nyuuzyou/wikis huggingface dataset:
Composite of Surjections is Surjection

== Theorem ==
A composite of surjections is a surjection.
That is:
:If $g$ and $f$ are surjections, then so is $g \circ f$.
== Proof ==
Let $f: S_1 \to S_2$ and $g: S_2 \to S_3$ be surjections.
Then:
By definition of a composite mapping:
:$\map {g \circ f} x = \map g {\map f x} = \map g y = z$
Hence $g \circ f$ is surjective.
== Sources ==
* : $\S 3.6$. Products of bijective mappings. Permutations: $\text{(i)}$
* : Chapter $\text I$: Sets and Functions: Factoring Functions
* : $\S 5$. Induced mappings; composition; injections; surjections; bijections: Theorem $5.10 \ (2)$
* : Appendix $\text{A}.4$: Functions: Theorem $\text{A}.4$
* : Chapter $1$: Sets and mappings: $\S 1.3$: Mappings: Exercise $2$
* : Chapter $2$: Maps and relations on sets: Exercise $2 \ \text{(b)}$
* : $1$: Set Theory and Logic: $\S 2$: Functions
* : $1$: Set Theory and Logic: $\S 2$: Functions: Exercise $2.4 \ \text{(d)}$
* : Appendix $\text A$: Set Theory: Functions
Category:Surjections
Category:Composite Mappings
Composite of Symmetric Relations is not necessarily Symmetric

== Theorem ==
Let $A$ be a set.
Let $\RR$ and $\SS$ be symmetric relations on $A$.
Then their composite $\RR \circ \SS$ is not necessarily symmetric.
== Proof ==
Proof by Counterexample:
Let:
We note that both $\RR$ and $\SS$ are symmetric relations on $A$.
We have by definition of composition of relations that:
:$\RR \circ \SS = \set {\tuple {x, z} \in A \times A: \exists y \in A: \tuple {x, y} \in \SS \land \tuple {y, z} \in \RR}$
By inspection, we see that:
:$\RR \circ \SS = \set {\tuple {3, 1} }$
demonstrating that $\RR \circ \SS$ is not symmetric.
Category:Composite Relations
Category:Symmetric Relations
Composite of Three Mappings in Cycle forming Injections and Surjection

== Theorem ==
Let $A$, $B$ and $C$ be non-empty sets.
Let $f: A \to B$, $g: B \to C$ and $h: C \to A$ be mappings.
Let the following hold:
:$h \circ g \circ f$ is an injection
:$f \circ h \circ g$ is an injection
:$g \circ f \circ h$ is a surjection.
where:
:$g \circ f$ (and so on) denote composition of mappings.
Then each of $f$, $g$ and $h$ are bijections.
== Proof ==
First note that from Composition of Mappings is Associative:
:$\paren {h \circ g} \circ f = h \circ \paren {g \circ f}$
and so on.
However, while there is no need to use parenthesis to establish the order of composition of mappings, in the following the technique will be used in order to clarify what is being done.
We have that $\paren {h \circ g} \circ f$ and $\paren {f \circ h} \circ g$ are injections.
From Injection if Composite is Injection it follows that:
:$f$ is an injection
:$g$ is an injection.
Similarly, because $h \circ \paren {g \circ f}$ and $f \circ \paren {h \circ g}$ are injections:
:$g \circ f$ is an injection
:$h \circ g$ is an injection.
We have that $g \circ \paren {f \circ h}$ is a surjection.
From Surjection if Composite is Surjection it follows that $g$ is a surjection.
Similarly, because $\paren {g \circ f} \circ h$ is a surjection:
:$g \circ f$ is a surjection.
Thus we have deduced that:
:$g$ is a bijection
and:
:$g \circ f$ is a bijection.
From Mapping Composed with Bijection forming Bijection is Bijection it follows that:
:$f$ is a bijection.
Thus from Inverse of Bijection is Bijection we have that $f^{-1}: B \to A$ and $g^{-1}: C \to B$ are both bijections.
Similarly, $\paren {g \circ f}^{-1}: C \to A$ is a bijection, which from Inverse of Composite Bijection is the same as $f^{-1} \circ g^{-1}$.
== Sources ==
* : $\S 5$. Induced mappings; composition; injections; surjections; bijections: Exercise $16$
Category:Injections
Category:Surjections
Category:Bijections
Category:Composite Mappings
Composite of Total Relations is Total

