nyuuzyou/wikis huggingface dataset:
Epic Equalizer is Isomorphism

== Theorem ==
Let $\mathbf C$ be a metacategory.
Let $e: E \to C$ be the equalizer of two morphisms $f, g: C \to D$.
Let $e$ be an epimorphism.
Then $e$ is an isomorphism.
== Proof ==
As $e$ equalises $f$ and $g$, $f \circ e = g \circ e$.
Since $e$ is epic, it follows that $f = g$.
Then in the equaliser diagram: 
:::$\begin{xy}\xymatrix{
 E 
 \ar[r]^*{e}
&
 A
 \ar[r]<2pt>^*{f}
 \ar[r]<-2pt>_*{g} 
&
 B
\\
 A
 \ar@{.>}[u]^*{k}
 \ar[ur]_*{\operatorname{id}_A}
}\end{xy}$
We have that:
:$f \circ 1_A = g \circ 1_A$
so there is a unique $k$ with $e \circ k = \operatorname{id}_A$.
Then:
By Equalizer is Monomorphism, it follows that $k \circ e = \operatorname{id}_E$.
This gives $k$ as an inverse to $e$.
Thus $e$ is an isomorphism.
== Sources ==
* : $\S 3.10$: Theorem $2$
Category:Category Theory
Epicycloid whose Ratio of Generating Circle Radii is Rational is Closed Curve

== Theorem ==
Consider the epicycloid $E$ generated by a epicycle $C_1$ of radius $b$ rolling within a deferent $C_2$ of radius $a$.
Let $k = \dfrac a b$ be a rational number.
Then $E$ is a closed curve.
== Proof ==
== Sources ==
* 
* 
Category:Epicycloids
Epimenides Paradox

== Paradox ==
 famously stated:
:All Cretans are liars.
As himself was a Cretan, then naturally he would have been telling a lie when he said it.
That would mean all Cretans are not liars, which would mean (as was a Cretan) he would have been telling the truth.
This would then mean was a liar.
Hence this statement contradicts itself.
== Resolution ==
Just because someone has been categorised as a "liar", does not necessarily mean that everything they say is a lie.
In a similar vein, just because someone is categorised as a "truth teller", does not necessarily mean that nothing they say is a lie.
Hence this is a false dichotomy.
Besides, the negation of All Cretans are liars is not All Cretans are not liars, but Not all Cretans are liars.
This is perfectly consistent with himself as being one of the Cretans who is a liar.
Thus the Epimenides Paradox can be classified as a falsidical paradox, as it is only faulty reasoning that renders this statement paradoxical.
== Also see ==
* Second Socratic Paradox
== Historical Note ==
Category:Falsidical Paradoxes
Epimenides Paradox/Historical Note

== Historical Note on Epimenides Paradox ==
 is quoted as having said:
:The Cretans, always liars, evil beasts, idle bellies!
This may have been in response to a Cretan stating that Zeus is mortal, which offended 's religious sensibilities.
Hence this statement is not an intrinsic paradox, but rather an opinion of the Cretans from .
In fact, Epimenides Paradox has been considered paradoxical only from the $19$th century, when the rules of logic had only recently begun to be explored.
Category:Historical Notes
Epimorphism Preserves Commutativity

== Theorem ==
Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an epimorphism.
Let $\circ$ be a commutative operation.
Then $*$ is also a commutative operation.
== Proof ==
Let $\struct {S, \circ}$ be an algebraic structure in which $\circ$ is commutative.
Let $\struct {S, \circ}$ be an algebraic structure in which $\circ$ is associative.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an epimorphism.
Suppose $S$ is the empty set.
It follows from the definition of an epimorphism that:
$\phi$ is a surjective homomorphism
By Empty Mapping to Empty Set is Bijective, the empty map is bijective
By definition of bijection, the empty map is an epimorphism.
Therefore, suppose $\phi$ is the empty map, which is indeed an epimorphism.
By definition of a homomorphism, $\phi$ can be defined as:
:$\forall \O \in S: \map \phi {\O \circ \O} = \map \phi \O * \map \phi \O$
By Image of Empty Set is Empty Set, $T$ is also the empty set.
It follows from the definition of the homomorphism that the binary operations $\circ$ and $*$ are also the empty map.
Hence, it is vacuously true that $\circ$ is commutative on $S$, when $S$ is empty, as required.
Let $\struct {S, \circ}$ be an algebraic structure in which $\circ$ is commutative.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an epimorphism.
Suppose $S$ is non-empty.
As an epimorphism is surjective, it follows that:
:$\forall u, v \in T: \exists x, y \in S: \map \phi x = u, \map \phi y = v$
So:
== Warning ==
== Also see ==
* Epimorphism Preserves Associativity
* Epimorphism Preserves Identity
* Epimorphism Preserves Inverses
* Epimorphism Preserves Semigroups
* Epimorphism Preserves Groups
* Epimorphism Preserves Distributivity
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Theorem $12.2$
* : Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Exercise $12.1$
Category:Epimorphisms (Abstract Algebra)
Category:Commutativity
Epimorphism Preserves Distributivity

