nyuuzyou/wikis huggingface dataset:
Composition of Inflationary and Idempotent Mappings

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Let $f$ and $g$ be inflationary and idempotent mappings on $S$.
Then the following are equivalent:
:$(1): \quad f \circ g$ and $g \circ f$ are both idempotent
:$(2): \quad f$ and $g$ commute (that is, $f \circ g = g \circ f$)
:$(3): \quad \Img {f \circ g} = \Img {g \circ f}$
where:
:$\circ$ represents composition
:$\Img f$ represents the image of a mapping $f$.
== Proof ==
=== $(2)$ implies $(1)$ ===
Follows from Composition of Commuting Idempotent Mappings is Idempotent.
=== $(1)$ implies $(2)$ ===
Suppose that $f \circ g$ and $g \circ f$ are idempotent.
Then $\paren {f \circ g} \circ \paren {f \circ g} = f \circ g$.
By Composition of Mappings is Associative and the definition of composition, we have for each $x \in S$:
:$\map f {\map g {\map f {\map g x} } } = \map f {\map g x}$
Because $\preceq$ is an ordering and hence reflexive:
:$\map f {\map g {\map f {\map g x} } } \preceq \map f {\map g x}$
Since $f$ is inflationary:
:$\map g {\map f {\map g x} } \preceq \map f {\map g {\map f {\map g x} } }$
Thus since $\preceq$ is an ordering and hence transitive:
:$\map g {\map f {\map g x} } \preceq \map f {\map g x}$
Since $g$ is inflationary:
:$\map f {\map g x} \preceq \map g {\map f {\map g x} }$
Thus since $\preceq$ is an ordering and hence antisymmetric:
:$\map g {\map f {\map g x} } = \map f {\map g x}$
Since this holds for all $x \in S$, Equality of Mappings shows that:
:$g \circ f \circ g = f \circ g$
 
The same argument, with the roles of $f$ and $g$ reversed, shows that:
:$f \circ g \circ f = g \circ f$
Combining everything, we obtain:
:$f \circ g = f \circ \paren {g \circ f \circ g} = f \circ \paren {g \circ f} = g \circ f$
Thus $f \circ g = g \circ f$, so $f$ and $g$ commute.
=== $(2)$ implies $(3)$ ===
Equals substitute for equals.
=== $(3)$ implies $(1)$ ===
Follows from Composition of Idempotent Mappings.
Category:Mapping Theory
Composition of Inverse Image Mappings of Mappings

== Theorem ==
Let $A, B, C$ be non-empty sets.
Let $f: A \to B, g: B \to C$ be mappings.
Let:
:$f^\gets: \powerset B \to \powerset A$
and
:$g^\gets: \powerset C \to \powerset B$
be the inverse image mappings of $f$ and $g$.
Then:
:$\paren {g \circ f}^\gets = f^\gets \circ g^\gets$
== Proof ==
Let $T \subseteq C$.
We have:
It remains to be shown that:
:$\Img f \cap \map {g^\gets} T = \O \iff \Img {g \circ f} \cap T = \O$
So:
== Sources ==
* : $\S 5$. Induced mappings; composition; injections; surjections; bijections: Theorem $5.7$
Category:Composite Relations
Category:Composite Mappings
Category:Inverse Image Mappings
Composition of Isometries is Isometry

== Theorem ==
Let:
:$\struct {X_1, d_1}$
:$\struct {X_2, d_2}$
:$\struct {X_3, d_3}$
be metric spaces.
Let:
:$\phi: \struct {X_1, d_1} \to \struct {X_2, d_2}$
:$\psi: \struct {X_2, d_2} \to \struct {X_3, d_3}$
be isometries.
Then the composite of $\phi$ and $\psi$ is also an isometry.
== Proof ==
An isometry is a distance-preserving mapping which is also a bijection.
From Composition of Distance-Preserving Mappings is Distance-Preserving, $\psi \circ \phi$ is a distance-preserving mapping.
From Composite of Bijections is Bijection, $\psi \circ \phi$ is a bijection.
Category:Metric Spaces
Composition of Left Regular Representation with Right

