nyuuzyou/wikis huggingface dataset:
Surface Area of Torus/Formulation 1

== Theorem ==
Let $\TT$ be a torus.
Let $r$ be the radius of the generating circle of $\TT$.
Let $R$ be the distance of the center of the generating circle from the axis of revolution of $\TT$.
Then the area $\AA$ of $\TT$ is given by:
:$\AA = 4 \pi^2 r R$
== Proof ==
Recall Pappus's Centroid Theorem for Surface Area:
In this context:
:$C$ is the generating circle of $\TT$, which has radius $r$
:the centroid of $C$ is the center of $C$, which is at a distance $R$ from the axis of revolution.
From Perimeter of Circle, the perimeter of $C$ is $2 \pi r$.
Also from Perimeter of Circle, the distance travelled by the center of $C$ is $2 \pi R$.
Hence:
:$\AA = \paren {2 \pi r} \paren {2 \pi R}$
Hence the result.
== Sources ==
* 
* 
Category:Surface Area of Torus
Surface Area of Torus/Formulation 2

== Theorem ==
Let $\TT$ be a torus.
Let $a$ and $b$ be the inner radius and outer radius respectively of $\TT$.
Then the surface area $\AA$ of $\TT$ is given by:
:$\AA = \pi^2 \paren {b^2 - a^2}$
== Proof ==
== Sources ==
* : $\S 4$: Geometric Formulas: Torus of Inner Radius $a$ and Outer Radius $b$: $4.46$
* : $\S 7$: Geometric Formulas: Torus of Inner Radius $a$ and Outer Radius $b$: $7.46.$
Category:Surface Area of Torus
Surface Integral/Examples/Electric Flux

== Example of Surface Integral ==
Let $\mathbf E$ be an electric field acting over a region of space $R$.
Let $S$ be a surface through which $\mathbf E$ acts.
Let $\d S$ be a small element of $S$ whose center is at a point $P$.
Then the electric flux through $S$ to which $\mathbf E$ gives rise is given by the surface integral:
:$\ds \iint_S \mathbf E \cdot \mathbf {\hat n} \rd S$
where $\mathbf {\hat n}$ denotes the unit normal to $S$ at $\d S$ in the direction of flow of $\mathbf E$.
== Sources ==
* : Chapter $\text {II}$: The Products of Vectors: $3$. Line and Surface Integrals
Category:Examples of Surface Integrals
Surface Integral/Examples/Flow of Heat

== Example of Surface Integral ==
Let $\mathbf h$ be the flow of heat within a body $B$ as a point-function..
Let $S$ be a surface through which $\mathbf h$ acts.
Let $\d S$ be a small element of $S$ whose center is at a point $P$.
Then the heat flow through $S$ to which $\mathbf h$ gives rise is given by the surface integral:
:$\ds \iint_S \mathbf h \cdot \mathbf {\hat n} \rd S$
where $\mathbf {\hat n}$ denotes the unit normal to $S$ at $\d S$ in the direction of flow of $\mathbf h$.
== Sources ==
* : Chapter $\text {II}$: The Products of Vectors: $3$. Line and Surface Integrals
Category:Examples of Surface Integrals
Surface Integral/Examples/Fluid in Motion

== Example of Surface Integral ==
Let $\mathbf v$ be the velocity within a body of fluid $B$ as a point-function.
Let $S$ be a surface through which $B$ is in motion.
Let $\d S$ be a small element of $S$ whose center is at a point $P$.
Then the flow rate of $B$ through $S$ is given by the surface integral:
:$\ds \iint_S \mathbf v \cdot \mathbf {\hat n} \rd S$
where $\mathbf {\hat n}$ denotes the unit normal to $S$ at $\d S$ in the direction of flow of $B$.
== Sources ==
* : Chapter $\text {II}$: The Products of Vectors: $3$. Line and Surface Integrals
Category:Examples of Surface Integrals
Surface Integral/Examples/Magnetic Flux

