nyuuzyou/wikis huggingface dataset:
Initial Value Theorem of Laplace Transform

== Theorem ==
Let $\laptrans {\map f t} = \map F s$ denote the Laplace transform of the real function $f$.
Then:
:$\ds \lim_{t \mathop \to 0} \map f t = \lim_{s \mathop \to \infty} s \, \map F s$
if those limits exist.
=== General Result ===
== Proof ==
We have that $\map {f'} t$ is piecewise continuous with one-sided limits and of exponential order.
Hence:
:$\ds \lim_{s \mathop \to \infty} \int_0^\infty e^{-s t} \map {f'} t \rd t = 0$
Suppose that $f$ is continuous at $t = 0$.
From Laplace Transform of Derivative:
:$(1): \quad \laptrans {\map {f'} t} = s \map F s - \map f 0$
Taking the limit as $s \to \infty$ in $(1)$, where it is assumed that $\map f t$ is continuous at $t = 0$:
:$0 = \ds \lim_{s \mathop \to \infty} s \map F s - \map f 0$
or:
:$\ds \lim_{s \mathop \to \infty} s \map F s = \map f 0 = \lim_{t \mathop \to 0} \map f t$
Suppose that $f$ is not continuous at $t = 0$.
From Laplace Transform of Derivative with Discontinuity at Zero:
:$\laptrans {\map {f'} t} = s \map F s - \map f {0^+}$
which means:
:$(2): \quad \laptrans {\map {f'} t} = s \map F s - \ds \lim_{u \mathop \to 0} \map f u$
Similarly taking the limit as $s \to \infty$ in $(2)$, where it is assumed that $\map f t$ is continuous at $t = 0$:
:$0 = \ds \lim_{s \mathop \to \infty} s \map F s - \lim_{u \mathop \to 0} \map f u$
and so:
:$\ds \lim_{s \mathop \to \infty} s \map F s = \lim_{u \mathop \to 0} \map f u = \lim_{t \mathop \to 0} \map f t$
== Examples ==
== Sources ==
* : Chapter $1$: The Laplace Transform: Some Important Properties of Laplace Transforms: $11$. Initial-value theorem: Theorem $1 \text{-} 16$
* : Chapter $1$: The Laplace Transform: Solved Problems: Initial and Final Value Theorems: $25$
Category:Laplace Transforms
Initial Value Theorem of Laplace Transform/Examples/Example 1

== Example of use of Initial Value Theorem of Laplace Transform ==
Consider the real function $f: \R \to \R$ defined as:
:$\map f t = 3 e^{-2 t}$
From Laplace Transform of Exponential:
:$\laptrans {\map f t} = \dfrac 3 {s + 2}$
Then by the Initial Value Theorem of Laplace Transform:
== Sources ==
* : Chapter $1$: The Laplace Transform: Solved Problems: Initial and Final Value Theorems: $27 \ \text{(a)}$
Category:Examples of Laplace Transforms
Initial Value Theorem of Laplace Transform/General Result

== Theorem ==
Let $f$ and $g$ be real functions.
Let $\laptrans {\map f t} = \map F s$ denote the Laplace transform of $f$.
Let $\ds \lim_{t \mathop \to 0} \dfrac {\map f t} {\map g t} = 1$.
Then:
:$\ds \lim_{s \mathop \to \infty} \dfrac {\map F s} {\map G s} = 1$
if those limits exist.
== Proof ==
== Sources ==
* : Chapter $1$: The Laplace Transform: Some Important Properties of Laplace Transforms: $13$. Generalization of initial-value theorem: Theorem $1 \text{-} 18$
Category:Laplace Transforms
Injection/Examples

== Examples of Injections ==
=== $2 x$ Function on Integers is Injective but not Surjective ===
=== $2 x + 1$ Function on Integers is Injective ===
=== $-x$ Function on Integers is Injective ===
=== Square Function on $\N$ is Injective ===
=== Cube Function is Injective ===
=== Arbitrary Example ===
== Examples of Mappings which are Not Injections ==
Category:Examples of Injections
Injection/Examples/2x+1 Function on Integers

== Example of Injection ==
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 an injection.
== Proof ==
Let $x_1$ and $x_2$ be integers.
Then:
Hence $f$ is an injection by definition.
== Sources ==
* : Chapter $3$. Mappings: Exercise $2 \ \text {(iv) (a)}$
Category:Examples of Injections
Injection/Examples/2x Function on Integers