== Theorem ==
Let $\struct {R_1, +_1, \circ_1}$ and $\struct {R_2, +_2, \circ_2}$ be algebraic structures.
Let $\phi: R_1 \to R_2$ be an epimorphism.
:If $\circ_1$ is left distributive over $+_1$, then $\circ_2$ is left distributive over $+_2$.
:If $\circ_1$ is right distributive over $+_1$, then $\circ_2$ is right distributive over $+_2$.
Consequently, if $\circ_1$ is distributive over $+_1$, then $\circ_2$ is distributive over $+_2$.
That is, epimorphism preserves distributivity.
== Proof ==
Throughout the following, we assume the morphism property holds for $\phi$ for both operations.
It remains to be shown that for non-empty $R_1$, $\paren {x +_1 y}, \paren {y +_1 z}, \paren {x \circ_1 y }, \paren {y \circ_1 z}, \paren {x \circ_1 z}, \paren { x \circ_1 \paren {y +_1 z} }, \paren { \paren {x +_1 y} \circ_1 z } \in \Dom \phi$.
=== Left Distributivity ===
Suppose $R_1$ is the empty set.
It follows from the definition of an epimorphism that
$\phi$ is a surjective homomorphism
By Empty Mapping to Empty Set is Bijective, the empty map is bijective
By definition of bijection, the empty map is an epimorphism.
Therefore, suppose $R_1$ is the empty map, which is indeed an epimorphism.
By Image of Empty Set is Empty Set, $R_2$ is also the empty set.
It follows from the definition of the homomorphism that the binary operations $+_1$, $+_2$, $\circ_1$ and $\circ_2$ are also the empty map.
If $\circ_1$ is left distributive over $+_1$, then it is vacuously true that $\circ_2$ is left distributive over $+_2$, as required.
Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be an epimorphism.
Suppose $R_1$ is non-empty.
Then: 
So $\circ_2$ is left distributive over $+_2$.
=== Right Distributivity ===
Suppose $\circ_1$ is right distributive over $+_1$. 
Suppose $R_1$ is the empty set.
By the similar reasoning shown for left distributivity, it follows that if $\circ_1$ is left distributive over $+_1$, then it is vacuously true that $\circ_2$ is left distributive over $+_2$, as required.
Suppose $R_1$ is non-empty.
Then:
So $\circ_2$ is right distributive over $+_2$.
=== Distributivity ===
If $\circ_1$ is distributive over $+_1$, then it is both right and left distributive over $+_1$.
Hence from the above, $\circ_2$ is both right and left distributive over $+_2$.
That is, $\circ_2$ is distributive over $+_2$.
== Warning ==
Category:Epimorphisms (Abstract Algebra)
Category:Distributive Operations
Epimorphism Preserves Groups

== Theorem ==
Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an epimorphism.
Let $\struct {S, \circ}$ be a group.
Then $\struct {T, *}$ is also a group.
== Proof ==
From Epimorphism Preserves Semigroups, if $\struct {S, \circ}$ is a semigroup then so is $\struct {T, *}$.
From Epimorphism Preserves Identity, if $\struct {S, \circ}$ has an identity $e_S$, then $\map \phi {e_S}$ is the identity for $*$.
From Epimorphism Preserves Inverses, if $x^{-1}$ is an inverse of $x$ for $\circ$, then $\map \phi {x^{-1} }$ is an inverse of $\map \phi x$ for $*$.
The result follows from the definition of group.
== Warning ==
== Also see ==
* Isomorphism Preserves Groups
* Epimorphism Preserves Associativity
* Epimorphism Preserves Commutativity
* Epimorphism Preserves Identity
* Epimorphism Preserves Inverses
* Epimorphism Preserves Semigroups
* Epimorphism Preserves Distributivity
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Theorem $12.2$: Corollary
* : Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Exercise $12.1$
Category:Group Epimorphisms
Epimorphism Preserves Identity

