nyuuzyou/wikis huggingface dataset:
Inverse Cosine of Imaginary Number

== Theorem ==
:$\cos^{-1} x = \pm \, i \cosh^{-1} x$
== Proof ==
== Sources ==
* : $\S 8$: Hyperbolic Functions: $8.94$: Relationship between Inverse Hyperbolic and Inverse Trigonometric Functions
Category:Inverse Cosine
Category:Inverse Hyperbolic Cosine
Inverse Cotangent of Imaginary Number

== Theorem ==
:$\map {\cot^{-1} } {i x} = - i \coth^{-1} x$
== Proof ==
== Sources ==
* : $\S 8$: Hyperbolic Functions: $8.98$: Relationship between Inverse Hyperbolic and Inverse Trigonometric Functions
Category:Inverse Cotangent
Category:Inverse Hyperbolic Cotangent
Inverse Element/Examples/Addition Modulo 6

== Examples of Inverse Elements ==
Consider the additive group of integers modulo $6$, whose Cayley table is given below:
Each element of this group is invertible.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 4$: Neutral Elements and Inverses
Category:Examples of Inverse Elements
Category:Modulo Addition
Inverse Element/Examples/Multiplication Modulo 6

== Examples of Inverse Elements ==
Consider the multiplicative monoid of integers modulo $6$ , whose Cayley table is given below:
The only invertible elements of this group are $\eqclass 1 6$ and $\eqclass 5 6$.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 4$: Neutral Elements and Inverses
Category:Examples of Inverse Elements
Category:Modulo Multiplication
Inverse Element/Examples/Rational Multiplication

== Examples of Inverse Elements ==
Consider the multiplicative group of positive rational numbers $\struct {\Q_{> 0}, \times}$.
:$\tfrac 4 {23}$ and $5 \tfrac 3 4$ are inverses of each other.
== Proof ==
Then we have:
== Sources ==
* 
* 
Category:Examples of Inverse Elements
Category:Rational Multiplication
Inverse Element/Examples/Square Root Function

== Examples of Inverse Elements ==
Let $\struct {\CC, \circ}$ be the monoid of all real functions $\CC$ under composition $\circ$ over the closed real interval $\closedint 0 1$.
Not all elements of $\CC$ have an inverse mapping, but in particular let $f$ be defined as:
:$\forall x \in \closedint 0 1: \map f x = x^2$
$\map f x$ is in fact a bijection and has inverse mapping $\inv f x$ defined as:
:$\forall x \in \closedint 0 1: \inv f x = \sqrt x$
== Proof ==
== Sources ==
* 
* 
Category:Examples of Inverse Elements
Category:Rational Multiplication
Inverse Element/Examples/Symmetry Group of Square

== Examples of Inverse Elements ==
Consider the symmetry group of the square:
Each element of this group is invertible, for example:
:$r^{-1} = r^3$
:${t_x}^{-1} = t_x$
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 4$: Neutral Elements and Inverses
Category:Examples of Inverse Elements
Inverse Element in Inverse Completion of Commutative Monoid

== Theorem ==
Let $\struct {S, \circ}$ be a commutative monoid.
Let $\struct {C, \circ} \subseteq \struct {S, \circ}$ be the subsemigroup of cancellable elements of $\struct {S, \circ}$.
Let $\struct {T, \circ'}$ be an inverse completion of $\struct {S, \circ}$.
Then the inverse of an element of $S$ which is invertible for $\circ$ is also its inverse for $\circ'$.
== Proof ==
Let the identity of $\struct {S, \circ}$ be $e$.
Let $z$ be the inverse of $y$ for $\circ$:
:$z \circ y = e$
:$y \circ z = e$
From Identity of Inverse Completion of Commutative Monoid:
:$z \circ' y = e$
:$y \circ' z = e$
Hence $z$ is the inverse of $y$ for $\circ'$.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $\S 20$: The Integers: Theorem $20.1: \ 5^\circ$
Category:Inverse Completions
Inverse Element is Power of Order Less 1

== Theorem ==
Let $G$ be a group whose identity is $e$.
Let $g \in G$ be of finite order.
Then:
:$\order g = n \implies g^{n - 1} = g^{-1}$
== Proof ==
== Sources ==
* : $\S 5.4$. Cyclic groups: Example $101$
* : Chapter $3$: Elementary consequences of the definitions: Definition $3.9$: Remark $2$
Category:Inverse Elements
Category:Order of Group Elements
Inverse Element of Bijection

