nyuuzyou/wikis huggingface dataset:
Composition of Open Mappings is Open Mapping

== Theorem ==
Let $\struct {X, \tau_X}$, $\struct {Y, \tau_Y}$ and $\struct {Z, \tau_Z}$ be topological spaces.
Let $f : X \to Y$ and $g : Y \to Z$ be open mappings.
Then $g \circ f : X \to Z$ is an open mapping.
== Proof ==
Let $U$ be an open set in $\struct {X, \tau_X}$.
Since $f : X \to Y$ is open, we have $f \sqbrk U \in \tau_Y$.
Since $g : Y \to Z$ is open, we have $g \sqbrk {f \sqbrk U} \in \tau_Z$.
That is, whenever $U \in \tau_X$, we have $\paren {g \circ f} \sqbrk U \in \tau_Z$.
So $g \circ f : X \to Z$ is an open mapping.
Category:Open Mappings
Composition of Permutations is not Commutative

== Theorem ==
Let $S$ be a set.
Let $\map \Gamma S$ denote the set of permutations on $S$.
Let $\pi, \rho$ be elements of $\map \Gamma S$
Then it is not necessarily the case that:
:$\pi \circ \rho = \rho \circ \pi$
where $\circ$ denotes composition.
== Proof ==
Proof by Counterexample:
Let $S := \set {1, 2, 3}$.
Let:
Then:
while:
== Sources ==
* : $\S 4.2$. Commutative and associative operations: Example $62$
Category:Permutations
Category:Composite Mappings
Category:Commutativity
Composition of Product Mappings on Natural Numbers

== Theorem ==
Let $a \in \N$ be a natural number.
Let $\mu_a: \N \to \N$ be the mapping defined as:
:$\forall x \in \N: \map {\mu_a} x = x a$
Then:
:$\mu_{a b} = \mu_b \circ \mu_a$
== Proof ==
== Sources ==
* : Chapter $1$: Sets and mappings: $\S 1.3$: Mappings: Exercise $7$
Category:Composite Mappings
Category:Natural Number Multiplication
Composition of Regular Representations

== Theorem ==
Let $\struct {S, *}$ be a semigroup.
Let $\lambda_x, \rho_x$ be the left and right regular representations of $\struct {S, *}$ $x$.
Let $\lambda_x \circ \lambda_y$, $\rho_x \circ \rho_y$ etc. be defined as the composition of the mappings $\lambda_x$ and $\lambda_y$ etc.
Then $\forall x, y \in S$, the following results hold:
=== Composition of Left Regular Representations ===
=== Composition of Right Regular Representations ===
=== Composition of Left Regular Representation with Right ===
Category:Semigroups
Category:Regular Representations
Composition of Relation with Inverse is Symmetric

== Theorem ==
Let $\RR \subseteq S \times T$ be a relation.
Then the composition of $\RR$ with its inverse $\RR^{-1}$ is symmetric:
:$(1): \quad \RR^{-1} \circ \RR$ is a symmetric relation on $S$
:$(2): \quad \RR \circ \RR^{-1}$ is a symmetric relation on $T$.
== Proof ==
Note that this result holds for any $\RR \subseteq S \times T$, and does not require that $\struct {S, \RR}$ necessarily be a relational structure.
Thus $\tuple {a, b} \in \RR^{-1} \circ \RR \implies \tuple {b, a} \in \RR^{-1} \circ \RR$ and thus $\RR^{-1} \circ \RR$ is symmetric.
As $\RR = \paren {\RR^{-1} }^{-1}$ from Inverse of Inverse Relation, it follows that $\RR \circ \RR^{-1} = \paren {\RR^{-1} }^{-1} \circ \RR^{-1}$ is likewise a symmetric relation.
The domain of $\RR^{-1} \circ \RR$ is $S$ from Domain of Composite Relation, as is its codomain from Codomain of Composite Relation and the definition of Inverse Relation.
Similarly, the codomain of $\RR \circ \RR^{-1}$ is $T$, as is its domain.
This completes the proof.
Category:Symmetric Relations
Category:Composite Relations
Composition of Relations Preserves Subsets

