nyuuzyou/wikis huggingface dataset:
Cancellable Semiring with Unity is Additive Semiring

== Theorem ==
Let $\struct {S, *, \circ}$ be a cancellable semiring with unity $1_S$.
Then the distributand $*$ is commutative.
That is to say, $\struct {S, *, \circ}$ is also an additive semiring.
== Proof ==
Let $\struct {S, *, \circ}$ be a semiring, all of whose elements of $S$ are cancellable for $*$.
We expand the expression $\paren {a * b} \circ \paren {c * d}$ using the distributive law in two ways:
So, by the fact that all elements of $\struct {S, *}$ are cancellable (and thus, a fortiori, are $a \circ c$ and $b \circ d$), we have:
As this is true for all $a, b, c, d \in S$, it is true in particular if $c = d = 1_S$.
Thus it is clear that $b * a = a * b$, which is exactly to say that $*$ is commutative.
Hence the result, by definition of additive semiring.
Category:Semirings
Cancellable iff Regular Representations Injective

== Theorem ==
Let $\struct {S, \circ}$ be an algebraic structure.
Then $a \in S$ is cancellable :
:the left regular representation $\map {\lambda_a} x$ is injective
and:
:the right regular representation $\map {\rho_a} x$ is injective.
== Proof ==
=== Left Cancellable ===
=== Right Cancellable ===
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups
Category:Regular Representations
Category:Cancellability
Cancellation Law for Field Product

== Theorem ==
Let $\struct {F, +, \times}$ be a field whose zero is $0_F$ and whose unity is $1_F$.
Let $a, b, c \in F$.
Then:
:$a \times b = a \times c \implies a = 0_F \text { or } b = c$
== Proof ==
Let $a \times b = a \times c$.
Then:
That is:
:$a \ne 0 \implies b = c$
Suppose $b \ne c$.
Then by Rule of Transposition:
:$\map \neg {a \ne 0}$
that is:
:$a = 0_F$
and we note that in this case:
:$a \times b = 0_F = a \times c$
== Sources ==
* : Chapter $1$: Axioms for the Real Numbers: $2$. Fields: Theorem $3 \ \text {(vi)}$
Category:Field Theory
Category:Cancellation Laws
Cancellation Law for Ring Product of Integral Domain

== Theorem ==
Let $\struct {D, +, \circ}$ be an integral domain whose zero is $0_D$.
Let $D^*$ denote $D \setminus \set {0_D}$, that is, $D$ without its zero.
Let $a \in D^*$.
Then:
:$\forall x, y \in D: a \circ x = a \circ y \implies x = y$
That is, all elements of $D^*$ are cancellable for the ring product.
== Proof ==
From the definition of integral domain, no elements of $D^*$ are zero divisors.
From Ring Element is Zero Divisor iff not Cancellable, it follows that all elements of $D^*$ are cancellable for the ring product $\circ$.
== Sources ==
* : Chapter $1$: Integral Domains: $\S 4$. Elementary Properties: Theorem $2 \ \text{(vii)}$
* : Chapter $1$: Rings - Definitions and Examples: $3$: Some special classes of rings: Lemma $1.4$
* : $\S 55.4$ Special types of ring and ring elements
Category:Integral Domains
Category:Cancellation Laws
Cancellation Laws

== Theorem ==
Let $G$ be a group.
Let $a, b, c \in G$.
Then the following hold:
;Right cancellation law
:$b a = c a \implies b = c$
;Left cancellation law
:$a b = a c \implies b = c$
That is, the group operation is cancellable.
Let $e$ be the identity element of $G$.
Then:
=== Corollary 1 ===
=== Corollary 2 ===
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Sources ==
* : $\text{I}$: Groups: $\S 1$ Semigroups, Monoids and Groups: Theorem $1.2 \text{(ii)}$
* : $\S 1$: Exercise $5$
Category:Cancellation Laws
Category:Group Theory
Category:Cancellability
Category:Named Theorems
Cancellation Laws/Corollary 2

== Corollary to Cancellation Laws ==
Let $g$ and $h$ be elements of a group $G$ whose identity element is $e$.
Then:
:$h g = g \implies h = e$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $\text{II}$: Groups: The Group Property: Theorem $1 \ \text {(ii)}$
Category:Cancellation Laws
Category:Identity Elements
Cancellation Laws/Proof 1

