nyuuzyou/wikis huggingface dataset:
Composition of Compatible Closure Operators

== Theorem ==
Let $S$ be a set.
Let $f, g: \powerset S \to \powerset S$ be closure operators on $S$.
Let $\CC_f$ and $\CC_g$ be the sets of closed sets of $S$ with respect to $f$ and $g$ respectively.
For each subset $T$ of $S$, let the following hold:
:$(1): \quad$ If $T$ is closed with respect to $g$, then $\map f T$ is closed with respect to $g$.
::That is, if $T \in \CC_g$ then $\map f T \in \CC_g$.
:$(2): \quad$ If $T$ is closed with respect to $f$, then $\map g T$ is closed with respect to $f$.
::That is, if $T \in \CC_f$ then $\map g T \in \CC_f$.
Let $\CC_h = \CC_f \cap \CC_g$.
Then:
:$\CC_h$ induces a closure operator $h$ on $S$
:$f \circ g = g \circ f = h$, where $\circ$ represents composition of mappings.
== Proof ==
First we show that $\CC_h$ induces a closure operator on $S$.
Let $\AA \subseteq \CC_h$.
By Intersection is Largest Subset:
:$\AA \subseteq \CC_f$
and:
:$\AA \subseteq \CC_g$
Thus by Intersection of Closed Sets is Closed/Closure Operator:
:$\ds \bigcap \AA \in \CC_f$
and
:$\ds \bigcap \AA \in \CC_g$
Thus by the definition of set intersection:
:$\ds \bigcap \AA \in \CC_h$
Thus by Closure Operator from Closed Sets, $C_h$ induces a closure operator $h$ on $S$.
Now we will show that $f \circ g = h$.
Having established that, it can be seen that $g \circ f = h$ will hold by reversing the variable names.
Let $T \subseteq S$.
By definition of closed set:
:$\map f {\map g T} \in \CC_f$
:$\map g T \in \CC_g$
By the premise:
:$\map f {\map g T} \in \CC_g$
Thus by definition of set intersection:
:$\map f {\map g T} \in \CC_f \cap \CC_g = C_h$
So $\map f {\map g T}$ is closed with respect to $h$.
By Set Closure is Smallest Closed Set/Closure Operator:
:$\map h T \subseteq f {\map g T}$
By definition of closed set:
:$\map h T \in C_h$
Thus by definition of set intersection:
:$\map h T \in C_f$
and
:$\map h T \in C_g$
By Set Closure is Smallest Closed Set/Closure Operator:
:$\map g T \subseteq \map h T$
By Closure Operator: Axiom $(2)$ $f$ is order-preserving:
:$\map f {\map g T} \subseteq \map f {\map h T}$
Recall that $\map h T \in C_f$.
By definition of closed set:
:$\map f {\map h T} = \map h T$
Thus:
:$\map f {\map g T} \subseteq \map h T$
We have that:
:$\map h T \subseteq \map f {\map g T}$
So by definition of set equality:
:$\map h T = \map f {\map g T}$
Category:Closure Operators
Composition of Computable Real-Valued Functions is Computable

== Theorem ==
Let $D \subseteq \R^n$ and $E \subseteq \R^m$ be subsets of real cartesian space.
Let $f : D \to \R$ be computable.
Let $g_1, g_2, \dotsc, g_n : E \to \R$ be computable.
Suppose that, for every $\bsx \in E$:
:$\tuple {\map {g_1} \bsx, \map {g_2} \bsx, \dotsc, \map {g_n} \bsx} \in D$
Then $h : E \to \R$ defined as:
:$\map h {x_1, \dotsc, x_m} = \map f {\map {g_1} {x_1, \dotsc, x_m}, \dotsc, \map {g_n} {x_1, \dotsc, x_m}}$
is computable.
== Proof ==
Follows immediately from:
* Composition of Sequentially Computable Real-Valued Functions is Sequentially Computable
* Composition of Computably Uniformly Continuous Real-Valued Functions is Computably Uniformly Continuous
Category:Computability Theory
Composition of Computable Real Functions is Computable

== Theorem ==
Let $f,g : \R \to \R$ be computable real functions.
Let $h : \R \to \R$ be defined as:
:$\map h x = \map f {\map g x}$
Then $h$ is computable.
== Proof ==
Follows immediately from:
* Composition of Sequentially Computable Real Functions is Sequentially Computable
* Composition of Computably Uniformly Continuous Real Functions is Computably Uniformly Continuous
Category:Computability Theory
Composition of Computably Uniformly Continuous Real-Valued Functions is Computably Uniformly Continuous