== Theorem ==
Let $A, B, S, T$ be relations as subsets of Cartesian products.
Let $A \subseteq B$ and $S \subset T$.
Then:
:$A \circ S \subseteq B \circ T$
== Proof ==
We have:
By definition of subset:
:$A \circ S \subseteq B \circ T$
Category:Relations
Category:Cartesian Product
Category:Composite Relations
Composition of Relations is Associative

== Theorem ==
The composition of relations is an associative binary operation:
:$\paren {\RR_3 \circ \RR_2} \circ \RR_1 = \RR_3 \circ \paren {\RR_2 \circ \RR_1}$
== Proof ==
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:
:$\Dom {\RR_2} = \Cdm {\RR_1}$
:$\Dom {\RR_3} = \Cdm {\RR_2}$
The two composite relations can be seen to have the same domain, thus:
and also the same codomain, thus:
So they are equal they have the same value at each point in their common domain, which this shows:
== Sources ==
* : Chapter $0$: Relations: Theorem $5 \ \text{(b)}$
* : $\S 10$: Inverses and Composites
* : Chapter $\text I$: Algebraic Structures: $\S 5$: Composites and Inverses of Functions: Exercise $5.8 \ \text{(b)}$
* : Chapter $\text I$: Sets and Functions: Problem $\text{AA}$: Relations
Category:Composite Relations
Composition of Relations is not Commutative

== Theorem ==
Composition of relations is, in general, not commutative.
That is, it is usually the case that:
:$\RR_1 \circ \RR_2 \ne \RR_2 \circ \RR_1$
for relations $\RR_1$ and $\RR_2$.
== Proof ==
Proof by Counterexample:
Let $\RR_1 := \struct {S, S, R_1}$ and $\RR_2 := \struct {S, S, R_2}$ be relations defined as:
Let:
:$S = \set {0, 1, 2}$
:$R_1 = \set {\tuple {0, 1} }$
:$R_2 = \set {\tuple {1, 2} }$
We have that:
:$\RR_1 \circ \RR_2 = \struct {S, S, \set {\tuple {0, 2} } }$
while:
:$\RR_2 \circ \RR_1 = \struct {S, S, \O}$
== Sources ==
* : Chapter $0$: Relations
Category:Relation Theory
Composition of Repeated Compositions of Injections

== Theorem ==
Let $S$ be a set.
Let $f: S \to S$ be an injection.
Let the sequence of mappings:
:$f^0, f^1, f^2, \ldots, f^n, \ldots$
be defined as:
:$\forall n \in \N: f^n \left({x}\right) = \begin{cases}
x & : n = 0 \\
f \left({x}\right) & : n = 1 \\
f \left({f^{n-1} \left({x}\right)}\right) & : n > 1
\end{cases}$
Then:
:$\forall m, n \in \Z_{\ge 0}: f^n \circ f^m = f^{m + n}$
where $f^n \circ f^m$ denotes composition of mappings.
== Proof ==
Proof by induction:
Let $m \in \Z_{\ge 0}$ be given.
For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the proposition:
:$f^n \circ f^m = f^{m + n}$
$P \left({0}\right)$ is true, as this is the case:
=== Basis for the Induction ===
$P \left({1}\right)$ is true, as this is the case:
:$f^{m+1} \left({x}\right) = f \left({f^m \left({x}\right)}\right)$
by definition.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k+1}\right)$ is true.
So this is our induction hypothesis:
:$f^k \circ f^m = f^{m + k}$
Then we need to show:
:$f^{k+1} \circ f^m = f^{m + k + 1}$
=== Induction Step ===
This is our induction step:
So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall m, n \in \Z_{\ge 0}: f^n \circ f^m = f^{m + n}$
Category:Injections
Category:Composite Mappings
Composition of Right Inverse with Mapping is Idempotent

== Theorem ==
Let $f: S \to T$ be a mapping.
Let $g: T \to S$ be a right inverse mapping of $f$.
Then:
:$\paren {g \circ f} \circ \paren {g \circ f} = g \circ f$
== Proof ==
== Sources ==
* : Chapter $3$. Mappings: Exercise $6$
Category:Composite Mappings
Composition of Right Regular Representations