== Example of Surface Integral ==
Let $\mathbf M$ be an magnetic field acting over a region of space $R$.
Let $S$ be a surface through which $\mathbf M$ acts.
Let $\d S$ be a small element of $S$ whose center is at a point $P$.
Then the magnetic flux through $S$ to which $\mathbf M$ gives rise is given by the surface integral:
:$\ds \iint_S \mathbf M \cdot \mathbf {\hat n} \rd S$
where $\mathbf {\hat n}$ denotes the unit normal to $S$ at $\d S$ in the direction of flow of $\mathbf M$.
== Sources ==
* : Chapter $\text {II}$: The Products of Vectors: $3$. Line and Surface Integrals
Category:Examples of Surface Integrals
Surface of Revolution as Warped Product Manifold

== Theorem ==
Let $H$ be the open upper half-plane.
Let $S_C \subseteq \R^3$ be the surface of revolution, where $C$ is its generating curve.
Endow $C$ with the Riemannian metric induced from the Euclidean metric on $H$.
Let $\Bbb S^1$ be the $1$-dimensional sphere, that is, a circle endowed with its standard metric.
Let $f : C \to \R$ be the distance to the axis of rotation, say the $x$-axis:
:$\map f {x, y} = y$
Then $S_C$ is isometric to the warped product manifold $C \times_f \Bbb S^1$.
== Proof ==
== Sources ==
* : $\S 2$: Riemannian Metrics. Methods for Constructing Riemannian Metrics
Category:Surfaces of Revolution
Category:Solid Geometry
Category:Riemannian Geometry
Surgery for Rings

== Theorem ==
Let $R$ and $S$ be commutative rings with unity, and $\phi: R \to S$ a ring monomorphism.
Then there is a ring $T$ isomorphic to $S$ that contains $R$ as a subring.
== Proof ==
Let $T$ be the disjoint union $T = R \cup \paren {S \setminus \Img \phi}$.
Define $\theta : T \to S$ as follows:
:If $x \in R$, then $\map \theta x = \map \phi x$
:If $x \in \paren {S \setminus \Img \phi}$ then $\map \theta x = x$
We claim that $\theta$ is an isomorphism.
Injectivity: Let $\map \theta x = \map \theta y$.
Since $\theta \sqbrk R = \phi \sqbrk R$, we have $\theta \sqbrk R \cap \paren {S \setminus \Img \phi} = \O$.
Therefore either $x, y \in R$ or $x, y \in \paren {S \setminus \Img \phi}$.
If $x, y \in R$ then $\map \phi x = \map \phi y$, so $\theta$ is injective because $\phi$ is.
If $x, y \in \paren {S \setminus \Img \phi}$ then $\theta$ acts as the identity and $x = y$.
Surjectivity: Under $\theta$, $R$ surjects onto $\Img \phi = \phi \sqbrk R$ by definition.
Moreover:
:$\theta \sqbrk {S \setminus \Img \phi} = I \sqbrk {S \setminus \Img \phi} = \paren {S \setminus \Img \phi}$
where $I$ denotes the identity mapping.
Therefore $\theta$ is surjective everywhere.
Now we endow $T$ with addition and multiplication: for $x, y \in T$, let
:$x + y = \map {\theta^{-1} } {\map \theta x + \map \theta y}$
:$x y = \map {\theta^{-1} } {\map \theta x \, \map \theta y}$
so trivially we have
:$\map \theta {x + y} = \map \theta x + \map \theta y$
:$\map \theta {x y} = \map \theta x \, \map \theta y$
Therefore $T$ is the isomorphic image of a ring, and an isomorphism is precisely a map that preserves the ring axioms.
Thus $T$ is a ring isomorphic to $S$ containing $R$ as a subring.
Category:Ring Isomorphisms
Category:Ring Monomorphisms
Surjection/Examples

== Examples of Surjections ==
=== Arbitrary Example $1$ ===
=== Arbitrary Example $2$ ===
=== Negative Function on Integers ===
=== Doubling Function on Reals ===
=== Floor Function of $\dfrac {x + 1} 2$ on $\Z$ ===
=== $\map f x = \dfrac x 2$ for $x$ Even, $0$ for $x$ Odd ===
=== Real Sine Function to $\closedint 1 1$ ===
=== Real Square Function to $\R_{\ge 0}$ ===
== Examples of Mappings which are Not Surjections ==
Category:Examples of Surjections
Surjection/Examples/Arbitrary Example 1

