nyuuzyou/wikis huggingface dataset:
Coproduct on Disjoint Union

== Theorem ==
Let $S_1$ and $S_2$ be sets.
Let $S_1 \sqcup S_2 := \paren {S_1 \times \set 1} \cup \paren {S_2 \times \set 2}$ be the disjoint union of $S_1$ and $S_2$.
Let $i_1: S_1 \to S_1 \sqcup S_2$ and $i_2: S_2 \to S_1 \sqcup S_2$ be the mappings defined as:
:$\forall s_1 \in S_1: \map {i_1} {s_1} = \tuple {s_1, 1}$
:$\forall s_2 \in S_2: \map {i_2} {s_2} = \tuple {s_2, 2}$
Then $\struct {S_1 \sqcup S_2, i_1, i_2}$ is a coproduct of $S_1$ and $S_2$.
== Proof ==
For $\struct {S_1 \sqcup S_2, i_1, i_2}$ to be a coproduct, it is necessary that:
:for all sets $X$ and all mappings $f_1: S_1 \to X$ and $f_2: S_2 \to X$
::there exists a unique mapping $h: S_1 \sqcup S_2 \to X$ such that:
:::$h \circ i_1 = f_1$
:::$h \circ i_2 = f_2$
Let $h$ be the mapping defined as:
:$\forall \tuple {s, t} \in S_1 \sqcup S_2: \map h {s, t} = \begin {cases} \map {f_1} s & : t = 1 \\ \map {f_2} s & : t = 2 \end {cases}$
Then:
The existence of $h$ is apparent.
It remains to prove that $h$ is unique.
Let $h_1: S_1 \sqcup S_2 \to X$ and $h_2: S_1 \sqcup S_2 \to X$ both be mappings with the properties of $h$.
Let $\tuple {s, t} \in S_1 \sqcup S_2$ be arbitrary.
Then:
and:
Hence $h_1 = h_2$ and so $h$ is unique
Hence $\struct {S_1 \sqcup S_2, i_1, i_2}$ is a coproduct of $S_1$ and $S_2$.
== Sources ==
* : $\S 5$. Induced mappings; composition; injections; surjections; bijections: Exercise $18$
Category:Set Coproducts
Category:Disjoint Unions
Coreflexive Relation is Subset of Diagonal Relation

== Theorem ==
A coreflexive relation is a subset of the diagonal relation.
== Proof ==
Let $\RR \subseteq S \times S$ be a coreflexive relation.
Let $\tuple {x, y} \in \RR$.
By definition of coreflexive, it follows that $x = y$, and hence $\tuple {x, y} = \tuple {x, x}$.
So by definition of the diagonal relation:
:$\tuple {x, y} \in \Delta_S$
Hence the result.
Category:Coreflexive Relations
Category:Diagonal Relation
Corollary of Gauss Lemma for Riemannian Manifolds

== Theorem ==
Let $\struct {M, g}$ be a Riemannian manifold.
Let $U = \map {\exp_p} {\map {B_\epsilon} 0 }$ be a geodesic ball centered at $p \in M$. 
Let $r$ be the radial distance function.
Let $\partial_r$ be the radial vector field.
Suppose $\grad$ is the gradient operator.
Then:
:$\forall x \in U \setminus \set p : \valueat {\grad r}x = \valueat {\partial_r} x$
where $\setminus$ denotes the set difference.
== Proof ==
== Sources ==
* : $\S 6$: Geodesics and Distance. Geodesics Are Locally Minimizing
Category:Riemannian Manifolds
Correctness of Definition of Increasing Mappings Satisfying Inclusion in Lower Closure