== Example of Injection which is Not a Surjection ==
Let $f: \Z \to \Z$ be the mapping defined on the set of integers as:
:$\forall x \in \Z: \map f x = 2 x$
Then $f$ is an injection, but not a surjection.
== Proof ==
Let $x_1$ and $x_2$ be integers.
Then:
Hence $f$ is an injection by definition.
Now consider $y = 2 n + 1$ 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 $8$
* : Appendix $\text{A}.4$: Functions: Problem Set $\text{A}.4$: $23$
* : Chapter $4$: Mappings: Exercise $5$
Category:Examples of Injections
Injection/Examples/Arbitrary Example 1

== Example of Injection ==
Let $S$ be the set $\set {3, 6}$.
Let $T$ be the set $\set {9, 36, 150$.
Let $f: S \to T$ be the square function:
:$\forall x \in S: \map f x = x^2$
Then $f$ is an injection.
== Proof ==
We have:
Hence by definition $f$ is an injection.
== Sources ==
* 
* 
Category:Examples of Injections
Injection/Examples/Cube Function

== Example of Injection ==
Let $f: \R \to \R$ be the real function defined as:
:$\forall x \in \R: \map f x = x^3$
Then $f$ is an injection.
== Proof ==
From Odd Power Function on Real Numbers is Strictly Increasing, $f$ is strictly increasing.
From Strictly Monotone Real Function is Bijective, it follows that $f$ is bijective.
Hence by definition $f$ is an injection.
== Sources ==
* : Chapter $\text I$: Sets and Functions: Composition of Functions
* : $\S 22$: Injections; bijections; inverse of a bijection
* 
Category:Examples of Injections
Injection/Examples/Negative Function on Integers

== Example of Mapping which is Not an Injection ==
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 an injection.
== Proof ==
Let $x_1$ and $x_2$ be integers.
Then:
Hence $f$ is an injection by definition.
== Sources ==
* : Chapter $3$. Mappings: Exercise $2 \ \text {(ii) (a)}$
Category:Examples of Injections
Injection/Examples/Non-Injection/Arbitrary Mapping on Sets

== Example of Mapping which is Not an Injection ==
Let $A = \set {a, b, c, d}$.
Let $B = \set {1, 2, 3, 4, 5}$.
Let $f \subseteq {A \times B}$ be the mapping defined as:
:$f = \set {\tuple {a, 1}, \tuple {b, 4}, \tuple {c, 5}, \tuple {d, 4} }$
Then $f$ is not a injection.
== Proof ==
For $f$ to be an injection, it would be necessary that:
:$\forall x_1, x_2 \in A: \map f {x_1} = \map f {x_2} \implies x_1 = x_2$
By definition of the squaring operation, we have:
:$\map f b = \map f d$
But unless $b \ne d$.
Hence $f$ is not an injection.
== Sources ==
* : Appendix $\text{A}.4$: Functions
Category:Examples of Injections
Injection/Examples/Non-Injection/Half Even Zero Odd

== Example of Mapping which is Not an Injection ==
Let $f: \Z \to Z$ be the real function defined 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 not an injection.
== Proof ==
For $f$ to be an injection, it would be necessary that:
:$\forall x_1, x_2 \in \Z: x_1 \ne x_2 \implies \map f {x_1} \ne \map f {x_2}$
But take, for example, $x_1 = 1, x_2 = 3$.
Then:
:$\map f {x_1} = 0 = \map f {x_2}$
and it follows that $f$ is not an injection.
== Sources ==
* : Chapter $3$. Mappings: Exercise $2 \ \text {(iii) (a)}$
Category:Examples of Injections
Injection/Examples/Square Function on Natural Numbers

== Example of Injection ==
Let $f: \N \to \N$ be the mapping defined as:
:$\forall n \in \N: \map f n = n^2$
Then $f$ is an injection, but not a surjection.
== Proof ==
Let $g: \R_{\ge 0} \to \R$ be the mapping defined as:
:$\forall x \in \R_{\ge 0}: \map g x = x^2$
We have that $f$ is a restriction of $g$.
From Positive Power Function on Non-negative Reals is Strictly Increasing, $g$ is strictly increasing.
From Strictly Monotone Real Function is Bijective, it follows that $g$ is bijective.
By definition of bijection, $g$ is an injection.
It follows from Restriction of Injection is Injection that $f$ is an injection.
Let $p \in \N$ be a prime number.
From Square Root of Prime is Irrational:
:$\nexists n \in \N: n^2 = p$
and so $f$ is not a surjection.
== Sources ==
* : $\S 2$: Sets and functions: Some special types of function
Category:Examples of Injections
Injection Induces Total Ordering