== Theorem ==
Let $D \subseteq \R^n$ and $E \subseteq \R^m$ be subsets of real cartesian space.
Let $f : D \to \R$ be computably uniformly continuous.
Let $g_1, g_2, \dotsc, g_n : E \to \R$ be computably uniformly continuous.
Suppose that, for every $\bsx \in E$:
:$\tuple {\map {g_1} \bsx, \map {g_2} \bsx, \dotsc, \map {g_n} \bsx} \in D$
Then $h : E \to \R$ defined as:
:$\map h {x_1, \dotsc, x_m} = \map f {\map {g_1} {x_1, \dotsc, x_m}, \dotsc, \map {g_n} {x_1, \dotsc, x_m}}$
is computably uniformly continuous.
== Proof ==
By definition of computably uniformly continuous, for each of:
:$f, g_1, \dotsc, g_n$
there exists a total recursive function $d_f : \N \to \N$ such that, for all $k \in \N$ and $\bsx, \bsy \in \Dom f$:
:$\norm {\bsx - \bsy} < \dfrac 1 {\map {d_f} k + 1} \implies \size {\map f \bsx - \map f \bsy} < \dfrac 1 {k + 1}$
Define $d_h : \N \to \N$ as:
:$\map {d_h} k = \map \max {\map {d_{g_1}} {n \map {d_f} k + \paren {n - 1}}, \dotsc, \map {d_{g_n}} {n \map {d_f} k + \paren {n - 1}}}$
By definition of general max operation, it is obtained by substitution using the binary max operation.
Thus, by:
* Maximum Function is Primitive Recursive
* Multiplication is Primitive Recursive
* Addition is Primitive Recursive
* Constant Function is Primitive Recursive
* Primitive Recursive Function is Total Recursive Function
we have that $d_h$ is a total recursive function.
Let $k \in \N$ and $\bsx, \bsy \in E$ satisfy:
:$\norm {\bsx - \bsy} < \dfrac 1 {\map {d_h} k + 1}$
By Max Operation Yields Supremum of Parameters/General Case, and the definition of supremum:
:$\map {d_h} k \ge \map {d_{g_i}} {n \map {d_f} k + \paren {n - 1}}$
for every $1 \le i \le n$.
Therefore:
:$\dfrac 1 {\map {d_h} k + 1} \le \dfrac 1 {\map {d_{g_i}} {n \map {d_f} k + \paren {n - 1}} + 1}$
Hence:
:$\norm {\bsx - \bsy} < \dfrac 1 {\map {d_{g_i}} {n \map {d_f} k + \paren {n - 1}} + 1}$
for every $1 \le i \le n$.
By assumption on $d_{g_i}$:
:$\size {\map {g_i} \bsx - \map {g_i} \bsy} < \dfrac 1 {n \map {d_f} k + \paren {n - 1} + 1} = \dfrac 1 {n \paren {\map {d_f} k + 1}}$
for every $1 \le i \le n$.
Then:
Additionally, by assumption:
:$\tuple {\map {g_1} \bsx, \dotsc, \map {g_n} \bsx} \in D$
:$\tuple {\map {g_1} \bsy, \dotsc, \map {g_n} \bsy} \in D$
Therefore, by assumption on $d_f$:
:$\size {\map f {\map {g_1} \bsx, \dotsc, \map {g_n} \bsx} - \map f {\map {g_1} \bsy, \dotsc, \map {g_n} \bsy}} < \dfrac 1 {k + 1}$
Thus, by definition of $h$:
:$\size {\map h \bsx - \map h \bsy} < \dfrac 1 {k + 1}$
As $k, \bsx, \bsy$ were arbitrary, $h$ is computably uniformly continuous by definition.
Category:Computability Theory
Composition of Computably Uniformly Continuous Real Functions is Computably Uniformly Continuous

