avewright/proofwiki-math huggingface dataset:
Theorem: Limit Inferior of Restriction Net is Supremum of Image of Directed Subset

Let $L = \struct {S, \vee_1, \wedge_1, \preceq_1}$ and $\struct {T, \vee_2, \wedge_2, \preceq_2}$ be up-complete lattices.
Let $f: S \to T$ be an increasing mapping.
Let $D \subseteq S$ be a directed subset of $S$.
Let $\struct {D, \preceq'}$ be a directed ordered subset of $L$.
Let $f \restriction D: D \to T$, the restriction of mapping, be a net in $T$.
Then $\map \liminf {f \restriction D} = \map \sup {f \sqbrk D}$

Proof:
We will prove that
:(lemma): $\forall j \in D: \map {\inf_L} {\paren {f \restriction D} \sqbrk {\map {\preceq'} j} } = \map f j$
Let $j \in D$.
By definitions of image of element and upper closure of element:
:$\map {\preceq'} j = j^{\succeq'}$
By Upper Closure in Ordered Subset is Intersection of Subset and Upper Closure:
:$j^{\succeq'} = D \cap j^{\succeq_1}$
By Intersection is Subset:
:$j^{\succeq'} \subseteq j^{\succeq_1}$
By Image of Subset under Mapping is Subset of Image:
:$\paren {f \restriction D} \sqbrk {j^{\succeq'} } \subseteq \paren {f \restriction D} \sqbrk {j^{\succeq_1} }$
By Infimum of Subset and Infimum of Image of Upper Closure of Element under Increasing Mapping:
:$\map f j \preceq_2 \map {\inf_L} {\paren {f \restriction D} \sqbrk {\map {\preceq'} j} }$
By definition of reflexivity:
:$j \preceq' j$
By definition of image of element:
:$j \in \map {\preceq'} j$
By definition of image of set:
:$\map f j \in \paren {f \restriction D} \sqbrk {\map {\preceq'} j}$
By definitions of infimum and lower bound:
:$\map {\inf_L} {\paren {f \restriction D} \sqbrk {\map {\preceq'} j} } \preceq_2 \map f j$
Thus by definition of antisymmetry:
:$\map {\inf_L} {\paren {f \restriction D} \sqbrk {\map {\preceq'} j} } = \map f j$
{{qed|lemma}}
Thus
{{begin-eqn}}
{{eqn | l = \map \liminf {f \restriction D}
      | r = \sup \set {\map \inf {\paren {f \restriction D} \sqbrk {\map {\preceq'} j} }: j \in D}
      | c = {{Defof|Limit Inferior of Net}}
}}
{{eqn | r = \sup \set {\map f j: j \in D}
      | c = (lemma)
}}
{{eqn | r = \map \sup {f \sqbrk D}
      | c = {{Defof|Image of Subset under Mapping}}
}}
{{end-eqn}}
{{qed}}
Theorem: Infimum of Image of Upper Closure of Element under Increasing Mapping

Let $\struct {S, \preceq}$ and $\struct {T, \precsim}$ be ordered set.
Let $f: S \to T$ be an increasing mapping.
Let $x \in S$.
Then $\map \inf {f \sqbrk {x^\succeq} } = \map f x$

Proof:
By Infimum of Upper Closure of Element:
:$\inf x^\succeq = x$
By definition of infimum:
:$x$ is lower bound for $x^\succeq$
Thus by Increasing Mapping Preserves Lower Bounds:
:$\map f x$ is lower bound for $f \sqbrk {x^\succeq}$
By definition of reflexivity:
:$x \preceq x$
By definition of upper closure of element:
:$x \in x^\succeq$
By definition of image of set:
:$\map f x \in f \sqbrk {x^\succeq}$
Thus by definition:
:$\forall y \in T: y$ is lower bound for $f \sqbrk {x^\succeq} \implies y \precsim \map f x$
Thus by definition of infimum:
:$\map \inf {f \sqbrk {x^\succeq} } = \map f x$
{{qed}}
Theorem: Poulet Numbers which are also Magic Constant for Magic Square

The sequence of Poulet numbers which are also the magic constant of a magic square begins:
:$1105, 2465, \ldots$