== Theorem ==
Let $\left({T, \le}\right)$ be a totally ordered set, and let $S$ be a set.
Let $f: S \to T$ be an injection.
Then $\le_f$, the ordering induced by $f$, is a total ordering.
== Proof ==
By Ordering Induced by Injection is Ordering, $\le_f$ is an ordering.
Let $s_1, s_2 \in S$.
Then as $\le$ is a total ordering, at least one of the following holds:
:$(1): \qquad f \left({s_1}\right) \le f \left({s_2}\right)$
:$(2): \qquad f \left({s_2}\right) \le f \left({s_1}\right)$
If $(1)$ holds, the definition of $\le_f$ implies that $s_1 \le s_2$.
Similarly, if $(2)$ holds, $s_2 \le s_1$.
This leads to the conclusion that at least one of $s_1 \le s_2$ and $s_2 \le s_1$ holds.
Hence $\le_f$ is a total ordering.
Category:Total Orderings
Category:Injections
Injection Induces Well-Ordering

== Theorem ==
Let $\left({T, \le}\right)$ be a well-ordered set.
Let $S$ be a set.
Let $f: S \to T$ be an injection.
Then $\le_f$, the ordering induced by $f$, is a well-ordering.
== Proof ==
By Ordering Induced by Injection is Ordering, $\le_f$ is an ordering.
Let $S' \subseteq S$ be non-empty.
Then as $\le$ is a well-ordering, the set:
:$f \left({S'}\right) = \left\{{f \left({s}\right): s \in S'}\right\}$
has a minimal element, say $f \left({s_0}\right)$.
That is, for all $s \in S', f \left({s_0}\right) \le f \left({s}\right)$.
By definition of $\le_f$, this implies:
:$\forall s \in S': s_0 \le_f s$
That is, $s_0$ is a minimal element of $S'$.
Hence $\le_f$ is a well-ordering, as $S'$ was arbitrary.
Category:Well-Orderings
Category:Injections
Injection from Finite Set to Itself is Surjection

== Theorem ==
Let $S$ be a finite set.
Let $f: S \to S$ be an injection.
Then $f$ is also a surjection.
=== Corollary ===
== Proof ==
Let $a \in S$.
We need to show that there exists $b \in S$ such that $a = \map f b$.
Consider what happens when $f$ is applied repeatedly on $S$.
Let $f^2$ denote $f \circ f$ and, generally, $f^n := f \circ f^{n-1}$.
Consider the sequence of elements of $S$:
:$a, \map f a, \map {f^2} a, \ldots$
Because $S$ is a finite set, there must be repetitions.
That is, there must exist $r, s \in \N$ such that:
:$\map {f^r} a = \map {f^s} a$
where $r \ne s$.
, assume $r > s$.
$f$ is an injection.
Therefore by Composite of Injections is Injection, $f^s$ is an injection.
By Injection iff Left Cancellable, $f^s$ is left cancellable.
Thus:
That is, $b = \map {f^{r - s - 1} } a$.
== Sources ==
* : Chapter $3$. Mappings: Exercise $8$
* : Chapter $1$: Sets and mappings: $\S 1.3$: Mappings: Lemma $3$
Category:Injections
Category:Surjections
Category:Finite Sets
Injection from Finite Set to Itself is Surjection/Corollary

== Theorem ==
Let $S$ be a finite set.
Let $f: S \to S$ be an injection.
Then $f$ is a permutation.
== Proof ==
From Injection from Finite Set to Itself is Surjection, $f$ is a surjection.
As $f$ is thus both an injection and a surjection, $f$ is a bijection by definition.
Thus as $f$ is a bijection to itself, it is by definition a permutation.
== Sources ==
* : $0$: Some Conventions and some Basic Facts
Category:Injections
Category:Permutations
Injection from Proper Class to Class

== Theorem ==
Let $A$ be a class.
Let $P$ be a proper class.
Let $f: P \to A$ be an injection.
Then $A$ is proper.
== Proof ==
 $A$ is not proper.
Then it must be a set.
By Injection to Image is Bijection, it follows that the restriction $f \restriction_{P \times f \sqbrk P}$ is a bijection.
By the corollary of Class is Proper iff Bijection from Class to Proper Class, $f \sqbrk P$ is proper.
But since $f \sqbrk P \subseteq A$, this contradicts Subclass of Set is Set.
And so by contradiction, $A$ cannot be a set.
Therefore $A$ is proper.
Category:Gödel-Bernays Class Theory
Category:Class Mappings
Injection from Set to Power Set