== Theorem ==
Let $f,g : \R \to \R$ be computably uniformly continuous real functions.
Let $h : \R \to \R$ be defined as:
:$\map h x = \map f {\map g x}$
Then $h$ is computably uniformly continuous.
== Proof ==
By definition of computably uniformly continuous, there exist total recursive functions $d_f, d_g : \N \to \N$ such that:
:For all $n \in \N$ and $x,y \in \R$ such that:
::$\size {x - y} < \dfrac 1 {\map {d_f} n + 1}$
:it holds that:
::$\size {\map f x - \map f y} < \dfrac 1 {n + 1}$
and:
:For all $n \in \N$ and $x,y \in \R$ such that:
::$\size {x - y} < \dfrac 1 {\map {d_g} n + 1}$
:it holds that:
::$\size {\map g x - \map g y} < \dfrac 1 {n + 1}$
Define:
:$\map {d_h} n = \map {d_g} {\map {d_f} n}$
As $d_h$ is obtained by substitution on recursive functions, it is also recursive.
As both of those functions are total, it follows that $d_h$ is also total.
Now, let $n \in \N$ and $x,y \in \R$ satisfy:
:$\size {x - y} < \dfrac 1 {\map {d_h} n + 1}$
By definition of $d_h$:
:$\size {x - y} < \dfrac 1 {\map {d_g} {\map {d_f} n} + 1}$
By assumption on $d_g$:
:$\size {\map g x - \map g y} < \dfrac 1 {\map {d_f} n + 1}$
By assumption on $d_f$:
:$\size {\map f {\map g x} - \map f {\map g y}} < \dfrac 1 {n + 1}$
But:
:$\map f {\map g x} = \map h x$
:$\map f {\map g y} = \map h y$
Thus:
:$\size {\map h x - \map h y} < \dfrac 1 {n + 1}$
As $n$, $x$, and $y$ were arbitrary, $h$ is computably uniformly continuous by definition.
Category:Computability Theory
Composition of Continuous Linear Transformations is Continuous Linear Transformation

== Theorem ==
Let $K$ be a field.
Let $\struct {X, \norm {\, \cdot \,}_X}$, $\struct {Y, \norm {\, \cdot \,}_Y}$, $\struct{Z, \norm {\, \cdot \,}_Z}$ be normed vector spaces over $K$.
$\map {CL} {X, Y}$ be the continuous linear transformation space.
Let $\norm {\, \cdot \,}$ be the supremum operator norm.
Let $S \circ T : X \to Z$ be the composition of mappings such that:
:$\forall T \in \map {CL} {X, Y} : \forall S \in \map {CL} {Y, Z} : \forall x \in X : \map {S \circ T} x := \map S {\map T x}$
Then $S \circ T \in \map {CL} {X, Z}$.
== Proof ==
=== Linearity ===
Follows from Composition of Linear Transformations is Linear Transformation.
=== Continuity ===
We have that:
By Continuity of Linear Transformation between Normed Vector Spaces, $S \circ T$ is continuous.
== Sources ==
* : Chapter $\S 2.4$: Composition of continuous linear transformations
Category:Functional Analysis
Category:Mapping Theory
Category:Continuity
Category:Linear Transformations
Composition of Continuous Mapping on Compact Space Preserves Uniform Convergence

== Theorem ==
Let $\struct{ X, d_X }$ be a compact metric space.
Let $\struct{ Y, d_Y }, \struct{ Z, d_Z }$ be metric spaces.
Let $\phi : Y \to Z$ be a continuous mapping.
For all $n \in \N$, let $f_n : X \to Y$ be a continuous mapping.
Let $f: X \to Y$ be a mapping such that $\sequence {f_n}_{n \in \N}$ converges to $f$ uniformly on $X$.
Then the sequence $\sequence {\phi \circ f_n}_{n \in \N}$ converges to $\phi \circ f$ uniformly on $X$.
== Proof ==
 that $\sequence {\phi \circ f_n}_{n \in \N}$ do not converge to $\phi \circ f$ uniformly.
This implies that there exists $\epsilon \in \R_{>0}$ such that for all $n \in \N$, there exist $N_n \ge n$ and $x_n \in X$ such that:
:$\map {d_Z}{ \map { \phi \circ f_{N_n} }{ x_n } , \map { \phi \circ f }{ x_n} } \ge \epsilon$
From Compact Subspace of Metric Space is Sequentially Compact in Itself, it follows that $\struct{ X, d_X }$ is sequentially compact in itself.
This implies that there exist $x' \in X$ and a subsequence $\sequence {x_{n_m} }_{m \in \N}$ such that $x_{n_m} \to x'$ for $m \to \infty$.
We calculate:
which is a contradiction, as $\epsilon > 0$.
Category:Uniform Convergence
Category:Sequentially Compact Spaces
Composition of Dirac Delta Distribution with Function with Simple Zero