== Theorem ==
Let $A$ be a set.
Let $\RR$ and $\SS$ be total relations on $A$.
Then their composite $\RR \circ \SS$ is also total.
== Proof ==
Recall the definition of composition of relations:
Hence in this particular context:
:$\RR \circ \SS := \set {\tuple {x, z} \in A \times A: \exists y \in A: \tuple {x, y} \in \SS \land \tuple {y, z} \in \RR}$
Let $x$ and $y$ in $A$ be arbitrary.
Because $\SS$ is total, either $x \mathrel \SS y$ or $y \mathrel \SS x$.
, suppose $x \mathrel \SS y$.
From Relation is Connected and Reflexive iff Total, $\RR$ is reflexive.
That is:
:$y \mathrel \RR y$
Thus we have:
:$\exists y \in A: \tuple {x, y} \in \SS \land \tuple {y, y} \in \RR$
and so by definition of composition of relations:
:$\tuple {x, y} \in \RR \circ \SS$
Similarly for $y \mathrel \SS x$:
:$\exists x \in A: \tuple {y, x} \in \SS \land \tuple {x, x} \in \RR$
giving:
:$\tuple {y, x} \in \RR \circ \SS$
So, by definition, $\RR \circ \SS$ is total.
Category:Composite Relations
Category:Total Relations
Composite of Transitive Relations is not necessarily Transitive

== Theorem ==
Let $A$ be a set.
Let $\RR$ and $\SS$ be transitive relations on $A$.
Then their composite $\RR \circ \SS$ is not necessarily also transitive.
== Proof ==
Recall the definition of composition of relations:
;Proof by Counterexample
:
Let $A = \set {a, b, c}$.
Let $\RR$ be defined as:
:$\RR = \set {\tuple {a, a}, \tuple {b, b}, \tuple {c, c}, \tuple {b, c} }$
Let $\SS$ be defined as:
:$\SS = \set {\tuple {a, a}, \tuple {b, b}, \tuple {c, c}, \tuple {a, b} }$
Both $\RR$ and $\SS$ can be seen to be transitive.
Then we have:
:$\RR \circ \SS = \set {\tuple {a, a}, \tuple {b, b}, \tuple {c, c}, \tuple {a, b}, \tuple {b, c} }$
We note that:
:$\tuple {a, b}, \tuple {b, c} \in \RR \circ \SS$
but:
:$\tuple {a, c} \notin \RR \circ \SS$
Hence, by definition, $\RR \circ \SS$ is not transitive.
Category:Composite Relations
Category:Transitive Relations
Composite with Constant Mapping is Constant Mapping

== Theorem ==
Let $f_c: S \to T$ be the constant mapping defined as:
:$\forall x \in S: \map {f_c} x = c$
where $c \in T$.
Then for all mappings $g: \Dom g \to S$:
:$f_c \circ g$ is a constant mapping
and for all mappings $h: T \to \Cdm h$:
:$h \circ f_c$ is a constant mapping
where:
:$\Dom g$ denotes the domain of $g$
:$\Cdm h$ denotes the codomain of $h$
:$\circ$ denotes composition of mappings.
== Proof ==
As $c$ is constant, so is $\map h c$.
Category:Constant Mappings
Category:Composite Mappings
Composition Series/Examples

== Examples of Composition Series ==
=== Cyclic Group $C_8$ ===
=== Quaternion Group $Q$ ===
=== Dihedral Group $D_4$ ===
=== Dihedral Group $D_6$ ===
=== Symmetric Group $S_2$ ===
=== Symmetric Group $S_3$ ===
=== Symmetric Group $S_4$ ===
=== Symmetric Group $S_n$ for $n > 2$ where $n \ne 4$ ===
Category:Examples of Composition Series
Composition Series/Examples/Cyclic Group C8

