nyuuzyou/wikis huggingface dataset:
Inverse of Injective and Surjective Mapping is Mapping/Proof 2

== Theorem ==
== Proof ==
Let $f: S \to T$ be a mapping such that:
::$(1): \quad f$ is an injection
::$(2): \quad f$ is a surjection.
Let $t \in T$.
Then as $f$ is a surjection:
:$\exists s \in S: t = \map f s$
As $f$ is an injection, there is only one $s \in S$ such that $t = \map f s$.
Define $\map g t = s$.
As $t \in T$ is arbitrary, it follows that:
:$\forall t \in T: \exists s \in S: \map g t = s$
such that $s$ is unique for a given $t$.
That is, $g: T \to S$ is a mapping.
By the definition of $g$:
:$(1): \quad \forall t \in T: \map f {\map g t} = t$
Let $s \in S$.
Let:
:$(2): \quad s' = \map g {\map f s}$
Then:
Thus $f: S \to T$ and $g: T \to S$ are inverse mappings of each other.
== Sources ==
* : Chapter $1$: Theory of Sets: $\S 9$: Inverse Functions, Extensions, and Restrictions
Category:Mapping is Injection and Surjection iff Inverse is Mapping
Inverse of Inner Automorphism

== Theorem ==
Let $G$ be a group.
Let $x \in G$.
Let $\kappa_x$ be the inner automorphism of $G$ given by $x$.
Then:
:$\paren {\kappa_x}^{-1} = \kappa_{x^{-1} }$
== Proof ==
Let $G$ be a group whose identity is $e$.
Let $x \in G$.
Let $\kappa_x \in \Inn G$.
Then from the definition of inner automorphism:
:$\forall g \in G: \map {\kappa_x} g = x g x^{-1}$
As $G$ is a group:
:$x \in G \implies x^{-1} \in G$
So:
:$\kappa_{x^{-1} } \in \Inn G$
and is defined as:
:$\forall g \in G: \map {\kappa_{x^{-1} } } g = x^{-1} g \paren {x^{-1} }^{-1} = x^{-1} g x$
Now we need to show that:
:$\kappa_x \circ \kappa_{x^{-1} } = I_G = \kappa_{x^{-1} } \circ \kappa_x$
where $I_G: G \to G$ is the identity mapping.
So:
Thus:
:$\forall g \in G: \map {\kappa_x \circ \kappa_{x^{-1} } } g = \map {I_G} g = \map {\kappa_{x^{-1} } \circ \kappa_x} g$
Hence the result.
Category:Inner Automorphisms
Inverse of Inverse/General Algebraic Structure

== Theorem ==
Let $\struct {S, \circ}$ be an algebraic structure with an identity element $e$.
Let $x \in S$ be invertible, and let $y$ be an inverse of $x$.
Then $x$ is also an inverse of $y$.
== Proof ==
Let $x \in S$ be invertible, where $y$ is an inverse of $x$.
Then:
:$x \circ y = e = y \circ x$
by definition.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 4$: Neutral Elements and Inverses: Theorem $4.3$
Category:Inverse Elements
Inverse of Inverse/Monoid

== Theorem ==
Let $\struct {S, \circ}$ be a monoid.
Let $x \in S$ be invertible, and let its inverse be $x^{-1}$.
Then $x^{-1}$ is also invertible, and:
:$\paren {x^{-1} }^{-1} = x$
== Proof ==
By Inverse in Monoid is Unique, any inverse of $x$ is unique, and can be denoted $x^{-1}$.
From Inverse of Inverse in General Algebraic Structure:
:$x^{-1}$ is invertible and its inverse is $x$.
That is:
:$\paren {x^{-1} }^{-1} = x$
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 4$: Neutral Elements and Inverses: Theorem $4.3$
* : $\S 3.1$: Monoids
Category:Monoids
Category:Inverse Elements
Inverse of Inverse Relation