== Example of Surjection ==
Let $S$ and $T$ be sets such that:
Let $f: S \to T$ be the mapping defined as:
Then $f$ is a surjection.
== Sources ==
* : $\S 5$. Induced mappings; composition; injections; surjections; bijections: Example $5.1$
Category:Examples of Surjections
Surjection/Examples/Doubling Function on Reals

== Example of Surjection ==
Let $f: \R \to \R$ be the real function defined as:
:$\forall x \in \R: \map f x = 2 x$
Then $f$ is a surjection.
== Proof ==
For $f$ to be a surjection, it is necessary that:
:$\forall y \in \R: \exists x \in \R: y = 2 x$
This is the case.
Thus $f$ is a surjection by definition.
== Sources ==
* : Chapter $2$: 'And you do addition?': $\S 2.1$: Functions
Category:Examples of Surjections
Surjection/Examples/Floor of Half x+1 on Integers

== Example of Surjection ==
Let $f: \Z \to \Z$ be the mapping defined on the set of integers as:
:$\forall x \in \Z: \map f x = \floor {\dfrac {x + 1} 2}$
where $\floor {\, \cdot \,}$ denotes the floor function.
Then $f$ is a surjection, but not an injection.
== Proof ==
Let $y \in \Z$ be an integer.
Consider the integer $x = 2 y$.
Then:
Thus:
:$\forall y \in \Z: \exists x \in \Z: \map f x = y$
Thus $f$ is a surjection by definition.
Let $n \in \Z$ be an integer.
Let $x_1 = 2 n - 1$ and $x_2 = 2 n$.
Then:
while:
Thus we have that $x_1 \ne x_2$ but $\map f {x_1} = \map f {x_2}$.
Hence by definition $f$ is not an injection.
== Sources ==
* : Chapter $4$: Mappings: Exercise $5$
Category:Examples of Surjections
Surjection/Examples/Half Even Zero Odd

== Example of Surjection ==
Let $f: \Z \to \Z$ be the mapping defined on the set of integers as:
:$\forall x \in \Z: \map f x = \begin{cases} \dfrac x 2 & : x \text { even} \\ 0 & : x \text { odd} \end{cases}$
Then $f$ is a surjection.
== Proof ==
Let $y \in \Z$ be an integer.
Consider the integer $x = 2 y$.
Then:
:$\exists x \in \Z: y = \dfrac x 2$
That is:
:$x = 2 y$
Thus $f$ is a surjection by definition.
== Sources ==
* : Chapter $3$. Mappings: Exercise $2 \ \text {(iii) (b)}$
Category:Examples of Surjections
Surjection/Examples/Negative Function on Integers

== Example of Surjection ==
Let $f: \Z \to \Z$ be the mapping defined on the set of integers as:
:$\forall x \in \Z: \map f x = -x$
Then $f$ is a surjection.
== Proof ==
For $f$ to be a surjection, it is necessary that:
:$\forall y \in \Z: \exists x \in \Z: -x = y$
This is the case by setting $ x = -y$, which is an integer since $y$ is an integer.
Thus $f$ is a surjection by definition.
== Sources ==
* : Chapter $3$. Mappings: Exercise $2 \ \text {(ii) (b)}$
Category:Examples of Surjections
Surjection/Examples/Non-Surjection/2x+1 Function on Integers

== Example of Mapping which is Not a Surjection ==
Let $f: \Z \to \Z$ be the mapping defined on the set of integers as:
:$\forall x \in \Z: \map f x = 2 x + 1$
Then $f$ is not a surjection.
== Proof ==
Consider $y = 2 n$ for some $n \in \Z$.
There exists no $x \in \Z$ such that $\map f x = y$.
Thus by definition $f$ is not a surjection.
== Sources ==
* : Chapter $3$. Mappings: Exercise $2 \ \text {(iv) (b)}$
Category:Examples of Injections
Surjection/Examples/Non-Surjection/Arbitrary Mapping on Sets