== Theorem ==
Let $\struct {S, *}$ be a semigroup.
Let $\lambda_x, \rho_y$ be the left and right regular representations of $\struct {S, *}$ with respect to $x$ and $y$ respectively.
Let $\lambda_x \circ \rho_y$, $\rho_y \circ \lambda_x$ etc. be defined as the composition of the mappings $\lambda_x$ and $\rho_y$.
Then $\forall x, y \in S$:
: $\lambda_x \circ \rho_y = \rho_y \circ \lambda_x$
== Proof ==
Let $z \in S$.
== Sources ==
* : Chapter $6$: An Introduction to Groups: Exercise $2$
Category:Semigroups
Category:Regular Representations
Composition of Left Regular Representations

== Theorem ==
Let $\struct {S, *}$ be a semigroup.
Let $\lambda_x$ be the left regular representation of $\struct {S, *}$ with respect to $x$.
Let $\lambda_x \circ \lambda_y$ be defined as the composition of the mappings $\lambda_x$ and $\lambda_y$.
Then $\forall x, y \in S$:
:$\lambda_x \circ \lambda_y = \lambda_{x * y}$
== Proof ==
Let $z \in S$.
== Sources ==
* : Chapter $6$: An Introduction to Groups: Exercise $2$
Category:Semigroups
Category:Regular Representations
Composition of Linear Isometries is Linear Isometry

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $\struct {X, \norm \cdot_X}$, $\struct {Y, \norm \cdot_Y}$ and $\struct {Z, \norm \cdot_Z}$ be normed vector spaces over $\GF$.
Let $T : X \to Y$ and $S : Y \to Z$ be linear isometries.
Then $S T$ is a linear isometry.
== Proof ==
From Composition of Linear Transformations is Linear Transformation, $S T$ is a linear transformation. 
For $x \in X$, we have:
So $S T$ is a linear isometry.
Category:Linear Isometries
Composition of Linear Real Functions

== Theorem ==
Let $a, b, c, d \in \R$ be real numbers.
Let $\theta_{a, b}: \R \to \R$ be the real function defined as:
:$\forall x \in \R: \map {\theta_{a, b} } x = a x + b$
Let $\theta_{c, d} \circ \theta_{a, b}$ denote the composition of $\theta_{c, d}$ with $\theta_{a, b}$.
Then:
:$\theta_{c, d} \circ \theta_{a, b} = \theta_{a c, b c + d}$
== Proof ==
== Sources ==
* : Chapter $3$. Mappings: Exercise $5$
Category:Linear Algebra
Composition of Linear Transformations is Isomorphic to Matrix Product

== Theorem ==
Let $R$ be a ring with unity.
Let $F$, $G$ and $H$ be free $R$-modules of finite dimension $p, n, m > 0$ respectively.
Let $\sequence {a_p}$, $\sequence {b_n}$ and $\sequence {c_m}$ be ordered bases 
Let $\map {\LL_R} {G, H}$ denote the set of all linear transformations from $G$ to $H$.
Let $\map {\MM_R} {m, n}$ be the $m \times n$ matrix space over $R$.
Let $\sqbrk {u; \sequence {c_m}, \sequence {b_n} }$ be the matrix of $u$ relative to $\sequence {b_n}$ and $\sequence {c_m}$.
Let $M: \map {\LL_R} {G, H} \to \map {\MM_R} {m, n}$ be defined as:
:$\forall u \in \map {\LL_R} {G, H}: \map M u = \sqbrk {u; \sequence {c_m}, \sequence {b_n} }$
Then:
:$\forall u \in \map {\LL_R} {F, G}, v \in \map {\LL_R} {G, H}: \sqbrk {v \circ u; \sequence {c_m}, \sequence {a_p} } = \sqbrk {v; \sequence {c_m}, \sequence {b_n} } \sqbrk {u; \sequence {b_n}, \sequence {a_p} }$
== Proof ==
Follows directly from Relative Matrix of Composition of Linear Transformations.
== Motivation ==
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 29$. Matrices: Theorem $29.1$
Category:Linear Transformations
Category:Matrix Algebra
Composition of Linear Transformations is Linear Transformation

== Theorem ==
Let $K$ be a field.
Let $X, Y, Z$ be vector spaces over $K$.
Let $T_1 : X \to Y$ and $T_2 : Y \to Z$ be linear transformations.
Then the composition $T_2 \circ T_1 : X \to Z$ is a linear transformation.
== Proof ==
Let $\lambda \in K$ and $u, v \in X$. 
Then, we have: 
so $T_2 \circ T_1 : X \to Z$ is a linear transformation.
Category:Linear Transformations
Composition of Mapping and Inclusion is Restriction of Mapping