== Theorem ==
The inverse of an inverse relation is the relation itself:
:$\paren {\RR^{-1} }^{-1} = \RR$
== Proof ==
== Sources ==
* : Chapter $0$: Relations: Theorem $5 \ \text{(a)}$
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings
* : Chapter $\text{I}$: Sets and Functions: Problem $\text{AA}$: Relations
* : $\S 1.11$: Relations
Category:Inverse Relations
Inverse of Inverse of Bijection

== Theorem ==
Let $f: S \to T$ be a bijection.
Then:
:$\paren {f^{-1} }^{-1} = f$
where $f^{-1}$ is the inverse of $f$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Exercise $1.3: \ 12 \ \text{(b)}$
* : $\S 3.3$. Injective, surjective, bijective; inverse mappings
* : Chapter $\text{I}$: Sets and Functions: Composition of Functions
* : Chapter $1$: Mappings: $\S 13$
* : $\S 1.11$: Relations: Theorem $11.10$
* : Chapter $1$: Sets and mappings: $\S 1.3$: Mappings: Exercise $3$
* : Chapter $6$: Order Isomorphism and Transfinite Recursion: $\S 1$ A few preliminaries
Category:Bijections
Category:Inverse Mappings
Category:Inverse of Inverse of Bijection
Inverse of Inverse of Bijection/Proof 1

== Theorem ==
== Proof ==
Let $f: S \to T$ be a bijection.
From Composite of Bijection with Inverse is Identity Mapping we have:
:$f^{-1} \circ f = I_S$
:$f \circ f^{-1} = I_T$
where $I_S$ and $I_T$ are the identity mappings on $S$ and $T$ respectively.
The result follows from Left and Right Inverses of Mapping are Inverse Mapping.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 5$: Composites and Inverses of Functions: Theorem $5.5$
* : $\S 25.3$: Some further results and examples on mappings
* : Appendix $\text{A}.7$: Inverses: Proposition $\text{A}.7.5 \ (2)$
Category:Inverse of Inverse of Bijection
Inverse of Inverse of Matrix

== Theorem ==
Let $\mathbf A$ be a nonsingular matrix.
Then:
:$\paren {\mathbf A^{-1} }^{-1} = \mathbf A$
That is, a nonsingular matrix equals the inverse of its inverse.
== Proof ==
By definition of inverse matrix:
:$\mathbf A^{-1} \mathbf A = \mathbf I$
where $\mathbf I$ is the unit matrix.
Thus the inverse of $\mathbf A^{-1}$ is $\mathbf A$.
Hence the result.
Category:Inverse Matrices
Inverse of Isometric Isomorphism

== Theorem ==
Let $\struct {R, \norm {\,\cdot\,}_R}$ and $\struct {S, \norm {\,\cdot\,}_S}$ be normed division rings.
Let $\phi:R \to S$ be a mapping.
Then $\phi:R \to S$ is an isometric isomorphism $\phi^{-1}: S \to R$ is also an isometric isomorphism.
== Proof ==
By Inverse of Algebraic Structure Isomorphism is Isomorphism then:
:$\phi: R \to S$ is an ring isomorphism $\phi^{-1}: S \to R$ is also an ring isomorphism.
Let $d_R$ and $d_S$ be the metric induced by the norms $\norm {\,\cdot\,}_R$ and $\norm {\,\cdot\,}_S$ respectively.
By Inverse of Isometry of Metric Spaces is Isometry then:
:$\phi: \struct {R, d_R} \to \struct {S, d_S}$ is an isometry $\phi^{-1}: \struct {S, d_S} \to \struct {R, d_R}$ is also an isometry.
The result follows.
Category:Normed Division Rings
Category:Isometric Isomorphisms (Normed Division Rings)
Inverse of Left-Total Relation is Right-Total