== Theorem ==
Let $\struct {S, *}$ be a semigroup.
Let $\rho_x$ be the right regular representation of $\struct {S, *}$ with respect to $x$.
Let $\rho_x \circ \rho_y$ be defined as the composition of the mappings $\rho_x$ and $\rho_y$.
Then $\forall x, y \in S$:
: $\rho_x \circ \rho_y = \rho_{y * x}$
== Proof ==
Let $z \in S$.
== Sources ==
* : Chapter $6$: An Introduction to Groups: Exercise $2$
Category:Semigroups
Category:Regular Representations
Composition of Ring Automorphisms is Ring Automorphism

== Theorem ==
Let $R$ be a set.
Let:
* $\left({R, +_1, \circ_1}\right)$
* $\left({R, +_2, \circ_2}\right)$
* $\left({R, +_3, \circ_3}\right)$
be rings.
Let:
* $\phi: \left({R, +_1, \circ_1}\right) \to \left({R, +_2, \circ_2}\right)$
* $\psi: \left({R, +_2, \circ_2}\right) \to \left({R, +_3, \circ_3}\right)$
be (ring) automorphisms.
Then the composite of $\phi$ and $\psi$ is also a (ring) automorphism.
== Proof ==
A ring automorphism is a ring isomorphism $f$ from a set to itself.
That is:
:$\operatorname{Dom} \left({\phi}\right) = \operatorname{Cdm} \left({\phi}\right)$
:$\operatorname{Dom} \left({\psi}\right) = \operatorname{Cdm} \left({\psi}\right)$
From Composition of Ring Isomorphisms is Ring Isomorphism, $\psi \circ \phi$ is a ring isomorphism.
By definition of composition of mappings:
:$\operatorname{Cdm} \left({\phi}\right) = \operatorname{Dom} \left({\psi}\right)$
Thus:
:$\operatorname{Dom} \left({\phi}\right) = \operatorname{Cdm} \left({\psi}\right) = R$
and so $\psi \circ \phi$ is a ring automorphism.
== Sources ==
* : $\S 2.2$: Homomorphisms
Category:Ring Automorphisms
Composition of Ring Endomorphisms is Ring Endomorphism

== Theorem ==
Let $R$ be a set.
Let:
: $\left({R, +_1, \circ_1}\right)$
: $\left({R, +_2, \circ_2}\right)$
: $\left({R, +_3, \circ_3}\right)$
be rings.
Let:
: $\phi: \left({R, +_1, \circ_1}\right) \to \left({R, +_2, \circ_2}\right)$
: $\psi: \left({R, +_2, \circ_2}\right) \to \left({R, +_3, \circ_3}\right)$
be (ring) endomorphisms.
Then the composite of $\phi$ and $\psi$ is also a (ring) endomorphism.
== Proof ==
A ring endomorphism is a ring homomorphism $f$ from a set to itself.
That is:
:$\operatorname{Dom} \left({\phi}\right) = \operatorname{Cdm} \left({\phi}\right)$
:$\operatorname{Dom} \left({\psi}\right) = \operatorname{Cdm} \left({\psi}\right)$
From Composition of Ring Homomorphisms is Ring Homomorphism, $\psi \circ \phi$ is a ring homomorphism.
By definition of composition of mappings:
:$\operatorname{Cdm} \left({\phi}\right) = \operatorname{Dom} \left({\psi}\right)$
Thus:
:$\operatorname{Dom} \left({\phi}\right) = \operatorname{Cdm} \left({\psi}\right) = R$
and so $\psi \circ \phi$ is a ring endomorphism.
== Sources ==
* : $\S 2.2$: Homomorphisms: Definition $2.4$
Category:Ring Endomorphisms
Composition of Ring Epimorphisms is Ring Epimorphism

