nyuuzyou/wikis huggingface dataset:
Inverse in Group is Unique/Proof 3

== Theorem ==
== Proof ==
Let $x, y \in G$.
We already have, from the definition of inverse element, that:
:$\forall x \in G: \exists x^{-1} \in G: x \circ x^{-1} = e = x^{-1} \circ x$
By Group has Latin Square Property, there exists exactly one $a \in G$ such that $a \circ x = y$.
Similarly, there exists exactly one $b \in G$ such that $x \circ b = y$.
Substituting $e$ for $y$, it follows that $x^{-1}$ as defined above is unique.
== Sources ==
* : Chapter $\text {I}$: The Group Concept: $\S 2$: The Axioms of Group Theory
* : $\S 4.6$. Elementary theorems on groups: Theorem $\text{(i)}$
* : Chapter $3$: Elementary consequences of the definitions: Proposition $3.3$: Remark $3$
Category:Inverse in Group is Unique
Inverse in Monoid is Unique

== Theorem ==
Let $\struct {S, \circ}$ be a monoid.
Then an element $x \in S$ can have at most one inverse for $\circ$.
== Proof ==
Let $e$ be the identity element of $\struct {S, \circ}$.
Suppose $x \in S$ has two inverses: $y$ and $z$.
Then:
Similarly:
So whichever way round you do it, $y = z$ and the inverse of $x$ is unique.
== Also see ==
* Inverse not always Unique for Non-Associative Operation
* Identity is Unique
* Inverse in Group is Unique
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 4$: Neutral Elements and Inverses: Theorem $4.2$
* : $\text{I}$: Groups: $\S 1$ Semigroups, Monoids and Groups: Theorem $1.2 \text{(iii)}$
* : $2$: Examples of Groups and Homomorphisms: $2.2$ Definitions $\text{(ii)}$
* : $\S 31.2$ Identity element and inverses
* : $\S 3.1$: Monoids
* : Chapter $1$: Basic Definitions and Results
Category:Monoids
Category:Inverse Elements
Inverse is Mapping implies Mapping is Injection

== Theorem ==
Let $S$ and $T$ be sets.
Let $f: S \to T$ be a mapping.
Let the inverse $f^{-1} \subseteq T \times S$ itself be a mapping.
Then $f$ is an injection.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Inverse is Mapping implies Mapping is Surjection
Category:Mapping is Injection and Surjection iff Inverse is Mapping
Inverse is Mapping implies Mapping is Injection/Proof 1

== Theorem ==
== Proof ==
Let $f^{-1}: T \to S$ be a mapping.
Let $\map f {x_a} = y$ and $\map f {x_b} = y$.
Then:
Thus, by definition, $f$ is an injection.
Category:Mapping is Injection and Surjection iff Inverse is Mapping
Inverse is Mapping implies Mapping is Injection/Proof 2

== Theorem ==
== Proof ==
Let $f^{-1}: T \to S$ be a mapping.
Thus $f$ is by definition an injection.
== Sources ==
* : Chapter $1$: Theory of Sets: $\S 9$: Inverse Functions, Extensions, and Restrictions
Category:Mapping is Injection and Surjection iff Inverse is Mapping
Inverse is Mapping implies Mapping is Injection and Surjection

== Theorem ==
Let $S$ and $T$ be sets.
Let $f: S \to T$ be a mapping.
Let the inverse $f^{-1} \subseteq T \times S$ itself be a mapping.
Then:
:$(1): \quad f$ is an injection
:$(2): \quad f$ is a surjection.
== Proof ==
This is divided into two parts:
=== Inverse is Mapping implies Mapping is Injection ===
=== Inverse is Mapping implies Mapping is Surjection ===
Category:Mapping is Injection and Surjection iff Inverse is Mapping
Inverse is Mapping implies Mapping is Surjection

== Theorem ==
Let $S$ and $T$ be sets.
Let $f: S \to T$ be a mapping.
Let the inverse $f^{-1} \subseteq T \times S$ itself be a mapping.
Then $f$ is a surjection.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Inverse is Mapping implies Mapping is Injection
Category:Mapping is Injection and Surjection iff Inverse is Mapping
Inverse is Mapping implies Mapping is Surjection/Proof 1