== Theorem ==
Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an epimorphism.
Let $\struct {S, \circ}$ have an identity element $e_S$.
Then $\struct {T, *}$ has the identity element $\map \phi {e_S}$.
== Proof ==
Let $\struct {S, \circ}$ be an algebraic structure in which $\circ$ has an identity $e_S$.
Then:
:$\forall x \in S: x \circ e_S = x = e_S \circ x$
Thus:
:$\forall x \in S: x \circ e_S, e_S \circ x \in \Dom \phi$
Hence:
and:
The result follows because every element $y \in T$ is of the form $\map \phi x$ with $x \in S$.
== Warning ==
== Also see ==
* Group Homomorphism Preserves Identity
* Epimorphism Preserves Associativity
* Epimorphism Preserves Commutativity
* Epimorphism Preserves Inverses
* Epimorphism Preserves Semigroups
* Epimorphism Preserves Groups
* Epimorphism Preserves Distributivity
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Theorem $12.2$
* : Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Exercise $12.1$
Category:Epimorphisms (Abstract Algebra)
Category:Identity Elements
Epimorphism Preserves Inverses

== Theorem ==
Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an epimorphism.
Let $\struct {S, \circ}$ have an identity $e_S$.
Let $x^{-1}$ be an inverse element of $x$ for $\circ$.
Then $\map \phi {x^{-1} }$ is an inverse element of $\map \phi x$ for $*$.
That is:
:$\map \phi {x^{-1} } = \paren {\map \phi x}^{-1}$
== Proof ==
Let $\struct {S, \circ}$ be an algebraic structure in which $\circ$ has an identity element $e_S$.
From Epimorphism Preserves Identity, it follows that $\struct {T, *}$ also has an identity element, which is $\map \phi {e_S}$.
Let $y$ be an inverse of $x$ in $\struct {S, \circ}$. 
By definition of inverse element:
:$x \circ y = e_S = y \circ x$
Then:
So $\map \phi y$ is an inverse of $\map \phi x$ in $\struct {T, *}$.
== Warning ==
== Also see ==
* Group Homomorphism Preserves Inverses
* Epimorphism Preserves Associativity
* Epimorphism Preserves Commutativity
* Epimorphism Preserves Identity
* Epimorphism Preserves Semigroups
* Epimorphism Preserves Groups
* Epimorphism Preserves Distributivity
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Theorem $12.2$
* : Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Exercise $12.1$
Category:Epimorphisms (Abstract Algebra)
Category:Inverse Elements
Epimorphism Preserves Modules

== Theorem ==
Let $\struct {G, +_G, \circ}_R$ be an $R$-module.
Let $\struct {H, +_H, \circ}_R$ be an $R$-algebraic structure.
Let $\phi: G \to H$ be an epimorphism.
Then $H$ is an $R$-module.
=== Corollary ===
If $G$ is a unitary $R$-module, then so is $H$.
== Proof ==
If $\struct {G, +_G, \circ}_R$ is an $R$-module, then:
$\forall x, y, \in G, \forall \lambda, \mu \in R$:
:$(1): \quad \lambda \circ \paren {x +_G y} = \paren {\lambda \circ x} +_G \paren {\lambda \circ y}$
:$(2): \quad \paren {\lambda +_R \mu} \circ x = \paren {\lambda \circ x} +_G \paren {\mu \circ x}$
:$(3): \quad \paren {\lambda \times_R \mu} \circ x = \lambda \circ \paren {\mu \circ x}$
If $\phi: G \to H$ is an epimorphism, then:
:$\forall x, y \in G: \map \phi {x +_G y} = \map \phi x +_H \map \phi y$
:$\forall x \in S: \forall \lambda \in R: \map \phi {\lambda \circ x} = \lambda \circ \map \phi x$
:$\forall y \in H: \exists x \in G: y = \map \phi x$
As $\phi$ is an epimorphism, we can accurately specify the behaviour of all elements of $H$, as they are the images of elements of $G$.
If $\phi$ were not an epimorphism, that is not surjective, we would have no way of knowing the behaviour of elements of $H$ outside of the image of $G$.
Hence the specification that $\phi$ needs to be an epimorphism.
Now we check the module axioms in turn.
=== ===
Thus is shown to hold for $H$.
=== ===
Thus is shown to hold for $H$.
=== ===
Thus is shown to hold for $H$.
So all the module axioms for $H$ are satisfied.
== Also see ==
* Homomorphic Image of R-Module is R-Module
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations
Category:Epimorphism Preserves Modules
Category:Epimorphisms (Abstract Algebra)
Category:Module Theory
Epimorphism Preserves Modules/Corollary