== Theorem ==
== Proof ==
Let $a, b, c \in G$ and let $a^{-1}$ be the inverse of $a$.
Suppose $b a = c a$.
Then:
Thus, the right cancellation law holds.
The proof of the left cancellation law is analogous.
== Sources ==
* : $\S 35.1$: Elementary consequences of the group axioms
Category:Cancellation Laws
Cancellation Laws/Proof 2

== Theorem ==
== Proof ==
From its definition, a group is a monoid, all of whose elements have inverses and thus are invertible.
From Invertible Element of Monoid is Cancellable, it follows that all its elements are therefore cancellable.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Theorem $7.1$ Corollary
Category:Cancellation Laws
Cancellation Laws/Proof 3

== Corollary to Group has Latin Square Property ==
== Proof ==
Suppose $x = b a = c a$.
By Group has Latin Square Property, there exists exactly one $y \in G$ such that $x = y a$.
That is, $x = b a = c a \implies b = c$.
Similarly, suppose $x = a b = a c$.
Again by Group has Latin Square Property, there exists exactly one $y \in G$ such that $x = a y$.
That is, $a b = a c \implies b = c$.
== Sources ==
* : $\S 1.4$: Theorem $1$ Corollary
Category:Cancellation Laws
Cancellation Property of Product Inverse Operator

== Theorem ==
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $\oplus: G \times G \to G$ be the product inverse of $\circ$ on $G$.
Then:
:$\forall x, y, z \in G: \paren {x \oplus z} \oplus \paren {y \oplus z} = x \oplus y$
== Proof ==
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Exercise $7.7 \ \text {(a)}: 4^\circ$
Category:Product Inverse Operation
Cancellation of Meet in Boolean Algebra

== Theorem ==
Let $\struct {S, \vee, \wedge, \neg}$ be a Boolean algebra.
Let $a, b, c \in S$.
Let:
Then:
: $a = b$
== Proof ==
Follows from Cancellation of Join in Boolean Algebra through the Duality Principle
== Also see ==
* Cancellation of Join in Boolean Algebra
Category:Boolean Algebras
Canonical Basis of Free Module on Set is Basis

== Theorem ==
Let $R$ be a ring with unity.
Let $I$ be a set.
Let $R^{\paren I}$ be the free $R$-module on $I$.
Let $B$ be its canonical basis.
Then $B$ is a basis of $R^{\paren I}$.
== Proof ==
Let $R$ be a ring with unity.
Let $I$ be a set.
Let $R^{\paren I}$ be the free $R$-module on $I$.
Let $B$ be its canonical basis.
Recall the definition of canonical basis:
It is to be shown that $B$ is a basis.
Hence the result.
Category:Free Modules
Canonical Bijection from Completely Prime Filters to Frame Homomorphisms