== Theorem ==
Let $\RR \subseteq S \times T$ be a relation on $S \times T$.
Let $\RR^{-1} \subseteq T \times S$ be the inverse of $\RR$.
Then:
:$\RR$ is left-total $\RR^{-1}$ is right-total.
== Proof ==
From Inverse of Inverse Relation, the inverse of $\RR^{-1}$ is $\RR$.
From Inverse of Right-Total Relation is Left-Total:
:$\RR^{-1}$ is right-total $\RR$ is left-total.
Hence the result.
Category:Inverse Relations
Category:Left-Total Relations
Category:Right-Total Relations
Inverse of Linear Function on Real Numbers

== Theorem ==
Let $a, b \in \R$ be real numbers such that $a \ne 0$.
Let $f: \R \to \R$ be the real function defined as:
:$\forall x \in \R: \map f x = a x + b$
Then the inverse of $f$ is given by:
:$\forall y \in \R: \map {f^{-1} } y = \dfrac {y - b} a$
== Proof ==
We have that Linear Function on Real Numbers is Bijection.
Let $y = \map f x$.
Then:
and so:
:$\forall y \in \R: \map {f^{-1} } y = \dfrac {y - b} a$
== Sources ==
* : Chapter $3$. Mappings: Exercise $4$
Category:Linear Algebra
Inverse of Linear Isometry is Linear Isometry

== Theorem ==
Let $\struct {X, \norm \cdot_X}$ and $\struct {Y, \norm \cdot_Y}$ be normed vector spaces.
Let $T : X \to Y$ be an invertible (in the sense of a mapping) linear isometry with inverse $T^{-1} : Y \to X$. 
Then $T^{-1}$ is a linear isometry.
== Proof ==
From Inverse of Linear Transformation is Linear Transformation, we have: 
:$T^{-1}$ is a linear transformation.
Since $T$ is a linear isometry, we have: 
:$\norm {T x}_Y = \norm x_X$
for each $x \in X$.
Note that for each $y \in Y$, we have $T^{-1} y \in X$. 
We then have: 
So $T$ is a linear transformation with: 
:$\norm {T^{-1} y}_X = \norm y_Y$ 
for each $y \in Y$.
So $T^{-1}$ is a linear isometry.
Category:Linear Isometries
Inverse of Linear Operator is Linear Operator

== Theorem ==
Let $X$ be a vector space.
Let $A : X \to X$ be an invertible (in the sense of a mapping) linear transformation with inverse mapping $A^{-1} : X \to X$.
Then $A^{-1}$ is a linear operator.
== Proof ==
Applying Inverse of Linear Transformation is Linear Transformation in the case $U = V = X$ we have: 
:$A^{-1}$ is a linear transformation.
Since $A^{-1}$ is a linear transformation $X \to X$, we have: 
:$A^{-1}$ is a linear operator.
Category:Linear Operators
Inverse of Linear Transformation is Linear Transformation

== Theorem ==
Let $K$ be a field. 
Let $V$ and $U$ be vector spaces over $K$.
Let $A : V \to U$ be an invertible (in the sense of a mapping) linear transformation with inverse mapping $A^{-1} : U \to V$.
Then $A^{-1}$ is a linear transformation.
== Proof ==
We aim to show that: 
:$\map {A^{-1} } {\alpha x + \beta y} = \alpha A^{-1} x + \beta A^{-1} y$
for all $x, y \in U$ and $\alpha, \beta \in K$.
Since $A$ is a linear transformation, we have: 
:$\map A {\alpha u + \beta v} = \alpha A u + \beta A v$
for each $u, v \in V$. 
Note that we have $A^{-1} x \in V$ and $A^{-1} y \in V$, so that: 
:$\map A {\alpha A^{-1} x + \beta A^{-1} y} = \alpha x + \beta y$
for each $x, y \in U$ and $\alpha, \beta \in K$.
Composing with $A^{-1}$, we then have: 
:$\map {\paren {A^{-1} \circ A} } {\alpha A^{-1} x + \beta A^{-1} y} = \map {A^{-1} } {\alpha x + \beta y}$
From the definition of an inverse mapping, we therefore have: 
:$\ds \map {A^{-1} } {\alpha x + \beta y} = \alpha A^{-1} x + \beta A^{-1} y$
for all $x, y \in U$ and $\alpha, \beta \in K$.
So:
:$A^{-1}$ is a linear transformation.
Category:Linear Transformations
Inverse of Many-to-One Relation is One-to-Many