== Theorem ==
Let $S, T$ be sets.
Let $f: S \to T$ be a mapping.
Let $A \subseteq S$.
Then $f \circ i_A = f \restriction A$
where
:$i_A$ denotes the inclusion mapping of $A$,
:$f \restriction A$ denotes the restriction of $f$ to $A$.
== Proof ==
By definition of inclusion mapping:
:$i_A: A \to S$
By definitions of composition of mappings and restriction of mapping:
:$f \circ i_A: A \to T$ and $f \restriction A: A \to T$
Let $a \in A$.
Thus
== Sources ==
* 
Category:Restrictions
Composition of Mapping with Inclusion is Restriction

== Theorem ==
Let $S$ and $T$ be sets.
Let $f: S \to T$ be a mapping.
Let $A \subseteq S$ be a subset of the domain of $S$.
Let $i_A: A \to S$ be the inclusion mapping from $A$ to $S$.
Then:
:$f \circ i_A = f \restriction_A$
where $f \restriction_A$ denotes the restriction of $f$ to $A$.
== Proof ==
=== Equality of Domains ===
=== Equality of Codomains ===
=== Equality of Graph ===
Let $x \in A$.
All three criteria are seen to be fulfilled.
The result follows from Equality of Mappings.
== Sources ==
* : Notation and Terminology
Category:Composite Mappings
Category:Inclusion Mappings
Composition of Mapping with Mapping Restricted to Image

== Theorem ==
Let $A, B, C$ be sets.
Let $f : A \to B, g: B \to C$ be mappings.
Let $g \circ f : A \to C$ denote the composite mapping of $f$ with $g$.
Let $f \sqbrk A$ denote the image of $A$ under $f$.
Let $f \restriction_{A \times f \sqbrk A} : A \to f \sqbrk A$ denote the restriction of $f$ to $A \times f \sqbrk A$.
Let $g \restriction_{f \sqbrk A} : f \sqbrk A \to C$ denote the restriction of $g$ to $f \sqbrk A$.
Let $g \restriction_{f \sqbrk A} \circ f \restriction_{A \mathop \times f \sqbrk A} \mathop : A \to C$ denote the composite mapping of $f \restriction_{A \times f \sqbrk A}$ with $g \restriction_{f \sqbrk A}$.
Then:
:$g \restriction_{f \sqbrk A} \circ f \restriction_{A \mathop \times f \sqbrk A} \mathop = g \circ f$
== Proof ==
From Restriction of Mapping is Mapping:
:$f \restriction_{A \times f \sqbrk A} : A \to f \sqbrk A$ is a mapping
and
:$g \restriction_{f \sqbrk A} : f \sqbrk A \to C$ is a mapping
By definition of composite mapping:
:$g \restriction_{f \sqbrk A} \circ f \restriction_{A \mathop \times f \sqbrk A} \mathop : A \to C$ is a well-defined mapping
We have:
From Equality of Mappings:
:$g \restriction_{f \sqbrk A} \circ f \restriction_{A \mathop \times f \sqbrk A} \mathop = g \circ f$
Category:Mapping Theory
Category:Composite Mappings
Category:Restrictions
Composition of Mappings/Examples

== Examples of Compositions of Mappings ==
=== Compositions of $x^2$ with $2 x + 1$ ===
=== Compositions of $\sin x$ with $2 x + 1$ ===
=== Compositions of $x^2 + 1$ with $x + 1$ ===
=== Arbitrary Finite Sets ===
=== Arbitrary Finite Set with Itself ===
Category:Examples of Compositions of Mappings
Composition of Mappings/Examples/Arbitrary Finite Set with Itself

== Example of Compositions of Mappings ==
Let $X = Y = \set {a, b}$.
Consider the mappings from $X$ to $Y$:
The Cayley table illustrating the compositions of these $4$ mappings is as follows:
:$\begin{array}{c|cccc}
\circ & f_1 & f_2 & f_3 & f_4 \\
\hline
f_1 & f_1 & f_2 & f_3 & f_4 \\
f_2 & f_2 & f_2 & f_2 & f_2 \\
f_3 & f_3 & f_3 & f_3 & f_3 \\
f_4 & f_4 & f_3 & f_2 & f_1 \\
\end{array}$
We have that $f_1$ is the identity mapping and is also the identity element in the algebraic structure under discussion.
== Sources ==
* : Chapter $2$: Maps and relations on sets: Example $2.9$
Category:Examples of Compositions of Mappings
Category:Examples of Cayley Tables
Composition of Mappings/Examples/Arbitrary Finite Sets