== Theorem ==
== Proof ==
Let $f^{-1}: T \to S$ be a mapping.
 $f$ is not a surjection.
That is:
:$\exists y \in T: \neg \exists x \in S: \tuple {x, y} \in f$
By definition of inverse of mapping:
:$\exists y \in T: \neg \exists x \in S: \tuple {y, x} \in f^{-1}$
which would mean that $f^{-1}$ is not a mapping.
From this contradiction it follows that $f$ is a surjection.
Category:Mapping is Injection and Surjection iff Inverse is Mapping
Inverse is Mapping implies Mapping is Surjection/Proof 2

== Theorem ==
== Proof ==
Let $f^{-1}: T \to S$ be a mapping.
We have:
Thus $f$ is by definition a surjection.
== Sources ==
* : Chapter $1$: Theory of Sets: $\S 9$: Inverse Functions, Extensions, and Restrictions
Category:Mapping is Injection and Surjection iff Inverse is Mapping
Inverse not always Unique for Non-Associative Operation

== Theorem ==
Let $\struct {S, \circ}$ be an algebraic structure.
Let $\circ$ be a non-associative operation.
Then for any $x \in S$, it is possible for $x$ to have more than one inverse element.
== Proof ==
Proof by Counterexample:
Consider the algebraic structure $\struct {S, \circ}$ consisting of:
:The set $S = \set {a, b, e}$
:The binary operation $\circ$
whose Cayley table is given as follows:
:$\begin {array} {c|cccc}
\circ & e & a & b \\
\hline
e & e & a & b \\
a & a & e & e \\
b & b & e & e \\
\end {array}$
By inspection, we see that $e$ is the identity element of $\struct {S, \circ}$.
We also note that:
and so $\circ$ is not associative.
Note further that:
and also:
So both $a$ and $b$ are inverses of $a$.
Hence the result.
== Also see ==
* Inverse in Monoid is Unique
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 4$: Neutral Elements and Inverses
* : $\S 3.1$: Monoids: Exercise $(12)$
Category:Inverse Elements
Inverse of Algebraic Structure Isomorphism is Isomorphism

== Theorem ==
Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be a mapping.
Then $\phi$ is an isomorphism $\phi^{-1}: \struct {T, *} \to \struct {S, \circ}$ is also an isomorphism.
=== General Result ===
== Proof ==
Let $\phi$ be an isomorphism.
Then by definition $\phi$ is a bijection.
Thus $\exists \phi^{-1}$ such that $\phi^{-1}$ is also a bijection from Bijection iff Inverse is Bijection.
That is:
:$\exists \phi^{-1}: \struct {T, *} \to \struct {S, \circ}$
It follows that:
So $\phi^{-1}: \struct {T, *} \to \struct {S, \circ}$ is a homomorphism.
$\phi^{-1}$ is also (from above) a bijection.
Thus, by definition, $\phi^{-1}$ is an isomorphism.
Let $\phi^{-1}: \struct {T, *} \to \struct {S, \circ}$ be an isomorphism.
Applying the same result as above in reverse, we have that $\paren {\phi^{-1} }^{-1}: \struct {S, \circ} \to \struct {T, *}$ is also an isomorphism.
But by Inverse of Inverse of Bijection:
:$\paren {\phi^{-1} }^{-1} = \phi$
and hence the result.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 6$: Isomorphisms of Algebraic Structures: Theorem $6.1: \ 2^\circ$
* : $\S 2.2$: Homomorphisms
Category:Inverse of Algebraic Structure Isomorphism is Isomorphism
Category:Isomorphisms (Abstract Algebra)
Inverse of Algebraic Structure Isomorphism is Isomorphism/General Result

== Theorem ==
Let $\phi: \struct {S, \circ_1, \circ_2, \ldots, \circ_n}$ and $\struct {T, *_1, *_2, \ldots, *_n}$ be algebraic structures.
Let $\phi: \struct {S, \circ_1, \circ_2, \ldots, \circ_n} \to \struct {T, *_1, *_2, \ldots, *_n}$ be a mapping.
Then:
:$\phi: \struct {S, \circ_1, \circ_2, \ldots, \circ_n} \to \struct {T, *_1, *_2, \ldots, *_n}$ is an isomorphism
:
:$\phi^{-1}: \struct {T, *_1, *_2, \ldots, *_n} \to \struct {S, \circ_1, \circ_2, \ldots, \circ_n}$ is also an isomorphism.
== Proof ==
As $\paren {\phi^{-1} }^{-1} = \phi$, it suffices to show the sufficient condition.
Suppose that $\phi$ is an isomorphism.
We shall show that $\phi^{-1}$ is also an isomorphism.
We have that $\phi$ is a bijection.
Hence from Inverse of Bijection is Bijection we have that $\phi^{-1}$ is also a bijection.
Then for all $x, y \in T$ and $i \in \set {1, 2, \ldots, n}$:
Therefore $\phi^{-1}$ is a homomorphism.
Category:Inverse of Algebraic Structure Isomorphism is Isomorphism
Inverse of Antireflexive Relation is Antireflexive