== Theorem ==
For every set $S$, there exists an injection from $S$ to its power set $\powerset S$.
== Proof ==
If $S = \O$, the empty mapping suffices, as it is vacuously an injection.
Let $f: S \to \powerset S$ be the mapping defined as:
:$\forall s \in S: \map f s = \set s$
Let $s, t \in S$ such that $\map f s = \map f t$.
Then $\set s = \set t$.
By definition of set equality it follows directly that $s = t$.
Hence $f$ is the required injection, by definition.
Category:Injections
Category:Power Set
Injection has Surjective Left Inverse Mapping

== Theorem ==
Let $S$ and $T$ be sets such that $S \ne \O$.
Let $f: S \to T$ be a injection.
Then there exists a surjection $g: T \to S$ such that:
:$g \circ f = I_S$
== Proof 1 ==
== Proof 2 ==
Category:Injections
Category:Surjections
Category:Left Inverse Mappings
Category:Injection has Surjective Left Inverse Mapping
Injection has Surjective Left Inverse Mapping/Proof 1

== Theorem ==
== Proof ==
Since $S$ is non-empty, we can choose an element $x \in S$.
Since $f$ is an injection, for each $t \in \Img f$ there exists a unique $s \in S$ such that:
:$\map f s = t$
Thus by Law of Excluded Middle there exists a well-defined mapping $T \to S$ given by:
:$\map g t = \begin {cases} s & : \paren {t \in \Img f} \land \paren {\map f s = t} \\ x & : t \notin \Img f \end {cases}$
By construction, for any given $s \in S$, the element $\map f s$ maps to $s$ under $g$.
Therefore $g: T \to S$ is a surjection.
Category:Injection has Surjective Left Inverse Mapping
Injection if Composite is Injection

== Theorem ==
Let $f$ and $g$ be mappings such that their composite $g \circ f$ is an injection.
Then $f$ is an injection.
== Proof ==
Let $g \circ f$ be injective.
We need to show that $\map f a = \map f b \implies a = b$.
So suppose $\map f a = \map f b$.
Then:
and so $a = b$ as $g \circ f$ is an injection.
So we have shown that:
:$\map f a = \map f b \implies a = b$
Hence the result from the definition of injection.
== Also see ==
* Surjection if Composite is Surjection
== Sources ==
* : $\S 3.6$. Products of bijective mappings. Permutations
* : Chapter $\text I$: Algebraic Structures: $\S 5$: Composites and Inverses of Functions: Theorem $5.3: \ 1^\circ$
* : Chapter $\text{I}$: Sets and Functions: Exercise $\text{H}$
* : Appendix: Elementary set and number theory
* : $\S 5$. Induced mappings; composition; injections; surjections; bijections: Exercise $14 \ \text{(a)}$
* : $\S 25$: Some further results and examples on mappings: Worked Example $2$
* : Chapter $1$: Sets and mappings: $\S 1.3$: Mappings: Exercise $2$
* : $1$: Set Theory and Logic: $\S 2$: Functions: Exercise $2.4 \ \text{(c)}$
Category:Injections
Category:Composite Mappings
Injection iff Cardinal Inequality

== Theorem ==
Let $\card T$ denote the cardinal number of $T$.
Let $S$ and $T$ be sets such that $S \sim \card S$ and $T \sim \card T$.
Then:
:$\card S \le \card T $ there exists an injection $f: S \to T$.
== Proof ==
=== Necessary Condition ===
Suppose that $\card S \le \card T$.
Let $g : S \to \card S$ be a bijection and $h: \card T \to T$ be a bijection.
It follows that $g: S \to \card T$ is an injection by the fact that $\card T \le \card S$.
Then from Composite of Injections is Injection, $h \circ g: S \to T$ is an injection.
=== Sufficient Condition ===
The other direction follows from Injection implies Cardinal Inequality.
Category:Cardinals
Injection iff Left Cancellable