== Example of Composition Series ==
There is $1$ composition series of the cyclic group $C_8$, up to isomorphism:
:$\set e \lhd C_2 \lhd C_4 \lhd C_8$
== Proof ==
From Cyclic Group is Abelian and Subgroup of Abelian Group is Normal, all subgroups of $C_n$ are normal in $C_n$.
This leads directly to the composition series:
:$\set e \lhd C_2 \lhd C_4 \lhd C_8$
== Sources ==
* : Chapter $2$: Normal and Composition Series: $\S 74 \ \beta$
Category:Examples of Composition Series
Category:Cyclic Group of Order 8
Composition Series/Examples/Dihedral Group D4

== Example of Composition Series ==
There are $2$ composition series of the dihedral group $D_4$, up to isomorphism:
:$\set e \lhd C_2 \lhd C_4 \lhd D_4$
:$\set e \lhd C_2 \lhd K_4 \lhd D_4$
where:
:$C_n$ denotes the cyclic group of order $n$.
:$K_4$ denotes the Kline $4$-group.
== Proof ==
Let $D_4$ be defined as by Group Presentation of Dihedral Group:
:$D_4 = \gen {\alpha, \beta: \alpha^4 = \beta^2 = e, \beta \alpha \beta = \alpha^{−1} }$
From Subgroups of Dihedral Group $D_4$, $D_4$ has $2$ subgroups of index $2$:
:$\set {e, \alpha, \alpha^2, \alpha^3}$
:$\set {e, \alpha^2, \beta, \beta \alpha^2}$
We have that:
:$\set {e, \alpha, \alpha^2, \alpha^3} = \gen \alpha = C_4$
and:
:$\set {e, \alpha^2, \beta, \beta \alpha^2} = D_2 = K_4$
that is, the Kline $4$-group.
By Subgroup of Index 2 is Normal, both of these are normal.
Thus we have so far:
:$\set e \lhd \cdots \lhd C_4 \lhd D_4$
:$\set e \lhd \cdots \lhd K_4 \lhd D_4$
We have from Cyclic Group is Abelian and Subgroup of Abelian Group is Normal that all subgroups of $C_4$ are normal in $C_4$.
This leads directly to the composition series:
:$\set e \lhd C_2 \lhd C_4 \lhd D_4$
Similarly, the Kline $4$-group is abelian, and it has subgroups of order $2$, that is, $C_2$.
This leads to the composition series:
:$\set e \lhd C_2 \lhd K_4 \lhd D_4$
== Sources ==
* : Chapter $2$: Normal and Composition Series: $\S 74 \ \beta$
Category:Examples of Composition Series
Category:Dihedral Group D4
Composition Series/Examples/Dihedral Group D6

