nyuuzyou/wikis huggingface dataset:
Inverse of Conditional is Contrapositive of Converse

== Theorem ==
Let $p \implies q$ be a conditional.
Then the inverse of $p \implies q$ is the contrapositive of its converse.
== Proof ==
The inverse of $p \implies q$ is:
:$\neg p \implies \neg q$
The converse of $p \implies q$ is:
:$q \implies p$
The contrapositive of $q \implies p$ is:
:$\neg p \implies \neg q$
The two are seen to be equal.
Category:Conditional
Inverse of Conditional is Converse of Contrapositive

== Theorem ==
Let $p \implies q$ be a conditional.
Then the inverse of $p \implies q$ is the converse of its contrapositive.
== Proof ==
The inverse of $p \implies q$ is:
:$\neg p \implies \neg q$
The contrapositive of $p \implies q$ is:
:$\neg q \implies \neg p$
The converse of $\neg q \implies \neg p$ is:
:$\neg p \implies \neg q$
The two are seen to be equal.
Category:Conditional
Inverse of Curve under Inversive Transformation

== Theorem ==
Let $\CC$ be a circle embedded in a Cartesian plane $\EE$ whose center $O$ is at the origin $\tuple {0, 0}$ and whose radius is $r$.
Let $f$ be the inversive transformation of $\EE$ $\CC$.
Let $P = \tuple {x, y}$ be an arbitrary point of $\CC$.
The inverse point $P' = \tuple {x', y'}$ of $P$ under $f$ is given by:
== Proof ==
== Sources ==
* 
* 
Category:Inversive Transformations
Inverse of Curve under Inversive Transformation/Mistake

== Source Work ==
:
:inversion 1.
== Mistake ==
:A curve $\map f {x, y} = 0$ has an inverse $\map f {x', y'} = 0$, where
::$x' = \dfrac {r^2 x} {x^2 + y^2} \qquad y' = \dfrac {r^2 y} {x^2 + y^2}$
== Correction ==
The above statement is true only when the inversion center is at the origin.
For a general inversion circle of radius $r$ and inversion center $\tuple {x_0, y_0}$, the inverse point $P' = \tuple {x', y'}$ of a general point $P$ under $f$ is given by:
== Sources ==
* 
* 
* 
Category:Mistakes/The Penguin Dictionary of Mathematics
Inverse of Diagonal Matrix

== Theorem ==
Let:
:$\mathbf D = \begin{bmatrix}
a_{11} & 0 & \cdots & 0 \\
0 & a_{22} & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & a_{nn} \\
\end{bmatrix}$
be an $n \times n$ diagonal matrix.
Then its inverse is given by:
:$\mathbf D^{-1} = \begin{bmatrix}
\dfrac 1 {a_{11}} & 0 & \cdots & 0 \\
0 & \dfrac 1 {a_{22}} & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & \dfrac 1 {a_{nn}} \\
\end{bmatrix}$
provided that none of the diagonal elements are zero.
If any of the diagonal elements are zero, $\mathbf D$ is singular. 
== Proof ==
, consider the right inverse of $\mathbf D$.
Suppose none of the diagonal elements are zero. 
Then by the definition of inverse, our assertion is that the matrix product of the two matrices in question is the unit matrix of order $n$.
Now, observe that:
Now suppose one of the diagonal elements is zero.
Then $\map \det {\mathbf D} = 0$, where $\det$ indicates the determinant of $\mathbf D$.
From Determinant of Inverse Matrix it would follow that:
:$\map \det {\mathbf D^{-1} } = \dfrac 1 {\map \det {\mathbf D} }$
But this equation has no solution, and so $\mathbf D$ cannot admit an inverse.
Category:Diagonal Matrices
Category:Inverse Matrices
Inverse of Diagonal Relation

== Theorem ==
Let $S$ be a set.
Let $\Delta_S$ denote the diagonal relation on $S$.
Let ${\Delta_S}^{-1}$ denote the inverse of $\Delta_S$.
Then:
:${\Delta_S}^{-1} = \Delta_S$
== Proof ==
By definition of diagonal relation:
:$\Delta_S = \set {\tuple {x, x} \in S \times S: x \in S}$
By definition of inverse relation:
:${\Delta_S}^{-1} = \set {\tuple {x, x} \in S \times S: x \in S}$
Hence it follows that:
:$\tuple {x, x} \in \Delta_S \iff \tuple {x, x} \in {\Delta_S}^{-1}$
Category:Diagonal Relation
Category:Inverse Relations
Inverse of Direct Image Mapping does not necessarily equal Inverse Image Mapping