== Theorem ==
Let:
:$\struct {R_1, +_1, \circ_1}$
:$\struct {R_2, +_2, \circ_2}$
:$\struct {R_3, +_3, \circ_3}$
be rings.
Let:
:$\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$
:$\psi: \struct {R_2, +_2, \circ_2} \to \struct {R_3, +_3, \circ_3}$
be (ring) epimorphisms.
Then the composite of $\phi$ and $\psi$ is also a (ring) epimorphism.
== Proof ==
A ring epimorphism is a ring homomorphism which is also a surection.
From Composition of Ring Homomorphisms is Ring Homomorphism, $\psi \circ \phi$ is a ring homomorphism.
From Composite of Surjections is Surjection, $\psi \circ \phi$ is a surection.
== Sources ==
* : $\S 2.2$: Homomorphisms: Definition $2.4$
Category:Ring Epimorphisms
Composition of Ring Homomorphisms is Ring Homomorphism

== Theorem ==
Let:
:$\struct {R_1, +_1, \odot_1}$
:$\struct {R_2, +_2, \odot_2}$
:$\struct {R_3, +_3, \odot_3}$
be rings.
Let:
:$\phi: \struct {R_1, +_1, \odot_1} \to \struct {R_2, +_2, \odot_2}$
:$\psi: \struct {R_2, +_2, \odot_2} \to \struct {R_3, +_3, \odot_3}$
be homomorphisms.
Then the composite of $\phi$ and $\psi$ is also a homomorphism.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Exercise $12.2$
* : $\S 2.2$: Homomorphisms: Definition $2.4$
Category:Ring Homomorphisms
Category:Composition of Ring Homomorphisms is Ring Homomorphism
Composition of Ring Homomorphisms is Ring Homomorphism/Proof 2

== Theorem ==
== Proof ==
Let $\psi \circ \phi$ denote the composite of $\phi$ and $\psi$.
Then what we are trying to prove is denoted:
:$\paren {\psi \circ \phi}: \struct {R_1, +_1, \odot_1} \to \struct {R_3, +_3, \odot_3}$ is a homomorphism.
To prove the above is the case, we need to demonstrate that the morphism property is held by $+_1$ and $\odot_1$ under $\psi \circ \phi$.
We take two elements $x, y \in R_1$, and put them through the following wringer with respect to $+_1$:
The same applies to $\odot_1$:
Disentangling the confusing and tortuous expressions above, we (eventually) see that this shows that the morphism property is indeed held by both $+_1$ and $\circ_1$ under $\psi \circ \phi$.
Thus $\paren {\psi \circ \phi}: \struct {R_1, +_1, \odot_1} \to \struct {R_3, +_3, \odot_3}$ is a homomorphism.
Category:Composition of Ring Homomorphisms is Ring Homomorphism
Composition of Ring Isomorphisms is Ring Isomorphism

== Theorem ==
Let:
* $\left({R_1, +_1, \circ_1}\right)$
* $\left({R_2, +_2, \circ_2}\right)$
* $\left({R_3, +_3, \circ_3}\right)$
be rings.
Let:
* $\phi: \left({R_1, +_1, \circ_1}\right) \to \left({R_2, +_2, \circ_2}\right)$
* $\psi: \left({R_2, +_2, \circ_2}\right) \to \left({R_3, +_3, \circ_3}\right)$
be (ring) isomorphisms.
Then the composite of $\phi$ and $\psi$ is also a (ring) isomorphism.
== Proof ==
A ring isomorphism is a ring homomorphism which is also a bijection.
From Composition of Ring Homomorphisms is Ring Homomorphism, $\psi \circ \phi$ is a ring homomorphism.
From Composite of Bijections is Bijection, $\psi \circ \phi$ is a bijection.
== Sources ==
* : $\S 2.2$: Homomorphisms: Definition $2.4$
Category:Ring Isomorphisms
Composition of Ring Monomorphisms is Ring Monomorphism

== Theorem ==
Let:
:$\struct {R_1, +_1, \circ_1}$
:$\struct {R_2, +_2, \circ_2}$
:$\struct {R_3, +_3, \circ_3}$
be rings.
Let:
:$\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$
:$\psi: \struct {R_2, +_2, \circ_2} \to \struct {R_3, +_3, \circ_3}$
be (ring) monomorphisms.
Then the composite of $\phi$ and $\psi$ is also a (ring) monomorphism.
== Proof ==
A ring monomorphism is a ring homomorphism which is also an injection.
From Composition of Ring Homomorphisms is Ring Homomorphism, $\psi \circ \phi$ is a ring homomorphism.
From Composite of Injections is Injection, $\psi \circ \phi$ is an injection.
== Sources ==
* : $\S 2.2$: Homomorphisms: Definition $2.4$
Category:Ring Monomorphisms
Composition of Sequence with Mapping