== Theorem ==
Let $\delta \in \map {\DD'} \R$ be the Dirac delta distribution.
Let $\sequence {\map {\delta_n} x}_{n \mathop \in \N}$ be a delta sequence.
Let $f : \R \to \R$ be a real function with a simple zero at $x_0$.
Let $f$ be strictly monotone.
Let $\phi \in \map \DD \R$ be a test function.
Then in the distributional sense it holds that:
:$\ds \map \delta {\map f x} = \frac {\map \delta {x - x_0}}{\size {\map {f'} {x_0}} }$
which can be interpreted as:
:$\ds \int_{-\infty}^\infty \map \delta {\map f x} \map \phi x \rd x = \frac {\map \phi {x_0}}{\size {\map {f'} {x_0}} }$
which more strictly means that:
:$\ds \lim_{n \mathop \to \infty} \int_{-\infty}^\infty \map {\delta_n} {\map f x} \map \phi x \rd x = \frac {\map \phi {x_0}}{\size {\map {f'} {x_0}} }$
=== Corollary ===
== Proof 1 ==
== Proof 2 ==
Category:Composition of Dirac Delta Distribution with Function with Simple Zero
Category:Dirac Delta Distribution
Category:Delta Sequence
Composition of Dirac Delta Distribution with Function with Simple Zero/Corollary

== Corollary to Composition of Dirac Delta Distribution with Function with Simple Zero ==
Dirac delta distribution is even: 
:$\ds \lim_{n \mathop \to \infty} \int_{-\infty}^\infty \map {\delta_n} {-x} \map \phi x \rd x = \lim_{n \mathop \to \infty} \int_{-\infty}^\infty \map {\delta_n} x \map \phi x \rd x$
which can be abbreviated to:
:$\map \delta {-x} = \map \delta x$
== Proof ==
Let $\map \phi x \in \map \DD \R$ be a test function.
Let $\sequence {\map {\delta_n} x}_{n \mathop \in \N}$ be a delta sequence.
By definition:
:$\ds \lim_{n \mathop \to \infty} \int_{-\infty}^\infty \map {\delta_n} x \map \phi x \rd x = \map \phi 0$
Consider a delta sequence $\sequence {\map {\delta_n} {-x} }_{n \mathop \in \N}$.
This is a composition of $\sequence {\map {\delta_n} x}_{n \mathop \in \N}$ with the function $\map f x = - x$ with a simple zero at $x_0 = 0$.
By Composition of Dirac Delta Distribution with Function with Simple Zero:
Abusing the notation, this could be understood as:
:$\ds \int_{-\infty}^\infty \map \delta {- x} \map \phi x \rd x = \int_{-\infty}^\infty \map \delta x \map \phi x \rd x$
or
:$\map \delta {- x} = \map \delta x$
== Sources ==
* : Chapter $2$. The Schwartz-Sobolev Theory of Distributions $2.5$ Algebraic Operations on Distributions
Category:Dirac Delta Distribution
Category:Composition of Dirac Delta Distribution with Function with Simple Zero
Composition of Dirac Delta Distribution with Function with Simple Zero/Proof 1

== Theorem ==
== Proof ==
Suppose $\map {f'} {x_0} > 0$.
Then:
Suppose $\map {f'} {x_0} < 0$.
Then:
Altogether:
== Sources ==
* : Chapter $3$. Additional Properties of Distributions $3.1$ Transformation Properties of the Delta Distribution
Category:Composition of Dirac Delta Distribution with Function with Simple Zero
Composition of Dirac Delta Distribution with Function with Simple Zero/Proof 2