Proof:
From the sequence of Poulet numbers, these are Poulet numbers:
:$1105, 2465, \ldots$
Then we have:
{{begin-eqn}}
{{eqn | l = 1105
      | r = \dfrac {13 \paren {13^2 + 1} } 2
      | c = so $1105$ is the magic constant of the order $13$ magic square
}}
{{eqn | l = 2465
      | r = \dfrac {17 \paren {17^2 + 1} } 2
      | c = so $2465$ is the magic constant of the order $17$ magic square
}}
{{end-eqn}}
{{expand|Further investigation needed. I have not seen this correspondence explored on the internet. I note though that both $1105$ and $2465$ are also Carmichael.}}
Category:Poulet Numbers
Category:Magic Squares
ahytsnqo4qufaah5amkvd28zlhygl7u
Theorem: Composition of Mapping and Inclusion is Restriction of Mapping

Let $S, T$ be sets.
Let $f: S \to T$ be a mapping.
Let $A \subseteq S$.
Then $f \circ i_A = f \restriction A$
where
:$i_A$ denotes the inclusion mapping of $A$,
:$f \restriction A$ denotes the restriction of $f$ to $A$.

Proof:
By definition of inclusion mapping:
:$i_A: A \to S$
By definitions of composition of mappings and restriction of mapping:
:$f \circ i_A: A \to T$ and $f \restriction A: A \to T$
Let $a \in A$.
Thus
{{begin-eqn}}
{{eqn | l = \map {\paren {f \circ i_A} } a
      | r = \map f {\map {i_A} a}
      | c = {{Defof|Composition of Mappings}}
}}
{{eqn | r = \map f a
      | c = {{Defof|Inclusion Mapping}}
}}
{{eqn | r = \map {\paren {f \restriction A} } a
      | c = {{Defof|Restriction of Mapping}}
}}
{{end-eqn}}
{{qed}}
Theorem: Mapping at Limit Inferior Precedes Limit Inferior of Composition Mapping and Sequence implies Mapping Preserves Directed Suprema

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 a mapping such that
:for all directed sets $\struct {D, \precsim}$ and nets $N:D \to S$ in $S$: $\map f {\liminf N} \preceq_2 \map \liminf {f \circ N}$
Then $f$ preserves directed suprema.

Proof:
Let $D$ be a directed subset of $S$.
Assume that
:$D$ admits a supremum.
Thus by definition of complete lattice:
:$f \sqbrk D$ admits a supremum.
Thus by Mapping at Limit Inferior Precedes Limit Inferior of Composition Mapping and Sequence implies Supremum of Image is Mapping at Supremum of Directed Subset:
:$\map \sup {f \sqbrk D} = \map f {\sup D}$
{{qed}}
Theorem: Thurston's Geometrization Conjecture

When a topological manifold of dimension $3$ has been split into its connected sum and the Jaco-Shalen-Johannson torus decomposition, the remaining components each admit exactly one of the following geometries:
:$(1): \quad$ Euclidean geometry
:$(2): \quad$ Hyperbolic geometry
:$(3): \quad$ Spherical geometry
:$(4): \quad$ The geometry of $\mathbb S^2 \times \R$
:$(5): \quad$ The geometry of $\mathbb H^2 \times \R$
:$(6): \quad$ The geometry of the universal cover $S L_2 \R^~$ of the Lie group $S L_2 \R$
:$(7): \quad$ Nil geometry
:$(8): \quad$ Sol geometry
where:
:$\mathbb S^2$ is the 2-sphere
:$\mathbb H^2$ is the hyperbolic plane.

Proof:
{{ProofWanted|Discuss {{AuthorRef|Grigori Perelman}}'s work -- it is generally believed his proof of this is valid, in which case this needs to be renamed to Theorem.}}
{{Namedfor|William Paul Thurston|cat = Thurston}}
Theorem: Mapping Preserves Directed Suprema implies Mapping is Continuous

Let $\struct {S, \preceq_1, \tau_1}$ and $\struct {T, \preceq_2, \tau_2}$ be up-complete ordered sets with Scott topologies.
Let $f: S \to T$ be a directed suprema preserving mapping.
Then $f$ is continuous.

