nyuuzyou/wikis huggingface dataset:
Convolution Operator is Continuous Linear Transformation

== Theorem ==
Let $\map {L^1} \R$ and $\map {L^\infty} \R$ be the real Lebesgue $1$- and real Lebesgue $\infty$-spaces respectively.
Let $f \in \map {L^1} \R$.
Let $f* : \map {L^\infty} \R \to \map {L^\infty} \R$ be the convolution operator such that:
:$\ds \forall t \in \R : \forall g \in \map {L^\infty} \R : \map {\paren {f * g} } t = \int_{-\infty}^\infty \map f {t - \tau} \map g \tau \rd \tau$
Then $f*$ is well defined and $f* \in \map {CL} {\map {L^\infty} \R}$, where $CL$ denotes the continuous linear transformation space.
== Proof ==
Thus, an element of the image of $f*$ is bounded.
Therefore, $f*$ is well-defined.
By definition, $f*$ is integral operator.
By Integral Operator is Linear, $f*$ is linear.
We have that $\norm f_1 \in \R$.
By Continuity of Linear Transformation between Normed Vector Spaces, $f*$ is continuous.
== Sources ==
* : Chapter $\S 2.3$: The normed space $\map {CL} {X,Y}$
Category:Operator Theory
Category:Improper Integrals
Category:Linear Transformations
Category:Continuous Transformations
Convolution Theorem

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $f: \R \to \GF$ and $g: \R \to \GF$ be functions.
Let their Laplace transforms $\laptrans {\map f t} = \map F s$ and $\laptrans {\map g t} = \map G s$ exist.
Then:
:$\map F s \map G s = \ds \laptrans {\int_0^t \map f u \map g {t - u} \rd u}$
== Proof 1 ==
== Proof 2 ==
== Also presented as ==
Some sources give this as:
:$\invlaptrans {\map F s \map G s} = \ds \int_0^t \map f u \map g {t - u} \rd u$
== Sources ==
* : $\S 32$: Table of General Properties of Laplace Transforms: $32.15$
Category:Laplace Transforms
Category:Named Theorems
Category:Convolution Theorem
Convolution Theorem/Proof 1

== Theorem ==
== Proof ==
 
 
where $s_M$ is defined to be:
:$\ds \int_{t \mathop = 0}^M \int_{u \mathop = 0}^t e^{-s t} \map f u \map g {t - u} \rd u \rd t$
The region in the plane over which $(1)$ is to be integrated is $\mathscr R_{t u}$ below:
:
Setting $t - u = v$, that is $t = u + v$, the shaded region above is transformed into the region $\mathscr R_{u v}$ the $u v$ plane:
:
Thus:
where $\dfrac {\map \partial {u, t} } {\map \partial {u, v} }$ is the Jacobian of the transformation:
 
Thus the of $(2)$ is:
:$(3): \quad \ds s_M = \int_{v \mathop = 0}^M \int_{u \mathop = 0}^{M - v} e^{-s \paren {u + v} } \map f u \map g v \rd u \rd v$
Let $\map K {u, v}$ be the function defined as:
:$\map K {u, v} = \begin{cases} e^{-s \paren {u + v} } \map f u \map g v & : u + v \le M \\ 0 & : u + v > M \end{cases}$
This function is defined over the square region in the diagram below:
:
but is zero over the lighter shaded portion.
Now we can write $(3)$ as:
Hence the result.
== Sources ==
* : Chapter $2$: The Inverse Laplace Transform: Solved Problems: The Convolution Theorem: $20$
Category:Convolution Theorem
Convolution of Integrable Function with Bounded Function

== Theorem ==
Let $f: \R^n \to \R$ be a Lebesgue integrable function.
Let $g: \R^n \to \R$ be an essentially bounded function under Lebesgue measure $\lambda^n$.
Then the convolution $f * g$ of $f$ and $g$ is bounded and continuous.
In particular, $f * g$ is again essentially bounded.
== Proof ==
== Sources ==
* : $14.8$
Category:Measure Theory
Convolution of Measurable Function and Measure is Bilinear