== Theorem ==
== Proof ==
Let $H$ be the Heaviside step function.
Let $T \in \map \DD \R$ be a distribution associated with $\map H {\map f x}$:
:$\ds T = T_{\map H {\map f x}}$
We have that:
Taking the derivative of the yields:
Taking the derivative of the yields:
:$\forall x \ne x_0 : \dfrac \d {\d x} \map H {\map f x} = 0$
Furthermore:
:$\forall x \in \R : \map {f'} x > 0 : \paren {\map f {x_0^+} = 1} \land \paren {\map f {x_0^-} = 0}$
:$\forall x \in \R : \map {f'} x < 0 : \paren {\map f {x_0^+} = 0} \land \paren {\map f {x_0^-} = 1}$
By Jump Rule the reads:
Define the composite Dirac delta distribution according to Distributional Derivative of Heaviside Step Function.
We have that:
:$T_{\map H x}' = \delta_0 = \delta_x$
Then:
Hence:
and:
Category:Composition of Dirac Delta Distribution with Function with Simple Zero
Composition of Direct Image Mappings of Mappings

== Theorem ==
Let $A, B, C$ be non-empty sets.
Let $f: A \to B$ and $g: B \to C$ be mappings.
Let:
:$f^\to: \powerset A \to \powerset B$
and
:$g^\to: \powerset B \to \powerset C$
be the direct image mappings of $f$ and $g$.
Then:
:$\paren {g \circ f}^\to = g^\to \circ f^\to$
== Proof 1 ==
== Proof 2 ==
Category:Composite Mappings
Category:Direct Image Mappings
Category:Composition of Direct Image Mappings of Mappings
Composition of Direct Image Mappings of Mappings/Proof 1

== Theorem ==
== Proof ==
Let $S \subseteq A$ such that $S \ne \O$.
Then:
Now we treat the case where $S = \O$:
and the result is complete.
== Sources ==
* : $\S 5$. Induced mappings; composition; injections; surjections; bijections: Theorem $5.7$
Category:Composition of Direct Image Mappings of Mappings
Composition of Direct Image Mappings of Relations

== Theorem ==
Let $A, B, C$ be non-empty sets.
Let $\RR_1 \subseteq A \times B, \RR_2 \subseteq B \times C$ be relations.
Let:
:${\RR_1}^\to: \powerset A \to \powerset B$
and
:${\RR_2}^\to: \powerset B \to \powerset C$
be the direct image mappings of $\RR_1$ and $\RR_2$.
Then:
:$\paren {\RR_2 \circ \RR_1}^\to = {\RR_2}^\to \circ {\RR_1}^\to$
== Proof ==
Let $S \subseteq A, S \ne \O$.
Then:
Now we treat the case where $S = \O$:
Category:Composite Relations
Category:Composite Mappings
Category:Direct Image Mappings
Composition of Direct Sums of Bounded Linear Operators on Hilbert Space

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $\sequence {\family {\HH_i, \innerprod \cdot \cdot_i} }_{i \mathop \in I}$ be a $I$-indexed family of Hilbert spaces over $\GF$.
For each $i \in I$, let $T_i : \HH_i \to \HH_i$ and $S_i : \HH_i \to \HH_i$ be bounded linear operators.
Suppose that:
:$\ds \sup_{i \mathop \in I} \norm {T_i}_{\map B {\HH_i} } < \infty$
and:
:$\ds \sup_{i \mathop \in I} \norm {S_i}_{\map B {\HH_i} } < \infty$
where $\norm {\, \cdot \,}_{\map B {\HH_i} }$ is the norm of a bounded linear operator on $\HH_i$.
Let:
:$\ds \HH = \bigoplus_{i \mathop \in I} \HH_i$
be the Hilbert space direct sum of $\sequence {\family {\HH_i, \innerprod \cdot \cdot_i} }_{i \mathop \in I}$ with inner product $\innerprod \cdot \cdot$ and norm $\norm {\, \cdot \,}$.
Let:
:$\ds T = \bigoplus_{i \mathop \in I} T_i$
:$\ds S = \bigoplus_{i \mathop \in I} S_i$
:$\ds \bigoplus_{i \mathop \in I} T_i S_i$ 
be the direct sum of $\family {T_i}_{i \mathop \in I}$, $\family {S_i}_{i \mathop \in I}$ and $\family {T_i S_i}_{i \mathop \in I}$ respectively.
Then we have:
:$\ds T S = \bigoplus_{i \mathop \in I} T_i S_i$ 
That is:
:$\ds \paren {\bigoplus_{i \mathop \in I} T_i} \paren {\bigoplus_{i \mathop \in I} S_i} = \bigoplus_{i \mathop \in I} T_i S_i$
== Proof ==
From Norm on Bounded Linear Transformation is Submultiplicative we have:
Hence we have:
:$\ds \sup_{i \mathop \in I} \norm {T_i S_i}_{\map B {\HH_i} } \le \paren {\sup_{i \mathop \in I} \norm {T_i}_{\map B {\HH_i} } } \paren {\sup_{i \mathop \in I} \norm {S_i}_{\map B {\HH_i} } } < \infty$
Hence:
:$\ds \bigoplus_{i \mathop \in I} T_i S_i$ is well-defined.
Let $f \in \HH$ and $i \in I$.
Then we have:
:$\ds \map {\paren {\paren {\bigoplus_{i \mathop \in I} T_i} \paren {\bigoplus_{i \mathop \in I} S_i} f} } i = T_i \map {\paren {\paren {\bigoplus_{i \mathop \in I} S_i} f} } i = T_i S_i \map f i = \map {\paren {\paren {\bigoplus_{i \mathop \in I} T_i S_i} f} } i$
by the definition of the direct sum.
Hence we have:
:$\ds \paren {\bigoplus_{i \mathop \in I} T_i} \paren {\bigoplus_{i \mathop \in I} S_i} f = \paren {\bigoplus_{i \mathop \in I} T_i S_i} f$
for each $f \in \HH$.
Hence:
:$\ds \paren {\bigoplus_{i \mathop \in I} T_i} \paren {\bigoplus_{i \mathop \in I} S_i} = \bigoplus_{i \mathop \in I} T_i S_i$
Category:Hilbert Space Direct Sums
Composition of Distance-Preserving Mappings is Distance-Preserving