== Theorem ==
Let $S$ and $T$ be sets.
Let $f: S \to T$ be a bijection.
Then:
:$\map {f^{-1} } y = x \iff \map f x = y$
where $f^{-1}$ is the inverse mapping of $f$.
== Proof ==
Suppose $f$ is a bijection.
Because $f^{-1}$ is a bijection from Bijection iff Inverse is Bijection, it is by definition a mapping.
The result follows directly from Image of Element under Inverse Mapping.
== Sources ==
* : $1$. Preliminary ideas; sets, vector spaces: $1.1$. Sets
* : Chapter $1$: Mappings: $\S 13$
* : Appendix $\text{A}.7$: Inverses: Proposition $\text{A}.7.4$
* : Chapter $2$: 'And you do addition?': $\S 2.1$: Functions: Definition $2.1.8$
* : Chapter $6$: Order Isomorphism and Transfinite Recursion: $\S 1$ A few preliminaries
Category:Bijections
Category:Inverse Mappings
Inverse Element of Injection

== Theorem ==
Let $S$ and $T$ be sets.
Let $f: S \to T$ be an injection.
Then:
:$\map {f^{-1} } y = x \iff \map f x = y$
== Proof ==
=== Necessary Condition ===
Let $y = \map f x$.
From the definition of the preimage of an element:
:$\map {f^{-1} } y = \set {x \in S: \tuple {y, x} \in f}$
Thus:
:$x \in \map {f^{-1} } y$
By definition of injection, $\map {f^{-1} } y$ is a singleton:
:$\map {f^{-1} } y = \set x$
which can be expressed as:
:$\map {f^{-1} } y = x$
=== Sufficient Condition ===
Let $\map {f^{-1} } y = x$.
Thus by definition of direct image mapping:
:$\map {f^\gets} {\set y} = \set x$
Then:
:$\map {f^\to} {\map {f^\gets} {\set y} } = \map {f^\to} {\set x}$
So from the :
:$\set y = \map {f^\to} {\set x}$
and so by definition of direct image mapping:
:$y = \map f x$
== Sources ==
* : $\S 10$: Inverses and Composites
Category:Injections
Category:Inverse Mappings
Inverse Elements of Right Transversal is Left Transversal

== Theorem ==
Let $G$ be a group.
Let $H$ be a subgroup of $G$.
Let $S \subseteq G$ be a right transversal for $H$ in $G$.
Let $T$ be the set defined as:
:$T := \set {x^{-1}: x \in S}$
where $x^{-1}$ is the inverse of $x$ in $G$.
Then $T$ is a left transversal for $H$ in $G$.
== Proof ==
Let $g \in G$.
We show that $g H$ contains exactly $1$ element of $T$.
Since $S$ is a right transversal:
:$\exists ! x \in S: x \in H g^{-1}$
By Right Cosets are Equal iff Element in Other Right Coset:
:$H x = H g^{-1}$
By Right Cosets are Equal iff Left Cosets by Inverse are Equal:
:$x^{-1} H = g H$
We have from definition $x^{-1} \in T$.
The result follows from uniqueness of $x$.
== Sources ==
* : Chapter $6$: Cosets: Exercise $4$
Category:Transversals (Group Theory)
Inverse Evaluation Isomorphism of Annihilator

== Theorem ==
Let $R$ be a commutative ring with unity.
Let $G$ be a module over $R$ whose dimension is finite.
Let $G^*$ be the algebraic dual of $G$.
Let $G^{**}$ be the algebraic dual of $G^*$.
Let $N$ be a submodule of $G^*$.
Let $J$ be the evaluation isomorphism from $G$ onto $G^{**}$.
Let $N^\circ$ be the annihilator of $N$.
Then:
:$J^{-1} \sqbrk {N^\circ} = \set {x \in G: \forall t \in N: \map t x = 0}$
where $J^{-1} \sqbrk {N^\circ}$ denotes the preimage of $N^\circ$ under $J$.
== Proof ==
We are given that $G$ is finite-dimensional.
Hence by Evaluation Isomorphism is Isomorphism, $J: G \to G^{**}$ is an isomorphism, and therefore a surjection.
Thus:
:$N^\circ = \set {x^\wedge \in G^{**}: \forall t \in N: \map {x^\wedge} t = 0}$
where $x^\wedge$ is as defined in the definition of the evaluation linear transformation.
The result follows.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations
Category:Linear Transformations
Inverse Function Theorem