== Example of Mapping which is Not a Surjection ==
Let $A = \set {a, b, c}$.
Let $B = \set {1, 2, 3}$.
Let $f \subseteq {A \times B}$ be the mapping defined as:
:$f = \set {\tuple {a, 2}, \tuple {b, 1}, \tuple {c, 1} }$
Then $f$ is not a surjection.
== Proof ==
For $f$ to be a surjection, it would be necessary that:
:$\forall y \in B: \exists x \in A: \map f x = y$
However $3 \in B$ does not fit this condition.
Hence $f$ is not a surjection.
== Sources ==
* : Appendix $\text{A}.4$: Functions
Category:Examples of Surjections
Surjection/Examples/Real Sine Function to Image

== Example of Surjection ==
Let $I$ denote the closed real interval $\closedint {-1} 1$.
Let $f: \R \to I$ be the mapping defined on the set of real numbers as:
:$\forall x \in \R: \map f x = \sin x$
where $\sin$ denotes the sine function.
Then $f$ is a surjection, but not an injection.
== Sources ==
* : Chapter $\text {II}$: Topological Spaces: $\S 9$. Functions: Example $1$
Category:Examples of Surjections
Surjection/Examples/Real Square Function to Non-Negative Reals

== Example of Surjection ==
Let $f: \R \to \R_{\ge 0}$ be the real square function whose codomain is the set of non-negative reals:
:$\forall x \in \R: \map f x = x^2$
Then $f$ is a surjection.
== Proof ==
Let $y \in \R_{\ge 0}$.
Let $x = +\sqrt y$.
From Existence of Square Roots of Positive Real Number, there exists such a $y$.
Then:
:$x^2 = y$
That is:
:$y = \map f x$
Hence:
:$\forall y \in \R_{\ge 0}: \exists x \in \R: y = \map f x$
and the result follows by definition of surjection.
== Sources ==
* : Chapter $\text{I}$: Sets and Functions: Composition of Functions
Category:Examples of Surjections
Category:Square Function
Surjection Induced by Powerset is Induced by Surjection

== Theorem ==
Let $\RR \subseteq S \times T$ be a relation.
Let $\RR^\to: \powerset S \to \powerset T$ be the direct image mapping $\RR$.
Let $\RR^\to$ be a surjection.
Let $X = \Preimg \RR$, that is, the preimage of $\RR$.
Then $\RR {\restriction_X} \subseteq X \times T$, that is, the restriction of $\RR$ to $X$, is a surjection.
== Proof ==
Let $X$ be the preimage of $\RR$.
Suppose $\RR {\restriction_X} \subseteq X \times T$ is a mapping, but not a surjection.
Then:
:$\exists y \in T: \neg \exists x \in S: \tuple {x, y} \in \RR$
Because no element of $S$ relates to $y$, no subset of $S$ contains any element of $S$ that relates to the subset $\set y \subseteq T$.
Thus:
:$\exists \set y \in \powerset T: \neg \exists X \in \powerset S: \map {\RR^\to} X = \set y$
So we see that $\RR^\to: \powerset S \to \powerset T$ is not a surjection.
Now, suppose $\RR {\restriction_X} \subseteq X \times T$ is not even a mapping.
This could happen, according to the definition of a mapping, in one of two ways:
:$(1): \quad \exists x \in X: \neg \exists y \in T: \tuple {x, y} \in \RR$
:$(2): \quad \exists x \in S: \tuple {x, y_1} \in \RR \land \tuple {x, y_2} \in \RR: y_1 \ne y_2$
Because $X$ is already the preimage of $\RR$, the first of these cannot happen here.
For the second, it can be seen that neither $\set {y_1}$ nor $\set {y_2}$ can be in $\Rng {\map {\RR^\to} {\powerset S} }$.
Therefore $\RR^\to: \powerset S \to \powerset T$ can not be a surjection.
Thus, by the Rule of Transposition, the result follows.
Category:Surjections
Category:Direct Image Mappings
Surjection by Free Module