Proof:
Let $P$ be a closed subset of $T$.
By Closed Set iff Lower and Closed under Directed Suprema in Scott Topological Ordered Set:
:$P$ is lower and closed under directed suprema.
We will prove that
:for all directed subset $D$ of $S$: $D \subseteq f^{-1} \sqbrk P \implies \sup D \in f^{-1} \sqbrk P$
Let $D$ be a directed subset of $S$ such that
:$D \subseteq f^{-1} \sqbrk P$
By definition of mapping preserves directed suprema:
:$f$ preserves the supremum of $D$.
By definition of up-complete:
:$D$ admits a supremum.
By definition of mapping preserves the supremum:
:$\map \sup {f \sqbrk D} = \map f {\sup D}$
By Directed Suprema Preserving Mapping is Increasing:
:$f$ is increasing.
By Image of Directed Subset under Increasing Mapping is Directed:
:$f \sqbrk D$ is directed.
By Preimage is Subset of Set implies Image of Set is Subset:
:$f \sqbrk D \subseteq P$
By definition of closed under directed suprema:
:$\map \sup {f \sqbrk D} \in P$
Thus by definition of preimage of set:
:$\sup D \in f^{-1} \sqbrk P$
{{qed|lemma}}
By definition:
:$f^{-1} \sqbrk P$ is closed under directed suprema.
By Preimage of Lower Section under Increasing Mapping is Lower:
:$f^{-1} \sqbrk P$ is lower.
Thus by Closed Set iff Lower and Closed under Directed Suprema in Scott Topological Ordered Set:
:$f^{-1} \sqbrk P$ is closed.
{{qed}}
Theorem: Secant of Complex Number

Let $a$ and $b$ be real numbers.
Let $i$ be the imaginary unit.
Then:
:$\sec \paren {a + b i} = \dfrac {\cos a \cosh b + i \sin a \sinh b} {\cos^2 a \cosh^2 b + \sin^2 a \sinh^2 b}$
where:
:$\sec$ denotes the complex secant function.
:$\sin$ denotes the real sine function
:$\cos$ denotes the real cosine function
:$\sinh$ denotes the hyperbolic sine function
:$\cosh$ denotes the hyperbolic cosine function

Proof:
{{begin-eqn}}
{{eqn | l = \sec \paren {a + b i}
      | r = \frac 1 {\cos \paren {a + b i} }
      | c = {{Defof|Complex Secant Function}}
}}
{{eqn | r = \dfrac 1 {\cos a \cosh b - i \sin a \sinh b}
      | c = Cosine of Complex Number
}}
{{eqn | r = \dfrac {\cos a \cosh b + i \sin a \sinh b} {\paren {\cos a \cosh b + i \sin a \sinh b} \paren {\cos a \cosh b - i \sin a \sinh b} }
      | c = multiplying denominator and numerator by $\cos a \cosh b + i \sin a \sinh b$
}}
{{eqn | r = \dfrac {\cos a \cosh b + i \sin a \sinh b} {\cos^2 a \cosh^2 b - i^2 \sin^2 a \sinh^2 b}
      | c = Difference of Two Squares
}}
{{eqn | r = \dfrac {\cos a \cosh b + i \sin a \sinh b} {\cos^2 a \cosh^2 b + \sin^2 a \sinh^2 b}
      | c = {{Defof|Imaginary Unit}}
}}
{{end-eqn}}
{{qed}}
Theorem: Tangent of Complex Number/Formulation 1

:$\tan \paren {a + b i} = \dfrac {\sin a \cosh b + i \cos a \sinh b} {\cos a \cosh b - i \sin a \sinh b}$

Proof:
{{begin-eqn}}
{{eqn | l = \tan \paren {a + b i}
      | r = \frac {\sin \paren {a + b i} } {\cos \paren {a + b i} }
      | c = {{Defof|Complex Tangent Function}}
}}
{{eqn | r = \dfrac {\sin a \cosh b + i \cos a \sinh b} {\cos a \cosh b - i \sin a \sinh b}
      | c = Sine of Complex Number and Cosine of Complex Number
}}
{{end-eqn}}
{{qed}}
Theorem: Tangent of Complex Number/Formulation 2

:$\tan \paren {a + b i} = \dfrac {\tan a + i \tanh b} {1 - i \tan a \tanh b}$

Proof:
{{begin-eqn}}
{{eqn | l = \tan \paren {a + b i}
      | r = \dfrac {\sin a \cosh b + i \cos a \sinh b} {\cos a \cosh b - i \sin a \sinh b}
      | c = Tangent of Complex Number: Formulation 1
}}
{{eqn | r = \dfrac {\tan a \cosh b + i \sinh b} {\cosh b - i \tan a \sinh b}
      | c = multiplying denominator and numerator by $\dfrac 1 {\cos a}$
}}
{{eqn | r = \frac {\tan a + i \tanh b} {1 - i \tan a \tanh b}
      | c = multiplying denominator and numerator by $\dfrac 1 {\cosh b}$
}}
{{end-eqn}}
{{qed}}
Theorem: Tangent of Complex Number/Formulation 3