== Theorem ==
Let $n \in \N$ be a natural number.
Let $f: \R^n \to \R^n$ be a mapping on the real Cartesian space of $n$ dimensions.
Let $\mathbf x \in \R^n$ be an element of $\R^n$.
Let the Jacobian matrix of $f$ be nonsingular in the locality of $\mathbf x$.
Then there exists a local single-valued differentiable inverse for $f$ at the locality of $\mathbf x$.
== Proof ==
== Examples ==
== Sources ==
* 
* 
Category:Inverse Function Theorem
Category:Jacobian Matrices
Category:Differential Calculus
Inverse Function Theorem/Examples/Square Function

== Example of Use of Inverse Function Theorem ==
The real function $f: \R \to \R$ defined as:
:$\forall x \in \R: \map f x = x^2$
does not have a local differentiable inverse around $x = 0$, because $\map f 0 = 0$.
However, it does have a local differentiable inverse around every $a \ne 0$, because $\map f a \ne 0$.
== Sources ==
* 
* 
Category:Inverse Function Theorem
Inverse Function Theorem for Real Functions

== Theorem ==
Let $n \ge 1$ and $k \ge 1$ be natural numbers.
Let $\Omega \subset \R^n$ be open.
Let $f: \Omega \to \R^n$ be a vector-valued function of class $C^k$.
Let $a \in \Omega$.
Let the differential $D \map f a$ of $f$ at $a$ be invertible.
Then there exist open sets $U \subset \Omega$ and $V \subset \R^n$ such that:
:$a \in U$
:the restriction of $f$ to $U$ is a $C^k$-diffeomorphism $f: U \to V$.
== Proof ==
Category:Implicit Functions
Inverse Hyperbolic Cosecant is Odd Function

== Theorem ==
Let $x \in \R$.
Then:
:$\map {\csch^{-1} } {-x} = -\csch^{-1} x$
where $\map {\csch^{-1} } {-x}$ denotes the inverse hyperbolic cosecant function.
== Proof ==
== Sources ==
* : $\S 8$: Hyperbolic Functions: $8.67$: Relations Between Inverse Hyperbolic Functions
Category:Inverse Hyperbolic Cosecant
Category:Examples of Odd Functions
Inverse Hyperbolic Cosecant of Imaginary Number

== Theorem ==
:$\map {\csch^{-1} } {i x} = -i \csc^{-1} x$
== Proof ==
== Sources ==
* : $\S 8$: Hyperbolic Functions: Relationship between Inverse Hyperbolic and Inverse Trigonometric Functions: $8.103$
* : $\S 14$: Hyperbolic Functions: Relationship between Inverse Hyperbolic and Inverse Trigonometric Functions: $14.103.$
Category:Inverse Cosecant
Category:Inverse Hyperbolic Cosecant
Inverse Hyperbolic Cosine of Imaginary Number

== Theorem ==
:$\cosh^{-1} x = \pm \, i \cos^{-1} x$
== Proof ==
== Sources ==
* : $\S 8$: Hyperbolic Functions: $8.95$: Relationship between Inverse Hyperbolic and Inverse Trigonometric Functions
Category:Inverse Cosine
Category:Inverse Hyperbolic Cosine
Inverse Hyperbolic Cotangent is Odd Function

== Theorem ==
Let $x \in \R$.
Then:
:$\map {\coth^{-1} } {-x} = -\coth^{-1} x$
where $\map {\coth^{-1} } {-x}$ denotes the inverse hyperbolic cotangent function.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 8$: Hyperbolic Functions: $8.66$: Relations Between Inverse Hyperbolic Functions
Category:Inverse Hyperbolic Cotangent
Category:Inverse Hyperbolic Cotangent is Odd Function
Category:Examples of Odd Functions
Inverse Hyperbolic Cotangent of Imaginary Number