== Example of Compositions of Mappings ==
Let:
Let $\theta: A \to B$ and $\phi: B \to C$ be defined in two-row notation as:
Then:
:$\phi \circ \theta = \dbinom {1 \ 2 \ 3} {w \ v \ w}$
== Sources ==
* : $\S 3.4$. Product of mappings: Example $49$
* : Appendix $\text{A}.4$: Functions: Theorem $\text{A}.4$
Category:Examples of Compositions of Mappings
Composition of Mappings/Examples/Compositions of sin x with 2x+1

== Example of Compositions of Mappings ==
Let $f: \R \to \R$ be the real function defined as:
:$\forall x \in \R: \map f x = \sin x$
Let $g: \R \to \R$ be the real function defined as:
:$\forall x \in \R: \map g x = 2 x + 1$
Then the compositions of $f$ with $g$ are:
$f \circ g: \R \to \R$:
:$\forall x \in \R: \map {\paren {f \circ g} } x = \map \sin {2 x + 1}$
$g \circ f: \R \to \R$:
:$\forall x \in \R: \map {\paren {g \circ f} } x = 2 \sin x + 1$
Note that:
:$\map {\paren {f \circ g} } 0 = \map \sin {2 \times 0 + 1} \approx 0 \cdotp 84$
:$\map {\paren {g \circ f} } 0 = 2 \times \sin 0 + 1 = 1$
demonstrating that composition of mappings is in general not commutative.
== Sources ==
* : Chapter $2$: Maps and relations on sets: Example $2.8$
Category:Examples of Compositions of Mappings
Composition of Mappings/Examples/Compositions of x^2+1 with x+1

== Example of Compositions of Mappings ==
Let $f: \R \to \R$ be the real function defined as:
:$\forall x \in \R: \map f x = x^2 + 1$
Let $g: \R \to \R$ be the real function defined as:
:$\forall x \in \R: \map g x = x + 1$
Then the compositions of $f$ with $g$ are:
== Sources ==
* 
* 
Category:Examples of Compositions of Mappings
Composition of Mappings/Examples/Compositions of x^2 with 2x+1

== Example of Compositions of Mappings ==
Let $f: \R \to \R$ be the real function defined as:
:$\forall x \in \R: \map f x = x^2$
Let $g: \R \to \R$ be the real function defined as:
:$\forall x \in \R: \map g x = 2 x + 1$
Then the compositions of $f$ with $g$ are:
$f \circ g: \R \to \R$:
:$\forall x \in \R: \map {\paren {f \circ g} } x = \paren {2 x + 1}^2$
$g \circ f: \R \to \R$:
:$\forall x \in \R: \map {\paren {g \circ f} } x = 2 x^2 + 1$
Note that:
:$\map {\paren {f \circ g} } 1 = \paren {2 \times 1 + 1}^2 = 9$
:$\map {\paren {g \circ f} } 1 = 2 \times 1^2 + 1 = 3$
demonstrating that composition of mappings is in general not commutative.
== Sources ==
* : $\S 24$: Composition of Mappings
Category:Examples of Compositions of Mappings
Composition of Mappings is Associative