== Theorem ==
Let $A$ be a ring.
Let $M$ be a left $A$-module.
Then there exists a free $A$-module $F$ and a surjective $A$-module homomorphism $f : F \to M$.
== Proof ==
Let $F = A^{\paren M}$ be the free $A$-module on the set $M$.
Let $c : M \to A^{\paren M}$ be the canonical mapping on $F$.
Let $f : F \to M$ be the $A$-module homomorphism induced the by the Universal Property of Free Modules applied to the identity $\operatorname {id}_M$ of $M$.
We have:
:$f \circ c = \operatorname {id}_M$
Thus $f$ is a split epimorphism in the category of sets.
By Split Epimorphism is Epic and Surjection iff Epimorphism in Category of Sets $f$ is surjective.
Category:Homological Algebra
Surjection from Aleph to Ordinal

== Theorem ==
Let $x$ and $y$ be ordinals.
Suppose that:
:$0 < y < \aleph_{x+1}$
Then there is a surjection:
:$f : \aleph_x \to y$
== Proof ==
:$y < \aleph_{x+1}$, then $y < \aleph_x \lor y \sim \aleph_x$ by Ordinal Less than Successor Aleph.
In either case, $\left|{ y }\right| \le \aleph_x$ by Ordinal in Aleph iff Cardinal in Aleph and Equivalent Sets have Equal Cardinal Numbers.
The existence of the surjection follows from Surjection iff Cardinal Inequality.
== Sources ==
* : $\S 10.46 \ (3)$
Category:Aleph Mapping
Surjection from Class to Proper Class

== Theorem ==
Let $A$ be a class.
Let $\mathrm P$ be a proper class.
Let $f: A \to \mathrm P$ be a surjection.
Then $A$ is proper.
== Proof ==
 $A$ is not proper.
Then $A$ must be a set.
By the Axiom of Powers, $\powerset A$ is also a set.
Let $g: f \sqbrk A \to \powerset A$ be defined as:
:$\map g {\map f a} = f^{-1} \sqbrk {\set {\map f a} }$
It should be noted that:
:$\forall a \in A: \map {\paren {g \circ f} } a \ne \O$:
Suppose that $\map g {\map f a} = \map g {\map f b}$.
Let $x \in f^{-1} \sqbrk {\set {\map f a} }$.
Then by the definition of a singleton:
:$x \in f^{-1} \sqbrk {\set {\map f a} } \implies \map f x \in \set {\map f a} \implies \map f x = \map f a$
By the same argument:
:$x \in f^{-1} \sqbrk {\set {\map f b} } \implies \map f x = \map f b$
Hhence by the properties of equality:
:$\map f a = \map f b$
It has been shown that $g$ is an injection.
Because $f$ is a surjection, it follows by definition of surjection that:
:$f \sqbrk A = \mathrm P$
But this contradicts Injection from Proper Class to Class.
Thus by contradiction $A$ must be proper.
Hence the result.
Category:Gödel-Bernays Class Theory
Category:Class Mappings
Surjection from Finite Set to Itself is Permutation

== Theorem ==
Let $S$ be a finite set.
Let $f: S \to S$ be an surjection.
Then $f$ is a permutation.
== Proof ==
From Surjection iff Right Inverse, $f$ has a right inverse $g: S \to S$.
From Right Inverse Mapping is Injection, $g$ is an injection.
From Injection from Finite Set to Itself is Permutation, $g$ is a permutation and so a bijection.
From Inverse of Bijection is Bijection, $f$ is also a bijection.
Thus as $f$ is a bijection to itself, it is by definition a permutation.
== Sources ==
* : Chapter $3$. Mappings: Exercise $8$
* : $0$: Some Conventions and some Basic Facts
Category:Surjections
Category:Permutations
Surjection from Natural Numbers iff Countable

== Theorem ==
Let $S$ be a non-empty set.
Then $S$ is countable there exists a surjection $f: \N \to S$.
=== Corollary 1 ===
=== Corollary 2 ===
== Proof ==
=== Necessary Condition ===
Suppose that $f: \N \to S$ is a surjection.
By Surjection from Natural Numbers iff Right Inverse, $f$ admits a right inverse $g: S \to \N$.
We have that $g$ is an injection by Right Inverse Mapping is Injection.
Hence the result, by the definition of a countable set.
=== Sufficient Condition ===
Suppose that $S$ is countable and non-empty.
Then there is an injection $g: S \to \N$.
By Injection has Surjective Left Inverse Mapping, there is a surjection $f: \N \to S$.
== Sources ==
* : $1$: Set Theory and Logic: $\S 7$: Countable and Uncountable Sets: Theorem $7.1$
Category:Countable Sets
Category:Natural Numbers
Category:Surjections
Surjection from Natural Numbers iff Countable/Corollary 1