== Theorem ==
Let $\RR$ be a relation on a set $S$.
If $\RR$ is antireflexive, then so is $\RR^{-1}$.
== Proof ==
Hence the result by definition of antireflexive relation.
Category:Antireflexive Relations
Category:Inverse Relations
Inverse of Antisymmetric Relation is Antisymmetric

== Theorem ==
Let $\RR$ be a relation on a set $S$.
If $\RR$ is antisymmetric, then so is $\RR^{-1}$.
== Proof ==
Let $\RR$ be antisymmetric.
Then:
:$\tuple {x, y} \land \tuple {y, x} \in \RR \implies x = y$
It follows that:
:$\tuple {y, x} \land \tuple {x, y} \in \RR^{-1} \implies x = y$
Thus it follows that $\RR^{-1}$ is also antisymmetric.
Category:Antisymmetric Relations
Category:Inverse Relations
Inverse of Antitransitive Relation is Antitransitive

== Theorem ==
Let $\RR$ be a relation on a set $S$.
If $\RR$ is antitransitive, then so is $\RR^{-1}$.
== Proof ==
Let $\RR$ be antitransitive.
Then:
:$\tuple {x, y}, \tuple {y, z} \in \RR \implies \tuple {x, z} \notin \RR$
Thus:
:$\tuple {y, x}, \tuple {z, y} \in \RR^{-1} \implies \tuple {z, x} \notin \RR^{-1}$
and so $\RR^{-1}$ is antitransitive.
Category:Antitransitive Relations
Category:Inverse Relations
Inverse of Asymmetric Relation is Asymmetric

== Theorem ==
Let $\RR$ be a relation on a set $S$.
If $\RR$ is asymmetric, then so is $\RR^{-1}$.
== Proof ==
Let $\RR$ be asymmetric.
Let $\tuple {x, y} \in \RR^{-1}$.
Then:
Thus it follows that $\RR^{-1}$ is also asymmetric.
Category:Asymmetric Relations
Category:Inverse Relations
Inverse of Bijection is Bijection

== Theorem ==
Let $f: S \to T$ be a bijection in the sense that:
:$(1): \quad f$ is an injection
:$(2): \quad f$ is a surjection.
Then the inverse $f^{-1}$ of $f$ is itself a bijection by the same definition.
== Proof ==
Let $f$ be both an injection and a surjection.
From Mapping is Injection and Surjection iff Inverse is Mapping it follows that its inverse $f^{-1}$ is a mapping.
From Inverse of Inverse Relation:
:$\paren {f^{-1} }^{-1} = f$
Thus the inverse of $f^{-1}$ is $f$.
But then $f$, being a bijection, is by definition a mapping.
So from Mapping is Injection and Surjection iff Inverse is Mapping it follows that $f^{-1}$ is a bijection.
== Also see ==
* Composite of Bijection with Inverse is Identity Mapping
== Sources ==
* : Introduction $\S 2$: Product sets, mappings
* : $\S 3.3$. Injective, surjective, bijective; inverse mappings
* : Chapter $\text I$: Algebraic Structures: $\S 5$: Composites and Inverses of Functions: Theorem $5.5$
* : Appendix: Elementary set and number theory
* : $\S 1.11$: Relations: Theorem $11.9$
* : Appendix $\text{A}.4$: Functions: Problem Set $\text{A}.4$: $25$
* : Chapter $1$: Sets and mappings: $\S 1.3$: Mappings: Exercise $3$
* : $\S 1$: Naive Set Theory: $\S 1.6$: Functions
* : Chapter $2$: Maps and relations on sets: Proposition $2.15$
* : Appendix $\text{A}.7$: Inverses: Proposition $\text{A}.7.4$
* : $1$: Set Theory and Logic: $\S 2$: Functions
* : Chapter $6$: Order Isomorphism and Transfinite Recursion: $\S 1$ A few preliminaries
* 
Category:Bijections
Category:Inverse Mappings
Inverse of Cantor Pairing Function