== Theorem ==
:$\map {\coth^{-1} } {i x} = i \cot^{-1} x$
== Proof ==
== Sources ==
* : $\S 8$: Hyperbolic Functions: $8.99$: Relationship between Inverse Hyperbolic and Inverse Trigonometric Functions
Category:Inverse Cotangent
Category:Inverse Hyperbolic Cotangent
Inverse Hyperbolic Secant of Imaginary Number

== Theorem ==
:$\sech^{-1} x = \pm \, i \sec^{-1} x$
== Proof ==
== Sources ==
* : $\S 8$: Hyperbolic Functions: $8.101$: Relationship between Inverse Hyperbolic and Inverse Trigonometric Functions
Category:Inverse Secant
Category:Inverse Hyperbolic Secant
Inverse Hyperbolic Sine is Odd Function

== Theorem ==
Let $x \in \R$.
Then:
:$\map \arsinh {-x} = -\arsinh x$
where $\arsinh$ denotes the inverse hyperbolic sine function.
== Proof ==
== Sources ==
* : $\S 8$: Hyperbolic Functions: $8.64$: Relations Between Inverse Hyperbolic Functions
Category:Inverse Hyperbolic Sine
Category:Examples of Odd Functions
Inverse Hyperbolic Sine of Imaginary Number

== Theorem ==
:$\map {\sinh^{-1} } {i x} = i \sin^{-1} x$
== Proof ==
== Sources ==
* : $\S 8$: Hyperbolic Functions: $8.93$: Relationship between Inverse Hyperbolic and Inverse Trigonometric Functions
Category:Inverse Sine
Category:Inverse Hyperbolic Sine
Inverse Hyperbolic Tangent/Examples/i

== Example of Inverse Hyperbolic Tangent ==
:$\inv \tanh i = \dfrac {\paren {4 k + 1} \pi i} 4$
== Proof ==
:$\inv \tanh i := \set {z \in \C: \tanh z = i}$
Thus:
== Sources ==
* : $\S 4$. Elementary Functions of a Complex Variable: Exercise $9 \ \text{(ii)}$
Category:Examples of Inverse Hyperbolic Tangent
Inverse Hyperbolic Tangent is Odd Function

== Theorem ==
Let $x \in \R$.
Then:
:$\map {\tanh^{-1} } {-x} = -\tanh^{-1} x$
where $\map {\tanh^{-1} } {-x}$ denotes the inverse hyperbolic tangent function.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 8$: Hyperbolic Functions: $8.65$: Relations Between Inverse Hyperbolic Functions
Category:Inverse Hyperbolic Tangent
Category:Inverse Hyperbolic Tangent is Odd Function
Category:Examples of Odd Functions
Inverse Hyperbolic Tangent of Imaginary Number

== Theorem ==
:$\map {\tanh^{-1} } {i x} = i \tan^{-1} x$
== Proof ==
== Sources ==
* : $\S 8$: Hyperbolic Functions: $8.97$: Relationship between Inverse Hyperbolic and Inverse Trigonometric Functions
Category:Inverse Tangent
Category:Inverse Hyperbolic Tangent
Inverse Image Mapping Induced by Projection

== Theorem ==
Let $\family {S_i}_{i \mathop \in I}$ be a family of sets.
Let $\ds S = \prod_{i \mathop \in I} S_i$ be the Cartesian product of $\family {S_i}_{i \mathop \in I}$.
For each $j \in I$, let $\pr_j: S \to S_j$ denote the $j$-th projection.
For each $j \in I$ let $\pr_j^\gets: \powerset {S_i} \to \powerset S$ denote the inverse image mapping induced by $\pr_j$
Then for all $j \in I$, $\pr_j^\gets$ is the mapping defined by:
:$\ds \forall T \subseteq S_i: \map {\pr_j^\gets} T = \prod_{i \mathop \in I} T_i$
where:
:$T_i = \begin {cases} T & i = j \\ S_i & i \ne j \end {cases}$
== Proof ==
Let $j \in I$.
Let $T \subseteq S_j$.
Let $\family {T_i}_{i \mathop \in I}$ be the family of sets defined by:
:$T_i = \begin {cases} T & : i = j \\ S_i & : i \ne j \end {cases}$
Then:
Category:Projections
Category:Inverse Image Mappings
Inverse Image Mapping is Lower Adjoint to Composite Involving Direct Image Mapping