== Theorem ==
Let $S$ and $T$ be sets.
Let $\RR \subseteq S \times T$ be a relation.
Let $\RR^\to$ be the direct image mapping of $\RR$.
Let $\RR^\gets$ be the inverse image mapping of $\RR$.
Then it is not necessarily the case that:
:$\paren {\RR^\to}^{-1} = \RR^\gets$
where $\paren {\RR^\to}^{-1}$ denote the inverse of $\RR^\to$.
That is, the inverse of the direct image mapping of $\RR$ does not always equal the inverse image mapping of $\RR$.
== Proof ==
Proof by Counterexample:
Let $S = T = \set {0, 1}$.
Let $\RR = \set {\tuple {0, 0}, \tuple {0, 1} }$.
We have that:
:$\RR^{-1} = \set {\tuple {0, 0}, \tuple {1, 0} }$
:$\powerset S = \powerset T = \set {\O, \set 0, \set 1, \set {0, 1} }$
Thus, by inspection:
Note that $\paren {\RR^\to}^{-1}$ is the inverse of a mapping which is neither an injection nor a surjection, and so is itself not a mapping from $\powerset T$ to $\powerset S$.
This can be seen to be completely different from $\RR^\gets$, which can be determined by inspection to be:
Category:Direct Image Mappings
Category:Inverse Image Mappings
Inverse of Division Product

== Theorem ==
Let $\struct {R, +, \circ}$ be a commutative ring with unity.
Let $\struct {U_R, \circ}$ be the group of units of $\struct {R, +, \circ}$.
Let $a, b \in U_R$.
Then:
:$\paren {\dfrac a b}^{-1} = \dfrac {1_R} {\paren {a / b}} = \dfrac b a$
where $\dfrac x z$ is defined as $x \circ \paren {z^{-1} }$, that is, $x$ divided by $z$.
== Proof ==
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $23$. The Field of Rational Numbers: Theorem $23.7 \ (5)$
Category:Commutative Algebra
Inverse of Element in Semidirect Product

== Theorem ==
Let $N$ and $H$ be groups.
Let $H$ act by automorphisms on $N$ via $\phi$.
Let $N \rtimes_\phi H$ be the corresponding (outer) semidirect product.
Let $\tuple {n, h} \in N \rtimes_\phi H$.
Then:
== Proof ==
Follows from Semidirect Product of Groups is Group.
The alternatives follow from the fact that $H$ acts by automorphisms.
Category:Group Theory
Category:Semidirect Products
Inverse of Field Product

== Theorem ==
Let $\struct {F, +, \times}$ be a field whose zero is $0_F$ and whose unity is $1_F$.
Let $a, b \in F$ such that $a \ne 0$ and $b \ne 0$.
Then:
:$\paren {a \times b}^{-1} = b^{-1} \times a^{-1}$
== Proof ==
We are given that $a \ne 0$ and $b \ne 0$.
From Field has no Proper Zero Divisors and Rule of Transposition, we have:
:$a \times b \ne 0$
By we have that $\paren {a \times b}^{-1}$ exists.
Then we have:
Hence the result by definition of multiplicative inverse.
== Sources ==
* : Chapter $1$: Axioms for the Real Numbers: $2$. Fields: Theorem $3 \ \text {(vii)}$
Category:Field Theory
Category:Inverse Elements
Inverse of Field Product with Inverse

== Theorem ==
Let $\struct {F, +, \times}$ be a field whose zero is $0$ and whose unity is $1$.
Let $a, b \in F$ such that $a \ne 0$.
Then:
:$\paren {a \times b^{-1} }^{-1} = b \times a^{-1}$
This can also be expressed using the notation of division as:
:$1 / \paren {a / b} = b / a$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $3$: Field Theory: Definition and Examples of Field Structure: $\S 87 \delta$
Category:Field Theory
Category:Inverse of Field Product with Inverse
Inverse of Field Product with Inverse/Proof 1

