nyuuzyou/wikis huggingface dataset:
Mapping is Continuous iff Inverse Images of Open Sets are Open/Corollary

== Theorem ==
Let $X$ and $Y$ be normed vector spaces.
Let $f : X \to Y$ be a mapping.
Then:
:$f$ is continuous on $X$
:
:for every $F$ closed in $Y$, $\map {f^{-1}} F$ is closed in $X$.
== Proof ==
=== Sufficient Condition ===
Suppose that for every closed $F$ in $Y$, $\map {f^{-1}} F$ is closed in $X$.
Let $V$ be open in $Y$.
By definition, $Y \setminus V$ is closed in $Y$.
By assumption, $\map {f^{-1}} {Y \setminus V}$ is closed in $X$.
We have that:
Thus, $X \setminus \map {f^{-1} } V$ is closed in $X$.
By definition, $X \setminus \map {f^{-1} } V$ is closed iff $\map {f^{-1} } V$ is open.
Hence, for every $V$ open in $Y$, $ \map {f^{-1}} V$ is open in $X$.
By Mapping is Continuous iff Inverse Images of Open Sets are Open, $f$ is continuous on $X$.
=== Necessary Condition ===
Suppose that $f$ is continuous on $X$.
Let $F$ be closed in $Y$.
By definition, $Y \setminus F$ is open in $Y$.
By Mapping is Continuous iff Inverse Images of Open Sets are Open, $\map {f^{-1} } {Y \setminus F}$ is open on $X$.
Hence:
and $X \setminus \map {f^{-1} } F$ is open in $X$.
By definition, $\map {f^{-1}} F$ is closed in $X$
== Sources ==
* : Chapter $\S 2.2$: Continuous and linear maps. Continuous maps
Category:Continuous Mappings on Normed Vector Spaces
Mapping is Continuous implies Mapping Preserves Filtered Infima in Lower Topological Lattice

== Theorem ==
Let $T = \struct {S, \preceq, \tau}$ and $Q = \struct {X, \preceq', \tau'}$ be complete topological lattices with lower topologies.
Let $f: S \to X$ be a mapping such that:
:$f$ is a continuous mapping.
Then $f$ preserves filtered infima.
== Proof ==
Define $B := \set {\relcomp S {x^\succeq}: x \in S}$
By definition of lower topology:
:$B$ is an analytic sub-basis.
Let $F$ be a filtered subset of $S$ such that:
:$F$ admits an infimum in $T$.
Thus by definition of complete lattice:
:$f \sqbrk F$ admits an infimum in $Q$.
We will prove that:
:$\forall A \in B: \inf F \in A \implies F \cap A \ne \O$
Let $A \in B$ be such that:
:$\inf F \in A$
By definition of $B$:
:$\exists x \in S: A = \relcomp S {x^\succeq}$
By definition of relative complement:
:$\inf F \notin x^\succeq$
By definition of upper closure of element:
:$x \npreceq \inf F$
By definition of infimum:
:$x$ is not a lower bound for $F$.
By definition of lower bound:
:$\exists y \in F: x \npreceq y$
By definition of upper closure of element:
:$y \notin x^\succeq$
By definition of relative complement:
:$y \in A$
By definitions of intersection and non-empty set:
:$F \cap A \ne \O$
Then by If Infimum of Filtered Subset belongs to Element of Sub-Basis then Subset and Element Intersect implies Infimum of Subset belongs to Closure of Subset:
:$\inf F \in F^-$
We will prove that:
:$f$ is an increasing mapping.
Let $x, y \in S$ be such that:
:$x \preceq y$
By definition of reflexivity:
:$\map f x \preceq' \map f x$
By definition of upper closure of element:
:$\map f x \in \paren {\map f x}^{\succeq'}$
By definition of preimage of set:
:$x \in f^{-1} \sqbrk {\paren {\map f x}^{\succeq'} }$
By Complement of Upper Closure of Element is Open in Lower Topology:
:$\paren {\map f x}^{\succeq'}$ is closed.
By Continuity by Closed Sets:
:$f^{-1} \sqbrk {\paren {\map f x}^{\succeq'} }$ is closed.
By Closed Subset is Upper Section in Lower Topology:
:$f^{-1} \sqbrk {\paren {\map f x}^{\succeq'} }$ isan upper section.
By definition of upper section:
:$y \in f^{-1} \sqbrk {\paren {\map f x}^{\succeq'} }$
By definition of preimage of set:
:$\map f y \in \paren {\map f x}^{\succeq'}$
Thus by definition of upper closure of element:
:$\map f x \preceq' \map f y$
We will prove that:
:$\map f {\inf F}$ is lower bound for $f \sqbrk F$
By definition of infimum:
:$\map f {\inf F} \preceq' \map \inf {f \sqbrk F}$
We will prove that:
:$F \subseteq f^{-1} \sqbrk {\paren {\map \inf {f \sqbrk F} }^{\succeq'} }$
== Sources ==
* 
Category:Topological Order Theory
Mapping is Extension iff Composite with Inclusion