== Theorem ==
Let $S, T$ be sets.
Let $f: S \to T$ be a mapping.
Let $f^\to : \powerset S \to \powerset T$ denote the direct image mapping.
Let $f^\gets : \powerset T \to \powerset S$ denote the inverse image mapping.
Let $\complement_S: \powerset S \to \powerset S$ denote the relative complement mapping on the power set of $S$.
Let $\complement_T: \powerset T \to \powerset T$ denote the relative complement mapping on the power set of $T$.
Then:
:$f^\gets : \powerset T \to \powerset S$ is the lower adjoint to $\complement_T \circ f^\to \circ \complement_S : \powerset S \to \powerset T$
That is:
:$\forall A \subseteq S, B \subseteq T : \map {f^\gets} B \subseteq A$ $B \subseteq T \setminus \map {f^\to} {S \setminus A}$ 
== Proof ==
We have:
By definition of Galois connection:
:$\tuple{\complement_T \circ f^\to \circ \complement_S, f^\gets}$ is a Galois connection
It follows that $f^\gets : \powerset T \to \powerset S$ is the lower adjoint to $\complement_T \circ f^\to \circ \complement_S : \powerset S \to \powerset T$
Category:Images
Category:Preimages
Inverse Image Mapping is Upper Adjoint to Direct Image Mapping

== Theorem ==
Let $S, T$ be sets.
Let $f: S \to T$ be a mapping.
Let $f^\to : \powerset S \to \powerset T$ denote the direct image mapping.
Let $f^\gets : \powerset T \to \powerset S$ denote the inverse image mapping.
Then:
:$f^\gets : \powerset T \to \powerset S$ is the upper adjoint to $f^\to : \powerset S \to \powerset T$
That is:
:$\forall A \subseteq S, B \subseteq T : A \subseteq \map {f^\gets} B$ $\map {f^\to} A \subseteq B$
== Proof ==
From Image is Subset Iff Subset of Preimage:
:$\forall A \subseteq S, B \subseteq T : f \sqbrk A \subseteq B$ $A \subseteq f^{-1} \sqbrk B$ 
By definition of direct image mapping:
:$\forall A \subseteq S : \map {f^\to} A = f \sqbrk A$
By definition of inverse image mapping:
:$\forall B \subseteq T : \map {f^\gets} B = f^{-1} \sqbrk B$
Hence:
:$\forall A \subseteq S, B \subseteq T : A \subseteq \map {f^\gets} B$ $\map {f^\to} A \subseteq B$
By definition of Galois connection:
:$\tuple{f^\gets, f^\to}$ is a Galois connection
It follows that $f^\gets : \powerset T \to \powerset S$ is the upper adjoint to $f^\to : \powerset S \to \powerset T$
Category:Direct Image Mappings
Category:Inverse Image Mappings
Category:Galois Connections
Inverse Image Mapping of Bijection is Inverse of Direct Image Mapping

== Theorem ==
Let $S$ and $T$ be sets.
Let $f :S \to T$ be a bijection.
Let $f^\to$ be the direct image mapping of $f$.
Let $f^\gets$ be the inverse image mapping of $f$.
Then:
:$\paren {f^\to}^{-1} = f^\gets$
where $\paren {f^\to}^{-1}$ denotes the inverse of $f^\to$.
== Proof ==
From Mapping is Bijection iff Direct Image Mapping is Bijection:
:$f^\to$ is a bijection
Let $\paren {f^\to}^{-1}$ denote the inverse of $f^\to$.
We have:
Similarly, we have:
It follows that $f^\gets$ is the inverse of $f^\to$ by definition.
From Inverse Mapping is Unique:
:$\paren {f^\to}^{-1} = f^\gets$
Category:Direct Image Mappings
Category:Inverse Image Mappings
Inverse Image Mapping of Codomain is Preimage Set of Mapping

== Theorem ==
Let $S$ and $T$ be sets.
Let $\powerset S$ and $\powerset T$ be their power sets.
Let $f \subseteq S \times T$ be a mapping from $S$ to $T$.
Let $f^\gets: \powerset T \to \powerset S$ be the inverse image mapping of $f$:
:$\forall Y \in \powerset T: \map {f^\gets} Y = \begin {cases} \set {s \in S: \exists t \in Y: \map f s = t} & : \Img f \cap Y \ne \O \\ \O & : \Img f \cap Y = \O \end {cases}$
Then:
:$\map {f^\gets} T = \Preimg f$
where $\Preimg f$ is the preimage set of $f$.
== Proof ==
Category:Inverse Image Mappings
Inverse Image Mapping of Codomain is Preimage Set of Relation