== Theorem ==
== Proof ==
By definition, a field is a non-trivial division ring whose ring product is commutative.
By definition, a division ring is a ring with unity such that every non-zero element is a unit. 
Hence we can use Inverse of Division Product:
:$\paren {\dfrac a b}^{-1} = \dfrac {1_R} {\paren {a / b}} = \dfrac b a$
which applies to the group of units of a comutative ring with unity.
Category:Inverse of Field Product with Inverse
Inverse of Generator of Cyclic Group is Generator

== Theorem ==
Let $\gen g = G$ be a cyclic group.
Then:
:$G = \gen {g^{-1} }$
where $g^{-1}$ denotes the inverse of $g$.
Thus, in general, a generator of a cyclic group is not unique.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $5$: Subgroups: Exercise $14$
* : Chapter $2$: Subgroups and Cosets: $\S 44 \alpha$
* : $\S 39$. Cyclic Groups
Category:Inverse of Generator of Cyclic Group is Generator
Category:Cyclic Groups
Inverse of Generator of Cyclic Group is Generator/Proof 1

== Theorem ==
== Proof ==
Let $\gen g = G$.
Then from Set of Words Generates Group:
:$\map W {\set {g, g^{-1} } } = G$
But:
:$\gen {g^{-1} } = \map W {\set {g, g^{-1} } }$
and the result follows.
Category:Inverse of Generator of Cyclic Group is Generator
Inverse of Generator of Cyclic Group is Generator/Proof 2

== Theorem ==
== Proof ==
Let $C_n = \gen g$ be the cyclic group of order $n$.
By definition, $g^n = e$.
We have that $n - 1$ is coprime to $n$.
So it follows from that Power of Generator of Cyclic Group is Generator iff Power is Coprime with Order that:
: $C_n = \gen {g^{n - 1} }$
But from Inverse Element is Power of Order Less 1:
:$g^{n - 1} = g^{-1}$
== Also see ==
* Inverse of Generator of Cyclic Group is Generator/Proof 1: note that from Inverse Element is Power of Order Less 1:
:$g^{n - 1} = g^{-1}$
Category:Inverse of Generator of Cyclic Group is Generator
Inverse of Group Commutator

== Theorem ==
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $\sqbrk {g, h}$ denote the commutator of $g$ and $h$.
Then $\sqbrk {g, h}$ is the inverse of $\sqbrk {h, g}$:
:$\sqbrk {g, h} = \sqbrk {h, g}^{-1}$
== Proof ==
Category:Group Commutators
Category:Inverse Elements
Inverse of Group Inverse

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Let $g \in G$, with inverse $g^{-1}$.
Then:
:$\paren {g^{-1} }^{-1} = g$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $\text{II}$: Groups: The Group Property
* : $\text{I}$: Groups: $\S 1$ Semigroups, Monoids and Groups: Theorem $1.2 \text{(iv)}$
Category:Group Theory
Category:Inverse Elements
Category:Inverse of Group Inverse
Inverse of Group Inverse/Proof 2

== Theorem ==
== Proof ==
Let $g \in G$.
Then:
== Sources ==
* : $\S 4.6$. Elementary theorems on groups: Theorem $\text{(iii)}$
* : Chapter $2$: The Definition of Group Structure: $\S 28 \ (2)$
* : $\S 35.5$: Elementary consequences of the group axioms
* : Chapter $3$: Elementary consequences of the definitions: Corollary $3.5$
* : $\text{I}$: Groups: $\S 1$ Semigroups, Monoids and Groups: Theorem $1.2 \text{(iv)}$
Category:Inverse of Group Inverse
Inverse of Group Isomorphism is Isomorphism

== Theorem ==
Let $\struct {G, \circ}$ and $\struct {H, *}$ be groups.
Let $\phi: \struct {G, \circ} \to \struct {H, *}$ be a mapping.
Then $\phi$ is an isomorphism $\phi^{-1}: \struct {H, *} \to \struct {G, \circ}$ is also an isomorphism.
== Proof 1 ==
== Proof 2 ==
Category:Inverse of Group Isomorphism is Isomorphism
Category:Group Isomorphisms
Inverse of Group Isomorphism is Isomorphism/Proof 2