== Theorem ==
Let $S$ and $T$ be sets.
Let $A \subseteq S$.
Let $f: S \to T$ and $g: A \to T$ be mappings.
Then $f$ is an extension of $g$ :
:$f = g \circ i_A$
where $i_A$ is the inclusion mapping on $A$.
This can be illustrated using a commutative diagram as follows:
::$\begin {xy} \xymatrix@L + 2mu@ + 1em {
 A \ar[r]^*{i_A}
 \ar@{-->}[rd]_*{f = g \circ i_A}
&
 S \ar[d]^*{g}
\\
&
 T
} \end {xy}$
== Proof ==
=== Necessary Condition ===
Let $f: S \to T$ be an extension of $g: A \to T$.
Then by definition:
That is:
:$f = g \circ i_A$
=== Sufficient Condition ===
Let $f = g \circ i_A$ where $i_A$ is the inclusion mapping on $A$.
Then:
So, by definition, $f$ is an extension of $g$.
== Sources ==
* : Chapter $1$: Theory of Sets: $\S 9$: Inverse Functions, Extensions, and Restrictions
Category:Inclusion Mappings
Category:Composite Mappings
Mapping is Idempotent iff Restriction to Image is Identity Mapping

== Theorem ==
Let $S$ be a set.
Let $S^S$ denote the set of mappings from $S$ to itself.
Let $f \in S^S$ be a mapping on $S$.
Then:
:$f$ is idempotent
:
:the restriction of $f$ to $\Img f$ is the identity mapping.
== Proof ==
Recall the definitions:
:$\Img f$ denotes the image set of $f$
:The identity mapping $I_S$ is defined as:
::$\forall x \in f: \map {I_S} x = x$
:An idempotent mapping is a mapping with the property:
::$f \circ f = f$
:where $\circ$ denotes composition of mappings.
=== Necessary Condition ===
Let the restriction of $f$ to $\Img f$ be the identity mapping.
Then:
That is:
:$f \circ f = f$
and by definition $f$ is idempotent.
=== Sufficient Condition ===
Let $f$ be an idempotent mapping.
Let $y \in \Img f$ be arbitrary.
Then:
As $y \in \Img f$ is arbitrary, it follows $\map f y = y$ for all $y \in \Img f$.
The result follows by definition of the identity mapping.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets: Exercise $8.10 \ \text{(b)}$
Category:Identity Mappings
Category:Idempotence
Category:Restrictions
Mapping is Increasing implies Mapping at Infimum for Sequence Precedes Infimum for Composition of Mapping and Sequence

== Theorem ==
Let $\struct {S, \vee_1, \wedge_1, \preceq_1}$ and $\struct {T, \vee_2, \wedge_2, \preceq_2}$ be complete lattices.
Let $f: S \to T$ be an increasing mapping.
Let $\struct {D, \precsim}$ be a directed set.
Let $N: D \to S$ be a net in $S$.
Let $j \in D$.
Then $\map f {\map \inf {N \sqbrk {\map \precsim j} } } \preceq_2 \map \inf {\paren {f \circ N} \sqbrk {\map \precsim j} }$
== Proof ==
By definitions of image of set and composition of mappings:
:$f \sqbrk {N \sqbrk {\map \precsim j} } = \paren {f \circ N} \sqbrk {\map \precsim j}$
By definition of complete lattice:
:$f \sqbrk {N \sqbrk {\map \precsim j} }$ and $N \sqbrk {\map \precsim j}$ admit infima.
Thus by Subset and Image Admit Infima and Mapping is Increasing implies Infimum of Image Succeeds Mapping at Infimum:
:$\map f{\map \inf {N \sqbrk {\map \precsim j} } } \preceq_2 \map \inf {\paren {f \circ N} \sqbrk {\map \precsim j} }$
== Sources ==
* 
* 
Category:Order Theory
Mapping is Injection and Surjection iff Inverse is Mapping