== Theorem ==
Let $R = \struct {S, \preceq}$ be an ordered set.
Let $\map {\it Ids} R$ be the set of all ideals in $R$.
Let $L = \struct {\map {\it Ids} R, \precsim}$ be an ordered set
where $\precsim \mathop = \subseteq \restriction_{\map {\it Ids} R \times \map {\it Ids} R}$
Let:
:$M = \struct {F, \preccurlyeq}$
where:
:$F = \leftset {f: S \to \map {\it Ids} R: f} \,$ is an increasing mapping and $\rightset {\forall x \in S: \map f x \subseteq x^\preceq}$
:$\preccurlyeq$ is ordering on mappings generated by $\precsim$
:$x^\preceq$ denotes the lower closure of $x$.
Then $M$ is an ordered set.
== Proof ==
=== Reflexivity ===
Let $f \in F$.
By definition of reflexivity:
:$\forall x \in S: \map f x \precsim \map f x$
Thus by definition of ordering on mappings:
:$f \preccurlyeq f$
=== Transitivity ===
Let $f, g, h \in F$ such that
:$f \preccurlyeq g \preccurlyeq h$
By definition of ordering on mappings:
:$\forall x \in S: \map f x \precsim \map g x \precsim \map h x$
By definition of transitivity:
:$\forall x \in S: \map f x \precsim \map h x$
Thus by definition of ordering on mappings:
:$f \preccurlyeq h$
=== Antisymmetry ===
Let $f, g \in F$ such that
:$f \preccurlyeq g$ and $g \preccurlyeq f$
By definition of ordering on mappings:
:$\forall x \in S: \map f x \precsim \map g x \land \map g x \precsim \map f x$
By definition of antisymmetry:
:$\forall x \in S: \map f x = \map g x$
Thus by equality of mappings:
:$f = g$
Thus by definition:
:$M$ is an ordered set.
== Sources ==
* 
* 
Category:Order Theory
Correlation/Examples/Car Ownership and Alcohol Sales

== Example of Correlation ==
In a survey of centers of population, it is noticed that the number of people owning cars is highly correlated with the average daily sales of alcohol.
This, however, does not imply that car owners are likely to be drinkers.
Instead, it may merely reflect he fact that towns with higher population are likely to have both more car owners and sales of alcohol purely through the fact that they have more people.
== Sources ==
* 
* 
Category:Examples of Correlation
Correspondence Between Group Actions and Permutation Representations

== Theorem ==
Let $G$ be a group.
Let $X$ be a set.
There is a one-to-one correspondence between group actions of $G$ on $X$ and permutation representations of $G$ in $X$, as follows:
Let $\phi : G \times X \to X$ be a group action.
Let $\rho : G \to \struct {\map \Gamma X, \circ}$ be a permutation representation.
The following are equivalent:
:$(1): \quad \rho$ is the permutation representation associated to $\phi$
:$(2): \quad \phi$ is the group action associated to $\rho$
== Proof ==
For $g\in G$, define the mapping $\phi_g : X \to X$ as:
:$\map {\phi_g} x = \map \phi {g, x}$
Then $\rho$ is the permutation representation associated to $\phi$ :
:$\forall g \in G : \map \rho g = \phi_g$
By Equality of Mappings, this is equivalent to:
:$\forall g \in G : \forall x \in X : \map {\map \rho g} x = \map {\phi_g} x$
$\phi$ is the group action associated to $\rho$ :
:$\forall g \in G : \forall x \in X : \map \phi {g, x} = \map {\map \rho g} x$.
Because $\map {\phi_g} x = \map \phi {g, x}$, they are equivalent.
Category:Group Actions
Category:Permutation Representations
Correspondence Theorem (Group Theory)