== Theorem ==
== Proof ==
=== Necessary Condition ===
Let $\phi: G \to H$ be an isomorphism.
Then by definition $\phi$ is a bijection.
From Bijection iff Inverse is Bijection it follows that:
:$\exists \phi^{-1}: \struct {H, *} \to \struct {G, \circ}$
such that $\phi^{-1}$ is also a bijection.
Thus:
So $\phi^{-1}: \struct {H, *} \to \struct {G, \circ}$ is a homomorphism.
$\phi^{-1}$ is also (from above) a bijection.
Thus, by definition, $\phi^{-1}$ is an isomorphism.
=== Sufficient Condition ===
Let $\phi^{-1}: \struct {H, *} \to \struct {G, \circ}$ be an isomorphism.
Applying the same result as above in reverse, we have that $\paren {\phi^{-1} }^{-1}: \struct {G, \circ} \to \struct {H, *}$ is also an isomorphism.
But by Inverse of Inverse of Bijection:
: $\paren {\phi^{-1} }^{-1} = \phi$
and hence the result.
== Sources ==
* : $\S 7.2$. Some lemmas on homomorphisms: Lemma $\text{(v)}$
* : Chapter $\text{II}$: Groups: Quotient Groups
* : Chapter $2$: Group Homomorphism and Isomorphism: $\S 62$
* : Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $3 \ \text{(ii)}$
Category:Inverse of Group Isomorphism is Isomorphism
Inverse of Group Product

== Theorem ==
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $a, b \in G$, with inverses $a^{-1}, b^{-1}$.
Then:
:$\paren {a \circ b}^{-1} = b^{-1} \circ a^{-1}$
=== General Result ===
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Sources ==
* : $\text{I}$: Groups: $\S 1$ Semigroups, Monoids and Groups: Theorem $1.2 \ \text{(v)}$
Category:Group Theory
Category:Inverse Elements
Category:Inverse of Product
Category:Inverse of Group Product
Inverse of Group Product/General Result

== Theorem ==
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $a_1, a_2, \ldots, a_n \in G$, with inverses $a_1^{-1}, a_2^{-1}, \ldots, a_n^{-1}$.
Then:
: $\paren {a_1 \circ a_2 \circ \cdots \circ a_n}^{-1} = a_n^{-1} \circ \cdots \circ a_2^{-1} \circ a_1^{-1}$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $\text {I}$: The Group Concept: $\S 2$: The Axioms of Group Theory: $(1.13)$
* : $\S 4.6$. Elementary theorems on groups: Theorem $\text{(iv)}$
* : Chapter $2$: The Definition of Group Structure: $\S 28 \alpha$
* : $\S 35.7$: Elementary consequences of the group axioms
Category:Group Theory
Category:Inverse of Group Product
Inverse of Group Product/General Result/Proof 2

== Theorem ==
== Proof ==
Proof by induction:
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$\paren {a_1 \circ a_2 \circ \cdots \circ a_n}^{-1} = a_n^{-1} \circ \cdots \circ a_2^{-1} \circ a_1^{-1}$
$\map P 1$ is (trivially) true, as this just says:
:$\paren {a_1}^{-1} = a_1^{-1}$
=== Basis for the Induction ===
$\map P 2$ is the case:
:$\paren {a_1 \circ a_2}^{-1} = a_2^{-1} \circ a_1^{-1}$
which has been proved in Inverse of Group Product.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:$\paren {a_1 \circ a_2 \circ \cdots \circ a_k}^{-1} = a_k^{-1} \circ \cdots \circ a_2^{-1} \circ a_1^{-1}$
Then we need to show:
:$\paren {a_1 \circ a_2 \circ \cdots \circ a_{k + 1} }^{-1} = a_{k + 1}^{-1} \circ \cdots \circ a_2^{-1} \circ a_1^{-1}$
=== Induction Step ===
This is our induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \N_{> 0}: \paren {a_1 \circ a_2 \circ \cdots \circ a_n}^{-1} = a_n^{-1} \circ \cdots \circ a_2^{-1} \circ a_1^{-1}$
== Sources ==
* : $\S 1.4$: Theorem $2$
Category:Inverse of Group Product
Inverse of Group Product/Proof 1