== Theorem ==
Let $T$ be a countably infinite set.
Let $S$ be a non-empty set.
Then $S$ is countable there exists a surjection $f: T \to S$.
== Proof ==
Let $g: T \to \N$ be a bijection from $T$ to $\N$.
By Inverse of Bijection is Bijection, $g^{-1}: \N \to T$ is a bijection from $\N$ to $T$.
=== Necessary Condition ===
Let $f: T \to S$ be a surjection.
By Composite of Surjections is Surjection, $f \circ g^{-1}:\N \to S$ is a surjection.
By Surjection from Natural Numbers iff Countable, $S$ is a countable set.
=== Sufficient Condition ===
Suppose that $S$ is countable and non-empty.
By Surjection from Natural Numbers iff Countable, there exists a surjection $f: \N \to S$.
By Composite of Surjections is Surjection, $f \circ g:T \to S$ is a surjection.
Category:Countable Sets
Category:Surjections
Surjection from Natural Numbers iff Countable/Corollary 2

== Theorem ==
Let $T$ be a countably infinite set.
Let $S$ be an uncountable set.
Let $f:T \to S$ be a mapping.
Then $f$ is not a surjection.
== Proof ==
By Corollary 1 no mapping from $T$ to $S$ is a surjection.
Category:Countable Sets
Category:Surjections
Surjection from Natural Numbers iff Right Inverse

== Theorem ==
Let $S$ be a set.
Let $f: \N \to S$ be a mapping, where $\N$ denotes the set of natural numbers.
Then $f$ is a surjection $f$ admits a right inverse.
== Proof ==
=== Necessary Condition ===
Suppose that $g: S \to \N$ is a right inverse of $f$.
That is, let $f \circ g = I_S$, the identity mapping on $S$.
We have that $I_S$ is a surjection.
By Surjection if Composite is Surjection, it follows that $f$ is a surjection.
=== Sufficient Condition ===
Now, suppose that $f: \N \to S$ is a surjection.
By the definition of a surjection, it follows that:
:$\forall x \in S: \map {f^{-1} } x$ is non-empty
where $\map {f^{-1} } x$ denotes the preimage of $x$ under $f$.
From the Well-Ordering Principle, $\N$ is well-ordered.
Hence, we can define the mapping $g: S \to \N$ as:
:$\forall x \in S: \map g x = \map \min {\map {f^{-1} } x}$
By the definition of a smallest element, it follows that:
:$\forall x \in S: \map g x \in \map {f^{-1} } x$
That is:
:$\forall x \in S: \map f {\map g x} = \map {\paren {f \circ g} } x = x$
In other words, $g$ is a right inverse of $f$.
== Also see ==
* Surjection iff Right Inverse, which depends on the axiom of choice.
Category:Surjections
Category:Natural Numbers
Surjection if Composite is Surjection

== Theorem ==
Let $f: S_1 \to S_2$ and $g: S_2 \to S_3$ be mappings such that $g \circ f$ is a surjection.
Then $g$ is a surjection.
== Proof ==
Let $g \circ f$ be surjective.
Fix $z \in S_3$. 
Now find an $x \in S_1: \map {g \circ f} x = z$.
The surjectivity of $g \circ f$ guarantees this can be done.
Now find an $y \in S_2: f \paren x = y$.
$f$ is a mapping and therefore a left-total relation; which guarantees this too can be done.
It follows that:
== Also see ==
* Injection if Composite is Injection
== Sources ==
* : $\S 3.6$. Products of bijective mappings. Permutations
* : Chapter $\text I$: Algebraic Structures: $\S 5$: Composites and Inverses of Functions: Theorem $5.3: \ 2^\circ$
* : Chapter $\text{I}$: Sets and Functions: Exercise $\text{H}$
* : Appendix: Elementary set and number theory
* : $\S 5$. Induced mappings; composition; injections; surjections; bijections: Exercise $14 \ \text{(b)}$
* : Chapter $4$: Mappings: Exercise $13$
* : Chapter $1$: Sets and mappings: $\S 1.3$: Mappings: Exercise $2$
* : $1$: Set Theory and Logic: $\S 2$: Functions: Exercise $2.4 \ \text{(e)}$
Category:Surjections
Category:Composite Mappings
Surjection iff Cardinal Inequality