== Corollary to Epimorphism Preserves Modules ==
Let $\struct {G, +_G, \circ}_R$ be an unitary $R$-module.
Let $\struct {H, +_H, \circ}_R$ be an $R$-algebraic structure.
Let $\phi: G \to H$ be an epimorphism.
Then $H$ is a unitary $R$-module.
== Proof ==
Let $G$ be a unitary $R$-module.
From Epimorphism Preserves Modules we have that $H$ is an $R$-module.
Then by :
:$\forall x \in G: 1_R \circ x = x$
So:
Thus $H$ is also a unitary module.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations
Category:Epimorphism Preserves Modules
Category:Unitary Modules
Epimorphism Preserves Properties

== Theorem ==
=== Epimorphism Preserves Associativity ===
=== Epimorphism Preserves Commutativity ===
=== Epimorphism Preserves Distributivity ===
=== Epimorphism Preserves Identity ===
=== Epimorphism Preserves Inverses ===
=== Epimorphism Preserves Semigroups ===
=== Epimorphism Preserves Groups ===
== Warning ==
Category:Epimorphism Preserves Properties
Category:Epimorphisms (Abstract Algebra)
Epimorphism Preserves Rings

== Theorem ==
Let $\struct {R_1, +_1, \circ_1}$ be a ring, and $\struct {R_2, +_2, \circ_2}$ be a closed algebraic structure.
Let $\phi: R_1 \to R_2$ be an epimorphism.
Then $\struct {R_2, +_2, \circ_2}$ is a ring.
== Proof ==
From Epimorphism Preserves Groups, we have that if $\struct {R_1, +_1}$ is a group then so is $\struct {R_2, +_2}$.
From Epimorphism Preserves Semigroups, we have that if $\struct {R_1, \circ_1}$ is a semigroup then so is $\struct {R_2, \circ_2}$.
From Epimorphism Preserves Distributivity, we have that if $\circ_1$ distributes over $+_1$ then $\circ_2$ distributes over $+_2$.
So it follows from the definition of a ring that if $\struct {R_1, +_1, \circ_1}$ is a ring then so is $\struct {R_2, +_2, \circ_2}$.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $22$. New Rings from Old: Theorem $22.1$
Category:Ring Theory
Category:Ring Epimorphisms
Epimorphism Preserves Semigroups

== Theorem ==
Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an epimorphism.
Let $\struct {S, \circ}$ be a semigroup.
Then $\struct {T, *}$ is also a semigroup.
== Proof ==
As $\struct {S, \circ}$ is a semigroup, then by definition it is closed.
As $\phi$ is an epimorphism, it is by definition surjective.
That is:
:$T = \phi \sqbrk S$
where $\phi \sqbrk S$ denotes the image of $S$ under $\phi$.
From Morphism Property Preserves Closure it follows that $\struct {T, *}$ is closed.
As $\struct {S, \circ}$ is a semigroup, then by definition $\circ$ is associative.
From Epimorphism Preserves Associativity, $*$ is therefore also associative.
So:
:$\struct {T, *}$ is closed
and:
:$*$ is associative.
Therefore, by definition, $\struct {T, *}$ is a semigroup.
== Warning ==
== Also see ==
* Isomorphism Preserves Semigroups
* Epimorphism Preserves Associativity
* Epimorphism Preserves Commutativity
* Epimorphism Preserves Identity
* Epimorphism Preserves Inverses
* Epimorphism Preserves Groups
* Epimorphism Preserves Distributivity
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Theorem $12.2$: Corollary
* : Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Exercise $12.1$
Category:Epimorphisms (Abstract Algebra)
Category:Semigroups
Epimorphism from Division Ring to Ring