== Theorem ==
Let $\mu$ and $\nu$ be measures on the Borel $\sigma$-algebra $\BB^n$ on $\R^n$.
Let $f, f': \R^n \to \R$ be $\BB^n$-measurable functions.
Then for all $\lambda \in \R$:
:$\paren {\lambda f + f'} * \mu = \lambda \paren {f * \mu} + f' * \mu$
:$f * \paren {\lambda \mu + \nu} = \lambda \paren {f * \mu} + f * \nu$
provided the convolutions in these expressions exist.
That is, convolution $*$ is a bilinear operation.
== Proof ==
== Sources ==
* : $14.5 \ \text{(i)}$
Category:Measure Theory
Convolution of Measurable Functions is Bilinear

== Theorem ==
Let $\BB^n$ be the Borel $\sigma$-algebra on $\R^n$.
Let $f, f', g, g': \R^n \to \R$ be $\BB^n$-measurable functions.
Then for all $\lambda \in \R$:
:$\paren {\lambda f + f'} * g = \lambda \paren {f * g} + f' * g$
:$f * \paren {\lambda g + g'} = \lambda \paren {f * g} + f * g'$
provided the convolutions in these expressions exist.
That is, convolution $*$ is a bilinear operation.
== Proof ==
== Sources ==
* : $14.5 \ \text{(i)}$
Category:Measure Theory
Convolution of Measures as Pushforward Measure

== Theorem ==
Let $\mu$ and $\nu$ be measures on the Borel $\sigma$-algebra $\BB^n$ on $\R^n$.
Let $\alpha: \R^n \times \R^n \to \R^n, \ \map \alpha {\mathbf x, \mathbf y} = \mathbf x + \mathbf y$ be vector addition on $\R^n$.
Then we have the following equality of measures on $\BB^n$:
:$\mu * \nu = \map {\alpha_*} {\mu \times \nu}$
where $\mu * \nu$ is the convolution of $\mu$ and $\nu$, and $\map {\alpha_*} {\mu \times \nu}$ is the pushforward of the product measure $\mu \times \nu$ under $\alpha$.
== Proof ==
== Sources ==
* : $14.5 \ \text{(ii)}$
Category:Convolution of Measures
Convolution of Measures is Bilinear

== Theorem ==
Let $\mu, \mu', \nu$ and $\nu'$ be measures on the Borel $\sigma$-algebra $\BB^n$ on $\R^n$.
Then for all $\lambda \in \R$:
:$\paren {\lambda \mu + \mu'} * \nu = \lambda \paren {\mu * \nu} + \mu' * \nu$
:$\mu * \paren {\lambda \nu + \nu'} = \lambda \paren {\mu * \nu} + \mu * \nu'$
where $*$ denotes convolution.
That is, convolution is a bilinear operation.
== Proof ==
== Sources ==
* : $14.5 \ \text{(i)}$
Category:Convolution of Measures
Convolution of Real Function with Rectangle Function

== Theorem ==
Let $f: \R \to \R$ be a bounded piecewise continuous real function.
Consider the rectangle function $\Pi: \R \to \R$.
Then:
:$\forall x \in \R: \map \Pi x * \map f x = \ds \int_{x \mathop - \frac 1 2}^{x \mathop + \frac 1 2} \map f u \rd u$
where $*$ denotes the convolution integral.
== Proof ==
By definition of convolution integral:
:$\ds \map \Pi x * \map f x = \int_{-\infty}^\infty \map \Pi {x - t} \map f x \rd t$
The above integral exists because $f$ is discontinuous on a countable set (see Lebesgue-Vitali Theorem).
This is equal to:
:$\ds \int_{-\infty}^{x - \frac 1 2} \map \Pi {x - t} \map f t \rd t + \int_{x - \frac 1 2}^{x + \frac 1 2} \map \Pi {x - t} \map f t \rd t + \int_{x + \frac 1 2}^\infty \map \Pi {x - t} \map f t \rd t$ 
By definition of the rectangle function we have:
:$\map \Pi {x - t} = 1 \iff \size {x - t} \le \dfrac 1 2$
:$\map \Pi {x - t} = 0 \iff \size {x - t} > \dfrac 1 2$
Therefore:
What remains is:
== Sources ==
* : Chapter $1$: Introduction
Category:Rectangle Function
Category:Convolution Integrals
Coordinate Representation of Divergence