== Theorem ==
The composition of mappings is an associative binary operation:
:$\paren {f_3 \circ f_2} \circ f_1 = f_3 \circ \paren {f_2 \circ f_1}$
where $f_1, f_2, f_3$ are arbitrary mappings which fulfil the conditions for the relevant compositions to be defined.
== Proof ==
$\quad\quad \begin{xy}\xymatrix@L + 2mu@ + 1em {
E \ar[ddd]^*{f} \ar[rrr]^*{g \circ f} \ar@{..>}[drdrdr]_*{h \circ g \circ f} & & & G \ar[ddd]^*{h} \\ \\ \\
F \ar[rrr]^*{h \circ g} \ar[rururu]_*{g} & & & H
}\end{xy}$
From the definition, we know that a mapping is a relation.
First, note that from the definition of composition of relations, the following must be the case before the above expression is even to be defined:
:$(1): \quad \Dom {f_2} = \Cdm {f_1}$
:$(2): \quad \Dom {f_3} = \Cdm {f_2}$
where $\Cdm f$ denotes the codomain of the mapping $f$.
The two composite relations can be seen to have the same domain, as follows:
Also they have the same codomain, as is seen by:
As a mapping is a relation, we can use that the Composition of Relations is Associative‎:
:$\forall x \in \Dom {f_1}: \map {\paren {f_3 \circ f_2} \circ f_1} x = \map {f_3 \circ \paren {f_2 \circ f_1} } x$
Hence the result.
== Sources ==
* : Chapter $\text{I}$: Semi-Groups and Groups: $1$: Definition and examples of semigroups
* : $\S 10$: Inverses and Composites
* : Exercise $1.4: \ 5$
* : $\S 3.4$. Product of mappings
* : $\S 4.2$. Commutative and associative operations: Example $65$
* : Chapter $\text I$: Algebraic Structures: $\S 5$: Composites and Inverses of Functions: Theorem $5.1$
* : $\S 0.5$: Theorem $7$
* : Chapter $\text{I}$: Sets and Functions: Composition of Functions: Theorem $6$
* : Appendix: Elementary set and number theory
* : Chapter $1$: Mappings: $\S 16$
* : $\S 1.14$: Composition of Functions: Theorem $14.4$
* : $\S 5$. Induced mappings; composition; injections; surjections; bijections: Theorem $5.2$
* : Chapter $1$: Theory of Sets: $\S 8$: Composition of Functions and Diagrams
* : $0$: Some Conventions and some Basic Facts
* : $\S 24.1$: Composition of Mappings
* : Chapter $1$: Sets and mappings: $\S 1.3$: Mappings
* : Chapter $2$: Maps and relations on sets: Proposition $2.10$
* : Appendix $\text{A}.4$: Composition and Restriction
* 
* 
* : $\S 1.2$
* 
Category:Composite Mappings
Category:Examples of Associative Operations‎
Composition of Mappings is Composition of Relations

== Theorem ==
Let $f_1: S_1 \to S_2$ and $f_2: S_2 \to S_3$ be mappings such that the domain of $f_2$ is the same as the codomain of $f_1$.
Let $f_2 \circ f_1$ be the composite of $f_1$ and $f_2$.
Let $f_1$ and $f_2$ be considered as relations on $S_1 \times S_2$ and $S_2 \times S_3$ respectively.
Then $f_2 \circ f_1$ defined as the composition of relations coincides with the definition of $f_2 \circ f_1$ as the composition of mappings.
== Proof ==
By definition of composition of mappings:
:$f_2 \circ f_1 = \set {\tuple {x, z} \in S_1 \times S_3: \exists y \in S_2: \tuple {x, y} \in f_1 \land \tuple {y, z} \in f_2}$
Let $\RR_1 \subseteq S_1 \times T_1$ and $\RR_2 \subseteq S_2 \times T_2$ be relations.
The composite of $\RR_1$ and $\RR_2$ is defined as:
:$\RR_2 \circ \RR_1 := \set {\tuple {x, z} \in S_1 \times T_2: \exists y \in S_2 \cap T_1: \tuple {x, y} \in \RR_1 \land \tuple {y, z} \in \RR_2}$
When $T_1 = S_2$ and $T_2 = S_3$, this becomes:
:$\RR_2 \circ \RR_1 := \set {\tuple {x, z} \in S_1 \times T_3: \exists y \in S_2: \tuple {x, y} \in \RR_1 \land \tuple {y, z} \in \RR_2}$
The definitions can be seen to be identical.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 5$: Composites and Inverses of Functions: Exercise $5.8 \ \text{(a)}$
Category:Composite Mappings
Category:Composite Relations
Composition of Mappings is Left Distributive over Homomorphism of Pointwise Operation