== Theorem ==
Let $S$ and $T$ be sets.
Let $\powerset S$ and $\powerset T$ be their power sets.
Let $\RR \subseteq S \times T$ be a relation on $S \times T$.
Let $\RR^\gets: \powerset T \to \powerset S$ be the inverse image mapping of $\RR$:
:$\forall X \in \powerset S: \map {\RR^\to} X = \begin {cases} \set {t \in T: \exists s \in X: \tuple {x, t} \in \RR} & : X \ne \O \\ \O & : X = \O \end {cases}$
Then:
:$\map {\RR^\gets} T = \Preimg \RR$
where $\Preimg \RR$ is the preimage of $\RR$.
== Proof ==
Category:Inverse Image Mappings
Inverse Image Mapping of Frame Homomorphism Preserves Completely Prime Filter

== Theorem ==
Let $L_1 = \struct{S_1, \preceq_1}$ and $L_2 = \struct{S_2, \preceq_2}$ be frames.
Let $\phi : L_1 \to L_2$ be a frame homomorphism.
Let $\phi^\gets$ denote the inverse image mapping of $\phi$.
Let $F$ be a completely prime filter of $L_2$.
Then:
:$\map {\phi^\gets} F$ is a completely prime filter of $L_1$
== Proof ==
From Characterization of Completely Prime Filter in Complete Lattice it is sufficient to show:
:$(1)\quad\forall A \subseteq L : \bigvee A \in \map {\phi^\gets} F \iff \paren{\exists a \in A : a \in \map {\phi^\gets} F}$
:$(2)\quad\forall $ finite $A \subseteq L : \bigwedge A \in \map {\phi^\gets} F \iff \paren{\forall a \in A : a \in \map {\phi^\gets} F}$
=== $\map {\phi^\gets} F$ satisfies Statement $(1)$ ===
Let $\set{a_i : i \in I}$ be an indexed subset of elements of $L$.
We have:
It follows that $\map {\phi^\gets} F$ satisfies statement $(1)$.
=== $\map {\phi^\gets} F$ satisfies Statement $(2)$ ===
Let $\set{a_i : i \in J}$ be a finite indexed subset of elements of $L$.
We have:
It follows that $\map {\phi^\gets} F$ satisfies statement $(2)$.
The result follows from Characterization of Completely Prime Filter in Complete Lattice.
Category:Locales
Category:Frame Homomorphisms
Category:Inverse Image Mappings
Category:Completely Prime Filters
Inverse Image Mapping of Injection is Surjection

== Theorem ==
Let $S$ and $T$ be sets.
Let $f: S \to T$ be a injection.
Let $f^\gets: \powerset T \to \powerset S$ be the inverse image mapping of $f$.
Then $f^\gets$ is a surjection.
== Proof ==
Let $f^\to: \powerset S \to \powerset T$ be the direct image mapping by $f$.
Let $X \in \powerset S$.
Let $Y = \map {f^\to} X$.
By Subset equals Preimage of Image iff Mapping is Injection:
:$\map {f^\gets} Y = X$
As such a $Y$ exists for each $X \in \powerset S$, $f^\gets$ is surjective.
Category:Injections
Category:Surjections
Category:Inverse Image Mappings
Inverse Image Mapping of Mapping is Mapping

== Theorem ==
Let $S$ and $T$ be sets.
Let $f: S \to T$ be a mapping from $S$ to $T$.
Let $f^\gets$ be the inverse image mapping of $f$:
:$f^\gets: \powerset T \to \powerset S: \map {f^\gets} Y = f^{-1} \sqbrk Y$
Then $f^\gets$ is indeed a mapping.
== Proof ==
$f^{-1}$ is a relation.
So Inverse Image Mapping of Relation is Mapping applies directly.
== Also see ==
* Inverse of Direct Image Mapping does not necessarily equal Inverse Image Mapping
== Sources ==
* : $\S 5$. Induced mappings; composition; injections; surjections; bijections
Category:Inverse Image Mappings
Inverse Image Mapping of Relation is Mapping