== Theorem ==
Let $\struct{L, \vee, \wedge, \preceq}$ be a locale.
Let $\map {\operatorname{Sp}} L = \struct{\map {\operatorname{pt}} L, \set{\Sigma_a : a \in L}}$ be the spectrum of $L$ as completely prime filters where:
:*$\quad\map {\operatorname{pt}} L$ denotes the set of points as completely prime filters of $L$.
:*$\quad$for each $a \in L$, $\Sigma_a = \set{p \in \map {\operatorname{pt}} L : a \in p}$
Let $\Sigma_L = \set{\Sigma_a : a \in L}$
Let $\map {\operatorname{Sp}'} L = \struct{\map {\operatorname{pt}'} L, \set{\Sigma'_a : a \in L}}$ be the spectrum of $L$ as frame homomorphisms where:
:*$\quad\map {\operatorname{pt}'} L$ denotes the set of points as frame homomorphisms of $L$.
:*$\quad$for each $a \in L$, $\Sigma'_a = \set{\phi \in \map {\operatorname{pt}'} L : \map \phi a = \top}$
Let $\Sigma'_L = \set{\Sigma'_a : a \in L}$
Let $f : \map {\operatorname{pt}} L \to \map {\operatorname{pt}'} L$ be the mapping defined by:
:$\forall p \in \map {\operatorname{pt}} L : \map f p$ is the frame homomorphim defined by:
::$\forall a \in L : \map {\map f p} a = \begin{cases}
\top & : a \in p\\
\bot & : a \notin p 
\end{cases}$
Then:
:$f$ is a bijection
:$f^\to \restriction_{\Sigma_L}$ is a surjection from $\Sigma_L$ to $\Sigma'_L$
== Proof ==
=== $f$ is a Bijection ===
From Frame Homomorphism Onto Two Induced by Completely Prime Filter:
:$f$ is a well-defined mapping
Let $g : \map {\operatorname{pt}'} L \to \map {\operatorname{pt}} L$ be the mapping defined by:
:$\forall \phi \in \map {\operatorname{pt}'} L : \map g \phi = \map {\phi^{-1}} \top$
where:
:$\map {\phi^{-1}} \top$ denotes the preimage of $\top \in \mathbf 2$ under the mapping $\phi$
From Completely Prime Filter Induced by Frame Homomorphism Onto Two:
:$g$ is a well-defined mapping
From Completely Prime Filter Induced by Frame Homomorphism Induced by Completely Prime Filter:
:$g \circ f = \operatorname {id}_{\map {\operatorname{pt}} L}$
where $\operatorname {id}_{\map {\operatorname{pt}} L}$ is the identity mapping on $\map {\operatorname{pt}} L$
From Frame Homomorphism Induced by Completely Prime Filter Induced by Frame Homomorphism:
:$f \circ g = \operatorname {id}_{\map {\operatorname{pt}'} L}$
where $\operatorname {id}_{\map {\operatorname{pt}} L}$ is the identity mapping on $\map {\operatorname{pt}'} L$
Hence $f$ is a bijection by definition.
=== $f^\to \restriction_{\Sigma_L}$ is a Surjection ===
We have:
It follows that $f^\to \restriction_{\Sigma_L}$ is a surjection onto $\Sigma'_L$ by definition.
The result follows.
Category:Spectra of Locales
Canonical Bijection from Completely Prime Filters to Meet Irreducible Elements

== Theorem ==
Let $\struct{L, \vee, \wedge, \preceq}$ be a locale.
Let $\map {\operatorname{Sp}} L = \struct{\map {\operatorname{pt}} L, \set{\Sigma_a : a \in L}}$ be the spectrum of $L$ as completely prime filters where:
:*$\quad\map {\operatorname{pt}} L$ denotes the set of points as completely prime filters of $L$.
:*$\quad$for each $a \in L$, $\Sigma_a = \set{p \in \map {\operatorname{pt}} L : a \in p}$
Let $\Sigma_L = \set{\Sigma_a : a \in L}$
Let $\map {\operatorname{Sp}'} L = \struct{\map {\operatorname{pt}'} L, \set{\Sigma'_a : a \in L}}$ be the spectrum of $L$ as meet-irreducible elements where:
:*$\quad\map {\operatorname{pt}'} L$ denotes the set of points as meet-irreducible elements of $L$.
:*$\quad$for each $a \in L$, $\Sigma'_a = \set{p \in \map {\operatorname{pt}'} L : a \npreceq p}$
Let $\Sigma'_L = \set{\Sigma'_a : a \in L}$
Let $f : \map {\operatorname{pt}} L \to \map {\operatorname{pt}'} L$ be the mapping defined by:
:$\forall p \in \map {\operatorname{pt}} L : \map f p = \bigvee \set{a \in L : a \notin p}$
where:
:$\bigvee \set{a \in L : a \notin p}$ denotes the supremum of the set $\set{a \in L : a \notin p}$
Then:
:$f$ is a bijection
:$f^\to \restriction_{\Sigma_L}$ is a surjection from $\Sigma_L$ to $\Sigma'_L$
== Proof ==
=== $f$ is a Bijection ===
From Meet Irreducible Element Induced by Completely Prime Filter:
:$f$ is a well-defined mapping
Let $g : \map {\operatorname{pt}'} L \to \map {\operatorname{pt}} L$ be the mapping defined by:
:$\forall m \in \map {\operatorname{pt}'} L : \map g m = \set{a \in L : a \npreceq m}$
From Completely Prime Filter Induced by Meet Irreducible Element:
:$g$ is a well-defined mapping
From Completely Prime Filter Induced by Meet Irreducible Induced by Completely Prime Filter:
:$g \circ f = \operatorname {id}_{\map {\operatorname{pt}} L}$
where $\operatorname {id}_{\map {\operatorname{pt}} L}$ is the identity mapping on $\map {\operatorname{pt}} L$
From Meet Irreducible Induced by Completely Prime Filter Induced by Meet Irreducible:
:$f \circ g = \operatorname {id}_{\map {\operatorname{pt}'} L}$
where $\operatorname {id}_{\map {\operatorname{pt}} L}$ is the identity mapping on $\map {\operatorname{pt}'} L$
Hence $f$ is a bijection by definition.
=== $f^\to \restriction_{\Sigma_L}$ is a Surjection ===
We have:
It follows that $f^\to \restriction_{\Sigma_L}$ is a surjection onto $\Sigma'_L$ by definition.
The result follows.
Category:Spectra of Locales
Canonical Bijection from Frame Homomorphisms to Continuous Maps