== Theorem ==
Let $G$ be a group.
Let $N \lhd G$ be a normal subgroup of $G$.
Then every subgroup of the quotient group $G / N$ is of the form $H / N = \set {h N: h \in H}$, where $N \le H \le G$.
Conversely, if $N \le H \le G$ then $H / N \le G / N$.
The correspondence between subgroups of $G / N$ and subgroups of $G$ containing $N$ is a bijection.
This bijection maps normal subgroups of $G / N$ onto normal subgroups of $G$ which contain $N$.
== Proof ==
Let $H'$ be a subgroup of $G / N$, so that it consists of a certain set $\set {h N}$ of left cosets of $N$ in $G$.
Let us define the subset $\map \beta {H'} \subseteq G$:
:$\map \beta {H'} = \set {g \in G: g N \in H'}$
Then clearly:
:$N \subseteq \map \beta {H'}$
Also:
:$e_G \in N$
so:
:$e_G \in \map \beta {H'}$
Let $x, y \in \map \beta {H'}$. Then:
We also have, from Quotient Group is Group:
:$\paren {x N}^{-1} = x^{-1} N \implies x^{-1} \in \map \beta {H'}$
Thus, by the Two-Step Subgroup Test, $\map \beta {H'} \le G$ that contains $N$.
Conversely, let $H$ be such that $N \le H \le G$.
Let $\map \alpha H = \set {h N: h \in H} \subseteq G / N$.
It is easily checked that $\map \alpha H \le G / N$.
Now, let $X$ be the set of subgroups of $G$ containing $N$ and $Y$ be the set of all subgroups of $G / N$.
We now need to show that $\alpha: X \to Y$ is a bijection.
We do this by checking that $\beta: Y \to X$ is the inverse of $\alpha$.
To do this, we show that $\alpha \circ \beta = I_Y$ and $\beta \circ \alpha = I_X$.
Suppose $N \le H \le G$. Then:
Thus $\beta \circ \alpha = I_X$.
Now let $H' \le G / N$. Then:
Thus $\alpha \circ \beta = I_Y$.
So, by Bijection iff Inverse is Bijection, $\alpha$ is a bijection.
Now let $H \lhd G$ such that $N \le H$.
We show that $\map \alpha H = H / N \lhd G / N$.
This follows by definition 3 of Normal Subgroup because: for any $h \in H, g \in G$
:$\paren {g N} \paren {h N} \paren {g N}^{-1} = g h g^{-1} N \in H / N$
:$\paren {g N}^{-1} \paren {h N} \paren {g N} = g^{-1} h g N \in H / N$
Conversely, let $H' \lhd G / N$.
Recall:
:$\map \beta {H'} = \set {g \in G : g N \in H'}$
Hence, for any $x \in G$ we have:
Now for any $h' \in \map \beta {H'}$, we have:
:$h'N \in H'$
From $H' \lhd G / N$:
For all $x \in G$:
:$\paren {x N}^{-1} \paren {h' N} \paren {x N} \in H'$
From $N \lhd G$:
:$x^{-1} h' x N \in H'$
This implies:
:$h' \in x \map \beta {H'} x^{-1}$
or:
:$\map \beta {H'} \subseteq x \map \beta {H'} x^{-1}$
Similarly, we can also show:
:$\map \beta {H'} \subseteq x^{-1} \map \beta {H'} x$
Hence by definition 4 of Normal Subgroup:
:$\map \beta {H'} \lhd G$
== Sources ==
* : Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 47 \eta$
* : Chapter $7$: Normal subgroups and quotient groups: Theorem $7.14$
Category:Normal Subgroups
Category:Quotient Groups
Category:Named Theorems
Correspondence Theorem (Set Theory)

== Theorem ==
Let $S$ be a set.
Let $\RR \subseteq S \times S$ be an equivalence relation on $S$.
Let $\mathscr A$ be the set of partitions of $S$ associated with equivalence relations $\RR'$ on $S$ such that:
:$\tuple {x, y} \in \RR \iff \tuple {x, y} \in \RR'$
Then there exists a bijection $\phi$ from $\mathscr A$ onto the set of partitions of the quotient set $S / \RR$.
== Proof ==
Denote the equivalence class of an element $x$ of $S$ by $\eqclass x {\RR}$ with respect to the relation $\RR$. 
Consider the relation on $S$:
:$\phi = \set {\tuple {\eqclass x {\RR'}, \eqclass x \RR}: x \in S}$
We prove that $\phi$ is a bijection. 
Let $\eqclass x \RR = \eqclass y \RR$.
Then:
:$\tuple {x, y} \in \RR$
Thus:
:$\tuple {x, y} \in \RR'$
Therefore:
:$\eqclass x {\RR'} = \eqclass y {\RR'}$
and so $\phi$ is one-to-many.
Let $\eqclass x {\RR'} = \eqclass y {\RR'}$.
Then:
:$\tuple {x, y} \in \RR'$
Thus:
:$\tuple {x, y} \in \RR$
Therefore $\eqclass x \RR = \eqclass y \RR$
Hence $\phi$ is many-to-one.
As $\phi$ is both one-to-many and many-to-one, it is by definition a one-to-one relation.
From the fact that $\Rng \phi = S / \RR$ we have that $\phi$ is both left-total and right-total.
We have demonstrated that $\phi$ is left-total and many-to-one, and so by definition is a mapping.
We have demonstrated that $\phi$ is right-total, and so by definition is a surjection.
We have demonstrated that $\phi$ is one-to-one, and so by definition is an injection.
Hence by definition $\phi$ is a bijection.
== Sources ==
* : $\S 6$: Exercise $18$
* P60
Category:Equivalence Relations
Category:Set Theory
Category:Quotient Sets
Category:Named Theorems
Correspondence Theorem for Localizations of Rings