== Theorem ==
Let $\struct {M, g}$ be a Riemannian manifold.
Let $U \subseteq M$ be an open set.
Let $\tuple {x^i}$ be local smooth coordinates.
Let $X$ be a smooth vector field on $M$.
Let $\operatorname {div}$ be the divergence operator.
Then:
:$\map {\operatorname {div} } {X^i \dfrac \partial {\partial x^i}} = \dfrac 1 {\sqrt g} \dfrac \partial {\partial x^i} \paren {X^i \sqrt g}$
== Proof ==
Let $\omega$ be the Riemannian volume form on $M$.
Recall that the divergence of $X$ is defined by the relation:
:$\map \d {\omega \rfloor X} = \map {\operatorname {div} } X \omega$, where $\rfloor$ denotes the interior product.
In local coordinates we have:
:$\omega = \sqrt g \rd x_1 \wedge \dots \wedge \d x_n$
Thus:
:$\omega \rfloor X = \paren {-1}^{i - 1} \sqrt g X^i \rd x_1 \wedge \cdots \wedge \widehat {\d x_i} \wedge \cdots \d x_n$
Thus:
:$\d \omega = \map {\dfrac \partial {\partial x_i} } {\sqrt g X^i} \rd x_1 \wedge \cdots \wedge \d x_n$
Thus:
:$\map {\operatorname {div} } X = \dfrac 1 {\sqrt g} \dfrac \partial {\partial x_i} \paren {X^i \sqrt g}$
== Sources ==
* : $\S 2$: Riemannian Metrics. Basic Constructions on Riemannian Manifolds
Category:Riemannian Geometry
Category:Divergence Operator
Coordinate Representation of Laplace-Beltrami Operator

== Theorem ==
Let $\struct {M, g}$ be a Riemannian manifold.
Let $U \subseteq M$ be an open set.
Let $\tuple {x^i}$ be local smooth coordinates.
Let $f \in \map {C^\infty} M : M \to \R$ be a smooth mapping on $M$.
Let $\nabla^2$ be the Laplace-Beltrami operator.
Then:
:$\nabla^2 f = \dfrac 1 {\sqrt {\det g} } \dfrac \partial {\partial x^i} \paren {g^{ij} \sqrt {\det g} \dfrac {\partial f} {\partial x^j}}$
== Proof ==
== Sources ==
* : $\S 2$: Riemannian Metrics. Basic Constructions on Riemannian Manifolds
Category:Riemannian Geometry
Category:Divergence Operator
Coordinates of Point of Reference of Normal Neighborhood

== Theorem ==
Let $\struct {M, g}$ be an $n$-dimensional Riemannian manifold.
Let $U_p$ be the normal neighborhood for $p \in M$.
Let $\struct {U_p, \tuple {x^i} }$ be any normal coordinate chart.
Then the coordinates of $p$ are $\tuple {0, \ldots, 0}$.
== Proof ==
== Sources ==
* : $\S 5$: The Levi-Civita Connection. Normal Neighborhoods and Normal Coordinates
Category:Normal Neighborhoods
Coordinates of Pole of Given Polar

== Theorem ==
Let $\CC$ be a circle of radius $r$ whose center is at the origin of a Cartesian plane.
Let $\LL$ be a straight line whose equation is given as:
:$l x + m y + n = 0$
Then the pole $P$ of $\LL$ with respect to $\CC$ is:
:$P = \tuple {-\dfrac l n r^2, -\dfrac m n r^2}$
=== Homogeneous Coordinates ===
== Proof ==
From Equation of Circle center Origin, we have that the equation of $\CC$ is:
:$x^2 + y^2 = r^2$
Let $P = \tuple {x_0, y_0}$.
By definition of polar:
:$x x_0 + y y_0 = r^2$
Comparing this with the equation for $\LL$:
:$\dfrac {x_0} l = \dfrac {y_0} m = \dfrac {r^2} {-n}$
The result follows.
== Sources ==
* : Chapter $\text {III}$. The Circle: $7$. To find the pole of the line $\ldots$ with respect to the circle $\ldots$
Category:Polars of Points
Coordinates of Pole of Given Polar/Homogeneous Coordinates