== Theorem ==
Let $f: S \to T$ be a mapping.
Then:
: $f: S \to T$ can be defined as a bijection in the sense that:
::$(1): \quad f$ is an injection
::$(2): \quad f$ is a surjection
:
:the inverse $f^{-1}$ of $f$ is itself a mapping.
== Proof ==
=== Necessary Condition ===
=== Sufficient Condition ===
== Sources ==
* : Chapter $1$: Mappings: $\S 13$
* : Chapter $1$: Theory of Sets: $\S 9$: Inverse Functions, Extensions, and Restrictions
Category:Injections
Category:Surjections
Category:Inverse Mappings
Category:Equivalence of Definitions of Bijection
Category:Mapping is Injection and Surjection iff Inverse is Mapping
Mapping is Injection and Surjection iff Inverse is Mapping/Proof 2

== Theorem ==
Let $S$ and $T$ be sets.
Let $f: S \to T$ be a mapping.
Then:
: $f: S \to T$ can be defined as a bijection in the sense that:
::$(1): \quad f$ is an injection
::$(2): \quad f$ is a surjection.
:
:the inverse $f^{-1}$ of $f$ is such that:
::for each $y \in T$, the preimage $\map {f^{-1} } y$ has exactly one element.
:That is, such that $f^{-1} \subseteq T \times S$ is itself a mapping.
== Proof ==
=== Necessary Condition ===
=== Sufficient Condition ===
Let $f^{-1}: T \to S$ be a mapping.
By Inverse Mapping is Bijection, both $f$ and $f^{-1}$ are bijections.
Hence, in particular, $f$ is a bijection.
Category:Mapping is Injection and Surjection iff Inverse is Mapping
Mapping is Injection iff Direct Image Mapping is Injection

== Theorem ==
Let $f: S \to T$ be a mapping.
Let $f^\to: \powerset S \to \powerset T$ be the direct image mapping of $f$.
Then:
:$f^\to$ is an injection
:$f: S \to T$ is also an injection.
== Proof ==
=== Necessary Condition ===
Suppose $f: S \to T$ is a mapping, but not injective.
Then:
:$\exists x_1 \ne x_2 \in S: \map f {x_1} = \map f {x_2} = y$
Let:
:$X_1 = \set {x_1}$
:$X_2 = \set {x_2}$
:$Y = \set y$
Then it follows that:
:$\map {f^\to} {X_1} = \map {f^\to} {X_2} = Y$
Thus $f^\to: \powerset S \to \powerset T$ is not injective.
So by the Rule of Transposition, the result follows.
=== Sufficient Condition ===
Suppose $f^\to: \powerset S \to \powerset T$ is not an injection.
Then:
:$\exists Y \in \powerset T: \exists X_1, X_2 \in \powerset S: X_1 \ne X_2 \land \map {f^\to} {X_1} = \map {f^\to} {X_2} = Y$
There are two cases to consider: 
:$(1): \quad$ Either $X_1$ or $X_2$ is the empty set
:$(2): \quad$ Neither $X_1$ nor $X_2$ is the empty set.
 assume that $X_1 = \O$.
Then, from Image of Empty Set is Empty Set:
:$Y = \map {f^\to} {X_1} = \O$
But that means:
:$\map {f^\to} {X_2} = \O$
Now $X_1 \ne X_2$, so $X_2 \ne \O$.
Thus:
:$\exists x_2 \in X_2: \neg \exists y \in T: \map f {x_2} = y$
which means that $f$ is not even a mapping, let alone an injection.
The same argument applies if $X_2$ is empty.
Now, assume that neither $X_1$ nor $X_2$ is the empty set.
As $X_1 \ne X_2$, there must be at least one element in either one which is not in the other.
 assume that $\exists x_1 \in X_1: x_1 \notin X_2$.
Now suppose $\map f {x_1} = y \in Y$.
However, as $\map {f^\to} {X_2} = Y$, there must be some $x_2 \in X_2$ such that $\map f {x_2} = y \in Y$.
So:
:$\exists x_1 \ne x_2 \in S: \map f {x_1} = \map f {x_2}$
which means, by definition, that $f$ is not injective.
Thus, if $f^\to: \powerset S \to \powerset T$ is not an injection then neither is $f: S \to T$.
Thus, by the Rule of Transposition, the result follows.
Category:Injections
Category:Direct Image Mappings
Mapping is Involution iff Bijective and Symmetric