== Theorem ==
Let $\sequence {a_j}_{j \mathop \in B}$ be a sequence.
Let $\sigma: A \to B$ be a mapping, where $A \subseteq \N$.
Then $\sequence {a_j} \circ \sigma$ is a sequence whose value at each $k \in A$ is $a_{\map \sigma k}$.
Thus $\sequence {a_j} \circ \sigma$ is denoted $\sequence {a_{\map \sigma k} }_{k \mathop \in A}$.
== Proof ==
By definition, a sequence is a mapping whose domain is a subset of $\N$.
Let the range of $\sequence {a_j}_{j \mathop \in B}$ be $S$.
Thus $\sequence {a_j}_{j \mathop \in B}$ can be expressed using the mapping $f: B \to S$ as:
:$\forall j \in B: \map f j = a_j$
Let $k \in A$.
Then $\map \sigma k \in B$.
By definition of composition of mappings:
Category:Sequences
Composition of Sequentially Computable Real-Valued Functions is Sequentially Computable

== Theorem ==
Let $D \subseteq \R^n$ and $E \subseteq \R^m$ be subsets of real cartesian space.
Let $f : D \to \R$ be sequentially computable.
Let $g_1, g_2, \dotsc, g_n : E \to \R$ be sequentially 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 sequentially computable.
== Proof ==
Let $\tuple {\sequence {x_{1,k}}_k, \dotsc, \sequence {x_{m,k}}_k}$ be an $m$-tuple of computable real sequences in $E$.
By definition of sequentially computable:
:$\sequence {\map {g_i} {x_{1,k}, \dotsc, x_{m,k}}}_k$ is sequentially computable
for every $1 \le i \le n$
Additionally, for every $k$:
:$\tuple {\map {g_1} {x_{1,k}, \dotsc, x_{m,k}}, \dotsc, \map {g_n} {x_{1,k}, \dotsc, x_{m,k}}} \in D$
by assumption.
Therefore:
:$\tuple {\sequence {\map {g_1} {x_{1,k}, \dotsc, x_{m,k}}}_k, \dotsc, \sequence {\map {g_n} {x_{1,k}, \dotsc, x_{m,k}}}_k}$
is an $n$-tuple of computable real sequences in $D$.
Hence, by definition of sequentially computable:
:$\sequence {\map f {\map {g_1} {x_{1,k}, \dotsc, x_{m,k}}, \dotsc, \map {g_n} {x_{1,k}, \dotsc, x_{m,k}}}}_k$
is computable.
But, for every $k \in \N$:
:$\map f {\map {g_1} {x_{1,k}, \dotsc, x_{m,k}}, \dotsc, \map {g_n} {x_{1,k}, \dotsc, x_{m,k}}} = \map h {x_{1,k}, \dotsc, x_{m,k}}$
Therefore:
:$\sequence {\map h {x_{1,k}, \dotsc, x_{m,k}}}_k$
is computable.
As $\tuple {\sequence {x_{1,k}}_k, \dotsc, \sequence {x_{m,k}}_k}$ was arbitrary, it follows that $h$ is sequentially computable by definition.
Category:Computability Theory
Composition of Sequentially Computable Real Functions is Sequentially Computable

== Theorem ==
Let $f,g : \R \to \R$ be sequentially computable real functions.
Let $h : \R \to \R$ be defined as:
:$\map h x = \map f {\map g x}$
Then $h$ is sequentially computable.
== Proof ==
Let $\sequence {x_n}$ be a computable real sequence.
As $g$ is sequentially computable:
:$\sequence {\map g {x_n}}$
is computable.
As $f$ is sequentially computable:
:$\sequence {\map f {\map g {x_n}}}$
is computable.
But:
:$\map f {\map g {x_n}} = \map h {x_n}$
Therefore:
:$\sequence {\map h {x_n}}$
is computable.
As $\sequence {x_n}$ was arbitrary, $h$ is sequentially computable by definition.
Category:Computability Theory
Composition of Symmetric Relation with Itself is Union of Products of Images