== Theorem ==
Let $\struct{L, \vee, \wedge, \preceq}$ be a locale.
Let $\map {\operatorname{Sp}} L = \struct{\map {\operatorname{pt}} L, \set{\Sigma_a : a \in L}}$ be the spectrum of $L$ as frame homomorphisms where:
:*$\quad\map {\operatorname{pt}} L$ denotes the set of points as frame homomorphisms of $L$.
:*$\quad$for each $a \in L$, $\Sigma_a = \set{\phi \in \map {\operatorname{pt}} L : \map \phi a = \top}$
Let $\Sigma_L = \set{\Sigma_a : a \in L}$
Let $\map {\operatorname{Sp}'} L = \struct{\map {\operatorname{pt}'} L, \set{\Sigma'_a : a \in L}}$ be the spectrum of $L$ as continuous maps where:
:*$\quad\map {\operatorname{pt}'} L$ denotes the set of points as continuous maps of $L$.
:*$\quad$for each $a \in L$, $\Sigma'_a = \set{g \in \map {\operatorname{pt}} L : \map {\loweradjoint g} a = \top}$ where:
::$\loweradjoint g : L \to 2$ is the frame homomorphism such that $\tuple{g,\loweradjoint g}$ is a Galois connection
Let $\Sigma'_L = \set{\Sigma'_a : a \in L}$
Let $f : \map {\operatorname{pt}} L \to \map {\operatorname{pt}'} L$ be the mapping defined by:
:$\forall \phi \in \map {\operatorname{pt}} L : \map f \phi = \upperadjoint \phi$
where $\tuple{\upperadjoint \phi, \phi}$ is a Galois connection
Then:
:$f$ is a bijection
:$f^\to \restriction_{\Sigma_L}$ is a surjection from $\Sigma_L$ to $\Sigma'_L$
== Proof ==
=== $f$ is a Bijection ===
From Frame Homomorphism is Lower Adjoint of Galois Connection and Galois Connection is Unique for Given Lower Adjoint:
:$\forall \phi \in \map {\operatorname{pt}} L: \exists !$ Galois connection $\tuple{\upperadjoint \phi, \phi}$
By definiton of continuous map:
:$\forall \phi \in \map {\operatorname{pt}} L: \exists !$ continuous map $\upperadjoint \phi : \tuple{\upperadjoint \phi, \phi}$ is a Galois connection
Hence $f$ is a well-defined mapping.
From Galois Connection is Unique for Given Lower Adjoint:
:$f$ is an injection
By definition of continuous map:
:$f$ is a surjection
Hence $f$ is a bijection by definition.
=== $f^\to \restriction_{\Sigma_L}$ is a Surjection ===
We have:
It follows that $f^\to \restriction_{\Sigma_L}$ is a surjection onto $\Sigma'_L$ by definition.
The result follows.
Category:Spectra of Locales
Canonical Form of Rational Number is Unique