== Theorem ==
Let $S$ be a set.
Let $f: S \to S$ be a mapping on $S$.
Then $f$ is an involution $f$ is both a bijection and a symmetric relation.
== Proof ==
By definition an involution on $S$ is a mapping such that:
:$\forall x \in S: \map f {\map f x} = x$
=== Necessary Condition ===
Let $f$ be an involution.
By Involution is Permutation, $f$ is a permutation and therefore by definition a bijection.
Then:
Thus $f$, considered as a relation, is symmetric.
Thus it has been shown that if $f$ is an involution, it is both a bijection and a symmetric relation.
=== Sufficient Condition ===
Let $f$ be a mapping which is both a bijection and a symmetric relation.
Then:
and so $f$ is shown to be an involution.
== Sources ==
* : $\S 5$. Induced mappings; composition; injections; surjections; bijections: Exercise $11$
Category:Involutions
Category:Bijections
Category:Symmetric Relations
Mapping is Surjection iff Direct Image Mapping is Surjection

== Theorem ==
Let $f: S \to T$ be a mapping.
Let $f^\to: \powerset S \to \powerset T$ be the direct image mapping of $f$.
Then:
:$f^\to$ is a surjection.
:$f: S \to T$ is also a surjection.
== Proof ==
=== Necessary Condition ===
Follows from Direct Image Mapping of Surjection is Surjection.
=== Sufficient Condition ===
Let $y \in S$.
Since $f^\to$ is a surjection:
:$\exists U \in \powerset S: \map {f^\to} U = \set y$
From definition of direct image mapping:
:$\set y \ne \O \implies U \ne \O$
Let $x \in U$.
Then from definition of direct image mapping:
:$\map f x = y$
Since $y$ is arbitrary, $f$ is a surjection.
== Also see ==
* Direct Image Mapping of Surjection is Surjection
Category:Surjections
Category:Direct Image Mappings
Mapping on Cartesian Product of Substructures is Restriction of Operation

== Theorem ==
Let $\struct {S, \circ}$ be an algebraic structure with $1$ operation.
Let $\struct {A, \circ {\restriction_A} }, \struct {B, \circ {\restriction_B} }$ be closed algebraic substructures of $\struct {S, \circ}$, where $\circ {\restriction_A}$ and $\circ {\restriction_B}$ are the operations induced by the restrictions of $\circ$ to $A$ and $B$ respectively.
Let the mapping $\phi: A \times B \to S$ be defined as:
:$\forall \tuple {a, b} \in A \times B: \map \phi {a, b} = a \circ b$
where $A \times B$ denotes the Cartesian product of $A$ and $B$.
Then $\phi$ is the restriction to $A \times B$ of the operation $\circ$ on $S \times S \to S$.
== Proof ==
Suppose the mapping $\phi: A \times B \to S$ is defined as:
:$\forall \tuple {a, b} \in A \times B: \map \phi {a, b} = a \circ b$
where $A \times B$ denotes the Cartesian product of $A$ and $B$.
Then:
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces
Category:Internal Direct Products
Mapping on Increasing Union