== Example of Composition Series ==
There are $3$ composition series of the dihedral group $D_6$, up to isomorphism:
:$\set e \lhd C_3 \lhd C_6 \lhd D_6$
:$\set e \lhd C_2 \lhd C_6 \lhd D_6$
:$\set e \lhd C_3 \lhd D_3 \lhd D_6$
where $C_n$ denotes the cyclic group of order $n$.
== Proof ==
Let $D_6$ be defined as by Group Presentation of Dihedral Group:
:$D_6 = \gen {\alpha, \beta: \alpha^6 = \beta^2 = e, \beta \alpha \beta = \alpha^{−1} }$
From Subgroups of Symmetry Group of Regular Hexagon, $D_6$ has $3$ subgroups of index $2$:
:$\set {e, \alpha, \alpha^2, \alpha^3, \alpha^4, \alpha^5}$
:$\set {e, \alpha^2, \alpha^4, \beta, \alpha^2 \beta, \alpha^4 \beta}$
:$\set {e, \alpha^2, \alpha^4, \alpha \beta, \alpha^3 \beta, \alpha^5 \beta}$
We have that:
:$\set {e, \alpha, \alpha^2, \alpha^3, \alpha^4, \alpha^5} = \gen \alpha = C_6$
and:
:$\set {e, \alpha^2, \alpha^4, \beta, \alpha^2 \beta, \alpha^4 \beta} = \gen {\alpha^2, \beta: \paren {\alpha^2}^3 = \beta^2 = e, \beta \alpha^2 \beta = \paren {\alpha^2}^{−1} } = D_3$
and:
:$\set {e, \alpha^2, \alpha^4, \alpha \beta, \alpha^3 \beta, \alpha^5 \beta} = \gen {\alpha^2, \alpha \beta: \paren {\alpha^2}^3 = \paren {\alpha \beta}^2 = e, \paren {\alpha \beta} \alpha^2 \paren {\alpha \beta} = \paren {\alpha^2}^{−1} } = D_3$
By Subgroup of Index 2 is Normal, all these are normal.
Thus we have so far:
:$\set e \lhd \cdots \lhd C_6 \lhd D_6$
:$\set e \lhd \cdots \lhd D_3 \lhd D_6$
We have from Cyclic Group is Abelian and Subgroup of Abelian Group is Normal that all subgroups of $C_6$ are normal in $C_6$.
This leads directly to the composition series:
:$\set e \lhd C_3 \lhd C_6 \lhd D_6$
:$\set e \lhd C_2 \lhd C_6 \lhd D_6$
We have that $D_3$ is isomorphic to $S_3$, the symmetric group on $3$ letters.
From Normal Subgroups of $S_3$, the only normal subgroup of $D_3$ is $C_3$.
This leads to the composition series:
:$\set e \lhd C_3 \lhd D_3 \lhd D_6$
As there are no more normal subgroups of $D_3$, there are no further composition series of $D_6$.
== Sources ==
* : Chapter $2$: Normal and Composition Series: $\S 74 \ \alpha$
Category:Examples of Composition Series
Category:Dihedral Group D6
Composition Series/Examples/Quaternion Group Q

== Example of Composition Series ==
There is $1$ composition series of the quaternion group $Q$, up to isomorphism:
:$\set e \lhd C_2 \lhd C_4 \lhd Q$
where:
:$C_n$ denotes the cyclic group of order $n$.
== Proof ==
Let $Q$ be defined as by Group Presentation of Quaternion Group:
:$Q = \gen {\alpha, \beta: \alpha^4 = e, \beta^2 = \alpha^2, \alpha \beta \alpha = \beta}$
From Subgroups of Quaternion Group, $Q$ has $3$ subgroups of index $2$:
Moreover from Subgroups of Quaternion Group, we see that all three of these subgroups are cyclic.
By Subgroup of Index 2 is Normal, these are normal.
Thus we have so far:
:$\set e \lhd \cdots \lhd C_4 \lhd Q$
We have from Cyclic Group is Abelian and Subgroup of Abelian Group is Normal that all subgroups of $C_4$ are normal in $C_4$.
This leads directly to the composition series:
:$\set e \lhd C_2 \lhd C_4 \lhd Q$
== Sources ==
* : Chapter $2$: Normal and Composition Series: $\S 74 \ \beta$
Category:Examples of Composition Series
Category:Quaternion Group
Composition Series/Examples/Symmetric Group S2

== Example of Composition Series ==
There is $1$ composition series of the symmetric group on $2$ letters $S_2$, up to isomorphism:
:$\set e = A_2 \lhd S_2$
where $A_2$ is the (degenerate) alternating group on $2$ letters.
Hence $S_2$ is (trivially) solvable.
== Proof ==
We have that $S_2$ is isomorphic to the parity group, which is the cyclic group $C_2$.
From Cyclic Group is Abelian and Subgroup of Abelian Group is Normal, all subgroups of $C_n$ are normal in $C_n$.
This leads directly to the composition series:
:$\set e \lhd C_2$
== Sources ==
* : Chapter $2$: The Symmetric Groups: $\S 84 \alpha$
Category:Examples of Composition Series
Category:Parity Group
Composition Series/Examples/Symmetric Group S3