== Theorem ==
The inverse of a many-to-one relation is a one-to-many relation, and vice versa.
== Proof ==
From the definition, a many-to-one relation $\RR \subseteq S \times T$ is such that:
:$\RR \subseteq S \times T: \forall x \in \Dom \RR: \tuple {x, y_1} \in \RR \land \tuple {x, y_2} \in \RR \implies y_1 = y_2$
Also from the definition, a one-to-many relation $\RR \subseteq S \times T$ is such that:
:$\RR \subseteq S \times T: \forall y \in \Img \RR: \tuple {x_1, y} \in \RR \land \tuple {x_2, y} \in \RR \implies x_1 = x_2$
The inverse of a relation $\RR$ is defined as:
:$\RR^{-1} = \set {\tuple {t, s}: \tuple {s, t} \in \RR}$
Also from the definition of inverse relation, we have that:
:$\Dom \RR = \Rng {\RR^{-1} }$
:$\Rng \RR = \Dom {\RR^{-1} }$
So, let $\RR$ be a many-to-one relation.
Putting the above all together, we have:
:$\RR^{-1} \subseteq T \times S: \forall x \in \Img {\RR^{-1} }: \tuple {y_1, x} \in \RR^{-1} \land \tuple {y_2, x} \in \RR^{-1} \implies y_1 = y_2$
and it can be seen that $\RR^{-1}$ is one-to-many.
Similarly, let $\RR$ be a one-to-many relation.
This gives us:
:$\RR^{-1} \subseteq T \times S: \forall y \in \Dom {\RR^{-1} }: \tuple {y, x_1} \in \RR^{-1} \land \tuple {y, x_2} \in \RR^{-1} \implies x_1 = x_2$
and it can be seen that $\RR^{-1}$ is many-to-one.
Category:Inverse Relations
Inverse of Mapping is One-to-Many Relation

== Theorem ==
Let $f$ be a mapping.
Then its inverse $f^{-1}$ is a one-to-many relation.
Hence $f^{-1}$ is not necessarily a mapping itself.
== Proof ==
We have that $f$ is a mapping.
Hence $f$ is a many-to-one relation.
Then from Inverse of Many-to-One Relation is One-to-Many, $f^{-1}$ is one-to-many.
== Sources ==
* : Chapter $6$: Order Isomorphism and Transfinite Recursion: $\S 1$ A few preliminaries
Category:Inverse Relations
Category:Inverse Mappings
Inverse of Mapping is Right-Total Relation

== Theorem ==
Let $f$ be a mapping.
Then its inverse $f^{-1}$ is a right-total relation.
== Proof ==
We have that $f$ is a mapping.
Hence $f$ is a left-total relation.
Then from Inverse of Left-Total Relation is Right-Total, $f^{-1}$ is right-total.
Category:Inverse Relations
Category:Inverse Mappings
Category:Right-Total Relations
Inverse of Mapping to Annihilator on Algebraic Dual is Bijection