== Theorem ==
Let $S_0, S_1, S_2, \ldots, S_i, \ldots$ be sets such that:
:$S_0 \subseteq S_1 \subseteq S_2 \subseteq \ldots \subseteq S_i \subseteq \ldots$
that is, each set is contained in the next as a subset.
Let $S$ be the increasing union of $S_0, S_1, S_2, \ldots, S_i, \ldots$:
:$\ds S = \bigcup_{i \mathop \in \N} S_i$
For each $i \in \N$, let $f_i: S_i \to T$ be a mapping such that:
:$\forall j < i: f_i \restriction_{S_j} = f_j$
where $f_i \restriction_{S_j}$ denotes the restriction of $f_i$ to $S_j$.
Then there is a unique mapping:
:$f: S \to T$
which extends each $f_i$ to $S$.
== Proof ==
Suppose $f: S \to T$ and $g: S \to T$ are both extensions of $f_i$ to $S$ for all $i \in \N$ such that $f \ne g$.
We have been given the domain and codomain of $f$.
Suppose there exists any $g: S \to T$ which is different from $f$.
This must be because $f$ and $g$ do not agree throughout their entire domain.
Thus:
:$\exists x \in S: \map f x \ne \map g x$
Then by definition of set union:
:$\exists S_k \subseteq S: x \in S_k$.
From the definition, both $f$ and $g$ are extensions of $S_k$.
But this means they must agree on $S_k$.
That is:
:$\forall x \in S_k: \map f x = \map g x$
This contradicts the supposition that $\map f x \ne \map g x$.
Thus:
:$\forall x \in S: \map f x = \map g x$
and $f$ has been shown to equal $g$.
Hence the result.
== Sources ==
* : Appendix $\text{A}.4$: Composition and Restriction: Problem $\text{A}.4.1$
Category:Mapping Theory
Category:Set Union
Mapping on Integers is Endomorphism of Max or Min Operation iff Increasing

== Theorem ==
Let $\Z$ denote the set of integers.
Let $f: \Z \to \Z$ be a mapping on $\Z$.
Let $\vee$ and $\wedge$ be the max operation and min operation on $\Z$ defined as:
Then:
:$f$ is an endomorphism of $\struct {\Z, \vee}$ or $\struct {\Z, \wedge}$
:
:$f$ is an increasing mapping.
== Proof ==
=== Necessary Condition ===
Suppose $f$ is an increasing mapping.
Let $x, y \in \Z$ and suppose $x \le y$.
By definition of an increasing mapping, we have $\map f x \le \map f y$.
Therefore:
:$\map f x \vee \map f y = \map f y = \map f {x \vee y}$
:$\map f x \wedge \map f y = \map f x = \map f {x \wedge y}$
Hence $f$ is an endomorphism of $\struct {\Z, \vee}$ and $\struct {\Z, \wedge}$.
As Conjunction implies Disjunction, $f$ is an endomorphism of $\struct {\Z, \vee}$ or $\struct {\Z, \wedge}$.
=== Sufficient Condition ===
Suppose $f$ is an endomorphism of $\struct {\Z, \vee}$.
Then for any $x, y \in \Z$, we have:
:$\map f x \vee \map f y = \map f {x \vee y}$
 suppose $x \le y$.
Then we have:
:$\map f y = \map f {x \vee y} = \map f x \vee \map f y$
and thus:
:$\map f x \le \map f y$
Hence $f$ is an increasing mapping.
Now suppose $f$ is an endomorphism of $\struct {\Z, \wedge}$.
Then for any $x, y \in \Z$, we have:
:$\map f x \wedge \map f y = \map f {x \wedge y}$
 suppose $x \le y$.
Then we have:
:$\map f x = \map f {x \wedge y} = \map f x \wedge \map f y$
and thus:
:$\map f x \le \map f y$
Hence $f$ is an increasing mapping.
Therefore if $f$ is an endomorphism of either $\struct {\Z, \vee}$ or $\struct {\Z, \wedge}$, $f$ is an increasing mapping.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Exercise $12.8$
Category:Endomorphisms
Category:Min Operation
Category:Max Operation
Category:Increasing Mappings
Mapping on Integers is Homomorphism between Max or Min Operation iff Decreasing

