbers
Category:Ordered Rings
Category:Identity Mappings
Identity Mapping is Permutation

== Theorem ==
The identity mapping $I_S: S \to S$ on the set $S$ is a permutation.
== Proof ==
The identity mapping $I_S$ is a bijection from $S$ to itself, by Identity Mapping is Bijection.
== Sources ==
* : $\S 3.6$. Products of bijective mappings. Permutations
* : Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Example $7.5$
* : $\S 24.2$: Composition of Mappings
Category:Identity Mappings
Identity Mapping is Relation Isomorphism

== Theorem ==
Let $\struct {S, \RR}$ be a relational structure.
Then the identity mapping $I_S: S \to S$ is a relation isomorphism from $\struct {S, \RR}$ to itself.
== Proof ==
By definition of identity mapping:
:$\forall x \in S: \map {I_S} x = x$
So:
:$x \mathrel \RR y \implies \map {I_S} x \mathrel \RR \map {I_S} y$
From Identity Mapping is Bijection, $I_S$ is a bijection.
Hence:
:$\map {I_S^{-1} } x = x$
So:
:$x \mathrel \RR y \implies \map {I_S^{-1} } x \mathrel \RR \map {I_S^{-1} } y$
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.9 \ \text{(b)}$
Category:Relation Isomorphisms
Category:Identity Mappings
Identity Mapping is Right Identity

== Theorem ==
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 1 ==
== Proof 2 ==
== Also see ==
* Identity Mapping is Left Identity
* Diagonal Relation is Right Identity
* Diagonal Relation is Left Identity
== Sources ==
* : Introduction $\S 2$: Product sets, mappings
* : Chapter $\text I$: Algebraic Structures: $\S 5$: Composites and Inverses of Functions
* : Chapter $\text{I}$: Sets and Functions: Composition of Functions
* : Chapter $1$: Mappings: $\S 16$
* : $\S 1.14$: Composition of Functions: Theorem $14.7$
* : Chapter $1$: Sets and mappings: $\S 1.3$: Mappings
* : Chapter $2$: Maps and relations on sets: Definition $2.11$
* : Appendix $\text{A}.5$: Identity, One-one, and Onto Functions
* : $\S 1.2$
Category:Identity Mappings
Category:Identity Mapping is Right Identity
Identity Mapping is Right Identity/Proof 1

== Theorem ==
== Proof ==
=== Equality of Codomains ===
The codomains of $f$ and $f \circ I_S$ are both equal to $T$ from Codomain of Composite Relation.
=== Equality of Domains ===
From Domain of Composite Relation:
:$\Dom {f \circ I_S} = \Dom {I_S}$
But from the definition of the identity mapping:
:$\Dom {I_S} = \Img {I_S} = S$
=== 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:
:$f \circ I_S = f$
Hence the result, by Equality of Mappings.
== Sources ==
* : $\S 3.5$. Identity mappings: Example $52$
* : Chapter $\text{I}$: Sets and Functions: Problem $\text{BB}$: Categorical Matters
* : $\S 24.3 \ \text{(i)}$: Composition of Mappings
Category:Identity Mapping is Right Identity
Identity Mapping is Semilattice Homomorphism

== Theorem ==
Let $S = \struct{A, \circ}$ be a semilattice.
Let $\operatorname{id}_A$ denote the identity mapping on $A$.
Then:
:$\operatorname{id}_A$ is a semilattice homomorphism of $S$ to $S$
== Proof ==
Follows immediately from Identity Mapping is Automorphism.
Category:Semilattice Homomorphisms
Identity Mapping is Surjection

== Theorem ==
On any set $S$, the identity mapping $I_S: S \to S$ is a surjection.
== Proof ==
The identity mapping is defined as:
:$\forall y \in S: \map {I_S} y = y$
Then we have:
Hence the result.
== Also see ==
* Identity Mapping is Injection
== Sources ==
* : $\S 5$. Induced mappings; composition; injections; surjections; bijections: Example $5.3$
Category:Surjections
Category:Identity Mappings
Identity Mapping on Metric Space is Homeomorphism