== Theorem ==
== Proof ==
The result follows from Group Product Identity therefore Inverses:
:$\paren {a \circ b} \circ \paren {b^{-1} \circ a^{-1} } = e \implies \paren {a \circ b}^{-1} = b^{-1} \circ a^{-1}$
== Sources ==
* : Chapter $\text {I}$: The Group Concept: $\S 2$: The Axioms of Group Theory
* : $\S 4.6$. Elementary theorems on groups: Theorem $\text{(iv)}$
* : $\S 1.4$: Theorem $2$
* : Chapter $\text{II}$: Groups: The Group Property
* : Chapter $2$: The Definition of Group Structure: $\S 28 \ (3)$
* : $\S 35.6$: Elementary consequences of the group axioms
* : Chapter $3$: Elementary consequences of the definitions: Corollary $3.4$
Category:Inverse of Group Product
Inverse of Hilbert Matrix

== Theorem ==
Let $H_n$ be the Hilbert matrix of order $n$:
:$\begin{bmatrix} a_{i j} \end{bmatrix} = \begin{bmatrix} \dfrac 1 {i + j - 1} \end{bmatrix}$
Then its inverse $H_n^{-1} = \sqbrk b_n$ can be specified as:
:$\begin{bmatrix} b_{i j} \end{bmatrix} = \begin{bmatrix} \dfrac {\paren {-1}^{i + j} \paren {i + n - 1}! \paren {j + n - 1}!} {\paren {\paren {i - 1}!}^2 \paren {\paren {j - 1}!}^2 \paren {n - j}! \paren {n - i}! \paren {i + j - 1} } \end{bmatrix}$
== Proof ==
From Hilbert Matrix is Cauchy Matrix, $H_n$ is a special case of a Cauchy matrix:
:$\begin{bmatrix} c_{i j} \end{bmatrix} = \begin{bmatrix} \dfrac 1 {x_i + y_j} \end{bmatrix}$
where:
:$x_i = i$
:$y_j = j - 1$
From Inverse of Cauchy Matrix, the inverse of the square Cauchy matrix of order $n$ is:
:$\begin{bmatrix} b_{i j} \end{bmatrix} = \begin{bmatrix} \dfrac {\ds \prod_{k \mathop = 1}^n \paren {x_j + y_k} \paren {x_k + y_i} } {\ds \paren {x_j + y_i} \paren {\prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {x_j - x_k} } \paren {\prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne i} } \paren {y_i - x_k} } } \end{bmatrix}$
Thus $H_n^{-1}$ can be specified as:
:$\begin{bmatrix} b_{i j} \end{bmatrix} = \begin{bmatrix} \dfrac {\ds \prod_{k \mathop = 1}^n \paren {i + k - 1} \paren {j + k - 1} } {\ds \paren {i + j - 1} \paren {\prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne i} } \paren {i - k} } \paren {\prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {j - k} } } \end{bmatrix}$
First, from Product of Products:
:$\ds \prod_{k \mathop = 1}^n \paren {i + k - 1} \paren {j + k - 1} = \prod_{k \mathop = 1}^n \paren {i + k - 1} \prod_{k \mathop = 1}^n \paren {j + k - 1}$
We address in turn the various factors of this expression for $b_{i j}$.
and similarly:
:$\ds \prod_{k \mathop = 1}^n \paren {j + k - 1} = \frac {\paren {j + n - 1}!} {\paren {j - 1}!}$
Then:
and similarly:
:$\ds \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {j - k} = \paren {j - 1}! \paren {-1}^{n - j} \paren {n - j}!$
Thus we can write:
== Sources ==
* : $\S 1.2.3$: Sums and Products: Exercise $45$
Category:Hilbert Matrices
Inverse of Homeomorphism between Metric Spaces is Homeomorphism

== Theorem ==
Let $M = \struct {A_1, d_1}, M_2 = \struct {A_2, d_2}$ be metric spaces.
Let $f: M_1 \to M_2$ be a homeomorphism.
Then $f^{-1}: M_2 \to M_1$ is also a homeomorphism.
== Proof ==
By definition, a homeomorphism is a bijection such that both $f$ and $f^{-1}$ are continuous.
As $f$ is a bijection then by Bijection iff Inverse is Bijection, so is $f^{-1}$.
So by definition $f^{-1}$ is a bijection such that both $f^{-1}$ and $\paren {f^{-1} }^{-1}$ are continuous.
The result follows from Inverse of Inverse of Bijection.
Category:Homeomorphisms (Metric Spaces)
Inverse of Homeomorphism is Homeomorphism