== Theorem ==
Let $A$ be a commutative ring with unity.
Let $S\subseteq A$ be a multiplicatively closed subset.
Let $A \overset \iota \to A_S$ be the localization at $S$.
Let $I$ be the set of saturated ideals of $A$ by $S$.
Let $J$ be the set of ideals of $A_S$.
=== Bijection ===
The direct image mapping $\iota^\to$ and the inverse image mapping $\iota^\gets$ induce reverse bijections between $I$ and $J$, specifically:
#For every ideal $\mathfrak a \in I$, its image $\map {\iota^{\to} } {\mathfrak a} = \map \iota {\mathfrak a} \in J$.
#For every ideal $\mathfrak b \in J$, its preimage $\map {\iota^{\gets} } {\mathfrak b} = \map {\iota^{-1} } {\mathfrak b} \in I$.
#The restrictions $\iota^\to: I \to J$ and $\iota^\gets: J \to I$ are reverse bijections.
=== Prime ideals ===
Let $\mathfrak p \subseteq A$ be an ideal.
# $\mathfrak p$ is a prime ideal of $A$ disjoint from $S$.
# $\mathfrak p$ is a prime ideal of $A$ saturated by $S$.
# $\map \iota {\mathfrak p}$ is a prime ideal of $A_S$.
=== Open embedding of prime spectrum ===
The induced map on spectra $\Spec \iota : \Spec {A_S} \to \Spec A$ is a topological open embedding.
== Proof ==
== Also see ==
* Correspondence Theorem for Quotient Rings
Category:Localization of Rings
Correspondence Theorem for Quotient Rings

== Theorem ==
Let $A$ be a commutative ring with unity.
Let $\mathfrak a \subseteq A$ be an ideal.
Let $A / \mathfrak a$ be the quotient ring and $\pi : A \to A / \mathfrak a$ the quotient ring epimorphism.
=== Bijection ===
== Also see ==
* First Ring Isomorphism Theorem
* Universal Property of Quotient Ring
* Correspondence Theorem for Localizations of Rings
=== Generalization ===
* Correspondence Theorem for Surjective Ring Homomorphisms
Category:Quotient Rings
Correspondence Theorem for Quotient Rings/Bijection

== Theorem ==
The direct image mapping $\pi^\to$ and the inverse image mapping $\pi^\gets$ induce reverse bijections between the ideals of $A$ containing $\mathfrak a$ and the ideals of $A/\mathfrak a$, specifically:
Let $I$ be the set of ideals of $A$ containing $\mathfrak a$.
Let $J$ be the set of ideals of $A / \mathfrak a$.
Then:
:$(1): \quad$ For every ideal $\mathfrak b \in I$, its image $\map {\pi^\to} {\mathfrak b} = \map \pi {\mathfrak b} \in J$
:$(2): \quad$ For every ideal $\mathfrak c \in J$, its preimage $\map {\pi^\gets} {\mathfrak c} = \map {\pi^{-1} } {\mathfrak c} \in I$
:$(3): \quad$ The restrictions $\pi^\to: I \to J$ and $\pi^\gets: J \to I$ are reverse bijections.
== Proof ==
Follows from Correspondence Theorem for Ring Epimorphisms/Bijection
Category:Quotient Rings
Correspondence Theorem for Ring Epimorphisms