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
The identity mapping $I_A: M \to M$ defined as:
:$\forall x \in A: \map {I_A} x = x$
is a homeomorphism.
== Proof ==
We have Identity Mapping is Bijection.
We also have Identity Mapping is Continuous.
Hence, by definition, $I_T$ is a homeomorphism.
Category:Homeomorphisms (Metric Spaces)
Category:Topologically Equivalent Metrics
Category:Identity Mappings
Identity Mapping on Normed Vector Space is Bounded Linear Operator

== Theorem ==
Let $\Bbb F$ be a subfield of $\C$. 
Let $\struct {X, \norm \cdot}$ be a normed vector space.
Let $I : X \to X$ be the identity mapping on $X$. 
Then $I$ is a bounded linear operator.
Further: 
Let $\map B X$, be the space of bounded linear operators on $X$.
Then:
:$\norm I_{\map B X} = 1$
where $\norm {\, \cdot \,}_{\map B X}$ denotes the norm on $\map B X$.
== Proof ==
Let $\lambda \in \Bbb F$ and $x, y \in X$.
Then, we have: 
:$\map I {\lambda x + y} = \lambda x + y = \lambda I x + I y$
So $I$ is a linear operator.
Further, we have: 
:$\norm {I x} = \norm x$ 
for each $x \in X$, so $I$ is bounded.
Further, for all $x \in X$ with $\norm x = 1$ we have that $\norm {I x} = 1$, so that: 
:$\set {\norm {I x} : \norm x = 1} = \set 1$
so that: 
:$\norm I_{\map B X} = 1$
Category:Bounded Linear Operators
Identity Mapping on Real Vector Space from Chebyshev to Euclidean Metric is Continuous

== Theorem ==
Let $\R^n$ be an $n$-dimensional real vector space.
Let $d_2$ be the Euclidean metric on $\R^n$.
Let $d_\infty$ be the Chebyshev distance on $\R^n$.
Let $I: \R^n \to \R^n$ be the identity mapping from $\R^n$ to itself.
Then the mapping:
:$I: \struct {\R^n, d_\infty} \to \struct {\R^n, d_2}$
is $\tuple {d_\infty, d_2}$-continuous.
== Proof ==
Let $a = \tuple {a_1, a_2, \ldots, a_n} \in \R^n$.
Let $\epsilon \in \R_{>0}$.
Let $\delta = \dfrac {\epsilon} {\sqrt n}$.
Let $x = \tuple {x_1, x_2, \ldots, x_n}$ be such that $\map {d_\infty} {x, a} < \delta$.
That is:
:$\ds \max_{i \mathop \le i \mathop \le n} \set {\size {a_i - x_i} } < \delta$
Then:
The result follows by definition of $\tuple {d_\infty, d_2}$-continuity.
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 3$: Continuity: Theorem $3.5$
Category:Continuous Mappings on Metric Spaces
Category:Identity Mappings
Category:Euclidean Metric
Category:Chebyshev Distance
Identity Mapping on Real Vector Space from Euclidean to Chebyshev Distance is Continuous

== Theorem ==
Let $\R^n$ be an $n$-dimensional real vector space.
Let $d_2$ be the Euclidean metric on $\R^n$.
Let $d_\infty$ be the Chebyshev distance on $\R^n$.
Let $I: \R^n \to \R^n$ be the identity mapping from $\R^n$ to itself.
Then the mapping:
:$I: \struct {\R^n, d_2} \to \struct {\R^n, d_\infty}$
is $\tuple {d_2, d_\infty}$-continuous.
== Proof ==
Let $a = \tuple {a_1, a_2, \ldots, a_n} \in \R^n$.
Let $\epsilon \in \R_{>0}$.
Let $\delta = \epsilon$.
Let $x = \tuple {x_1, x_2, \ldots, x_n}$ be such that $\map {d_2} {x, a} < \delta$.
That is:
:$\ds \sqrt {\sum_{i \mathop = i}^n \paren {a_i - x_i} } < \delta$
Then:
The result follows by definition of $\tuple {d_2, d_\infty}$-continuity.
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 3$: Continuity: Theorem $3.5$
Category:Continuous Mappings on Metric Spaces
Category:Identity Mappings
Category:Euclidean Metric
Category:Chebyshev Distance
Identity Mapping on Symmetric Group is Even Permutation