== Theorem ==
Let $\CC$ be a circle of radius $r$ whose center is at the origin of a Cartesian plane.
Let $\LL$ be a straight line whose equation is given as:
:$l x + m y + n = 0$
The pole $P$ of $\LL$ with respect to $\CC$, given in homogeneous Cartesian coordinates is:
:$P = \tuple {l, m, -\dfrac n {r^2} }$
== Proof ==
From Coordinates of Pole of Given Polar, $P$ can be expressed in conventional Cartesian coordinates as:
:$P = \tuple {-\dfrac l n r^2, -\dfrac m n r^2}$
Hence in homogeneous Cartesian coordinates:
:$P = \tuple {-\dfrac l n r^2, -\dfrac m n r^2, 1}$
From Multiples of Homogeneous Cartesian Coordinates represent Same Point, we multiply each coordinate by $-\dfrac n {r^2}$:
:$P = \tuple {l, m, -\dfrac n {r^2} }$
== Sources ==
* : Chapter $\text {III}$. The Circle: $7$. To find the pole of the line $\ldots$ with respect to the circle $\ldots$
Category:Polars of Points
Copeland-Erdős Constant is Normal

== Theorem ==
The Copeland-Erdős constant, whose decimal expansion is formed by concatenating the prime numbers in ascending order:
:$0 \cdotp 23571 \, 11317 \, 1923 \ldots$
is normal base $10$.
== Proof ==
== Sources ==
* 
* : $0,23571 11317 1923 \ldots$
* 
Category:Normal Numbers
Coprimality Criterion

== Theorem ==
== Proof ==
Let the less of two unequal numbers $AB, CD$ be continually subtracted from the greater, such that the number which is left over never measure the one before it till a unit is left.
We need to show that $AB$ and $CD$ are coprime.
:
Suppose $AB, CD$ are not coprime.
Then some natural number $E > 1$ will divide them both.
Let some multiple of $CD$ be subtracted from $AB$ such that the remainder $AF$ is less than $CD$.
Then let some multiple of $AF$ be subtracted from $CD$ such that the remainder $CG$ is less than $AF$.
Then let some multiple of $CG$ be subtracted from $FA$ such that the remainder $AH$ is a unit.
Since, then, $E$ divides $CD$, and $CD$ divides $BF$, then $E$ also divides $BF$.
But $E$ also divides $BA$.
Therefore $E$ also divides $AF$.
But $AF$ divides $DG$.
Therefore $E$ also divides $DG$.
But $E$ also divides the whole $DC$.
Therefore $E$ also divides the remainder $GC$.
But $CG$ divides $FH$.
Therefore $E$ also divides $FH$.
But $E$ also divides the whole $FA$.
Therefore $E$ also divides the remainder, that is, the unit $AH$.
But $E > 1$ so this is impossible.
Therefore, from , $AB$ and $CD$ are relatively prime.
== Sources ==
* : Book $\text{VII}$. Propositions
* : Chapter $\text {A}.4$: Euclid (flourished ca. $300$ B.C.)
Category:Number Theory
Category:Euclidean Algorithm
Coprimality Relation is Non-Reflexive

== Theorem ==
Consider the coprimality relation on the set of integers:
:$\forall x, y \in \Z: x \perp y \iff \gcd \set {x, y} = 1$
where $\gcd \set {x, y}$ denotes the greatest common divisor of $x$ and $y$.
Then:
:$\perp$ is non-reflexive.
== Proof ==
Proof by Counterexample:
We have from GCD of Integer and Divisor:
:$\gcd \set {n, n} = n$
and so, for example:
:$\gcd \set {2, 2} = 2$
and so:
:$2 \not \perp 2$
Hence $\perp$ is not reflexive.
But we also note that:
:$\gcd \set {1, 1} = 1$
and so:
:$1 \perp 1$
demonstrating that $\perp$ is not antireflexive either.
The result follows by definition of non-reflexive relation.
== Sources ==
* : $\S 6$. Indexed families; partitions; equivalence relations: Exercise $4 \ \text{(a)}$
Category:Coprime Integers
Coprimality Relation is Non-Transitive