== Theorem ==
Let $\struct {K, +, \circ}$ be a division ring whose zero is $0_K$.
Let $\struct {R, +, \circ}$ be a ring whose zero is $0_R$.
Let $\phi: K \to R$ be a ring epimorphism.
Then one of the following applies:
:$(1): \quad R$ is a null ring
:$(2): \quad R$ is a division ring and $\phi$ is a ring isomorphism.
== Proof ==
We have that the kernel of $K$ is an ideal.
From Ideals of Division Ring, $\map \ker K$ must therefore either be $0_K$ or $K$.
Let $\map \ker K = 0_K$.
Then by Ring Epimorphism with Trivial Kernel is Isomorphism $\phi$ is an ring isomorphism.
Thus $R$ a division ring like $K$.
Let $\map \ker K = K$.
Then:
:$\forall x \in K: \map \phi x = 0_R$
As $\phi$ is an epimorphism, it is surjective.
Therefore:
:$R = \set {0_R}$
Hence the result.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $23$. The Field of Rational Numbers: Theorem $23.3$
Category:Ring Theory
Category:Division Rings
Category:Ring Epimorphisms
Epimorphism from Integers to Cyclic Group

== Theorem ==
Let $\gen a = \struct {G, \circ}$ be a cyclic group.
Let $f: \Z \to G$ be a mapping defined as:
$\forall n \in \Z: \map f n = a^n$.
Then $f$ is a (group) epimorphism from $\struct {\Z, +}$ onto $\gen a$.
== Proof ==
By Powers of Element form Subgroup:
:$\forall n \in \N: a^n \in \gen a$
Hence by the Index Law for Monoids: Negative Index:
:$\forall n \in \Z: a^n \in \gen a$
Also, by Index Law for Monoids: Sum of Indices, $f$ is a homomorphism from $\struct {\Z, +}$ into $\struct {G, \circ}$.
By Homomorphism Preserves Subsemigroups, its codomain $f \sqbrk \Z$ is therefore a subgroup of $\gen a$ containing $a$.
By Existence of Unique Subgroup Generated by Subset: Singleton Generator, $\gen a$ is the smallest subgroup of $G$ containing $a$.
So:
:$f \sqbrk \Z = \gen a$
Therefore $f$ is an epimorphism from $\struct {\Z, +}$ onto $\struct {G, \circ}$.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $25$. Cyclic Groups and Lagrange's Theorem: Theorem $25.1$
Category:Integers
Category:Group Epimorphisms
Category:Cyclic Groups
Epimorphism from Polynomial Forms to Polynomial Functions

== Theorem ==
Let $D$ be an integral domain.
Let $D \sqbrk X$ be the ring of polynomial forms in $X$ over $D$.
Let $\map P D$ be the ring of polynomial functions over $D$.
The mapping $\kappa: D \sqbrk X \to \map P D$ given by:
:$\ds \map \kappa {\sum_{k \mathop = 0}^n {a_k \circ X^k} } = f$
where $\ds f = \sum_{k \mathop = 0}^n {a_k \circ x^k}, x \in D$
is a ring epimorphism.
== Proof ==
== Sources ==
* : $\S 64$ Polynomial rings over an integral domain
Category:Polynomial Theory
Epimorphism from Real Numbers to Circle Group

== Theorem ==
Let $\struct {K, \times}$ be the circle group, that is:
:$K = \set {z \in \C: \cmod z = 1}$
under complex multiplication.
Let $f: \R \to K$ be the mapping from the real numbers to $K$ defined as:
:$\forall x \in \R: \map f x = \cos x + i \sin x$
Then $f: \struct {\R, +} \to \struct {K, \times}$ is a group epimorphism.
Its kernel is:
:$\map \ker f = \set {2 \pi n: n \in \Z}$
== Proof ==
$f$ is a surjection from ...
Then:
So $f$ is a (group) homomorphism.
Thus $f$ is seen to be a surjective homomorphism.
Hence, by definition, it is a (group) epimorphism.
From Cosine of Multiple of Pi:
:$\forall n \in \Z: \cos n \pi = \paren {-1}^n$
and from Sine of Multiple of Pi:
:$\forall n \in \Z: \sin n \pi = 0$
From Sine and Cosine are Periodic on Reals, it follows that these are the only values of $\Z$ for which this holds.
For $\cos x + i \sin x = 1 + 0 i$ it is necessary that:
:$\cos x = 1$
:$\sin x = 0$
and it can be seen that the only values of $x$ for this to happen is:
:$x \in \set {2 \pi n: n \in \Z}$
Hence, by definition of kernel:
:$\map \ker f = \set {2 \pi n: n \in \Z}$
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Example $12.1$
Category:Circle Group
Category:Group Epimorphisms
Epimorphism into Projective Object Splits