== Theorem ==
Let $A$ be a set.
Let $\struct {S, \odot}$ and $\struct {T, \otimes}$ be algebraic structures.
Let:
:$S^A$ denote the set of mappings from $A$ to $S$.
Let $f$ be a homomorphism from $S$ to $T$.
Let $g, h \in S^A$ be mappings from $A$ to $S$.
Then:
:$f \circ \paren {g \odot h} = \paren {f \circ g} \otimes \paren {f \circ h}$
where:
:$g \odot h$ denotes the pointwise operation on $S^A$ induced by $\odot$
:$f \circ h$ denotes the composition of $f$ with $h$.
== Proof ==
First we establish:
The domain of $g$ and $h$ is $A$.
The codomain of $g$ and $h$ is $S$.
The domain of $f$ is $S$.
The codomain of $f$ is $T$.
Hence:
:the Domain of $g \odot h$ is $A$
:the codomain of $g \odot h$ is $S$
:the domain of $f \circ \paren {g \odot h}$ is $A$
:the codomain of $f \circ \paren {g \odot h}$ is $T$.
Then:
:the domain of $f \circ g$ is $A$
:the domain of $f \circ h$ is $A$
:the codomain of $f \circ g$ is $S$
:the codomain of $f \circ h$ is $S$
:the domain of $\paren {f \circ g} \otimes \paren {f \circ h}$ is $A$
:the codomain of $\paren {f \circ g} \otimes \paren {f \circ h}$ is $T$.
Hence both $f \circ \paren {g \odot h}$ and $\paren {f \circ g} \otimes \paren {f \circ h}$ have the same domain and codomain.
Let $a \in A$ be arbitrary.
Let $b \in B$ be such that $b = \map g a$. 
Let $c \in B$ be such that $c = \map h a$.
Let $d \in T$ such that $d = b \odot c$.
We have:
Hence the result by Equality of Mappings.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Exercise $13.11 \ \text{(b)}$
Category:Pointwise Operations
Category:Composite Mappings
Category:Homomorphisms (Abstract Algebra)
Category:Examples of Distributive Operations
Composition of Mappings is Right Distributive over Pointwise Operation

== Theorem ==
Let $A$ and $B$ be sets.
Let $\struct {S, \odot}$ be an algebraic structure.
Let:
:$B^A$ denote the set of mappings from $A$ to $B$
:$S^B$ denote the set of mappings from $B$ to $S$.
Let $f, g \in S^B$ be mappings from $B$ to $S$.
Let $h \in B^A$ be a mapping from $A$ to $B$.
Then:
:$\paren {f \odot g} \circ h = \paren {f \circ h} \odot \paren {g \circ h}$
where:
:$f \odot g$ denotes the pointwise operation on $S^B$ induced by $\odot$
:$g \circ h$ denotes the composition of $g$ with $h$.
== Proof ==
First we establish:
The domain of $h$ is $A$.
The codomain of $h$ is $B$
The domain of both $f$ and $g$ is $B$.
The codomain of both $f$ and $g$ is $S$.
Hence:
:the domain of $\paren {f \odot g} \circ h$ is $A$
:the Domain of $f \odot g$ is $B$
:the codomain of $f \odot g$ is $S$
:the codomain of $\paren {f \odot g} \circ h$ is $S$.
Then:
:the domain of $f \circ h$ is $A$
:the domain of $g \circ h$ is $A$
:the codomain of $f \circ h$ is $S$
:the codomain of $g \circ h$ is $S$
:the domain of $\paren {f \circ h} \odot \paren {g \circ h}$ is $A$
:the codomain of $\paren {f \circ h} \odot \paren {g \circ h}$ is $S$.
Hence both $\paren {f \odot g} \circ h$ and $\paren {f \circ h} \odot \paren {g \circ h}$ have the same domain and codomain.
Let $a \in A$ be arbitrary.
Let $b \in B$ be such that $b = \map h a$. 
We have:
Hence the result by Equality of Mappings.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Exercise $13.11 \ \text{(a)}$
Category:Pointwise Operations
Category:Composite Mappings
Category:Examples of Distributive Operations
Composition of Mappings is not Commutative