== Theorem ==
Consider the coprimality relation on the set of integers:
:$\forall x, y \in \Z: x \perp y \iff \gcd \set {x, y} = 1$
where $\gcd \set {x, y}$ denotes the greatest common divisor of $x$ and $y$.
Then:
:$\perp$ is non-transitive.
== Proof ==
Proof by Counterexample:
We have:
Hence we have:
:$2 \perp 3$ and $3 \perp 4$
However, it is not the case that $2 \perp 4$.
Thus $\perp$ is not transitive.
Then we have:
:$2 \perp 3$ and $3 \perp 5$
and also:
:$2 \perp 5$
Thus $\perp$ is not antitransitive either.
The result follows by definition of non-transitive relation.
== Sources ==
* : $\S 6$. Indexed families; partitions; equivalence relations: Exercise $4 \ \text{(a)}$
Category:Coprime Integers
Category:Examples of Non-Transitive Relations
Coprimality Relation is Symmetric

== Theorem ==
Consider the coprimality relation on the set of integers:
:$\forall x, y \in \Z: x \perp y \iff \gcd \set {x, y} = 1$
where $\gcd \set {x, y}$ denotes the greatest common divisor of $x$ and $y$.
Then:
:$\perp$ is symmetric.
== Proof ==
Hence the result by definition of symmetric relation.
== Sources ==
* : $\S 6$. Indexed families; partitions; equivalence relations: Exercise $4 \ \text{(a)}$
Category:Coprime Integers
Coprimality Relation is not Antisymmetric

== Theorem ==
Consider the coprimality relation on the set of integers:
:$\forall x, y \in \Z: x \perp y \iff \gcd \set {x, y} = 1$
where $\gcd \set {x, y}$ denotes the greatest common divisor of $x$ and $y$.
Then:
:$\perp$ is not antisymmetric.
== Proof ==
Proof by Counterexample:
We have:
:$\gcd \set {3, 5} = 1 = \gcd \set {5, 3}$
and so:
:$3 \perp 5$ and $5 \perp 3$
However, it is not the case that $3 = 5$.
The result follows by definition of antisymmetric relation.
Category:Coprime Integers
Coprime Divisors of Square Number are Square

== Theorem ==
Let $r$ be a square number.
Let $r = s t$ where $s$ and $t$ are coprime.
Then both $s$ and $t$ are square.
== Proof ==
Let $p$ be a prime factor of $s$.
Then for some $n \in \Z_{>0}$:
:$p^{2 n} \divides s t$
where $\divides$ denotes the divisibility relation.
But we have that $s \perp t$ and so $p \nmid t$.
Thus $p^{2 n} \divides s$.
This holds for all prime factors of $s$.
Thus $s$ is the product of squares of primes.
Thus $s$ is square.
The same argument holds for $t$.
Hence the result.
== Sources ==
* : Chapter $2$: Some Properties of $\Z$: Exercise $2.13$
Category:Square Numbers
Coprime Integers/Examples

== Examples of Coprime Integers ==
=== $2$ and $5$ ===
=== $3$ and $8$ ===
=== $5$ and $12$ ===
=== $7$ and $27$ ===
=== $-9$ and $16$ ===
=== $-18$ and $35$ ===
=== $-27$ and $-35$ ===
=== $72$ and $91$ ===
=== $6$ and $9$ ===
Category:Examples of Coprime Integers
Coprime Integers/Examples/-18 and 35

== Example of Coprime Integers ==
$-18$ and $35$ are coprime.
== Proof ==
The positive divisors of $-18$ are $1$, $2$, $3$, $6$, $9$ and $18$.
The positive divisors of $35$ are $1$, $5$, $7$ and $35$.
The only integer in common between them is $1$.
Hence the result by definition of coprime.
== Sources ==
* 
* 
Category:Examples of Coprime Integers
Coprime Integers/Examples/-27 and -35