== Theorem ==
A mapping $f$ is an injection $f$ is left cancellable.
== Proof ==
From the definition: a mapping $f: Y \to Z$ is left cancellable :
:$\forall X: \forall g_1: X \to Y, g_2: X \to Y: f \circ g_1 = f \circ g_2 \implies g_1 = g_2$
=== Necessary Condition ===
=== Sufficient Condition ===
== Also see ==
* Surjection iff Right Cancellable
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets: Exercise $8.10 \ \text{(a)}$
* : Chapter $\text{I}$: Sets and Functions: Problem $\text{BB}$: Categorical Matters
* : Appendix: Elementary set and number theory
* : $\S 5$. Induced mappings; composition; injections; surjections; bijections: Theorem $5.4$
* : Appendix $\text{A}.5$: Identity, One-one, and Onto Functions: Proposition $\text{A}.5.1: 1 \ \text{(d)}$
Category:Injections
Category:Cancellability
Category:Injection iff Left Cancellable
Injection iff Left Cancellable/Necessary Condition

== Theorem ==
Let $f: Y \to Z$ be an injection.
Then $f$ is left cancellable.
== Proof ==
From the definition: a mapping $f: Y \to Z$ is left cancellable :
:$\forall X: \forall g_1: X \to Y, g_2: X \to Y: f \circ g_1 = f \circ g_2 \implies g_1 = g_2$
Let $f: Y \to Z$ be an injection.
Let $X$ be a set
Let $g_1: X \to Y, g_2: X \to Y$ be mappings such that:
: $f \circ g_1 = f \circ g_2$
Then $\forall x \in X$:
As $f$ is an injection, $\map {g_1} x = \map {g_2} x$ and thus the condition for left cancellability holds.
== Sources ==
* : Chapter $4$: Mappings: Exercise $14$
Category:Injection iff Left Cancellable
Injection iff Left Cancellable/Sufficient Condition

== Theorem ==
Let $f: Y \to Z$ be a mapping which is left cancellable.
Then $f$ is an injection.
== Proof ==
From the definition: a mapping $f: Y \to Z$ is left cancellable :
:$\forall X: \forall g_1: X \to Y, g_2: X \to Y: f \circ g_1 = f \circ g_2 \implies g_1 = g_2$
We use a Proof by Contraposition.
That is, we show that if $f: Y \to Z$ is not injective, then $f$ is not left cancellable.
Hence, suppose $f: Y \to Z$ is not injective.
Then:
: $\exists y_1 \ne y_2 \in Y: f \left({y_1}\right) = f \left({y_2}\right)$
Let the two mappings $g_1: Y \to Y, g_2: Y \to Y$ be defined as follows:
:$\forall y \in Y: g_1 \left({y}\right) = y$
:$\forall y \in Y: g_2 \left({y}\right) = \begin{cases}
y_2 & : y = y_1 \\
y & : y \ne y_1
\end{cases}$
Thus we have $g_1 \ne g_2$ such that $f \circ g_1 = f \circ g_2$.
That is, $f$ is not left cancellable.
From Rule of Transposition it follows that if $f$ is left cancellable, it is injective.
== Sources ==
* : $\S 1.14$: Exercise $19 \ \text{(a)}$
Category:Injection iff Left Cancellable
Category:Proofs by Contraposition
Injection iff Left Inverse

== Theorem ==
A mapping $f: S \to T, S \ne \O$ is an injection :
:$\exists g: T \to S: g \circ f = I_S$
where $g$ is a mapping.
That is, $f$ has a left inverse.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Also see ==
* Surjection iff Right Inverse
* Conditions for Uniqueness of Left Inverse Mapping: in general the left inverse is not unique.
== Sources ==
* : $\S 10$: Inverses and Composites: Exercise $\text{(i)}$
* : Chapter $\text I$: Algebraic Structures: $\S 5$: Composites and Inverses of Functions: Exercise $5.6$
* : $\S 0.4$: Theorem $6$
* : $\S 1.3$: Exercise $5$
* : Appendix: Elementary set and number theory
* : $0$: Some Conventions and some Basic Facts
* : $\S 1.3$: Mappings: Exercise $5$
* : $1$: Set Theory and Logic: $\S 2$: Functions: Exercise $2.5 \ \text{(a)}$
Category:Injections
Category:Left Inverse Mappings
Category:Injection iff Left Inverse
Injection iff Left Inverse/Proof 1