== Theorem ==
Let $A$ and $B$ be commutative rings with unity.
Let $\pi : A \to B$ be a ring epimorphism.
Let $I$ be the set of ideals of $A$ containing the kernel $\ker \pi$.
Let $J$ be the set of ideals of $B$.
=== Bijection ===
=== Inclusion-preserving ===
The mappings $\pi^\to: I \to J$ and $\pi^\gets: J \to I$ are inclusion-preserving.
=== Isomorphism between Quotient Rings ===
Let $\mathfrak a \in I$ be an ideal of $A$.
Then the composition $A \overset \pi \to B \to B / \map \pi {\mathfrak a}$ with the quotient ring epimorphism has kernel $\mathfrak a$ and there is an isomorphism $A / \mathfrak a \to B / \map \pi {\mathfrak a}$.
=== Prime ideals ===
Let $\mathfrak a \in I$ be an ideal of $A$.
Then $\mathfrak a$ is a prime ideal $\map \pi {\mathfrak a}$ is.
=== Maximal ideals ===
Let $\mathfrak a \in I$ be an ideal of $A$.
Then $\mathfrak a$ is a maximal ideal $\map \pi {\mathfrak a}$ is.
=== Closed embedding of prime spectrum ===
The induced map on spectra $\Spec \pi : \Spec B \to \Spec A$ is a topological closed embedding.
== Also see ==
* Correspondence Theorem for Quotient Rings
* Correspondence Theorem for Module Epimorphisms
Category:Ring Theory
Correspondence Theorem for Ring Epimorphisms/Bijection

== Theorem ==
The direct image mapping $\pi^\to$ and the inverse image mapping $\pi^\gets$ induce reverse bijections between $I$ and $J$, specifically:
#For every ideal $\mathfrak a \in I$, its image $\pi^{\to}(\mathfrak a) = \pi(\mathfrak a) \in J$.
#For every ideal $\mathfrak b \in J$, its preimage $\pi^{\gets}(\mathfrak b) = \pi^{-1}(\mathfrak b) \in I$.
#The restrictions $\pi^\to : I \to J$ and $\pi^\gets : J \to I$ are reverse bijections.
== Proof ==
The first statements follow from:
* Preimage of Ideal under Ring Homomorphism is Ideal
* Image of Ideal under Ring Epimorphism is Ideal
The last statement follows from:
* Image of Preimage of Ideal under Ring Epimorphism
* Preimage of Image of Ideal under Ring Homomorphism
Category:Ring Homomorphisms
Correspondence between Abelian Groups and Z-Modules

== Theorem ==
=== Bijection ===
Let $\Z$ be the ring of integers.
Let $G$ be an abelian group.
Let $M$ be a unitary module over $\Z$.
# $G$ is the underlying group of $M$.
# $M$ is the $\Z$-module associated with $G$.
=== Homomorphisms ===
=== Isomorphism of categories ===
Category:Abelian Groups
Category:Module Theory
Correspondence between Abelian Groups and Z-Modules/Homomorphisms

== Theorem ==
Let $G, H$ be abelian groups.
Let $f : G \to H$ be a mapping.
# $f$ is a group homomorphism.
# $f$ is a $\Z$-module homomorphism between the $\Z$-modules associated with $G$ and $H$.
== Proof ==
Category:Abelian Groups
Category:Module Theory
Correspondence between Abelian Groups and Z-Modules/Isomorphism of Categories

== Theorem ==
Let $\Z$ be the ring of integers.
Let $\mathbf{Ab}$ be the category of abelian groups.
Let $\mathbf{\mathbb Z-Mod}$ be the category of unitary $\Z$-modules.
Then the:
* forgetful functor $\mathbf{\mathbb Z-Mod} \to \mathbf{Ab}$
* associated Z-module functor $\mathbf{Ab} \to \mathbf{\mathbb Z-Mod}$
are strict inverse functors.
In particular, $\mathbf{Ab}$ and $\mathbf{\mathbb Z-Mod}$ are isomorphic.
== Proof ==
Category:Examples of Isomorphisms of Categories
Category:Category of Modules
Category:Category of Abelian Groups
Correspondence between Irrational Numbers and Simple Infinite Continued Fractions