== Example of Coprime Integers ==
$-27$ and $-35$ are coprime.
== Proof ==
The positive divisors of $-27$ are $1$, $3$, $9$ and $27$.
The positive divisors of $-35$ are $1$, $5$, $7$ and $35$.
The only integer in common between them is $1$.
Hence the result by definition of coprime.
== Sources ==
* : Chapter $2$: Divisibility Theory in the Integers: $2.2$ The Greatest Common Divisor: Theorem $2 \text{-} 3$: Corollary
Category:Examples of Coprime Integers
Coprime Integers/Examples/-9 and 16

== Example of Coprime Integers ==
$-9$ and $16$ are coprime.
== Proof ==
The positive divisors of $-9$ are $1$, $3$ and $9$.
The positive divisors of $16$ are $1$, $2$, $4$, $8$ and $16$.
The only integer in common between them is $1$.
Hence the result by definition of coprime.
== Sources ==
* : Chapter $2$: Divisibility Theory in the Integers: $2.2$ The Greatest Common Divisor
Category:Examples of Coprime Integers
Coprime Integers/Examples/2 and 5

== Example of Coprime Integers ==
$2$ and $5$ are coprime.
== Proof ==
The divisors of $2$ are $1$ and $2$.
The divisors of $5$ are $1$ and $5$.
The only integer in common between them is $1$.
Hence the result by definition of coprime.
== Sources ==
* : Chapter $2$: Divisibility Theory in the Integers: $2.2$ The Greatest Common Divisor
Category:Examples of Coprime Integers
Coprime Integers/Examples/3 and 8

== Example of Coprime Integers ==
$3$ and $8$ are coprime.
== Proof ==
The divisors of $3$ are $1$ and $3$.
From Divisors of Power of Prime, the divisors of $8$ are $1$, $2$, $4$ and $8$.
The only integer in common between them is $1$.
Hence the result by definition of coprime.
Category:Examples of Coprime Integers
Coprime Integers/Examples/5 and 12

== Example of Coprime Integers ==
$5$ and $12$ are coprime.
== Proof ==
The positive divisors of $5$ are $1$ and $5$.
The positive divisors of $12$ are $1$, $2$, $3$, $4$, $6$ and $12$.
The only integer in common between them is $1$.
Hence the result by definition of coprime.
== Sources ==
* 
* 
Category:Examples of Coprime Integers
Coprime Integers/Examples/6 and 9

== Example of Integers which are not Coprime ==
$6$ and $9$ are not coprime.
== Proof ==
The positive divisors of $6$ are $1$, $2$, $3$ and $6$.
The positive divisors of $9$ are $1$, $3$ and $9$.
They have $3$ in common.
Hence the result by definition of coprime.
== Sources ==
* 
* 
Category:Examples of Coprime Integers
Coprime Integers/Examples/72 and 91

== Example of Coprime Integers ==
$72$ and $91$ are coprime.
== Proof ==
The positive divisors of $91$ are $1$, $7$, $13$ and $91$.
Apart from $1$, none of these integers is also a divisor of $72$.
Hence the result by definition of coprime.
== Sources ==
* 
* 
Category:Examples of Coprime Integers
Coprime Integers/Examples/7 and 27

== Example of Coprime Integers ==
$7$ and $27$ are coprime.
== Proof ==
The divisors of $7$ are $1$ and $7$.
The divisors of $27$ are $1$, $3$, $9$ and $27$.
The only integer in common between them is $1$.
Hence the result by definition of coprime.
== Sources ==
* : $\text {2-2}$ Divisibility: Example $\text {2-9}$
Category:Examples of Coprime Integers
Coprime Integers cannot Both be Zero

== Theorem ==
Let $a$ and $b$ be integers.
Let $a$ and $b$ be coprime.
Then it cannot be the case that $a = b = 0$.
== Proof ==
Let $a$ and $b$ be coprime.
Then by definition:
:$\gcd \set {a, b} = 1$
 $a = b = 0$.
Then $\gcd \set {a, b}$ is undefined.
But it is not possible both:
:for $\gcd \set {a, b}$ to be undefined
:for $\gcd \set {a, b} = 1$.
Hence by Proof by Contradiction it follows that it cannot be the case that $a = b = 0$.
Category:Coprime Integers
Category:Zero
Coprime Numbers form Fraction in Lowest Terms