:$\tan \paren {a + b i} = \dfrac {\tan a - \tan a \tanh ^2 b} {1 + \tan ^2 a \tanh ^2 b} + \dfrac {\tanh b + \tan ^2 a \tanh b} {1 + \tan ^2 a \tanh ^2 b} i$

Proof:
{{begin-eqn}}
{{eqn | l = \tan \paren {a + b i}
      | r = \frac {\tan a + i \tanh b} {1 - i \tan a \tanh b}
      | c = Tangent of Complex Number: Formulation 2
}}
{{eqn | r = \frac {\paren {\tan a + i \tanh b} \paren {1 + i \tan a \tanh b} } {1 + \tan ^2 a \tanh ^2 b}
      | c = multiplying denominator and numerator by $1 + i \tan a \tanh b$
}}
{{eqn | r = \frac {\tan a + i \tanh b + i \tan ^2 a \tanh b - \tan a \tanh ^2 b} {1 + \tan ^2 a \tanh ^2 b}
      | c = 
}}
{{eqn | r = \frac {\tan a - \tan a \tanh ^2 b} {1 + \tan ^2 a \tanh ^2 b} + \frac {\tanh b + \tan ^2 a \tanh b} {1 + \tan ^2 a \tanh ^2 b} i
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Cosecant of Complex Number

Let $a$ and $b$ be real numbers.
Let $i$ be the imaginary unit.
Then:
:$\csc \paren {a + b i} = \dfrac {\sin a \cosh b - i \cos a \sinh b} {\sin^2 a \cosh^2 b + \cos^2 a \sinh^2 b}$
where:
:$\csc$ denotes the complex cosecant function.
:$\sin$ denotes the real sine function
:$\cos$ denotes the real cosine function
:$\sinh$ denotes the hyperbolic sine function
:$\cosh$ denotes the hyperbolic cosine function

Proof:
{{begin-eqn}}
{{eqn | l = \csc \paren {a + b i}
      | r = \frac 1 {\sin \paren {a + b i} }
      | c = {{Defof|Complex Cosecant Function}}
}}
{{eqn | r = \dfrac 1 {\sin a \cosh b + i \cos a \sinh b}
      | c = Sine of Complex Number
}}
{{eqn | r = \dfrac {\sin a \cosh b - i \cos a \sinh b} {\paren {\sin a \cosh b - i \cos a \sinh b} \paren {\sin a \cosh b + i \cos a \sinh b} }
      | c = multiplying denominator and numerator by $\sin a \cosh b - i \cos a \sinh b$
}}
{{eqn | r = \dfrac {\sin a \cosh b - i \cos a \sinh b} {\sin^2 a \cosh^2 b - i^2 \cos^2 a \sinh^2 b}
      | c = Difference of Two Squares
}}
{{eqn | r = \dfrac {\sin a \cosh b - i \cos a \sinh b} {\sin^2 a \cosh^2 b + \cos^2 a \sinh^2 b}
      | c = {{Defof|Imaginary Unit}}
}}
{{end-eqn}}
{{qed}}
Theorem: Cotangent of Complex Number/Formulation 1

:$\cot \paren {a + b i} = \dfrac {\cos a \cosh b - i \sin a \sinh b} {\sin a \cosh b + i \cos a \sinh b}$

Proof:
{{begin-eqn}}
{{eqn | l = \cot \paren {a + b i}
      | r = \frac {\cos \paren {a + b i} } {\sin \paren {a + b i} }
      | c = {{Defof|Complex Cotangent Function}}
}}
{{eqn | r = \dfrac {\cos a \cosh b - i \sin a \sinh b} {\sin a \cosh b + i \cos a \sinh b}
      | c = Sine of Complex Number and Cosine of Complex Number
}}
{{end-eqn}}
{{qed}}
Theorem: Cotangent of Complex Number/Formulation 2