== Theorem ==
Let $T, T'$ be topological spaces.
Let $f: T \to T'$ be a homeomorphism.
Then $f^{-1}: T' \to T$ is also a homeomorphism.
== Proof ==
By definition, a homeomorphism is a bijection such that both $f$ and $f^{-1}$ are continuous.
As $f$ is a bijection then by Bijection iff Inverse is Bijection, so is $f^{-1}$.
So by definition $f^{-1}$ is a bijection such that both $f^{-1}$ and $\left({f^{-1}}\right)^{-1}$ are continuous.
The result follows from Inverse of Inverse of Bijection.
Category:Homeomorphisms (Topological Spaces)
Inverse of Identity Element is Itself

== Theorem ==
Let $\struct {S, \circ}$ be an algebraic structure with an identity element $e$.
Let the inverse of $e$ be $e^{-1}$.
Then:
:$e^{-1} = e$
That is, $e$ is self-inverse.
== Proof ==
From Identity Element is Idempotent:
:$e \circ e = e$
Hence the result by definition of identity element.
== Sources ==
* : $\S 4.6$. Elementary theorems on groups: Example $85$
* : $\S 31$ Identity element and inverses
* : $\S 35.4$: Elementary consequences of the group axioms
Category:Inverse Elements
Category:Identity Elements
Inverse of Identity Mapping

== Theorem ==
Let $S$ be a set.
Let $I_S: S \to S$ be the identity mapping on $S$.
Then the inverse of $I_S$ is itself:
:$\paren {I_S}^{-1} = I_S$
== Proof ==
From the nature of the identity mapping, we have:
:$I_S \circ I_S = I_S$
from which it follows by definition that $I_S$ is the inverse of itself.
== Sources ==
* : $\S 3.5$. Identity mappings
* : Chapter $\text{I}$: Sets and Functions: Composition of Functions
* : Appendix $\text{A}.7$: Inverses: Proposition $\text{A}.7.5 \ (1)$
Category:Inverse Mappings
Category:Identity Mappings
Inverse of Increasing Bijection need not be Increasing

== Theorem ==
Let $\struct {S, \preceq_1}$ and $\struct {T, \preceq_2}$ be ordered sets.
Let $\phi: S \to T$ be a bijection which is increasing.
Then $\phi^{-1}: T \to S$ is not necessarily also increasing.
== Proof ==
Proof by Counterexample:
Let $S := \powerset {\set {a, b} }$.
Let $T := \set {1, 2, 3, 4}$.
From Subset Relation on Power Set is Partial Ordering, $\struct {S, \subseteq}$ is an ordered set.
Clearly so is $\struct {T, \le}$ (although its ordering is in fact total, it is still technically an ordered set).
Let $\phi: S \to T$ be defined as:
By inspection it can be seen that $\phi$ is a bijection.
Also by inspection it can be seen that $\phi$ is increasing.
But note that while $2 \le 3$, it is not the case that $\set a \subseteq \set b$.
That is, $\phi^{-1}$ is not increasing.
== Sources ==
* : $\S 7$
Category:Increasing Mappings
Category:Bijections
Inverse of Infimum in Ordered Group is Supremum of Inverses