== Theorem ==
The composition of mappings is not in general a commutative binary operation:
:$f_2 \circ f_1 \ne f_1 \circ f_2$
== Proof ==
;Proof by Counterexample:
Let $f_1: S_1 \to S_2$ and $f_2: S_2 \to S_3$ be mappings.
First note that unless $S_1 = S_3$ then $f_2 \circ f_1$ is not even defined.
So in that case $f_2 \circ f_1$ is definitely not the same thing as $f_1 \circ f_2$.
So, let us suppose $S_1 = S_3$ and so we define $f_1: S_1 \to S_2$ and $f_2: S_2 \to S_1$.
If $S_1 \ne S_2$ then:
:$f_2 \circ f_1: S_1 \to S_1$
:$f_1 \circ f_2: S_2 \to S_2$
and so by Equality of Mappings they are unequal because their domains and codomains are different.
Finally, suppose $S_1 = S_2$, and consider the following.
:$S_1 = S_2 = \set {a, b}$
:$f_1 = \set {\tuple {a, a}, \tuple {b, a} }$
:$f_2 = \set {\tuple {a, b}, \tuple {b, b} }$
It is straightforward to check that $f_1$ and $f_2$ are mappings, and that:
:$f_1 \circ f_2 = \set {\tuple {a, b}, \tuple {b, b} }$
:$f_2 \circ f_1 = \set {\tuple {a, a}, \tuple {b, a} }$
Thus, even in this limitingly simple case, we see that:
:$f_2 \circ f_1 \ne f_1 \circ f_2$
== Examples ==
== Sources ==
* : $\S 10$: Inverses and Composites
* : $\S 3.4$. Product of mappings: Example $50$
* : $\S 5$. Induced mappings; composition; injections; surjections; bijections: Remark $3$
* : $\S 7.16 \ (3)$
* : $\S 24$: Composition of Mappings
* : Chapter $1$: Sets and mappings: $\S 1.3$: Mappings
* 
* 
Category:Composite Mappings
Category:Composition of Mappings is not Commutative
Composition of Mappings is not Commutative/Examples/Arbitrary Example 1

== Example of Compositions of Mappings ==
Let $f: \R \to \R$ be the real function defined as:
:$\forall x \in \R: \map f x = x^2 + 1$
Let $g: \R \to \R$ be the real function defined as:
:$\forall x \in \R: \map g x = x + 1$
Then the compositions of $f$ with $g$ are:
and it is immediately seen that:
:$g \circ f \ne f \circ g$
== Sources ==
* 
* 
Category:Examples of Compositions of Mappings
Composition of Mappings is not Commutative/Examples/Sum of Squares not Square of Sum

== Examples of Use of Composition of Mappings is not Commutative ==
Let $f: \R \times \R \to \R$ be the real-valued function defined as:
:$\forall \tuple {x, y} \in \R \times \R: \map f {x, y} = x^2 + y^2$
Let $g: \R \times \R \to \R$ be the real-valued function defined as:
:$\forall \tuple {x, y} \in \R \times \R: \map g {x, y} = x + y$
Let $h: \R \to \R$ be the real function defined as:
:$\forall x \in \R: \map g x = x^2$
Then we have that:
:$\map h {\map g {x, y} } = \paren {x + y}^2$
while:
:$\map g {\map h x, \map h y} = x^2 + y^2 = \map f {x, y}$
Hence the diagram:
$\quad\quad \begin{xy} \xymatrix@L+2mu@+1em{
\R \times \R \ar[r]^*{g}
\ar@{-->}[rd]_*{f}
&
\R \ar[d]^*{h}
\\ &
\R
}\end{xy}$
is not a commutative diagram.
== Sources ==
* : Chapter $1$: Theory of Sets: $\S 8$: Composition of Functions and Diagrams: Exercise $2$
Category:Composition of Mappings is not Commutative
Composition of Measurable Mappings is Measurable

== Theorem ==
Let $\struct {X_1, \Sigma_1}$, $\struct {X_2, \Sigma_2}$ and $\struct {X_3, \Sigma_3}$ be measurable spaces.
Let $f: X_1 \to X_2$ be a $\Sigma_1 \, / \, \Sigma_2$-measurable mapping.
Let $g: X_2 \to X_3$ be a $\Sigma_2 \, / \, \Sigma_3$-measurable mapping.
Then their composition $g \circ f: X_1 \to X_3$ is $\Sigma_1 \, / \, \Sigma_3$-measurable.
== Proof ==
Let $E_3 \in \Sigma_3$.
Then $\map {g^{-1} } {E_3} \in \Sigma_2$, and $\map {f^{-1} } {\map {g^{-1} } {E_3} } \in \Sigma_1$ as $f, g$ are measurable.
That is, $\map {\paren {g \circ f}^{-1} } {E_3} \in \Sigma_1$ for all $E_3 \in \Sigma_3$.
Hence, $g \circ f$ is $\Sigma_1 \, / \, \Sigma_3$-measurable.
== Sources ==
* : $7.4$
Category:Measure Theory
Category:Composite Mappings