== Theorem ==
Let $G$ be an $n$-dimensional vector space over a field.
Let $J: G \to G^{**}$ be the evaluation isomorphism.
Let $G^*$ be the algebraic dual of $G$.
Let $M$ be an $m$-dimensional subspace of $G$.
Let $M^\circ$ be the annihilator of $M$.
Let $N$ be a $p$-dimensional subspace of $G^*$.
Let $G_m$ denote the set of all $m$-dimensional subspaces of $G$.
Let ${G^*}_{n - m}$ denote the set of all $n - m$-dimensional subspaces of $G^*$.
Let $\phi: G_m \to {G^*}_{n - m}$ be the mapping from $G_m$ to the power set of ${G^*}_{n - m}$ defined as:
:$\forall M \in \powerset G: \map \phi M = M^\circ$
Then $\phi$ is a bijection.
The inverse of $\phi$ is the bijection:
:$N \to \map {J^\gets} {N^\circ}$
where $J^\gets$ is the inverse image mapping of $J$.
== Proof ==
By definition::
:$\paren {\map {J^\gets} {N^\circ} }^\circ = \set {z \in G^*: \forall x \in G: \forall t \in N: \map t x = 0: \map z x = 0}$
Thus:
:$N \subseteq \paren {\map {J^\gets} {N^\circ} }^\circ$
But as $\paren {\map {J^\gets} {N^\circ} }^\circ$ has dimension $n - \paren {n - p} = p$, it follows that $N = \paren {\map {J^\gets} {N^\circ} }^\circ$ by Dimension of Proper Subspace is Less Than its Superspace.
This follows by the definition of an inverse mapping.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations: Theorem $28.10$
Category:Annihilators
Category:Vector Subspaces
Inverse of Matrix Exponential

== Theorem ==
Let $\mathbf A$ be a square matrix.
Let $t \in \R$ be a real number.
Let $e^{\mathbf A t}$ denote the matrix exponential of $\mathbf A$.
Then:
:$\paren {e^{\mathbf A t} }^{-1} = e^{-\mathbf A t}$
where $\paren {e^{\mathbf A t} }^{-1}$ denotes the inverse of $e^{\mathbf A t}$.
== Proof ==
where:
:$\mathbf 0$ denotes the zero matrix of the appropriate order
:$\mathbf I$ denotes the identity matrix of the appropriate order.
Similarly:
:$e^{-\mathbf A t} e^{\mathbf A t} = \mathbf I$
Hence the result by definition of inverse matrix.
Category:Matrix Exponential
Category:Inverse Matrices
Inverse of Matrix Product

== Theorem ==
Let $\mathbf {A, B}$ be square matrices of order $n$
Let $\mathbf I$ be the $n \times n$ unit matrix.
Let $\mathbf A$ and $\mathbf B$ be nonsingular.
Then the matrix product $\mathbf {AB}$ is also nonsingular, and:
:$\paren {\mathbf A \mathbf B}^{-1} = \mathbf B^{-1} \mathbf A^{-1}$
== Proof ==
We are given that $\mathbf A$ and $\mathbf B$ are nonsingular.
From Product of Matrices is Nonsingular iff Matrices are Nonsingular, $\mathbf A \mathbf B$ is also nonsingular.
By the definition of inverse matrix:
:$\mathbf A \mathbf A^{-1} = \mathbf A^{-1} \mathbf A = \mathbf I$
and 
:$\mathbf B \mathbf B^{-1} = \mathbf B^{-1} \mathbf B = \mathbf I$
Now, observe that:
Similarly:
The result follows from the definition of inverse.
== Also see ==
* Transpose of Matrix Product
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.3$ The inverse of a matrix: Proposition $1.2$
Category:Inverse Matrices
Category:Conventional Matrix Multiplication
Inverse of Matrix is Scalar Product of Adjugate by Reciprocal of Determinant

== Theorem ==
Let $\mathbf A = \sqbrk a_n$ be a nonsingular square matrix of order $n$.
Let $\map \det {\mathbf A}$ be the determinant of $\mathbf A$.
Let $\adj {\mathbf A}$ be the adjugate of $\mathbf A$.
Then:
:$\mathbf A^{-1} = \dfrac 1 {\map \det {\mathbf A} } \cdot \adj {\mathbf A}$
where $\mathbf A^{-1}$ denotes the inverse of $\mathbf A$
== Proof 1 ==
== Proof 2 ==
== Examples ==
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.7$ Minors and cofactors
* 
* 
* 
* 
Category:Inverse of Matrix is Scalar Product of Adjugate by Reciprocal of Determinant
Category:Adjugate Matrices
Category:Determinants
Inverse of Matrix is Scalar Product of Adjugate by Reciprocal of Determinant/Examples/Order 2