== Theorem ==
The canonical form of a rational number is unique.
== Proof ==
Let $r \in \Q$ be a rational number.
Let $\dfrac p q$ and $\dfrac {p'} {q'}$ be two canonical forms of $r$.
, assume $q \le q'$.
By Equality of Rational Numbers:
:$p q' = p' q$
Therefore, $q'$ divides $p' q$.
As $\dfrac {p'} {q'}$ is a canonical form of $r$, by definition $q'$ is coprime to $p'$.
By Euclid's Lemma $q'$ divides $q$.
$q$ and $q'$ are both positive.
So from Absolute Value of Integer is not less than Divisors: Corollary:
:$q' \le q$
As $q' \le q$ and $q \le q'$ it follows that $q = q'$.
It then follows that $p = p'$.
Hence $\dfrac p q$ and $\dfrac {p'} {q'}$ are the same unique canonical form of $r$.
== Also see ==
* Existence of Canonical Form of Rational Number
== Sources ==
* : $\S 2.4$: The rational numbers and some finite fields: Exercise $1$
Category:Canonical Form of Rational Number
Canonical Form of Underdamped Oscillatory System

== Theorem ==
Consider a physical system $S$ whose behaviour can be described with the second order ODE in the form:
:$(1): \quad \dfrac {\d^2 x} {\d t^2} + 2 b \dfrac {\d x} {\d t} + a^2 x = 0$
for $a, b \in \R_{>0}$.
Let $b < a$, so as to make $S$ underdamped.
Then the value of $x$ can be expressed in the form:
:$x = \dfrac {x_0 a} \alpha e^{-b t} \map \cos {\alpha t - \theta}$
where:
:$\alpha = \sqrt {a^2 - b^2}$
:$\theta = \map \arctan {\dfrac b \alpha}$
This can be referred to as the canonical form of the solution of $(1)$.
== Proof ==
From Solution of Constant Coefficient Homogeneous LSOODE: Complex Roots of Auxiliary Equation, the general solution of $(1)$ is:
:$x = e^{-b t} \paren {C_1 \cos \alpha t + C_2 \sin \alpha t}$
where:
:$\alpha = \sqrt {a^2 - b^2}$
This is a homogeneous linear second order ODE with constant coefficients.
Let $m_1$ and $m_2$ be the roots of the auxiliary equation:
:$m^2 + 2 b + a^2 = 0$
From Solution to Quadratic Equation with Real Coefficients:
So from Solution of Constant Coefficient Homogeneous LSOODE: Complex Roots of Auxiliary Equation:
:$x = e^{-b t} \paren {C_1 \cos \alpha t + C_2 \sin \alpha t}$
where:
:$\alpha = \sqrt {a^2 - b^2}$
The following assumptions are made:
:We may label a particular point in time $t = 0$ at which the derivative of $x$ $t$ is itself zero.
:We allow that at this arbitrary $t = 0$, the value of $x$ is assigned the value $x = x_0$.
This corresponds, for example, with a physical system in which the moving body is pulled from its equilibrium position and released from stationary at time zero.
Differentiating $(1)$ $t$ gives:
:$\quad x' = -b e^{-b t} \paren {C_1 \cos \alpha t + C_2 \sin \alpha t} + e^{-b t} \paren {-\alpha C_1 \sin \alpha t + \alpha C_2 \cos \alpha t}$
Setting the initial condition $x = x_0$ when $t = 0$ in $(1)$:
Setting the initial condition $x' = 0$ when $t = 0$:
Hence:
From Multiple of Sine plus Multiple of Cosine: Cosine Form, $(2)$ can be expressed as:
== Also presented as ==
This can also be seen presented as:
:$x = \dfrac {x_0 \sqrt {\alpha^2 + b^2} } \alpha e^{-b t} \cos {\alpha t - \theta}$
== Sources ==
* : $\S 3.20$: Vibrations in Mechanical Systems: $(20)$
Category:Mechanics
Canonical Homomorphism to Polynomial Ring is Ring Monomorphism