== Example of Composition Series ==
There is $1$ composition series of the symmetric group on $3$ letters $S_3$, up to isomorphism:
:$\set e \lhd A_3 \lhd S_3$
where $A_3$ is the alternating group on $3$ letters.
Hence $S_3$ is (trivially) solvable.
== Sources ==
* : Chapter $2$: The Symmetric Groups: $\S 84 \alpha$
Category:Examples of Composition Series
Category:Symmetric Group on 3 Letters
Composition Series/Examples/Symmetric Group S4

== Example of Composition Series ==
There is $1$ composition series of the symmetric group on $4$ letters $S_4$, up to isomorphism:
:$\set e \lhd C_2 \lhd K_4 \lhd A_4 \lhd S_4$
where:
:$A_4$ is the alternating group on $4$ letters
:$K_4$ is the Klein four-group
:$C_n$ is the cyclic group of order $n$
Hence $S_4$ is solvable.
== Sources ==
* : Chapter $2$: The Symmetric Groups: $\S 84 \alpha$
Category:Examples of Composition Series
Category:Symmetric Group on 4 Letters
Composition Series/Examples/Symmetric Group Sn for n gt 2 where n ne 4

== Example of Composition Series ==
Let $n \in \Z$ such that $n > 2$ but $n \ne 4$.
There is $1$ composition series of the symmetric group on $n$ letters $S_n$, up to isomorphism:
:$\set e \lhd A_n \lhd S_n$
where $A_n$ is the alternating group on $n$ letters.
== Proof ==
First we note that from Alternating Group is Normal Subgroup of Symmetric Group:
:$A_n \lhd S_n$
By Quotient of Symmetric Group by Alternating Group is Parity Group:
:$\dfrac {S_n} {A_n}$ is a group of order $2$.
It follows that $A_n$ is the maximal normal subgroup of $S_n$.
$S_1$ is the trivial group whose composition series is simply:
:$\set e = S_1$
From Composition Series of Symmetric Group $S_2$:
:$\set e = A_2 \lhd S_2$
where $A_2$ is the (degenerate) alternating group on $2$ letters.
From Alternating Group is Simple except on 4 Letters, $A_n$ is a simple group for all $n \in \Z_{>0}$ except $n = 4$.
Indeed, we note that from Composition Series of Symmetric Group $S_4$:
:$\set e \lhd K_4 \lhd A_4 \lhd S_4$
where $K_4$ is the Klein four-group.
By definition of simple group, $A_n$ for $n \ne 4$ has only itself and the trivial group as normal subgroups.
Hence $\set e$ is the maximal normal subgroup of $A_n$ for $n \ne 4$.
Hence the result:
:$\set e \lhd A_n \lhd S_n$
== Sources ==
* 
Category:Examples of Composition Series
Category:Symmetric Groups
Composition Series of Group of Prime Power Order

== Theorem ==
Let $G$ be a group whose identity is $e$, and whose order is a prime power:
:$\order G = p^n, p \in \mathbb P, n \ge 1$
Then $G$ has a composition series:
:$\set e = G_0 \subset G_1 \subset \ldots \subset G_n = G$
such that:
:$\order {G_k} = p^k$
where:
:$G_k \lhd G_{k + 1}$
:$G_{k + 1} / G_k$ is cyclic and of order $p$.
== Proof ==
To be proved by induction on $n$.
Let $P_n$ be the proposition for $\order G = p^n$.
=== Basis for the Induction ===
$P_1$ is trivially true because:
:$\set e = G_0 \subset G_1 = G$
From Prime Group is Cyclic, a group whose order is prime is cyclic.
=== Induction Hypothesis ===
Suppose $P_k$ is true for all groups of order $p^k$ for all $k < n$.
We need to show that this implies $P_{k + 1}$ is true.
=== Induction Step ===
Let $G$ be a group of order $p^n$.
By Prime Power Group has Non-Trivial Proper Normal Subgroup, $G$ has a proper non-trivial normal subgroup.
There will be a finite number of these, so we are free to pick one of maximal order.
We call this $H$, such that $\order H = p^t, t < n$.
We need to show that $t = n - 1$.
Suppose $t < n - 1$.
Then $G / H$ is a group of order $p^{n - t} \ge p^2$.
Again by Prime Power Group has Non-Trivial Proper Normal Subgroup, $G / H$ has a proper non-trivial normal subgroup, which we will call $N$.
Let $H' = \set {g \in G: g H \in N}$.
We now show that $H \lhd G$.
Let $g, g' \in H'$.
Then $g H, g' H \in N$.
Since $N < G / H$:
:$\paren {g H} \paren {g' H} = g g' H \in N$
and so $g g' \in N$.
If $g \in H$, then $g H \in N$.
Since $N < G / H$:
:$\paren {g H}^{-1} = g^{-1} H \in N$
and so $g^{-1} \in H'$.
Next:
So clearly $H' / H = N$.
Therefore:
:$\dfrac {\order {H'} } {\order H} = \index {H'} N = \order N \ge p$
So:
:$\order {H'} \ge p \order H$
contradicting the maximality of $\order H$.
It follows that $t = n - 1$.
Finally, we set $G_{n - 1} = H$ and use induction to show that $P_{n - 1}$ holds.
We have that $G / H = G / G_{n - 1}$ is a group of order $p$
From Prime Group is Cyclic, it follows that $G / H$ is cyclic.
== Sources ==
* : Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 52$
Category:Cyclic Groups
Category:Composition Series
Category:P-Groups
Composition of 3 Mappings where Pairs of Mappings are Bijections