== Theorem ==
Let $\pi : \N^2 \to \N$ be the Cantor pairing function.
Define $k : \N \to \N$ as:
:$\map k z$ is the largest $k$ such that $T_k \le z$
where $T_k$ is the $k$-th triangular number.
Let $\pi_1 : \N \to \N$ be defined as:
:$\ds \map {\pi_1} z = z - T_{\map k z}$
Let $\pi_2 : \N \to \N$ be defined as:
:$\map {\pi_2} z = \map k z - \map {\pi_1} z$
Then:
* $\pi_1$ and $\pi_2$ are well-defined
* For every $z \in \N$:
::$\map \pi {\map {\pi_1} z, \map {\pi_2} z} = z$
* For every $x, y \in \N$:
::$\map {\pi_1} {\map \pi {x, y}} = x$
::$\map {\pi_2} {\map \pi {x, y}} = y$
== Proof ==
By definition of $\map k z$, we have $T_{\map k z} \le z$.
Thus, $z - T_{\map k z} \ge 0$.
Therefore, $\pi_1$ is well-defined.
 $\map {\pi_1} z > \map k z$.
That is:
:$z > T_{\map k z} + \map k z$
Or:
:$z \ge T_{\map k z} + \map k z + 1 = T_{\map k z + 1}$
which contradicts the maximality of $\map k z$.
Thus, by Proof by Contradiction:
:$\map {\pi_1} z \le \map k z$
Therefore:
:$\map k z - \map {\pi_1} z \ge 0$
and $\pi_2$ is well-defined.
Let $z \in \N$ be arbitrary.
We have:
Let $x, y \in \N$ be arbitrary.
We have:
Since:
:$T_{x + y} + x \ge T_{x + y}$
we have:
:$\map k {\map \pi {x, y}} \ge x + y$
Since:
:$x \le x + y$
we have:
:$T_{x + y} + x \le T_{x + y} + \paren {x + y + 1} = T_{x + y + 1}$
Hence:
:$\map k {\map \pi {x, y}} < x + y + 1$
Therefore:
:$\map k {\map \pi {x, y}} = x + y$
It follows:
And:
Category:Cantor Pairing Function
Inverse of Cantor Pairing Function is Primitive Recursive

== Theorem ==
Define $k : \N \to \N$ as:
:$\map k z$ is the largest $k$ such that $T_k \le z$
where $T_k$ is the $k$-th triangular number.
Let $\pi_1 : \N \to \N$ be defined as:
:$\ds \map {\pi_1} z = z - T_{\map k z}$
Let $\pi_2 : \N \to \N$ be defined as:
:$\map {\pi_2} z = \map k z - \map {\pi_1} z$
Then, $\pi_1$ and $\pi_2$ are primitive recursive functions.
== Proof ==
As $n = \map k z$ is the largest $n$ such that:
:$T_n \le z$
holds, it follows that $n = \map k z + 1$ is the smallest $n$ such that:
:$T_n \le z$
fails.
Or, in other words, $n = \map k z$ the smallest $n$ such that:
:$T_{n + 1} > z$
holds.
It follows that we can define $k : \N \to \N$ as:
:$\map k z = \map {\mu n} {T_{n + 1} > z}$
In order for $k$ to be primitive recursive in this way, we need the minimization to be bounded.
We will define:
:$\map k z = \map {\mu n \le z} {T_{n + 1} > z}$
 for $z \in \N$, every $n \le z$ fails $T_{n + 1} > z$.
Then, in particular, $T_{z + 1} > z$ fails.
But by definition of triangular number:
:$T_{z + 1} = z + 1 + T_z \ge z + 1 > z$
which is a contradiction.
Therefore, for every $z \in \N$:
:$\map {\mu n \le z} {T_{n + 1} > z} = \map {\mu n} {T_{n + 1} > z}$
Define $t : \N \to \N$ as:
:$\map t k = T_k$
which is primitive recursive by Triangular Numbers are Primitive Recursive.
Thus, we have:
:$\map k z = \map {\mu n \le z} {\map t {n + 1} > z}$
which is primitive recursive by:
* Bounded Minimization is Primitive Recursive
* Ordering Relations are Primitive Recursive
* Addition is Primitive Recursive
* Constant Function is Primitive Recursive
By Inverse of Cantor Pairing Function, $\pi_1$ and $\pi_2$ are well-defined.
That is, for every $z$:
:$\map {\pi_1} z \ge 0$
:$\map {\pi_2} z \ge 0$
or:
:$z \ge T_{\map k z}$
:$\map k z \ge \map {\pi_1} z$
Since these always hold, we have:
:$\map {\pi_1} z = z \mathop {\dot -} T_{\map k z}$
:$\map {\pi_2} z = \map k z \mathop {\dot -} \map {\pi_1} z$
where $\dot -$ denotes partial subtraction.
Thus, we have:
:$\map {\pi_1} z = z \mathop {\dot -} \map t {\map k z}$
From Cut-Off Subtraction is Primitive Recursive, this is primitive recursive.
By the same reasoning, $\pi_2$ is primitive recursive as well.
Category:Cantor Pairing Function
Category:Primitive Recursive Functions
Inverse of Central Unit of Ring is in Center