== Theorem ==
Let $\Z$ denote the set of integers.
Let $f: \Z \to \Z$ be a mapping on $\Z$.
Let $\vee$ and $\wedge$ be the operations on $\Z$ defined as:
Then:
:$f$ is a homomorphism either from $\struct {\Z, \vee}$ to $\struct {\Z, \wedge}$ or from $\struct {\Z, \wedge}$ to $\struct {\Z, \vee}$
:
:$f$ is a decreasing mapping.
== Proof ==
=== Necessary Condition ===
Let $f$ be a decreasing mapping.
Let $x, y \in \Z$ such that $x \le y$.
By definition of a decreasing mapping, we have $\map f x \ge \map f y$.
Therefore:
:$\map f x \vee \map f y = \map f x = \map f {x \wedge y}$
:$\map f x \wedge \map f y = \map f y = \map f {x \vee y}$
Hence:
:$f$ is a homomorphism from $\struct {\Z, \vee}$ to $\struct {\Z, \wedge}$
and also:
:$f$ is a homomorphism from $\struct {\Z, \wedge}$ to $\struct {\Z, \vee}$
As Conjunction implies Disjunction, $f$ is a homomorphism from $\struct {\Z, \vee}$ to $\struct {\Z, \wedge}$ or from $\struct {\Z, \wedge}$ to $\struct {\Z, \vee}$.
=== Sufficient Condition ===
Suppose $f$ is a homomorphism from $\struct {\Z, \wedge}$ to $\struct {\Z, \vee}$.
Then for any $x, y \in \Z$, we have:
:$\map f {x \wedge y} = \map f x \vee \map f y$
 suppose $x \le y$.
Then we have:
:$\map f x = \map f {x \wedge y} = \map f x \vee \map f y$
and thus:
:$\map f x \ge \map f y$
Hence $f$ is a decreasing mapping.
Now suppose $f$ is a homomorphism from $\struct {\Z, \vee}$ to $\struct {\Z, \wedge}$.
Then for any $x, y \in \Z$, we have:
:$\map f {x \vee y} = \map f x \wedge \map f y$
 suppose $x \le y$.
Then we have:
:$\map f y = \map f {x \vee y} = \map f x \wedge \map f y$
and thus:
:$\map f x \ge \map f y$
Hence $f$ is a decreasing mapping.
Therefore if $f$ is a homomorphism either from $\struct {\Z, \vee}$ to $\struct {\Z, \wedge}$ or from $\struct {\Z, \wedge}$ to $\struct {\Z, \vee}$, $f$ is a decreasing mapping.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Exercise $12.8$
Category:Homomorphisms (Abstract Algebra)
Category:Min Operation
Category:Max Operation
Category:Decreasing Mappings
Mapping on Quadratic Integers over 2 to Conjugate is Automorphism

== Theorem ==
Let $\Z \sqbrk {\sqrt 2}$ denote the set of quadratic integers over $2$:
:$\Z \sqbrk {\sqrt 2} := \set {a + b \sqrt 2: a, b \in \Z}$
that is, all numbers of the form $a + b \sqrt 2$ where $a$ and $b$ are integers.
Then the mapping $\phi: \Z \sqbrk {\sqrt 2} \to \Z \sqbrk {\sqrt 2}$ defined as:
:$\forall x = a + b \sqrt 2 \in \Z \sqbrk {\sqrt 2}: \map \phi {a + b \sqrt 2} = a - b \sqrt 2$
is a ring automorphism.
== Proof ==
We have Quadratic Integers over 2 form Subdomain of Reals.
First we note that:
:$\forall x \in \Z \sqbrk {\sqrt 2}: \map \phi x \in \Z \sqbrk {\sqrt 2}$
=== Proof of Bijection ===
Let $\map \phi {a + b \sqrt 2} = \map \phi {a' + b' \sqrt 2}$.
Then:
:$a - b \sqrt 2 = a' - b' \sqrt 2$
and so:
:$a + b \sqrt 2 = a' + b' \sqrt 2$
So $\phi$ is injective.
Now let $y = c + d \sqrt 2 \in \Z \sqbrk {\sqrt 2}$.
We have that:
Hence:
:$\forall y \in \Z \sqbrk {\sqrt 2}: \exists x \in \Z \sqbrk {\sqrt 2}: \map \phi x = y$
and so $\phi$ is surjective.
So $\phi$ is a bijection.
=== Proof of Morphism ===
Now consider $x = a + b \sqrt 2, y = c + d \sqrt 2$.
Then:
So $\phi$ is a ring homomorphism which is also a bijection, whose image equals its domain.
Hence the result by definition of ring automorphism.
== Sources ==
* : Chapter $5$: Rings: $\S 24$. Homomorphisms: Example $44$
Category:Examples of Ring Automorphisms
Category:Examples of Integral Domains
Category:Quadratic Integers
Mapping on Quadratic Integers over 3 to Conjugate is Automorphism