== Theorem ==
Let $A$, $B$, $C$ and $D$ be sets.
Let:
:$f: A \to B$
:$g: B \to C$
:$h: C \to D$
be mappings.
Let $g \circ f$ and $h \circ g$ be bijections.
Then $f$, $g$ and $h$ are all bijections.
== Proof ==
We note that both $g \circ f$ and $h \circ g$ are both injections and surjections by definition of bijection.
First it is shown that $g$ is a bijection.
We are given that:
:$g \circ f$ is a bijection.
From Injection if Composite is Injection it follows that $f$ is an injection.
From Surjection if Composite is Surjection it follows that $g$ is a surjection.
Similarly, we are given that:
:$h \circ g$ is a bijection.
From Injection if Composite is Injection it follows that $g$ is an injection.
From Surjection if Composite is Surjection it follows that $h$ is a surjection.
Thus we have that $g$ is a bijection.
Next it is shown that $f$ is a bijection.
 $f$ is not a surjection.
Then:
:$\exists y \in B: \neg \exists x \in A: \map f x = y$
But then $g$ is a surjection and so:
:$\forall z \in C: \exists y \in B: z = \map g y$
and so:
:$\exists z \in C: \neg \exists x \in A: \map g {\map f x} = z$
and so $g \circ f$ is not a surjection.
This contradicts our assertion that $g \circ f$ is a bijection and so a surjection.
Hence by Proof by Contradiction $f$ must be a surjection
So $f$ is both an injection and a surjection, and so a bijection.
Next it is shown that $h$ is a bijection.
 $h$ is not an injection.
Then:
:$\exists y_1, y_2 \in C: \map h {y_1} = \map h {y_2}$
But then $g$ is a bijection and so:
:$\exists x_1, x_2 \in B: y_1 = \map g {x_1} \ne \map g {x_2} = y_2$
and so:
:$\exists x_1, x_2 \in B: \map h {\map g {x_1} } = \map h {\map g {x_2} }$
and so $h \circ g$ is not an injection.
This contradicts our assertion that $h \circ g$ is a bijection and so an injection.
Hence by Proof by Contradiction $h$ must be an injection
So $h$ is both an injection and a surjection, and so a bijection.
Hence the result.
== Sources ==
* : $\S 5$. Induced mappings; composition; injections; surjections; bijections: Exercise $15$
Category:Bijections
Category:Composite Mappings
Composition of Addition Mappings on Natural Numbers

== Theorem ==
Let $a \in \N$ be a natural number.
Let $\alpha_a: \N \to \N$ be the mapping defined as:
:$\forall x \in \N: \map {\alpha_a} x = x + a$
Then:
:$\alpha_{a + b} = \alpha_b \circ \alpha_a$
== Proof ==
== Sources ==
* : Chapter $1$: Sets and mappings: $\S 1.3$: Mappings: Exercise $7$
Category:Composite Mappings
Category:Natural Number Addition
Composition of Affine Transformations is Affine Transformation