== Examples of Use of Inverse of Matrix is Scalar Product of Adjugate by Reciprocal of Determinant ==
:$\begin {pmatrix} a & b \\ c & d \end {pmatrix}^{-1} = \dfrac 1 {a d - b c} \begin {pmatrix} d & -b \\ -c & a \end {pmatrix}$
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.7$ Minors and cofactors
Category:Inverse of Matrix is Scalar Product of Adjugate by Reciprocal of Determinant
Inverse of Matrix is Scalar Product of Adjugate by Reciprocal of Determinant/Proof 2

== Theorem ==
== Proof ==
Let:
:$\mathbf A = \begin {bmatrix} a_{11} & \cdots & a_{1n} \\ \vdots & \ddots & \vdots \\ a_{n1} & \cdots & a_{nn} \end {bmatrix}$
:$\mathbf A^{-1} = \begin {bmatrix} b_{11} & \cdots & b_{1n} \\ \vdots & \ddots & \vdots \\ b_{n1} & \cdots & b_{nn} \end {bmatrix}$
Let $\tuple {\mathbf e_1, \mathbf e_2, \cdots, \mathbf e_n}$ be the standard ordered basis of $\R^n$.
Let $T: \R^n \to \R^n, \mathbf x \mapsto \map T {\mathbf x}$ be a linear transformation.
From Linear Transformation as Matrix Product, let:
Let $p, q \in \set {1, \dots, n}$.
Let $\mathbf I_n$ be the unit matrix of order $n$.
Let $\mathbf {A_p}$ be the matrix obtained by replacing the $p$th column of $\mathbf A$ with $\mathbf e_q$.
Let $C_{q p}$ be the cofactor of $a_{q p}$ in $\map \det {\mathbf A_p}$.
Hence:
Category:Inverse of Matrix is Scalar Product of Adjugate by Reciprocal of Determinant
Inverse of Multiplicative Inverse

== Theorem ==
Let $\struct {F, +, \times}$ be a field whose zero is $0_F$.
Let $a \in F$ such that $a \ne 0_F$.
Let $a^{-1}$ be the multiplicative inverse of $a$.
Then $\paren {a^{-1} }^{-1} = a$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $\text {I}$: Elementary Definitions: $\S 2$. Elementary Properties
Category:Field Theory
Category:Inverse of Multiplicative Inverse
Inverse of Neighborhood of Diagonal Point is Neighborhood

== Theorem ==
Let $T = \struct{X, \tau}$ be a topological Space.
Let $\tau_{X \times X}$ denote the product topology on the cartesian product $X \times X$.
Let $x \in X$.
Let $R$ be a neighborhood of $\tuple{x,x}$ in the product space $\struct{X \times X, \tau_{X \times X}}$.
Let $R^{-1}$ denote the inverse relation of $R$ where $R$ is viewed as a relation on $X \times X$. 
Then:
:$R^{-1}$ is a neighborhood of $\tuple{x,x}$ in $\struct{X \times X, \tau_{X \times X}}$.
== Proof ==
By definition of neighborhood:
:$\tuple{x, x} \in R$
By definition of product topology:
:$\BB = \set {V_1 \times V_2: V_1, V_2 \in \tau}$ is a basis for $\tau_{X \times X}$
From Characterization of Neighborhood by Basis:
:$\exists V_1, V_2 \in \tau : \tuple{x, x} \in V_1 \times V_2 : V_1 \times V_2 \subseteq R$
By definition of symmetric relation:
:$\tuple{x, x} \in V_2 \times V_1 : V_2 \times V_1 \subseteq R^{-1}$
By definition of product topology:
:$V_2 \times V_1 \in \tau_{X \times X}$
Hence $R^{-1}$ is a neighborhood $\tuple{x, x}$ by definition.
Category:Neighborhoods
Category:Inverse Relations
Inverse of Non-Reflexive Relation is Non-Reflexive