== Theorem ==
Let $\R \setminus \Q$ be the set of irrational numbers.
Let $S$ be the set of all simple infinite continued fractions in $\R$.
The mappings:
:$\R \setminus \Q \to S$ that sends an irrational number to its continued fraction expansion
:$S \to \R \setminus \Q$ that sends a simple infinite continued fractions to its value
are inverses of each other.
== Proof ==
Note that indeed a Simple Infinite Continued Fraction Converges to Irrational Number.
The result follows from:
: Continued Fraction Expansion of Irrational Number Converges to Number Itself
: Continued Fraction Expansion of Limit of Simple Infinite Continued Fraction equals Expansion Itself
== Also see ==
* Correspondence between Rational Numbers and Simple Finite Continued Fractions
Category:Simple Continued Fractions
Correspondence between Linear Group Actions and Linear Representations

== Theorem ==
Let $\struct {k, +, \cdot}$ be a field.
Let $V$ be a vector space over $k$ of finite dimension.
Let $\struct {G, *}$ be a finite group.
There is a one-to-one correspondence between linear group actions of $G$ on $V$ and linear representations of $G$ in $V$, as follows:
Let $\phi: G \times V \to V$ be a group action.
Let $\rho: G \to \map {\operatorname {Sym} } V$ be a permutation representation of $G$ on $V$.
The following are equivalent:
:$(1): \quad$ $\rho$ is the permutation representation associated to $\phi$
:$(2): \quad$ $\phi$ is the group action associated to $\rho$
If this is the case, the following are equivalent:
:$(1): \quad$ $\rho$ is a linear representation
:$(2): \quad$ $\phi$ is a linear group action
== Proof ==
The first equivalence follows from Correspondence Between Group Actions and Permutation Representations.
=== 1 implies 2 ===
Let $\rho : G \to \GL V$ be a linear representation of $G$ on $V$.
Then for all $g \in G$, $v_1, v_2 \in V$:
and for all $g \in G$, $v \in V$, $\lambda \in k$:
Therefore $\phi$ is a linear group action of $G$ on $V$.
=== 2 implies 1 ===
Let $\phi: G \times V \to V$ be a linear action of $G$ on $V$.
Then for all $g \in G$, $v_1, v_2 \in V$:
and for all $g \in G$, $v \in V$, $\lambda \in k$:
Therefore $\rho$ is a linear representation of $G$ on $V$.
Now let $g_1, g_2 \in G$.
We have for all $v \in V$:
Thus $\rho$ satisfies the homomorphism property.
Therefore:
:$\hat{}$ : (linear representations) $\to$ (linear actions)
:$\tilde{}$ : (linear actions) $\to$ (linear representations)
give a bijection.
== Also see ==
* Correspondence Between Group Actions and Permutation Representations
Category:Group Actions
Category:Representation Theory
Correspondence between Neighborhood Space and Topological Space

== Theorem ==
Let $S$ be a set.
Let $\struct {S, \tau}$ be a topological space.
Let $\struct {S, \NN}$ be the neighborhood space induced by $\tau$ on $S$.
Let $\phi: \struct {S, \tau} \to \struct {S, \NN}$ be the mapping defined as:
:$\forall x \in S: \map \phi x = x$
:$\forall U \in \tau: \phi \sqbrk U = U \in \NN$
Let $\struct {T, \NN'}$ be a neighborhood space.
Let $\struct {T, \tau'}$ be the topological space induced by $\NN$ on $S$.
Let $\psi: \struct {T, \NN'} \to \struct {T, \tau'}$ be the mapping defined as:
:$\forall y \in T: \map \psi y = y$
:$\forall V \in \NN': \psi \sqbrk V = V \in \tau'$
Then:
:$(1): \quad \phi^{-1} \sqbrk {\phi \sqbrk {\struct {S, \tau} } } = \struct {S, \tau}$
:$(2): \quad \psi^{-1} \sqbrk {\psi \sqbrk {\struct {T, \NN'} } } = \struct {T, \NN'}$
== Proof ==
From the construction of:
:the neighborhood space induced by $\tau$ on $S$
:the topological space induced by $\NN$ on $S$
the mappings $\phi$ and $\psi$ are well-defined mappings.
From Topological Space induced by Neighborhood Space induced by Topological Space, $\phi$ is a bijection.
From Neighborhood Space induced by Topological Space induced by Neighborhood Space, $\psi$ is a bijection.
Hence the result.
== Sources ==
* : Chapter $3$: Topological Spaces: $\S 3$: Neighborhoods and Neighborhood Spaces: Theorem $3.8$
Category:Neighborhood Spaces
Category:Topology
Correspondence between Rational Numbers and Simple Finite Continued Fractions