:$\map \cot {a + b i} = \dfrac {-1 - i \cot a \coth b} {\cot a - i \coth b}$

Proof:
{{begin-eqn}}
{{eqn | l = \map \cot {a + b i}
      | r = \dfrac {\cos a \cosh b - i \sin a \sinh b} {\sin a \cosh b + i \cos a \sinh b}
      | c = Cotangent of Complex Number: Formulation 1
}}
{{eqn | r = \dfrac {\cot a \cosh b - i \sinh b} {\cosh b + i \cot a \sinh b}
      | c = multiplying denominator and numerator by $\dfrac 1 {\sin a}$
}}
{{eqn | r = \dfrac {\cot a \coth b - i} {\coth b + i \cot a}
      | c = multiplying denominator and numerator by $\dfrac 1 {\sinh b}$
}}
{{eqn | r = \dfrac {-1 -i \cot a \coth b} {\cot a - i \coth b}
      | c = multiplying denominator and numerator by $-i$
}}
{{end-eqn}}
{{qed}}
Theorem: Cotangent of Complex Number/Formulation 3

:$\map \cot {a + b i} = \dfrac {\cot a \coth^2 b - \cot a} {\cot^2 a + \coth^2 b} + \dfrac {-\cot^2 a \coth b - \coth b} {\cot^2 a + \coth^2 b} i$

Proof:
{{begin-eqn}}
{{eqn | l = \map \cot {a + b i}
      | r = \dfrac {1 + i \cot a \coth b} {\cot a - i \coth b}
      | c = Cotangent of Complex Number: Formulation 2
}}
{{eqn | r = \dfrac {\paren {1 + i \cot a \coth b} \paren {\cot a + i \coth b} } {\paren {\cot a - i \coth b} \paren {\cot a + i \coth b} }
      | c = multiplying denominator and numerator by $\cot a + i \coth b$
}}
{{eqn | r = \dfrac {\paren {1 + i \cot a \coth b} \paren {\cot a + i \coth b} } {\cot^2 a + \coth^2 b}
      | c = Difference of Two Squares
}}
{{eqn | r = \dfrac {i \cot^2 a \coth b + \cot a - \cot a \coth^2 b + i \coth b} {\cot^2 a + \coth^2 b}
      | c = 
}}
{{eqn | r = \dfrac {\cot a \coth^2 b - \cot a} {\cot^2 a + \coth^2 b} + \dfrac {-\cot^2 a \coth b - \coth b} {\cot^2 a + \coth^2 b} i
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Mapping is Increasing implies Mapping at Infimum for Sequence Precedes Infimum for Composition of Mapping and Sequence

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:
:$\map f{\map \inf {N \sqbrk {\map \precsim j} } } \preceq_2 \map \inf {\paren {f \circ N} \sqbrk {\map \precsim j} }$
{{qed}}
Theorem: Set of Infima for Sequence is Directed

Let $\struct {S, \vee, \wedge, \preceq}$ be a complete lattice.
Let $\struct {A, \precsim}$ be a non-empty directed set.
Let $Z: A \to S$ be a net.
Let $D = \set {\map \inf {Z \sqbrk {\map \precsim j} }: j \in A}$ be a subset of $S$.
Then $D$ is directed.

Proof:
By definition of non-empty set:
:$\exists j: j \in A$
By definition of $D$:
:$\inf \left({Z \sqbrk {\map \precsim j} }\right) \in D$
Hence by definition:
:$D$ is a non-empty set.
Let $x, y \in D$.
By definition of $D$:
:$\exists j_1 \in A: x = \map \inf {Z \sqbrk {\map \precsim {j_1} } }$
and
:$\exists j_2 \in A: y = \map \inf {Z \sqbrk {\map \precsim {j_2} } }$
By definition of directed set:
:$\exists j \in A: j_1 \precsim j \land j_2 \precsim j$
By Preceding implies Image is Subset of Image:
:$\map \precsim j \subseteq \map {\mathord \precsim} {j_1}$ and $\map \precsim j \subseteq \map {\mathord \precsim} {j_2}$
By Image of Subset under Mapping is Subset of Image:
:$Z \sqbrk {\map \precsim j} \subseteq Z \sqbrk {\map \precsim {j_1} }$ and $Z \sqbrk {\map \precsim j} \subseteq Z \sqbrk {\map \precsim {j_2} }$
Thus by definition of $D$:
:$z := \map \inf {Z \sqbrk {\map \precsim j} } \in D$
Thus by Infimum of Subset:
:$x \preceq z$ and $y \preceq z$
{{qed}}
Theorem: Preceding implies Image is Subset of Image

Let $\struct {S, \precsim}$ be a preordered set.
Let $x, y \in S$ such that
:$x \precsim y$
Then $\map \precsim y \subseteq \mathord {\map \precsim x}$
where $\map \precsim y$ denotes the image of $y$ under $\precsim$.