== Theorem ==
Let $R$ be a commutative ring with unity.
Let $\struct {R \sqbrk X, \iota, X}$ be a polynomial ring over $R$ in one indeterminate $X$.
Then the canonical homomorphism $\iota : R \to R \sqbrk X$ is a ring monomorphism.
== Proof using explicit construction ==
== Proof using universal property ==
=== Outline of Proof ===
We apply the Universal Property of Polynomial Ring to construct a left inverse of $\iota$.
=== Proof ===
Let $\operatorname{id} : R \to R$ be the identity mapping.
Let $1$ be the unity of $R$.
By Identity Mapping is Ring Automorphism, $\operatorname{id}$ is a ring homomorphism.
By Universal Property of Polynomial Ring, there exists a ring homomorphism $h : R \sqbrk X \to R$ with $h \circ \iota = \operatorname{id}$.
By Identity Mapping is Injection, $\operatorname{id}$ is an injection.
By Injection if Composite is Injection, $\iota$ is an injection.
Thus $\iota$ is a ring monomorphism.
Category:Polynomial Theory
Canonical Injection from Ideal of External Direct Sum of Rings

== Theorem ==
Let $\struct {R_1, +_1, \circ_1}, \struct {R_2, +_2, \circ_2}, \dotsc, \struct {R_n, +_n, \circ_n}$ be rings.
Let $\ds \struct {R, +, \circ} = \prod_{k \mathop = 1}^n \struct {R_k, +_k, \circ_k}$ be their external direct product.
For each $k \in \closedint 1 n$, let:
:$R'_k = \set {\tuple {x_1, \dotsc, x_n} \in R: \forall j \ne k: x_j = 0}$
Let $\inj_k: R_k \to R$ be the canonical injection on the $k$th coordinate from $R_k$ into $\struct {R, +, \circ}$.
Let $\pr_k: R \to R'_k$ be the projection on the $k$th coordinate of $\struct {R, +, \circ}$ onto $R'_k$.
Then:
:$\inj_k: R_k \to R'_k$ is an isomorphism
:Its inverse is the restriction of $\pr_k$ to $R'_k$.
== Proof ==
From Ideal of External Direct Sum of Rings we have that $R'_k$ is an ideal of $R$, and thus a subring of $R$.
The result follows by application of Canonical Injection is Monomorphism.
== Also see ==
* Projection onto Ideal of External Direct Sum of Rings
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $22$. New Rings from Old: Theorem $22.8$
Category:Ideal Theory
Canonical Injection into Cartesian Product of Modules

== Theorem ==
Let $\struct {R, +_R, \times_R}$ be a ring.
Let $\struct {G, +, \circ}_R$ be the cartesian product of a sequence $\sequence {\struct {G_n, +_n, \circ_n}_R}$ of $R$-modules.
Then for each $j \in \closedint 1 n$, the canonical injection $\inj_j$ from $\struct {G_j, +_j, \circ_j}_R$ into $\struct {G, +, \circ}_R$ is a monomorphism.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations: Example $28.7$
Category:Module Theory
Category:Canonical Injections
Category:Canonical Injection into Cartesian Product of Modules
Canonical Injection into Cartesian Product of Modules/Proof 1

== Theorem ==
== Proof ==
To demonstrate that $\inj_j$ is an epimorphism, we need to show that:
:$(1): \quad \inj_j$ is an injection
:$(2): \quad \forall x, y \in G_j: \map {\inj_j} {x +_j y} = \map {\inj_j} x + \map {\inj_j} y$
:$(3): \quad \forall x_j \in G_j: \forall \lambda \in R: \map {\inj_j} {\lambda \circ_j x_j} = \lambda \circ \map {\inj_j} {x_j}$
Criteria $(1)$ and $(2)$ are a direct application of Canonical Injection is Monomorphism.
Let $x_j \in G_j$ be arbitrary.
Then we have:
Category:Canonical Injection into Cartesian Product of Modules
Canonical Injection into Cartesian Product of Modules/Proof 2