== Theorem ==
== Proof ==
Let $A, B$ be (natural) numbers which are prime to one another.
We need to show that $A$ and $B$ are the least of those which have the same ratio with them.
:
Suppose this were not the case.
Then there would be some numbers $C, D$ which are less than $A, B$ such that $A : B = C : D$.
From we have that $C$ measures $A$ the same number of times that $D$ measures $B$.
Now, as many times as $C$ measures $A$, let there be so many units in $E$.
Then from $E$ also measures $A$ according to the units in $C$.
For the same reason $E$ also measures $B$ according to the units in $D$.
Therefore $E$ measures $A$ and $B$, which are prime to one another, which is contrary to our initial hypothesis.
Hence the result.
== Sources ==
* : Book $\text{VII}$. Propositions
Category:Coprime Integers
Coproduct of Free Monoids

== Theorem ==
Let $\mathbf {Mon}$ be the category of monoids.
Let $\map M A$ and $\map M B$ be free monoids on sets $A$ and $B$, respectively.
Let $A \sqcup B$ be the disjoint union of $A$ and $B$.
Then the free monoid $\map M {A \sqcup B}$ on $A \sqcup B$ is the coproduct of $\map M A$ and $\map M B$ in $\mathbf {Mon}$.
== Proof ==
By Coproduct is Unique, it suffices to verify that $\map M {A \sqcup B}$ is a coproduct for $\map M A$ and $\map M B$.
By the UMP of $\map M A$, $\map M B$ and $\map M {A \sqcup B}$, we have the following commutative diagram:
$\quad\quad \begin {xy}
<0em, 5em>*+{N} = "N",
<-5em,0em>*+{\map M A} = "MA",
<0em,0em>*+{\map M {A \sqcup B} } = "MAB",
<5em,0em>*+{\map M B} = "MB",
<-5em,-5em>*+{A} = "A",
<0em,-5em>*+{A \sqcup B} = "AB",
<5em,-5em>*+{B} = "B",
"A";"MA" **@{-} ?>*@{>} ?*!/_.8em/{i_A},
"B";"MB" **@{-} ?>*@{>} ?*!/^.8em/{i_B},
"AB";"MAB" **@{-} ?>*@{>} ?*!/_.8em/{i_{A \mathop \sqcup B} },
"A";"AB" **@{-} ?>*@{>} ?*!/^.8em/{i_1},
"B";"AB" **@{-} ?>*@{>} ?*!/_.8em/{i_2},
"MA";"MAB" **@{-} ?>*@{>} ?*!/_.8em/{j_1},
"MB";"MAB" **@{-} ?>*@{>} ?*!/^.8em/{j_2},
"MA";"N" **@{-} ?>*@{>} ?*!/_.8em/{\bar f},
"MB";"N" **@{-} ?>*@{>} ?*!/^.8em/{\bar g},
"MAB";"N" **@{--} ?>*@{>} ?*!/_.8em/{\bar h},
\end{xy}$
Here (in the notation for free monoids):
:$j_1 = \overline {\paren {i_{A \mathop \sqcup B} \circ i_1} }$
:$j_2 = \overline {\paren {i_{A \mathop \sqcup B} \circ i_2}}$
and $i_1$, $i_2$ are the injections for the coproduct.
== Sources ==
* : $\S 3.2$: Example $3.5$
Category:Category of Monoids
Coproduct of Ordered Sets

== Theorem ==
Let $\mathbf{OrdSet}$ be the category of ordered sets.
Let $\left({P, \preceq_1}\right)$ and $\left({Q, \preceq_2}\right)$ be ordered sets.
Let $P \sqcup Q$ be the disjoint union of $P$ and $Q$.
Let $\preceq$ be the ordering on $P \sqcup Q$ defined by:
:$\left({x, i}\right) \preceq \left({y, j}\right)$ $i = j$ and $x \preceq_i y$
where $i = 1$ or $i = 2$ depending on whether $x,y \in P$ or $x,y \in Q$.
Then $\left({P \sqcup Q, \preceq}\right)$ is the coproduct of $P$ and $Q$ in $\mathbf{OrdSet}$.
== Proof ==
== Sources ==
* : $\S 3.2$
Category:Category of Ordered Sets