== 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 distance-preserving mappings.
Then the composite of $\phi$ and $\psi$ is also a distance-preserving mapping.
== Proof ==
Let $x,y \in X_1$.
Then:
By definition of a distance-preserving mapping, $\psi \circ \phi$ is distance-preserving.
Category:Metric Spaces
Composition of Elliptic Integral

== Theorem ==
An elliptic integral in $x$ can be reduced to the sum of:
:an elementary function
and constant multiples of integrals in three standard forms involving $x$ and:
:the (elliptic) modulus $k$
:the parameter $n$.
Those three standard forms are known as Legendre's standard elliptic integrals.
== Proof ==
== Also see ==
* Definition:Legendre's Standard Elliptic Integrals
== Sources ==
* 
* 
Category:Elliptic Integrals
Composition of Functors is Associative

== Theorem ==
Let $\mathbf A$, $\mathbf B$, $\mathbf C$ and $\mathbf D$ be metacategories.
Let $F: \mathbf A \to \mathbf B$, $G: \mathbf B \to \mathbf C$ and $H: \mathbf C \to \mathbf D$ be functors.
Then composition of functors is associative:
:$H \paren {G F} = \paren {H G} F$
== Proof ==
Let $A$ be an object of $\mathbf A$.
Then, solely by the definition of composite functor:
Then, , the same proof works for a morphism $f$ of $\mathbf A$ as well.
Hence the result.
Category:Category Theory
Composition of Galois Connections is Galois Connection