== Theorem ==
Let $R$ be a ring.
Let $\map Z R$ denote the center of $R$.
Let $u \in R$ be a unit of $R$.
Then:
:$u \in \map Z R \implies u^{-1} \in \map Z R$
== Proof ==
Follows directly from the definition of center and Inverse of Unit in Centralizer of Ring is in Centralizer.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $21$. Rings and Integral Domains: Theorem $21.5$: Corollary
Category:Ring Theory
Category:Subrings
Inverse of Circle Not Through Inversion Center

== Theorem ==
Let $\CC$ be a circle in the plane on center $O$.
Let $T : X \to Y$ be an inversive transformation with $\CC$ as the inversion circle. 
Then $O$ is the inversion center.
Let $K$ be an arbitrary circle distinct from $\CC$ and not through $O$.
Then the image under $T$ of all the points on $K$ lie on the same circle, distinct from both $K$ and $\CC$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $\text{III}$ / $\text{II}$ Section $4$: "Geometrical Transformations. Inversion."
* 
* 
Category:Inverse of Circle Not Through Inversion Center
Category:Inversive Transformations
Inverse of Circle Not Through Inversion Center/Proof 1

== Theorem ==
== Proof ==
Let $AB$ be the diameter of $K$, drawn so that $OAB$ are collinear points.
Let $A'$ and $B'$ be the image of $A$ and $B$ under $T$.
Let $K'$ be the circle with diameter $A'B'$
Let $C$ be an otherwise arbitrary point, lying on $K$.
:
By converse to Thales' Theorem:
:$C'$ is on circle with diameter $A'B'$
We note two other arrangements both covered by this proof.
* $K$ outside and tangent to $\CC$
* $K$ cuts $\CC$ but not through $O$
The third possibility is simply the inverse of an inversive transformation.
* $K$ inside $\CC$ and not through $O$
But this follows, since $T$ is an involution
== Sources ==
* 
Category:Inverse of Circle Not Through Inversion Center
Inverse of Circle Not Through Inversion Center/Proof 2

== Theorem ==
== Proof ==
:
Let the radius of $K$ be $k$.
Draw an arbitrary straight line from $O$ cutting $K$ at $A$ and $B$. 
Let $A'$ be the image of $A$ under $T$.
Let $B'$ be the image of $B$ under $T$.
There are two cases: 
:$(1): \quad$ The inversion circle $C$ may pass through straight line $OAB$ outside $K$
:$(2): \quad$ $C$ may cut $K$ and pass inside it.
The second possibility is illustrated using a dotted line.
The proof is the same for both.
Note: the inversion circle is not shown, for clarity.
Let:
:$OA = a$
:$OB = b$
:$OA' = a'$
:$OB' = b'$
:$OM = m$
Let $t$ be the length of the tangent line to $K$ from $O$.
Let the constant:
:$\dfrac {r^2} {t^2} = c^2$
Draw $A'Q \parallel BM$.
Let $OQ = q$.
Let $A'Q = p$.
By Equiangular Triangles are Similar:
:$\triangle OQA' \sim \triangle OMB$
:$\leadsto \dfrac q m = \dfrac {a'} b = c^2$
Also
:$\leadsto \dfrac p k = \dfrac {a'} b = c^2$
Substituting:
:$q = m c^2$
and
:$p = k c^2$
Since the of both of these are constant, for all positions of $A$ and $B$:
:$Q$ is the same point on $OM$
The length $A'Q = p$ is constant.
$A'$ is always the same distance from $Q$.
Recall:
:$\dfrac {b'} a = c^2$
So :
:$B'Q = p$
$B'$ is also always the same distance from $Q$.
So $B'Q = A'Q$.
It follows that $A'$ and $B'$ both lie on the same circle.
== Sources ==
* : Chapter $\text{III}$ / $\text{II}$ Section $4$: "Geometrical Transformations. Inversion."
Category:Inverse of Circle Not Through Inversion Center
Inverse of Circle Through Inversion Center is Straight Line Not Through Inversion Center