== Theorem ==
Let $\RR$ be a relation on a set $S$.
If $\RR$ is non-reflexive, then so is $\RR^{-1}$.
== Proof ==
Let $\RR$ be non-reflexive.
Then:
Thus $\RR^{-1}$ is not antireflexive.
Also:
Thus $\RR^{-1}$ is not reflexive.
Hence the result, by definition of non-reflexive relation.
Category:Non-Reflexive Relations
Category:Inverse Relations
Inverse of Non-Symmetric Relation is Non-Symmetric

== Theorem ==
Let $\RR$ be a relation on a set $S$.
If $\RR$ is non-symmetric, then so is $\RR^{-1}$.
== Proof ==
Let $\RR$ be non-symmetric.
Then:
:$\exists \tuple {x_1, y_1} \in \RR \implies \tuple {y_1, x_1} \in \RR$
and also:
:$\exists \tuple {x_2, y_2} \in \RR \implies \tuple {y_2, x_2} \notin \RR$
Thus:
:$\exists \tuple {y_1, x_1} \in \RR^{-1} \implies \tuple {x_1, y_1} \in \RR^{-1}$
and also:
:$\exists \tuple {y_2, x_2} \in \RR^{-1} \implies \tuple {x_2, y_2} \notin \RR^{-1}$
and so $\RR^{-1}$ is non-symmetric.
Category:Non-Symmetric Relations
Category:Inverse Relations
Inverse of Non-Transitive Relation is Non-Transitive

== Theorem ==
Let $\RR$ be a relation on a set $S$.
If $\RR$ is non-transitive, then so is $\RR^{-1}$.
== Proof ==
Let $\RR$ be non-transitive.
Then:
:$\exists x_1, y_1, z_1 \in S: \tuple {x_1, y_1}, \tuple {y_1, z_1} \in \RR, \tuple {x_1, z_1} \in \RR$
:$\exists x_2, y_2, z_2 \in S: \tuple {x_2, y_2}, \tuple {y_2, z_2} \in \RR, \tuple {x_2, z_2} \notin \RR$
So:
:$\exists x_1, y_1, z_1 \in S: \tuple {y_1, x_1}, \tuple {z_1, y_1} \in \RR^{-1}, \tuple {z_1, x_1} \in \RR^{-1}$
:$\exists x_2, y_2, z_2 \in S: \tuple {y_2, x_2}, \tuple {z_2, y_2} \in \RR^{-1}, \tuple {z_2, x_2} \notin \RR^{-1}$
So $\RR^{-1}$ is non-transitive.
Category:Non-Transitive Relations
Category:Inverse Relations
Inverse of Nonsingular 2 x 2 Real Square Matrix

== Theorem ==
Let $\mathbf A$ be an nonsingular $2 \times 2$ real square matrix defined as:
:$\mathbf A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$
Then its inverse matrix $\mathbf A^{-1}$ is:
:$\mathbf A^{-1} = \dfrac 1 {\map \det {\mathbf A} } \begin {pmatrix} d & -b \\ -c & a \end {pmatrix} = \dfrac 1 {a d - b c} \begin {pmatrix} d & -b \\ -c & a \end {pmatrix}$
== Proof ==
We construct $\begin {pmatrix} \mathbf A & \mathbf I \end {pmatrix}$:
:$\begin {pmatrix} \mathbf A & \mathbf I \end {pmatrix} = \paren {\begin {array} {cc|cc} a & b & 1 & 0 \\ c & d & 0 & 1 \\ \end {array} }$
In the following, $\sequence {e_n}_{n \mathop \ge 1}$ denotes the sequence of elementary row operations that are to be applied to $\begin {pmatrix} \mathbf A & \mathbf I \end {pmatrix}$.
The matrix that results from having applied $e_1$ to $e_k$ in order is denoted $\begin {pmatrix} \mathbf A_k & \mathbf B_k \end {pmatrix}$.
It is seen that $\begin {pmatrix} \mathbf A_4 & \mathbf B_4 \end {pmatrix}$ is the required reduced echelon form:
:$\mathbf A_4 = \mathbf I$
Hence by the Matrix Inverse Algorithm:
Hence the result.
== Sources ==
* 
* 
Category:Matrix Inverse Algorithm
Category:Inverse Matrices
Inverse of One-to-One Relation is One-to-One