== Theorem ==
Let $L_1 = \struct {S_1, \preceq_1}$, $L_2 = \struct {S_2, \preceq_2}$ and $L_3 = \struct {S_3, \preceq_3}$ be ordered sets.
Let:
:$g_1: S_1 \to S_2, g_2: S_2 \to S_3, d_1: S_2 \to S_1, d_2: S_3 \to S_2$
be mappings such that:
:$\tuple {g_1, d_1}$ and $\tuple {g_2, d_2}$
are Galois connections.
Then $\tuple {g_2 \circ g_1, d_1 \circ d_2}$ is also a Galois connection.
== Proof ==
By definition of Galois connection:
:$g_1$, $g_2$, $d_2$, and $d_1$ are increasing mappings.
Thus by Composite of Increasing Mappings is Increasing:
:$g_2 \circ g_1$ and $d_1 \circ d_2$ are increasing mappings.
Let $s \in S_3, t \in S_1$.
We will prove that
:$s \preceq_3 \map {\paren {g_2 \circ g_1} } t \implies \map {\paren {d_1 \circ d_2} } s \preceq_1 t$
Assume that
:$s \preceq_3 \map {\paren {g_2 \circ g_1} } t$
By definition of composition of mappings:
:$s \preceq_3 \map {g_2} {\map {g_1} t}$
By definition of Galois connection:
:$\map {d_2} s \preceq_2 \map {g_1} t$
By definition of increasing mapping:
:$\map {d_1} {\map {d_2} s} \preceq_1 \map {d_1} {\map {g_1} t}$
By Galois Connection Implies Order on Mappings
:$d_1 \circ g_1 \preceq_1 I_{S_1}$
By definitions of ordering on mappings and composition of mappings:
:$\map {g_1} {\map {d_1} t} \preceq_1 \map {I_{S_1} } t$
By definition of identity mapping:
:$\map {g_1} {\map {d_1} t} \preceq_1 t$
Because an ordering is a transitive relation:
:$\map {d_1} {\map {d_2} s} \preceq_1 t$
Thus by definition of composition of mappings:
:$\map {\paren {d_1 \circ d_2} } s \preceq_1 t$
Assume that
:$\map {\paren {d_1 \circ d_2} } s \preceq_1 t$
By definition of composition of mappings:
:$\map {d_1} {\map {d_2} s} \preceq_1 t$
By definition of Galois connection:
:$\map {d_2} s \preceq_2 \map {g_1} t$
By definition of increasing mapping:
:$\map {g_2} {\map {d_2} s} \preceq_1 \map {g_2} {\map {g_1} t}$
By Galois Connection Implies Order on Mappings
:$I_{S_3} \preceq_3 g_2 \circ d_2$
By definitions of ordering on mappings and composition of mappings:
:$\map {I_{S_3} } s \preceq_3 \map {g_2} {\map {d_2} s}$
By definition of identity mapping:
:$s \preceq_3 \map {g_2} {\map {d_2} s}$
Because an ordering is a transitive relation:
:$s \preceq_3 \map {g_2} {\map {g_1} t}$
Thus by definition of composition of mappings:
:$s \preceq_3 \map {\paren {g_2 \circ g_1} } t$
== Sources ==
* 
Category:Galois Connections
Composition of Idempotent Mappings

== Theorem ==
Let $S$ be a set.
Let $f, g: S \to S$ be idempotent mappings.
Suppose that $f \circ g$ and $g \circ f$ have the same images.
That is, suppose that $f \sqbrk {g \sqbrk S} = g \sqbrk {f \sqbrk S}$.
Then $f \circ g$ and $g \circ f$ are idempotent.
== Proof ==
Let $x \in S$.
By the premise:
:$\map f {\map g x} \in g \sqbrk {f \sqbrk S}$
Since $f \sqbrk S \subseteq S$:
:$\map f {\map g x} \in g \sqbrk S$
Thus for some $y \in S$:
:$\map f {\map g x} = \map g y$
Since $g$ is idempotent:
:$\map g {\map g y} = \map g y$
By the choice of $y$:
:$\map g {\map f {\map g x} } = \map g {\map g y} = \map g y = \map f {\map g x}$
Thus:
:$\map f {\map g {\map f {\map g x} } } = \map f {\map f {\map g x} } = \map f {\map g x}$
That is, $f \circ g$ is idempotent.
Category:Idempotent Mappings
Category:Composite Mappings
Composition of Identification Mappings is Identification Mapping

== Theorem ==
Let $T_1 = \struct {S_1, \tau_1}$ be a topological space.
Let $S_2$ and $S_3$ be sets.
Let $f_1: S_1 \to S_2$ and $f_2: S_2 \to S_3$ be mappings.
Let $\tau_2$ be the identification topology on $S_2$ with respect to $f_1$ and $\tau_1$.
Let $\tau_3$ be the identification topology on $S_3$ with respect to $f_2$ and $\tau_2$.
Let $\phi: S_1 \to S_3$ be the composition of $f_2$ with $f_1$:
:$\phi = f_2 \circ f_1$
Then $\phi$ is itself an identification mapping. 
== Proof ==
Suppose $V \subseteq S_3$ is an arbitrary subset.
We have the following chain of equivalences:
Thus, $\phi$ is an identification mapping.
== Sources ==
* : $3$: Continuity generalized: topological spaces: Exercise $3.9: 41$
Category:Identification Topology
Category:Composite Mappings
Composition of Inflationary Mappings is Inflationary

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Let $f, g: S \to S$ be inflationary mappings.
Then $f \circ g$, the composition of $f$ and $g$, is also inflationary.
== Proof ==
Let $x \in S$.
Since this holds for all $x \in S$, $f \circ g$ is inflationary.
Category:Order Theory