== Theorem ==
Let $S$ and $T$ be sets.
Let $\RR \subseteq S \times T$ be a relation on $S \times T$.
Let $\RR^\gets$ be the inverse image mapping of $\RR$:
:$\RR^\gets: \powerset T \to \powerset S: \map {\RR^\gets} Y = \RR^{-1} \sqbrk Y$
Then $\RR^\gets$ is indeed a mapping.
== Proof ==
$\RR^{-1}$, being a relation, obeys the same laws as $\RR$.
So Direct Image Mapping of Relation is Mapping applies directly.
== Also see ==
* Inverse of Direct Image Mapping does not necessarily equal Inverse Image Mapping
Category:Inverse Image Mappings
Inverse Image of Convex Set under Monotone Mapping is Convex

== Theorem ==
Let $\left({X, \le}\right)$ and $\left({Y, \preceq}\right)$ be ordered sets.
Let $f: X \to Y$ be a monotone mapping.
Let $C$ be a convex subset of $Y$.
Then $f^{-1} \left[{C}\right]$ is convex in $X$.
== Proof ==
Suppose $f$ is increasing.
Let $a, b, c \in X$ such that $a \le b \le c$.
Let $a, c \in f^{-1} \left[{C}\right]$.
By definition of inverse image:
: $f \left({a}\right), f \left({c}\right) \in C$
By definition of increasing mapping:
:$f \left({a}\right) \preceq f \left({b}\right) \preceq f \left({c}\right)$
Thus by definition of convex set:
:$f \left({b}\right) \in C$
Then by definition of inverse image:
:$b \in f^{-1} \left[{C}\right]$
Since this holds for all such triples, $f^{-1} \left[{C}\right]$ is convex in $X$.
A similar argument applies if $f$ is decreasing.
Category:Order Theory
Inverse Image of Direct Image of Inverse Image equals Inverse Image Mapping

== Theorem ==
Let $f: S \to T$ be a mapping.
Let:
:$f^\to: \powerset S \to \powerset T$ denote the direct image mapping of $f$
:$f^\gets: \powerset T \to \powerset S$ denote the inverse image mapping of $f$
where $\powerset S$ denotes the power set of $S$.
Then:
:$f^\gets \circ f^\to \circ f^\gets = f^\gets$
where $\circ$ denotes composition of mappings.
== Proof ==
Thus we have:
:$\forall B \in \powerset T: \map {f^\gets} B \subseteq \map {\paren {f^\gets \circ f^\to \circ f^\gets} } B \subseteq \map {f^\gets} B$
and the result follows.
== Also see ==
* Direct Image of Inverse Image of Direct Image equals Direct Image Mapping
== Sources ==
* : $\S 5$. Induced mappings; composition; injections; surjections; bijections: Exercise $4$
Category:Direct Image Mappings
Category:Inverse Image Mappings
Inverse Image of Set under Set-Like Relation is Set

== Theorem ==
Let $A$ be a class.
Let $\RR$ be a set-like endorelation on $A$.
Let $B \subseteq A$ be a set.
Then $\map {\RR^{-1} } B$, the inverse image of $B$ under $\RR$, is also a set.
== Proof ==
Since $\RR$ is set-like, $\map {\RR^{-1} } {\set x}$ is a set for each $x$ in $A$.
As $B \subseteq A$, this holds also for each $x \in B$.
But then $\ds \map {\RR^{-1} } B = \bigcup_{x \mathop \in B} \map {\RR^{-1} } {\set x}$, which is a set by the Axiom of Unions.
Category:Set Theory
Inverse Image under Embedding of Image under Relation of Image of Point

== Theorem ==
Let $S$ and $T$ be sets.
Let $\RR_S$ and $\RR_t$ be relations on $S$ and $T$, respectively.
Let $\phi: S \to T$ be a mapping with the property that:
:$\forall p, q \in S: \paren {p \mathrel {\RR_S} q \iff \map \phi p \mathrel {\RR_T} \map \phi q}$
Then for each $p \in S$:
:$\map {\RR_S} p = \phi^{-1} \sqbrk {\map {\RR_T} {\map \phi p} }$
== Proof ==
Let $p \in S$.
Thus by definition of set equality:
:$\map {\RR_S} p = \phi^{-1} \sqbrk {\map {\RR_T} {\map \phi p} }$
Category:Relation Theory