== Theorem ==
The inverse of a one-to-one relation is a one-to-one relation.
== Proof ==
Let $\RR$ be a one-to-one relation.
Let $\RR^{-1}$ denote its inverse
By definition, $\RR$ is a relation which is both many-to-one and one-to-many.
From Inverse of Many-to-One Relation is One-to-Many:
:$\RR^{-1}$ is both one-to-many and many-to-one.
Hence the result by definition of one-to-one relation.
Category:Inverse Relations
Inverse of Open Set in Product Space is Open in Inverse Product Space

== Theorem ==
Let $\struct{S_1, \tau_1}$ and $\struct{S_2, \tau_2}$ be topological spaces.
Let $\tau_{S_1 \times S_2}$ be the product topology on the Cartesian product $S_1 \times S_2$.
Let $\tau_{S_2 \times S_1}$ be the product topology on the Cartesian product $S_2 \times S_1$.
Let $V \subseteq S_1 \times S_2$.
Let $V^{-1} \subseteq S_2 \times S_1$ denote the inverse of $V$ viewed as a relation on $S_1 \times S_2$.
Then:
:$V$ is open in $\struct{S_1 \times S_2, \tau_{S_1 \times S_2}}$ $V^{-1}$ is open in $\struct{S_2 \times S_1, \tau_{S_2 \times S_1}}$
== Proof ==
=== Necessary Condition ===
Let $V$ be open in $\struct{S_1 \times S_2, \tau_{S_1 \times S_2}}$.
Let $\tuple{s_2, s_1} \in V^{-1}$
By definition of inverse:
:$\tuple{s_1, s_2} \in V$
By definition of product topology:
:$\BB_{12} = \set{U \times W : U \in \tau_1, W \in \tau_2}$ is a basis for $\tau_{S_1 \times S_2}$
By definition of basis:
:$\exists U \in \tau_1, W \in \tau_2 : \tuple{s_1, s_2} \in U \times W \subseteq V$
By definition of inverse:
:$\tuple{s_2, s_1} \in W \times U \subseteq V^{-1}$
By definition of product topology:
:$\BB_{21} = \set{W \times U : W \in \tau_2, U \in \tau_1}$ is a basis for $\tau_{S_2 \times S_1}$
Since $\tuple{s_2, s_1}$ was arbitrary, we have:
:$\forall \tuple{s_2, s_1} \in V^{-1} : \exists W \times U \in \BB_{21} : \tuple{s_2, s_1} \in W \times U \subseteq V^{-1}$
From Characterization of Set Equals Union of Sets:
:$V^{-1} = \bigcup \set{W \times U \in \BB_{21} : W \times U \subseteq V^{-1}}$
By :
:$V^{-1}$ is open in $\struct{S_2 \times S_1, \tau_{S_2 \times S_1}}$
=== Sufficient Condition ===
Let $V^{-1}$ be open in $\struct{S_2 \times S_1, \tau_{S_2 \times S_1}}$.
From Necessary Condition above:
:$\paren{V^{-1}}^{-1}$ is open in $\struct{S_1 \times S_2, \tau_{S_1 \times S_2}}$
From Inverse of Inverse Relation:
:$\paren{V^{-1}}^{-1} = V$
Hence:
:$V$ is open in $\struct{S_1 \times S_2, \tau_{S_1 \times S_2}}$
Category:Product Spaces