== Theorem ==
== Proof ==
$G$ can be seen as functions:
:$\ds f: A \to \bigcup_{a \mathop \in A} G_a$
Let $a \in A$.
Let $x, y \in G_a$.
Let $r \in R$.
So both $x + y \in G_a$ and $r x \in G_a$.
Let $b \in A$.
=== Case 1 ===
Let $b = a$.
Then:
:$\map {\map {\inj_a} {x + y} } b = x + y = \map {\map {\inj_a} x} b + \map {\map {\inj_a} y} b$
and:
:$\map {\map {\inj_a} {r x} } b = r x = r \, \map {\map {\inj_a} x} b$
=== Case 2 ===
Let $b \ne a$.
Then:
:$\map {\map {\inj_a} {x + y} } b = 0 + 0 = \map {\map {\inj_a} x} b + \map {\map {\inj_a} y} b$
and:
:$\map {\map {\inj_a} {r x} } b = 0 = \map r 0 = r \, \map {\map {\inj_a} x} b$
Therefore:
:$\map {\inj_a} {x + y} = \map {\inj_a} x + \map {\inj_a} y$
and:
:$\map {\inj_a} {r x} = r \, \map {\inj_a} x$
So $\inj_a$ is a homomorphism.
Combined with Canonical Injection is Injection gives that $\inj_a$ is a monomorphism.
Category:Canonical Injection into Cartesian Product of Modules
Canonical Injection into Metric Space Product with P-Product Metric is Continuous

== Theorem ==
Let $M_1 = \struct {A_1, d}$ and $M_2 = \struct {A_2, d'}$ be metric spaces.
Let $\AA := A_1 \times A_2$ be the cartesian product of $A_1$ and $A_2$.
Let $\MM = \struct {\AA, d_p}$ denote the metric space on $\AA$ where $d_p: \AA \to \R$ is one of the $p$-product metrics on $\AA$:
:$\map {d_p} {x, y} := \begin {cases}
\paren {\paren {\map d {x_1, y_1} }^p + \paren {\map {d'} {x_2, y_2} }^p}^{1/p} & : p \in \Z_{>0} \\
\ds \max_{i \mathop = 1}^n \set {\map d {x_1, y_1}, \map {d'} {x_2, y_2} } & : p = \infty
\end {cases}$
where:
:$x = \tuple {x_1, x_2}$
:$y = \tuple {y_1, y_2}$
Let $a \in A_1$ and $b \in A_2$ be fixed and arbitrary.
Let:
:$i_b: A_1 \to \AA$ be the mapping defined as:
::$\forall x \in A_1: \map {i_b} x = \tuple {x, b}$
:$i_a: A_2 \to \AA$ be the mapping defined as:
::$\forall y \in A_2: \map {i_a} y = \tuple {a, y}$
Then $i_b$ and $i_a$ are continuous in $M_1$ and $M_2$ respectively.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $2$: Continuity generalized: metric spaces: Exercise $2.6: 16$
Category:P-Product Metrics
Category:Canonical Injections
Category:Continuous Mappings on Metric Spaces
Category:Canonical Injection into Metric Space Product with P-Product Metric is Continuous
Canonical Injection into Metric Space Product with P-Product Metric is Continuous/Proof 1

== Theorem ==
== Proof ==
Let $\pr_1: \MM \to M_1$ and $\pr_2: \MM \to T_2$ be the first and second projections from $\MM$ onto its factors.
From Projection from Metric Space Product with P-Product Metric is Continuous, both $\pr_1$ and $\pr_2$ are continuous
The result follows from Continuous Mapping to Product Space.
Category:Canonical Injection into Metric Space Product with P-Product Metric is Continuous
Canonical Injection into Metric Space Product with P-Product Metric is Continuous/Proof 2

== Theorem ==
== Proof ==
We want to show that:
:$\forall c_1 \in A_1: \forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall x \in A_1: \map d {x, c} < \delta \implies \map {d_p} {\map {i_b} x, \map {i_b} c} < \epsilon$
and:
:$\forall c_2 \in A_2: \forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall x \in A_1: \map {d'} {x, c} < \delta \implies \map {d_p} {\map {i_a} c, \map {i_a} x} < \epsilon$
Let $c_1$ and $c_2$ in $A_1$ and $A_2$ respectively be arbitrary.
Let $\epsilon \in \R_{>0}$ also be arbitrary.
Let $\delta = \epsilon$.
Let $x \in A_1$ such that $\map d {x, c_1} < \delta$.
We have:
We have that $c_1$ and $\epsilon$ are arbitrary.
Hence, by definition, $i_b$ is continuous in $M_1$.
Let $y \in A_2$ such that $\map d {y, c_2} < \delta$.
We have:
We have that $c_2$ and $\epsilon$ are arbitrary.
Hence, by definition, $i_a$ is continuous in $M_2$.
Category:Canonical Injection into Metric Space Product with P-Product Metric is Continuous