== Theorem ==
Let an arbitrary circle $K$ be drawn in the plane.
Let $A'$ and $P'$ be arbitrary points on $K$.
Let $T$ be an inversive transformation such that:
:the inversion center of $T$ is $O$
:the inversion circle $O$ for $T$ is chosen such that $OA'$ is a diameter of $K$
:the radius of $O$ is $r$.
Let $A$ and $P$ be the images of $A'$ and $P'$ under $T$ respectively.
Let $L$ be a straight line drawn perpendicular to the extension of $OA'$ at point $A$.
Then $P$ lies on $L$.
== Proof ==
The diagrams show the two cases:
:$K$ completely inside the circle on $O$
:or with some points outside
The proof is the same for both cases.
$\angle AOP$ is shared.
By Perpendicular from Point to Straight Line in Plane is Unique:
:there is only one straight line perpendicular to a straight line at a given point.
So:
:$AP$ is the same straight line as $L$
That is, $P$ lies on $L$.
$P$ is the image under $T$ of an arbitrary point $P'$ on $K$, and $P$ lies on $L$.
== Sources ==
* : Chapter $\text{III}$ / $\text{II}$ Section $4$: "Geometrical Transformations. Inversion."
* 
* 
Category:Inversive Transformations
Inverse of Combinatorial Matrix

== Theorem ==
Let $C_n$ be the combinatorial matrix of order $n$ given by:
:$C_n = \begin{bmatrix}
x + y & y & \cdots & y \\
y & x + y & \cdots & y \\
\vdots & \vdots & \ddots & \vdots \\
y & y & \cdots & x + y
\end{bmatrix}$
Then its inverse $C_n^{-1} = \sqbrk b_n$ can be specified as:
:$b_{i j} = \dfrac {-y + \delta_{i j} \paren {x + n y} } {x \paren {x + n y} }$
where $\delta_{i j}$ is the Kronecker delta.
== Proof ==
From the definition of the combinatorial matrix:
:$C_n = x \mathbf I_n + y \mathbf J_n$
where:
:$\mathbf I_n$ is the unit matrix of order $n$
:$\mathbf J_n$ is the square ones matrix of order $n$.
From Square of Ones Matrix we have $\mathbf J_n^2 = n \mathbf J_n$.
Hence:
So we have found a specification for the matrix which, when multiplied by $C_n$, yields $\mathbf I_n$.
By using the identities $\mathbf I_n = \sqbrk {\delta_{i j} }_n$ and $\mathbf J_n = \sqbrk 1_n$ we obtain the stated result:
:$b_{i j} = \dfrac {-y + \delta_{i j} \paren {x + n y} } {x \paren {x + n y} }$
== Sources ==
* : $\S 1.2.3$: Sums and Products: Exercise $39$
Category:Combinatorial Matrix
Inverse of Commuting Pair

== Theorem ==
Let $\struct {S, \circ}$ be a monoid whose identity is $e_S$.
Let $x, y \in S$ such that $x$ and $y$ are both invertible.
Then $x$ commutes with $y$ :
: $\struct {x \circ y}^{-1} = x^{-1} \circ y^{-1}$
== Proof ==
== Examples ==
Category:Monoids
Category:Group Theory
Category:Commutativity
Category:Inverse of Commuting Pair
Inverse of Commuting Pair/Examples/Elements of Symmetric Group S3