Proof:
Let $z \in \mathord {\map \precsim y}$
By definition of image of element:
:$y \precsim z$
By definition of transitivity:
:$x \precsim z$
Thus by definition of image of element:
:$z \in \mathord {\map \precsim x}$
{{qed}}
Theorem: Universal Property of Group Ring

Let $R$ be a commutative ring with unity.
Let $G$ be a group.
Let $R \sqbrk G$ be the corresponding group ring.
Let $S$ be a commutative ring with unity.
Let $\phi: R \to S$ be a ring homomorphism.
Let $\beta : G \to S^\times$ be a group homomorphism, where $S^\times$ is the multiplicative group of $S$.
Then there exists a unique ring homomorphism from $R \sqbrk G$ to $S$ which extends $\phi$ and $\beta$.

Proof:
{{ProofWanted}}
Category:Universal Properties
Category:Group Rings
67dm0aljrwzkn75c8os68bnp1hssab4
Theorem: Continuous iff Directed Suprema Preserving

Let $\struct {S, \preceq_1, \tau_1}$ and $\struct {T, \preceq_2, \tau_2}$ be complete topological lattices with Scott topologies.
Let $f: S \to T$ be a mapping.
Then $f$ is continuous {{iff}} $f$ preserves directed suprema.

Proof:
=== Sufficient Condition ===
Assume that
:$f$ is continuous.
By Continuous iff Mapping at Limit Inferior Precedes Limit Inferior of Composition of Mapping and Sequence:
:for all directed set $\struct {D, \precsim}$ and net $N:D \to S$ in $S$: $\map f {\liminf N} \preceq_2 \map \liminf {f \circ N}$
Thus by Mapping at Limit Inferior Precedes Limit Inferior of Composition Mapping and Sequence implies Mapping Preserves Directed Suprema:
:$f$ preserves directed suprema.
{{qed|lemma}}
Theorem: Universal Property of Quotient of Topological Group

Let $G$ and $H$ be topological groups.
Let $N$ be a normal subgroup of $G$.
Let $\pi : G \to G/N$ be the quotient mapping.
Let $f : G \to H$ be a continuous group homomorphism whose kernel contains $N$.
Then there exists a unique continuous group homomorphism $\overline f: G / N \to H$ such that:
:$f = \overline f \circ \pi$

Proof:
Because $N \subset \ker f$, $f$ is constant on the cosets of $N$.
Thus $f$ is invariant under congruence modulo $N$.
By Universal Property of Quotient Set, there exists a unique mapping $\overline f: G / N \to H$ such that:
:$f = \overline f \circ \pi$
It suffices to verify that it is a continuous group homomorphism.
By Universal Property of Quotient Space, there exists a continuous mapping $\overline g: G / N \to H$ such that:
:$f = \overline g \circ \pi$
By uniqueness of $f$:
:$g = f$
By Universal Property of Quotient Group, there exists a group homomorphism $\overline h: G / N \to H$ such that:
:$f = \overline h \circ \pi$
By uniqueness of $f$:
:$h = f$
{{qed}}
Category:Topological Groups
Category:Universal Properties
sl31ilchz5brwtpmogo8gnhifh1enw1
Theorem: Universal Property of Quotient Set

Let $X$ and $Y$ be sets.
Let $\sim$ be an equivalence relation on $X$.
Let $\pi: X \to X / {\sim}$ be the quotient mapping to the quotient set.
Let $f: X \to Y$ be $\sim$-invariant.
Then there exists a unique mapping $\overline f : X / {\sim} \to Y$ such that $f = \overline f \circ \pi$.
:<nowiki>$\xymatrix{
X \ar[d]_\pi \ar[r]^f & Y\\
X/\sim \ar@{.>}[ru]_{\overline f} }$</nowiki>