== Theorem ==
Let $S_n$ denote the symmetric group on $n$ letters.
Let $e \in S_n$ be the identity permutation on $S_n$.
Then $e$ is an even permutation on $S_n$.
== Proof ==
By definition of the identity permutation:
:$\forall i \in \N_{>0}: \map e i = i$
Thus $e$ fixes all elements of $S_n$.
We have that a Transposition is of Odd Parity.
Hence a permutation is odd if is the composition of an odd number of transpositions.
The identity permutation is the composition of $0$ (zero) transpositions.
Hence the result.
== Sources ==
* : Chapter $2$: The Symmetric Groups: $\S 82$
Category:Symmetric Groups
Category:Identity Mappings
Category:Even Permutations
Identity Mapping to Coarser Topology is Continuous

== Theorem ==
Let $S$ be a set.
Let $\tau_1$ and $\tau_2$ be topologies on $S$.
That is, let $T_1 = \struct {S, \tau_1}$ and $T_2 = \struct {S, \tau_2}$ be topological spaces.
Let $I_S: \struct {S, \tau_1} \to \struct {S, \tau_2}$ denote the identity mapping on $S$:
:$\forall x \in S: \map {I_S} x = x$
Then:
:$I_S: T_1 \to T_2$ is a continuous mapping
:
:$\tau_2$ is coarser than $\tau_1$.
== Proof 1 ==
But $\tau_2 \subseteq \tau_1$ is the definition of $\tau_2$ being coarser than $\tau_1$.
== Proof 2 ==
=== Sufficient Condition ===
Suppose:
:$I_S: T_1 \to T_2$ is a continuous mapping.
Then, by definition of everywhere continuous mapping between $\struct {S, \tau_1}$ and $\struct {S, \tau_2}$:
:$U_2 \in \tau_2 \implies I_S^{-1} \sqbrk {U_2} \in \tau_1$
where $I_S^{-1} \sqbrk {U_2}$ denotes the preimage of $U$ under $I_S$.
By Preimage of Subset under Identity Mapping, we have:
:$U_2 \in \tau_2 \implies U_2 \in \tau_1$
So:
:$\tau_2 \subseteq \tau_1$
Hence, by definition of coarser topology:
:$\tau_2$ is coarser than $\tau_1$.
=== Necessary Condition ===
Suppose:
:$\tau_2$ is coarser than $\tau_1$.
That is:
:$U_2 \in \tau_2 \implies U_2 \in \tau_1$
By Preimage of Subset under Identity Mapping, we have:
:$U_2 \in \tau_2 \implies I_S^{-1} \sqbrk {U_2} \in \tau_1$
where $I_S^{-1} \sqbrk {U_2}$ denotes the preimage of $U_2$ under $I_S$.
So the mapping $I_S$ is by definition, everywhere continuous mapping between $\struct {S, \tau_1}$ and $\struct {S, \tau_2}$.
Hence:
:$I_S: T_1 \to T_2$ is a continuous mapping.
Hence the result.
Category:Continuous Mappings
Category:Identity Mappings
Identity Mapping to Expansion is Closed

== Theorem ==
Let $S$ be a set on which $\tau_1$ and $\tau_2$ are topologies such that:
:$\tau_1 \subseteq \tau_2$
that is, such that $\tau_2$ is an expansion of $\tau_1$.
Let $I_X: \struct {S, \tau_1} \to \struct {S, \tau_2}$ be the identity mapping from $\struct {S, \tau_1}$ to $\struct {S, \tau_2}$.
Then $I_S$ is closed.
== Proof ==
$\tau_1 \subseteq \tau_2$ means that every open set of $\struct {S, \tau_1}$ is also an open set of $\struct {S, \tau_2}$.
Let $A \subseteq S$ be closed in $\struct {S, \tau_1}$
Then by definition $S \setminus A$ is open in $\struct {S, \tau_1}$.
Then $\map {I_S} {S \setminus A} = S \setminus A$ is open in $\struct {S, \tau_2}$.
So, by definition, $\map {I_S} A = A$ is closed in $\struct {S, \tau_2}$.
So for all $A \subseteq S$ closed in $\struct {S, \tau_1}$, it holds that $\map {I_S} A$ is closed in $\struct {S, \tau_2}$.
So by definition of closed mapping, the result follows.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Functions, Products, and Subspaces
Category:Closed Mappings
Category:Identity Mappings