== Theorem ==
Let $S$ and $T$ be sets.
Let $S$ be non-empty.
Then:
:$0 < \card T \le \card S$
:
:there exists a surjection $f: S \to T$
where $\card S$ denotes the cardinality of $S$.
== Proof ==
=== Necessary Condition ===
Suppose $f: S \to T$ is a surjection.
Then $\Img f = T$ by definition.
From Cardinality of Image of Mapping not greater than Cardinality of Domain:
:$\card T \le \card S$
Furthermore, if $S$ is non-empty, then $\map f x \in T$ for some $x \in S$.
Thus, $T$ is non-empty and $0 < \card T$ by Cardinality of Empty Set.
=== Sufficient Condition ===
Suppose that $0 < \card T \le \card S$.
By Injection iff Cardinal Inequality, it follows that $g: T \to S$ for some injection $g$.
Take an arbitrary $y \in T$.
Define the function $f: S \to T$ as follows:
:$\map f x = \begin{cases} \map {g^{-1} } x & : x \in \Img g \\ y & : x \notin \Img g \end{cases}$
For any $z \in T$, $\map g z \in \Img g$.
Thus, $\map f x = z$ for some $x \in S$.
It follows that $f: S \to T$ is a surjection.
== Sources ==
* : $\S 10.27$
Category:Surjections
Category:Cardinals
Surjection iff Epimorphism in Category of Sets

== Theorem ==
Let $\mathbf{Set}$ be the category of sets.
Let $f: X \to Y$ be a morphism in $\mathbf{Set}$, i.e. a mapping.
Then $f$ is a surjection it is an epimorphism.
== Proof ==
=== Necessary Condition (proof 1 - contrapositive) ===
Suppose that $f$ is surjective.
Suppose further that we have mappings $g, h: Y \to Z$ such that $g \ne h$.
Then necessarily there exists some $y \in Y$ such that $\map g y \ne \map h y$ by Equality of Mappings.
As $f$ is surjective, it follows that there exists an $x \in X$ such that $\map f x = y$.
Hence, we conclude that:
:$\map g {\map f x} \ne \map h {\map f x}$
which, again by Equality of Mappings, means that $g \circ f \ne h \circ f$.
Therefore, $f$ is epic, by the Rule of Transposition.
=== Necessary Condition (proof 2 - direct) ===
Suppose that $f$ is surjective.
Suppose further that we have mappings $g, h: Y \to Z$ such that $g \circ f = h \circ f$.
Suppose that we have an arbitrary element $y \in Y$.
As $f$ is surjective, it follows that there is an $x \in X$ such that $\map f x = y$.
Combined with the assumption on $g$ and $h$, we obtain:
:$\map g y = \map g {\map f x} = \map h {\map f x} = \map h y$
Discharging $y$, we have $g = h$.
Discharging $g$ and $h$, we conclude that $f$ is epic by definition.
=== Sufficient Condition ===
Suppose that $f: X \twoheadrightarrow Y$ is an epimorphism.
By definition of surjection, it will suffice to show that:
:$\forall y \in Y: \exists x \in X: \map f x = y$
Let us reason by contradiction.
So suppose $f$ were not surjective.
Then there would be an $y_0 \in Y$ such that:
:$\forall x \in X: \map f x \ne y_0$
Consider the mappings defined by:
:$g: Y \to Y \cup \set Y, \map g y := y$
:$h: Y \to Y \cup \set Y, \map h y := \begin{cases} y & \text{if } y \ne y_0 \\ Y & \text{if } y = y_0 \end{cases}$
The assumption on $f$ yields that $g \circ f = h \circ f$.
Since $h \ne g$, it follows that $f$ cannot be epic.
This contradiction shows that $f$ is necessarily surjective.
== Sources ==
* : $\S 2.1$
* : $\S 2.8$: Exercise $2.1$
Category:Surjections
Category:Category of Sets