Proof:
=== Existence ===
For $x \in X$, let $\eqclass x {} \in X / {\sim}$ denote its equivalence class under $\sim$.
Define the relation $\overline f: X / {\sim} \to Y$ by:
:$\tuple {\eqclass x {}, y} \in \overline f \iff \map f x = y$
We check that this is a well-defined mapping.
By Quotient Mapping is Surjection, $f$ is left-total.
Let $\eqclass x {} \in X / {\sim}$ and $y, z \in Y$ with:
:$\tuple {\eqclass x {}, y} \in \overline f$
:$\tuple {\eqclass x {}, z} \in \overline f$
We show that $y = z$.
By definition of $\overline f$, there exist $u, v \in X$ with $\eqclass u {} = \eqclass v {} = \eqclass x {}$ and $\map f u = y$ and $\map f v = z$.
By definition of equivalence class, $u \sim v$.
Because $f$ is $\sim$-invariant, $y = z$.
Thus $\overline f$ is many-to-one.
Thus $\overline f$ is a well-defined mapping.
{{qed|lemma}}
Theorem: Continuous iff Way Below iff There Exists Element that Way Below and Way Below

Let $\struct {S, \preceq_1, \tau_1}$ and $\struct {T, \preceq_2, \tau_2}$ be complete continuous topological lattices with Scott topologies.
Let $f: S \to T$ be a mapping.
Then $f$ is continuous {{iff}}
:$\forall x \in S, y \in T: y \ll \map f x \iff \exists w \in S: w \ll x \land y \ll \map f w$
{{explain|link to definition of $\ll$ and $\gg$}}

Proof:
=== Sufficient Condition ===
Assume that
:$f$ is continuous.
By Continuous iff Directed Suprema Preserving:
:$f$ preserves directed suprema.
By Directed Suprema Preserving Mapping at Element is Supremum:
:$\forall x \in S: \map f x = \sup \set {\map f w: w \in S \land w \ll x}$
Thus by Mapping at Element is Supremum implies Way Below iff There Exists Element that Way Below and Way Below:
:$\forall x \in S, y \in T: y \ll \map f x \iff \exists w \in S: w \ll x \land y \ll \map f w$
{{qed|lemma}}
Theorem: Finite Symmetric Group is Ambivalent

Let $n$ be a natural number.
Let $S_n$ be a symmetric group of order $n$.
Then $S_n$ is ambivalent.

Proof:
The Conjugacy Classes of Symmetric Group are determined uniquely by the cycle type of the elements.
Since any element in $S_n$ is of the same cycle type with its inverse, they are in the same conjugacy class.
Hence they are conjugates of each other.
This implies that $S_n$ is ambivalent.
{{qed}}
Category:Ambivalent Groups
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Theorem: Alternating Groups that are Ambivalent

Let $n$ be a natural number.
Then the $n$th alternating group $A_n$ is ambivalent {{iff}} $n \in \set {1, 2, 5, 6, 10, 14}$.
{{OEIS|A115200}}

Proof:
{{ProofWanted|Seems related to the representation theory of alternating group.}}
Category:Alternating Groups
Category:Ambivalent Groups
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Theorem: Transitivity of Big-O Estimates/General

Let $X$ be a topological space.
Let $V$ be a normed vector space over $\R$ or $\C$ with norm $\norm {\,\cdot\,}$.
Let $f, g, h: X \to V$ be functions.
Let $x_0 \in X$.
Let $f = \map \OO g$ and $g = \map \OO h$ as $x \to x_0$, where $\OO$ denotes big-$\OO$ notation.
Then $f = \map \OO h$ as $x \to x_0$.

Proof:
Because $f = \map \OO g$ and $g = \map \OO h$, there exist neighborhoods $U$ and $V$ of $x_0$ and real numbers $c, d \ge 0$ such that:
:$\norm {\map f x} \le c \cdot \norm {\map g x}$ for all $x \in U$
:$\norm {\map g x} \le d \cdot \norm {\map h x}$ for all $x \in V$.
By Intersection of Neighborhoods in Topological Space is Neighborhood, $U\cap V$ is a neighborhood of $x_0$.
For $x \in U \cap V$, we have:
:$\norm {\map f x} \le c \cdot \norm {\map g x} \le c d \cdot \norm {\map h x}$
Thus $f = \map \OO h$ for $x \to x_0$.
{{qed}}
Category:Big-O Notation
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