== Theorem ==
Let $\Q$ be the set of rational numbers.
Let $S$ be the set of all simple finite continued fractions in $\Q$, whose last partial denominators is not $1$.
The mappings:
:$\Q \to S$ that sends an rational number to its continued fraction expansion
:$S \to \Q$ that sends a simple finite continued fractions to its value
are inverses of each other.
=== Even length ===
Let $S_0$ be the set of all simple finite continued fractions in $\Q$ of even length.
The mappings:
:$\Q \to S$ that sends an rational number to:
::its continued fraction expansion $\sqbrk {a_0, \ldots, a_n}$ if $n$ is even
::$\sqbrk {a_0, \ldots, a_n - 1, 1}$ if $n$ is odd
:$S \to \Q$ that sends a simple finite continued fractions to its value
are inverses of each other.
=== Odd length ===
Let $S_1$ be the set of all simple finite continued fractions in $\Q$ of odd length.
The mappings:
:$\Q \to S$ that sends an rational number to:
::its continued fraction expansion $\sqbrk {a_0, \ldots, a_n}$ if $n$ is odd
::$\sqbrk {a_0, \ldots, a_n - 1, 1}$ if $n$ is even
:$S \to \Q$ that sends a simple finite continued fractions to its value
are inverses of each other.
== Proof ==
Note that indeed Simple Finite Continued Fraction has Rational Value.
The result follows from:
:Value of Continued Fraction Expansion of Rational Number equals Number Itself
:Continued Fraction Expansion of Value of Simple Finite Continued Fraction equals Expansion Itself
== Also see ==
* Correspondence between Irrational Numbers and Simple Infinite Continued Fractions
Category:Simple Continued Fractions
Category:Rational Numbers
Correspondence between Set and Ordinate of Cartesian Product is Mapping

== Theorem ==
Let $S$ and $T$ be sets such that $T \ne \O$.
Let $S \times T$ denote their cartesian product.
Let $t \in T$ be given.
Let $j_t \subseteq S \times \paren {S \times T}$ be the relation on $S \times {S \times T}$ defined as:
:$\forall s \in S: \map {j_t} s = \tuple {s, t}$
Then $j_t$ is a mapping.
== Proof ==
First it is to be shown that $j_t$ is left-total.
This follows from the fact that $j_t$ is defined for all $s$:
:$\map {j_t} s = \tuple {s, t}$
Next it is to be shown that $j_t$ is many-to-one, that is:
:$\forall s_1, s_2 \in S: \map {j_t} {s_1} \ne \map {j_t} {s_2} \implies s_1 \ne s_2$
We have that:
Hence the result.
== Also see ==
* Definition:Canonical Injection (Abstract Algebra) for an instance of this construct in the context of algebraic structures
== Sources ==
* : Chapter $1$: Theory of Sets: $\S 6$: Functions: Exercise $6$
Category:Cartesian Product
Cosecant Function in terms of Gaussian Hypergeometric Function

== Theorem ==
:$\csc x = \dfrac 1 x \map F {\dfrac 1 2, \dfrac 1 2; \dfrac 3 2; \sin^2 x}$
where:
:$x$ is a real number with $x \ne \paren {k \pi}$ with $k \in \Z$
:$\csc$ denotes the cosecant function
:$\map F {a, b; c; x}$ is the Gaussian hypergeometric function of $x$: $\ds \sum_{k \mathop = 0}^\infty \dfrac { a^{\overline k} b^{\overline k} } { c^{\overline k} } \dfrac {x^k} {k!}$
== Proof ==
== Also see ==
* Arcsine Function in terms of Gaussian Hypergeometric Function
* Arctangent Function in terms of Gaussian Hypergeometric Function
* Secant Function in terms of Gaussian Hypergeometric Function
Category:Cosecant Function
Category:Gaussian Hypergeometric Function
Category:Hypergeometric Functions