== Theorem ==
Let $\RR$ be a symmetric relation on a set $S$.
Then:
:$\RR \circ \RR = \ds \map {\bigcup_{s \mathop \in S} } {\map \RR s \times \map \RR s}$
where
:$\RR \circ \RR$ is the composition of $\RR$ with itself
:$\map \RR s$ is the image of $s$ under $\RR$
:$\map \RR s \times \map \RR s$ is the Cartesian product of $\map \RR s$ with itself
== Proof ==
We have:
Category:Set Union
Category:Cartesian Product
Category:Symmetric Relations
Composition of Symmetries is Associative

== Theorem ==
Let $P$ be a geometric figure.
Let $S_P$ be the set of all symmetries of $P$.
Let $\circ$ denote composition of mappings.
Let $\phi, \psi, \chi$ be symmetries of $P$.
Then:
:$\paren {\phi \circ \psi} \circ \chi = \phi \circ \paren {\psi \circ \chi}$
That is, composition of symmetries is associative.
== Proof ==
From Composition of Symmetries is Symmetry:
:$\paren {\phi \circ \psi} \circ \chi$ is a symmetry
and:
:$\phi \circ \paren {\psi \circ \chi}$ is a symmetry.
It follows from Composition of Mappings is Associative that:
:$\paren {\phi \circ \psi} \circ \chi = \phi \circ \paren {\psi \circ \chi}$
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 2$: Compositions: Example $2.5$
Category:Symmetries (Geometry)
Composition of Symmetries is Symmetry

== Theorem ==
Let $P$ be a geometric figure.
Let $S_P$ be the set of all symmetries of $P$.
Let $\circ$ denote composition of mappings.
Let $\phi$ and $\psi$ be symmetries of $P$.
Then $\phi \circ \psi$ is also a symmetry of $P$.
== Proof ==
By definition of composition of mappings:
:$\phi \circ \psi$ is a mapping.
We have by definition of symmetry that:
:$\map \phi P$ is congruent to $P$
and:
:$\map \psi {\map \phi P}$ is congruent to $\map \phi P$
Therefore:
:$\phi \circ \psi$ is congruent to $P$
Thus $\phi \circ \psi$ is a symmetry of $P$.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 2$: Compositions: Example $2.5$
Category:Symmetries (Geometry)
Composition of Three Mappings which form Identity Mapping

== 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:
where:
:$g \circ f$ (and so on) denote composition of mappings
:$I_A$ (and so on) denote the identity mappings.
Then each of $f$, $g$ and $h$ are bijections, and:
where $f^{-1}$ (and so on) denote the inverse mappings.
== 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:
:$\paren {h \circ g} \circ f = I_A$
It follows from Injection iff Left Inverse that $f$ is an injection.
We also have that:
:$f \circ \paren {h \circ g} = I_B$
It follows from Surjection iff Right Inverse that $f$ is a surjection.
So $f$ is a bijection.
It follows from the corollary to Bijection iff Left and Right Inverse that $h \circ g$ is also a bijection.
Thus we have that $h \circ g$ is both a left inverse and a right inverse of $f$.
It follows by definition that $h \circ g$ is the inverse of $f$:
:$f^{-1} = h \circ g$
The same argument, mutatis mutandis, can be used to show that:
:$g$ and $h$ are bijections
:the inverse of $g$ is $f \circ h$
:the inverse of $h$ is $g \circ f$.
== Sources ==
* : Chapter $1$: Sets and mappings: $\S 1.3$: Mappings: Exercise $4$
Category:Bijections
Category:Composite Mappings
Composition with Zero Morphism is Zero Morphism