== Examples of Use of Inverse of Commuting Pair ==
Consider the Symmetric Group on $3$ Letters $S_3$, whose Cayley table is given as:
Let $x = \tuple {1 2 3}$ and $y = \tuple {1 3}$.
We have:
However:
== Sources ==
* : Chapter $3$: Elementary consequences of the definitions: Exercise $5$
Category:Inverse of Commuting Pair
Inverse of Composite Bijection

== Theorem ==
Let $f$ and $g$ be bijections such that $\Dom g = \Cdm f$.
Then:
:$\paren {g \circ f}^{-1} = f^{-1} \circ g^{-1}$
and $f^{-1} \circ g^{-1}$ is itself a bijection.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $\text{I}$: Sets and Functions: Exercise $\text{J}$
* : $\S 1.1$: Sets and Functions
Category:Bijections
Category:Inverse Mappings
Category:Composite Mappings
Category:Inverse of Composite Bijection
Inverse of Composite Bijection/Proof 1

== Theorem ==
== Proof ==
$\left({g \circ f}\right)^{-1} = f^{-1} \circ g^{-1}$ is a specific example of Inverse of Composite Relation.
As $f$ and $g$ are bijections then by Bijection iff Inverse is Bijection, so are both $f^{-1}$ and $g^{-1}$.
By Composite of Bijections is Bijection, it follows that $f^{-1} \circ g^{-1}$ is a bijection.
== Sources ==
Category:Inverse of Composite Bijection
Inverse of Composite Bijection/Proof 2

== Theorem ==
Let $f$ and $g$ be bijections.
Then:
:$\paren {g \circ f}^{-1} = f^{-1} \circ g^{-1}$
and $f^{-1} \circ g^{-1}$ is itself a bijection.
== Proof ==
Let $f: X \to Y$ and $g: Y \to Z$ be bijections.
Then:
Hence the result.
== Sources ==
* : $\S 3.6$. Products of bijective mappings. Permutations: Example $53$
* : Chapter $\text I$: Algebraic Structures: $\S 5$: Composites and Inverses of Functions: Theorem $5.6$
* : $\S 5$. Induced mappings; composition; injections; surjections; bijections: Theorem $5.10 \ (3)$
* : $\S 25.2$: Some further results and examples on mappings
* : Appendix $\text{A}.7$: Inverses: Proposition $\text{A}.7.5 \ (3)$
Category:Inverse of Composite Bijection
Inverse of Composite Relation

== Theorem ==
Let $\RR_2 \circ \RR_1 \subseteq S_1 \times S_3$ be the composite of the two relations $\RR_1 \subseteq S_1 \times S_2$ and $\RR_2 \subseteq S_2 \times S_3$.
Then:
:$\paren {\RR_2 \circ \RR_1}^{-1} = \RR_1^{-1} \circ \RR_2^{-1}$
== Proof ==
Let $\RR_1 \subseteq S_1 \times S_2$ and $\RR_2 \subseteq S_2 \times S_3$ be relations.
We assume that:
:$\Dom {\RR_2} = \Cdm {\RR_1}$
where $\Dom \RR$ denotes domain and $\Cdm \RR$ denotes codomain of a relation $\RR$.
This is necessary for $\RR_2 \circ \RR_1$ to exist.
From the definition of an inverse relation, we have:
:$\Dom {\RR_2} = \Cdm {\RR_2^{-1} }$
:$\Cdm {\RR_1} = \Dom {\RR_1^{-1} }$
So we confirm that $\RR_1^{-1} \circ \RR_2^{-1}$ is defined.
== Sources ==
* : Chapter $0$: Relations: Theorem $5 \ \text{(a)}$
* : $\S 10$: Inverses and Composites
* : Chapter $\text I$: Algebraic Structures: $\S 5$: Composites and Inverses of Functions: Exercise $5.8 \ \text{(c)}$
* : Chapter $\text{I}$: Sets and Functions: Problem $\text{AA}$: Relations
Category:Inverse Relations
Category:Composite Relations
Inverse of Composition of Subsets of Topological Group

== Theorem ==
Let $\struct {G, \odot, \tau}$ be a topological group.
Let $A, B \subseteq G$.
Then: 
:$\paren {A \odot B}^{-1} = B^{-1} \odot A^{-1}$
== Proof ==
The proof is identical to Inverse of Product of Subsets of Group.
== Sources ==
* : $1.1.2$: Examples of Topological Groups
Category:Topological Groups