== Theorem ==
Let $\Z \sqbrk {\sqrt 3}$ denote the set of quadratic integers over $3$:
:$\Z \sqbrk {\sqrt 3} := \set {a + b \sqrt 3: a, b \in \Z}$
that is, all numbers of the form $a + b \sqrt 3$ where $a$ and $b$ are integers.
Then the mapping $\phi: \Z \sqbrk {\sqrt 3} \to \Z \sqbrk {\sqrt 3}$ defined as:
:$\forall x = a + b \sqrt 3 \in \Z \sqbrk {\sqrt 3}: \map \phi {a + b \sqrt 3} = a - b \sqrt 3$
is a ring automorphism.
== Proof ==
We have Quadratic Integers over 2 form Subdomain of Reals.
First we note that:
:$\forall x \in \Z \sqbrk {\sqrt 3}: \map \phi x \in \Z \sqbrk {\sqrt 3}$
=== Proof of Bijection ===
Let $\map \phi {a + b \sqrt 3} = \map \phi {a' + b' \sqrt 3}$.
Then:
:$a - b \sqrt 3 = a' - b' \sqrt 3$
and so:
:$a + b \sqrt 3 = a' + b' \sqrt 3$
So $\phi$ is injective.
Now let $y = c + d \sqrt 3 \in \Z \sqbrk {\sqrt 3}$.
We have that:
Hence:
:$\forall y \in \Z \sqbrk {\sqrt 3}: \exists x \in \Z \sqbrk {\sqrt 3}: \map \phi x = y$
and so $\phi$ is surjective.
So $\phi$ is a bijection.
=== Proof of Morphism ===
Now consider $x = a + b \sqrt 3, y = c + d \sqrt 3$.
Then:
So $\phi$ is a ring homomorphism which is also a bijection, whose image equals its domain.
Hence the result by definition of ring automorphism.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 6$: Isomorphisms of Algebraic Structures: Exercise $6.7 \ \text {(b)}$
Category:Examples of Ring Automorphisms
Category:Examples of Integral Domains
Category:Quadratic Integers
Mapping on Total Ordering reflects Transitivity

== Theorem ==
Let $\struct {S, \preccurlyeq}$ be a totally ordered set.
Let $f: S \to T$ be a mapping to an arbitrary set $T$.
Let $\RR$ be a relation on $T$ defined such that:
:$\RR: = \set {\tuple {\map f x, \map f y}: x \preccurlyeq y}$
That is, $a$ is related to $b$ in $T$ they have preimages $x$ and $y$ under $f$ such that $x$ precedes $y$.
Then $\RR$ is transitive.
== Proof ==
Let $a, b, c \in T$ such that:
:$a \mathrel \RR b$
:$b \mathrel \RR c$
Then there exist $x, y, z \in S$ such that:
:$a = \map f x$
:$b = \map f y$
:$c = \map f z$
such that:
:$x \preccurlyeq y$
:$y \preccurlyeq z$
As $\preccurlyeq$ is a total ordering, it follows that:
:$x \preccurlyeq z$
and so by definition of $\RR$:
:$\map f x \mathrel \RR \map f z$
That is:
:$a \mathrel \RR c$
As $a$, $b$ and $c$ were arbitrary, it follows that $\RR$ is transitive.
== Sources ==
* : Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations: Exercise $41 \ \text {(a)}$
Category:Total Orderings
Category:Transitive Relations
Mapping reflects Preordering

== Theorem ==
Let $S$ and $T$ be sets.
Let $f: S \to T$ be a mapping.
Let ${\precsim} \subseteq T \times T$ be a preordering on $T$.
Let $\RR$ be the relation defined on $S$ by the rule:
:$x \mathrel \RR y \iff \map f x \precsim \map f y$
Then $\RR$ is a preordering on $S$.
== Proof ==
=== Reflexivity ===
Thus $\RR$ is reflexive.
=== Transitivity ===
Thus $\RR$ is transitive.
So, by definition, $\RR$ is a preordering.
== Sources ==
* : Chapter $1$: Sets and mappings: Further exercises: $6$
Category:Preorder Theory
Category:Mapping Theory