== Theorem ==
Let $\mathbf C$ be a category.
Let $0$ be a zero object in $\mathbf C$.
Let $A, B, C, D$ be objects in $\mathbf C$.
Let $f : A \to B$, $g : B \to C$ and $h : C \to D$ be morphisms in $\mathbf C$.
Let $g$ be a zero morphism.
Then $g \circ f$ and $h \circ g$ are zero morphisms.
== Proof ==
By definition of zero morphism, $g$ is the composition of the unique morphism $\alpha : B \to 0$ with the unique morphism $\beta : 0 \to C$.
Since $\alpha \circ f : A \to 0$ is a morphism with codomain $0$ and $0$ is a terminal object, $\alpha \circ f$ is the unique morphism $A \to 0$.
It follows that $g \circ f = \beta \circ \alpha \circ f$ is a zero morphism.
Since $h \circ \beta : 0 \to D$ is a morphism with domain $0$ and $0$ is an initial object, $h \circ \beta$ is the unique morphism $0 \to D$.
It follows that $h \circ g = h \circ \beta \circ \alpha$ is a zero morphism.
Category:Category Theory
Compositions of Closure Operators are both Closure Operators iff Operators Commute

== Theorem ==
Let $\left({S, \preceq}\right)$ be an ordered set.
Let $f$ and $g$ be closure operators on $S$.
Then the following are equivalent:
:$(1): \quad f \circ g$ and $g \circ f$ are both closure operators.
:$(2): \quad f$ and $g$ commute (that is, $f \circ g = g \circ f$).
:$(3): \quad \Img {f \circ g} = \Img {g \circ f}$
where $\Img \cdot$ denotes the image of a mapping. 
== Proof ==
By Composition of Inflationary Mappings is Inflationary:
:$f \circ g$ and $g \circ f$ are inflationary.
By Composite of Increasing Mappings is Increasing:
:$f \circ g$ and $g \circ f$ are increasing.
Thus each of the two composite mappings will be a closure operator it is idempotent.
Therefore the equivalences follow from Composition of Inflationary and Idempotent Mappings.
Category:Closure Operators
Compound Distribution of Poisson Distributed Bernoulli Trials has Poisson Distribution

== Theorem ==
Let $N$ be a discrete random variable with a Poisson distribution with expectation $\lambda$.
Let $X_1, X_2, \ldots, X_N$ be pairwise independent discrete random variables each with a Bernoulli distribution with parameter $P$.
Let $S_N : X_1 + X_2 + \cdots + X_N$ be the resulting compound distribution.
Then $S_N$ has a Poisson distribution with expectation $\lambda p$.
== Proof ==
== Sources ==
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Category:Compound Distributions
Category:Poisson Distribution
Category:Bernoulli Distribution
Compound Interest/Examples/Arbitrary Example 1

== Example of Compound Interest ==
Let $\pounds 1000$ be invested for $2$ years at $8 \%$ per annum.
Let interest be compounded half-yearly.
At the end of the $1$st $6$ months, the compound interest will be:
:$I_1 = \dfrac 8 {100} \times \dfrac 1 2 \times \pounds 1000 = \pounds 40$
At the end of the $2$nd $6$ months, the compound interest will be:
:$I_2 = \dfrac 8 {100} \times \dfrac 1 2 \times \pounds 1040 = \pounds 41 \cdotp 60$
At the end of the $3$rd $6$ months, the compound interest will be:
:$I_3 = \dfrac 8 {100} \times \dfrac 1 2 \times \pounds 1081 \cdotp 60 = \pounds 43 \cdotp 26$
At the end of the $4$th $6$ months, the compound interest will be:
:$I_4 = \dfrac 8 {100} \times \dfrac 1 2 \times \pounds 1124 \cdotp 86 = \pounds 44 \cdotp 99$
Hence the total interest earned will be $\pounds 169 \cdotp 85$.
== Proof ==
Using Formula for Compound Interest:
:$I = P \paren {\paren {1 + r}^n - 1}$
In this case:
:$P = 1000$
:$n = 4$
:$r = \dfrac {8 \%} 2 = 4 \% = 0.04$
Hence:
:$I = 1000 \paren {\paren {1 + 0 \cdotp 04}^4 - 1} \approx 1000 \times \paren {1 \cdotp 16985 - 1} = 169 \cdotp 85$
== Sources ==
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Category:Examples of Compound Interest