== Theorem ==
Let $\EE$, $\FF$ and $\GG$ be affine spaces with difference spaces $E$, $F$ and $G$ respectively.
Let $\LL: \EE \to \FF$ and $\MM: \FF \to \GG$ be affine transformations.
Let $L$ and $M$ be the tangent maps of $\LL$ and $\MM$ respectively.
Then the composition $\MM \circ \LL: \EE \to \FF$ is an affine transformation with tangent map $M \circ L$.
== Proof ==
Let $\NN = \MM \circ \LL : \EE \to \GG$ be the composition.
We want to show that for any $p, q \in \EE$
:$\map \GG Q = \map \GG p + \map {M \circ L} {\vec {p q} }$
We find that:
Now let:
:$p' = \map \LL p$
and:
:$q' = \map \LL p + \map L {\vec {p q} }$
so:
:$\vec {p' q'} = \map L {\vec {p q} }$
Then:
as required.
Category:Affine Geometry
Composition of Cartesian Products of Mappings

== Theorem ==
Let $I$ be an indexing set.
Let $\family {S_\alpha}_{\alpha \mathop \in I}$, $\family {T_\alpha}_{\alpha \mathop \in I}$ and $\family {U_\alpha}_{\alpha \mathop \in I}$ be families of sets all indexed by $I$.
For each $\alpha \in I$, let:
:$f_\alpha: S_\alpha \to T_\alpha$ be a mapping
:$g_\alpha: T_\alpha \to U_\alpha$ be a mapping.
Let:
:$\ds S = \prod_{\alpha \mathop \in I} S_\alpha$
:$\ds T = \prod_{\alpha \mathop \in I} T_\alpha$
:$\ds U = \prod_{\alpha \mathop \in I} U_\alpha$
Let $f: S \to T$ and $\ds g: T \to U$ be defined as:
:$\ds f = \prod_{\alpha \mathop \in I} f_\alpha$
:$\ds g = \prod_{\alpha \mathop \in I} g_\alpha$
Then their composition $g \circ f: S \to U$ is: 
:$\ds g \circ f: \prod_{\alpha \mathop \in I} g_\alpha \circ f_\alpha$
== Proof ==
First note that for all $\alpha \in I$:
:$\Dom {g_\alpha} = \Cdm {f_\alpha} = T_\alpha$
where $\Dom {g_\alpha}$ denotes the domain of $g_\alpha$ and $\Cdm {f_\alpha}$ denotes the codomain of $f_\alpha$.
So $g_\alpha \circ f_\alpha$ is defined for all $\alpha \in I$.
Similarly:
:$\Cdm f = \Dom g = T$
and so $g \circ f$ is defined.
Let $\mathbf x \in S$ be arbitrary:
:$\mathbf x = \family {x_\alpha \in S_\alpha}_{\alpha \mathop \in I}$
We have that:
== Sources ==
* : Chapter $1$: Theory of Sets: $\S 10$: Arbitrary Products: Exercise $2$
Category:Cartesian Product
Category:Composite Mappings
Composition of Commuting Idempotent Mappings is Idempotent

== Theorem ==
Let $S$ be a set.
Let $f, g: S \to S$ be idempotent mappings from $S$ to $S$.
Let:
:$f \circ g = g \circ f$
where $\circ$ denotes composition.
Then $f \circ g$ is idempotent.
== Proof 1 ==
== Proof 2 ==
Category:Idempotence
Category:Mapping Theory
Category:Composite Mappings
Category:Composition of Commuting Idempotent Mappings is Idempotent
Composition of Commuting Idempotent Mappings is Idempotent/Proof 2

== Theorem ==
== Proof ==
By Set of all Self-Maps under Composition forms Semigroup, the set of all self-maps on $S$ forms a semigroup under composition.
The result follows from Product of Commuting Idempotent Elements is Idempotent.
Category:Composition of Commuting Idempotent Mappings is Idempotent