== Theorem ==
Let $\struct {G, \circ, \preccurlyeq}$ be an ordered group.
Let $x, y \in G$.
Then:
:$\set {x, y}$ admits an infimum in $G$
:
:$\set {x^{-1}, y^{-1} }$ admits a supremum in $G$
in which case:
:$\paren {\inf \set {x, y} }^{-1} = \sup \set {x^{-1}, y^{-1} }$
== Proof ==
=== Sufficient Condition ===
Let $\set {x, y}$ admits an infimum $c$ in $G$.
Then:
and so $c^{-1}$ is an upper bound of $\set {x^{-1}, y^{-1} }$.
Suppose $d$ is an upper bound of $\set {x^{-1}, y^{-1} }$.
Then:
That is, $d^{-1}$ is a lower bound of $\set {x, y}$.
But because $c$ is an infimum of $\set {x, y}$:
That is, for an arbitrary upper bound $d$ of $\set {x^{-1}, y^{-1} }$:
:$c^{-1} \preccurlyeq d$
and so $c^{-1}$ is a supremum of $\set {x^{-1}, y^{-1} }$ by definition.
That is:
:$\paren {\inf \set {x, y} }^{-1} = \sup \set {x^{-1}, y^{-1} }$
=== Necessary Condition ===
Let $\set {x^{-1}, y^{-1} }$ admit a supremum $c$ in $G$.
Then:
and so $c^{-1}$ is a lower bound of $\set {x, y}$.
Suppose $d$ is a lower bound of $\set {x, y}$.
Then:
That is, $d^{-1}$ is an upper bound of $\set {x, y}$.
But because $c$ is a supremum of $\set {x^{-1}, y^{-1} }$:
That is, for an arbitrary lower bound $d$ of $\set {x, y}$:
:$d \preccurlyeq c^{-1}$
and so $c^{-1}$ is an infimum of $\set {x, y}$ by definition.
That is:
:$\paren {\inf \set {x, y} } = \sup \set {x^{-1}, y^{-1} }^{-1}$
from which it follows from that:
:$\paren {\inf \set {x, y} }^{-1} = \sup \set {x^{-1}, y^{-1} }$
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 15$: Ordered Semigroups: Exercise $15.10 \ \text {(c)}$
Category:Ordered Groups
Category:Suprema
Category:Infima
Inverse of Injection is Many-to-One Relation

== Theorem ==
Let $f: S \to T$ be an injection.
Let $f^{-1}: T \to S$ be the inverse relation of $f$.
Then $f^{-1}$ is many-to-one.
== Proof ==
Let $f: S \to T$ be an injection.
We have by definition of inverse relation that:
:$f^{-1} = \set {\tuple {t, s}: t = \map f s}$
Let $f: S \to T$ be an injection.
Let $\tuple {t, s_1} \in f^{-1}$ and $\tuple {t, s_2} \in f^{-1}$.
By definition, we have that $\map f {s_1} = t = \map f {s_2}$.
But as $f$ is an injection:
:$\map f {s_1} = \map f {s_2} \implies s_1 = s_2$
So we have that:
:$\tuple {t, s_1} \in f^{-1} \land \tuple {t, s_2} \in f^{-1} \implies s_1 = s_2$
and so by definition, $f^{-1}$ is a many-to-one relation.
Category:Injections
Category:Inverse Relations
Inverse of Injection is One-to-One Relation

== Theorem ==
Let $f$ be an injective mapping.
Then its inverse $f^{-1}$ is a one-to-one relation.
== Proof ==
We are given that $f$ is an injective mapping.
Hence by definition $f$ is a one-to-one relation.
The result follows from from Inverse of One-to-One Relation is One-to-One.
== Sources ==
* 
Category:Inverse Relations
Category:Injections
Category:Inverse Mappings
Inverse of Injective and Surjective Mapping is Mapping

== Theorem ==
Let $f: S \to T$ be a mapping such that:
:$(1): \quad f$ is an injection
:$(2): \quad f$ is a surjection.
Then the inverse $f^{-1}$ of $f$ is itself a mapping.
== Proof 1 ==
== Proof 2 ==
Category:Mapping is Injection and Surjection iff Inverse is Mapping
Inverse of Injective and Surjective Mapping is Mapping/Proof 1

== Theorem ==
== Proof ==
Recall the definition of the inverse of $f$:
$f^{-1} \subseteq T \times S$ is the relation defined as:
:$f^{-1} = \set {\tuple {t, s}: t = \map f s}$
Let $f: S \to T$ be a mapping such that:
::$(1): \quad f$ is an injection
::$(2): \quad f$ is a surjection.
By Inverse of Injection is Many-to-One Relation, $f^{-1}$ is many-to-one.
From Inverse of Surjection is Relation both Left-Total and Right-Total $\map {f^{-1} } y$ is left-total.
Thus $f^{-1}$ is:
:many-to-one
and
:left-total.
Hence, by definition, $f^{-1}$ is a mapping.
Category:Mapping is Injection and Surjection iff Inverse is Mapping