== Theorem ==
Let $\mathbf C$ be a metacategory.
Let $P \in \mathbf C_0$ be a projective object of $\mathbf C$.
Let $e: E \twoheadrightarrow P$ be an epimorphism.
Then $e$ is a split epimorphism, i.e. it admits a retraction $f: P \to E$.
== Proof ==
Consider the identity morphism $\operatorname{id}_P: P \to P$.
By definition of projective object, we obtain the following commutative diagram:
$\quad\quad \begin{xy}
<0em,0em>*+{P} = "P",
<4em,0em>*+{P} = "P2",
<4em,4em>*+{E} = "E",
"P";"E" **@{-} ?>*@{>} ?*!/_.6em/{f},
"P";"P2" **@{-} ?>*@{>} ?*!/^.6em/{\operatorname{id}_P},
"E";"P2" **@{-} ?>*@2{>} ?<>(.7)*{\vee} ?*!/_.6em/{e},
\end{xy}$
where $f = \bar{\operatorname{id}_P}$.
It follows that $e \circ f = \operatorname{id}_P$, and so $e$ is a split epimorphism.
== Sources ==
* : $\S 2.1.1$
Category:Category Theory
Epimorphism that is Split Monomorphism is Split Epimorphism

== Theorem ==
Let $\mathbf C$ be a metacategory.
Let $f: C \to D$ be a epimorphism and a split monomorphism.
Then $f: C \to D$ is a split epimorphism.
== Proof ==
Dual proof of Monomorphism that is Split Epimorphism is Split Monomorphism.
Category:Category Theory
Epitaph of Diophantus

== Classic Problem ==
:This tomb holds . Ah, how great a marvel!
:the tomb tells scientifically the measure of his life.
:God granted him to be a boy for the sixth part of his life,
:and adding a twelfth part to this, he clothed his cheeks with down;
:He lit him the light of wedlock after a seventh part,
:and five years after his marriage He granted him a son.
:Alas! late-born wretched child;
:after attaining the measure of half his father's life, chill Fate took him.
:After consoling his grief by this science of numbers for four years he ended his life.
== Solution ==
 died at the age of $84$.
Let $x$ be the number of years achieved by at his death.
His boyhood took up $\dfrac x 6$ years.
His adolescence took up another $\dfrac x {12}$ years.
After another another $\dfrac x 7$ years he married.
A son was born to him after another $5$ years.
After another $\dfrac x 2$ years, that son died.
(The assumption being made here is the conventional one: that the age of the son at his death is half the age of at the death of himself, not of his son, which was $4$ years earlier.)
After another $4$ years, himself died.
Thus we have:
Hence :
:was a boy for $14$ years
:was a youth for $7$ years
:married after another $12$ years at the age of $33$
:had a son born $5$ years later at the age of $38$
:who died $42$ years later when was $80$
:and died $4$ years later at the age of $84$.
== Also known as ==
This problem is often referred to as Diophantus's riddle.
== Historical Note ==
== Sources ==
* : Metrodorus' Arithmetical Epigrams: $126$
* : Chapter $2$: Divisibility Theory in the Integers: $2.4$ The Diophantine Equation $a x + b y = c$
* : $1$. Say it with Letters: Algebraic Amusements: The Ages of Man
* : $84$
* : Metrodorus and the Greek Anthology: $35$
* : $84$
* 
Category:Classic Problems
Category:Metrodorus' Arithmetical Epigrams
Category:Greek Anthology Book XIV
Category:Named Theorems/Metrodorus
Category:84
Epitaph of Diophantus/Historical Note

== Historical Note on Epitaph of Diophantus ==
Whether the Epitaph of Diophantus was actually posed by himself is unlikely.
The puzzle seems first to have appeared in the .
There are a number of translations that can be found. The one given here is that provided by .
== Sources ==
* : Metrodorus' Arithmetical Epigrams: $126$
* : $84$
* : $84$
Category:Historical Notes