== Theorem ==
== Proof ==
Let:
:$\exists g: T \to S: g \circ f = I_S$
From Identity Mapping is Injection, $I_S$ is injective, so $g \circ f$ is injective.
So from Injection if Composite is Injection, $f$ is an injection.
Note that the existence of such a $g$ requires that $S \ne \O$.
Now, assume $f$ is an injection.
We now define a mapping $g: T \to S$ as follows.
As $S \ne \O$, we choose $x_0 \in S$.
By definition of injection:
:$f^{-1} {\restriction_{\Img f} } \to S$ is a mapping
so it is possible to define:
:$\map g y = \begin{cases}
x_0: & y \in T \setminus \Img f \\
\map {f^{-1} } y: & y \in \Img f \end{cases}$
:
It does not matter what the elements of $T \setminus \Img f$ are.
Using the construction given, the equation $g \circ f = I_S$ holds whatever value (or values) we choose for $g \sqbrk {T \setminus \Img f}$.
The remaining elements of $T$ can be mapped arbitrarily, and will not affect the image of any $x \in S$ under the mapping $g \circ f$.
So, for all $x \in S$:
:$\map {g \circ f} x = \map g {\map f x}$
is the unique element of $S$ which $f$ maps to $\map f x$.
This unique element is $x$.
Thus $g \circ f = I_S$.
== Sources ==
* : Chapter $\text{I}$: Sets and Functions: Composition of Functions: Theorem $7$
* : Chapter $4$: Mappings: Exercise $17$
* : Appendix $\text{A}.7$: Inverses: Proposition $\text{A}.7.1$
Category:Injection iff Left Inverse
Injection iff Left Inverse/Proof 2

== Theorem ==
== Proof ==
Take the result Condition for Composite Mapping on Left:
Let $A, B, C$ be sets.
Let $f: A \to B$ and $g: A \to C$ be mappings.
Then:
:$\forall x, y \in A: \map f x = \map f y \implies \map g x = \map g y$
:
:$\exists h: B \to C$ such that $h$ is a mapping and $h \circ f = g$.
Let $C = A = S$, let $B = T$ and let $g = I_S$.
Then the above translates into:
:$\exists h: T \to S$ such that $h$ is a mapping and $h \circ f = g$
:
:$\forall x, y \in S: \map f x = \map f y \implies \map {I_S} x = \map {I_S} y$
But as $\map {I_S} x = x$ and $\map {I_S} y = y$ by definition of identity mapping, it follows that:
:$\exists h: T \to S$ such that $h$ is a mapping and $h \circ f = g$
:
:$\forall x, y \in S: \map f x = \map f y \implies x = y$
which is our result.
== Sources ==
* : $\S 5$. Induced mappings; composition; injections; surjections; bijections: Theorem $5.4$
Category:Injection iff Left Inverse
Injection iff Left Inverse/Proof 3

== Theorem ==
== Proof ==
Let $f: S \to T$ be an injection.
Then $f$ is a one-to-many relation.
By Inverse of Many-to-One Relation is One-to-Many, $f^{-1}: T \to S$ is many-to-one.
By Many-to-One Relation Extends to Mapping, there is a mapping $g: T \to S$ such that:
:$f^{-1} \subseteq g$
Let $\tuple {x, y} \in g \circ f$.
Then:
:$\exists z \in T: \tuple {x, z} \in f, \tuple {z, y} \in g$
Since $\tuple {x, z} \in f$:
:$\tuple {z, x} \in f^{-1} \subseteq g$
Since $\tuple {z, y} \in g$, $\tuple {z, x} \in g$ and $g$ is a mapping:
:$x = y$
so:
:$\tuple {x, y} \in I_S$
So we see that:
:$g \circ f \subseteq I_S$
Let $x \in S$.
Then:
:$\tuple {x, \map f x} \in f$
and:
:$\tuple {\map f x, x} \in f^{-1} \subseteq g$
So:
:$\tuple {x, x} \in g \circ f$
Thus:
:$I_S \subseteq g \circ f$
By definition of set equality:
:$I_S = g \circ f$
Category:Injection iff Left Inverse
Injection iff Monomorphism 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 an injection it is a monomorphism.
== Proof ==
=== Necessary Condition ===
Suppose that $f$ is injective.
Suppose further that we have mappings $g, h: Z \to X$ such that $g \ne h$.
Then necessarily there exists some $z \in Z$ such that $g \left({z}\right) \ne h \left({z}\right)$ by Equality of Mappings.
As $f$ is injective, it follows that:
:$\map f {\map g z} \ne \map f {\map h z}$
which, again by Equality of Mappings, means that $f \circ g \ne f \circ h$.
Hence $f$ is monic, by the Rule of Transposition.
=== Sufficient Condition ===
Suppose that $f: X \rightarrowtail Y$ is a monomorphism.
By definition of injection, it will suffice to show that:
:$x \ne x' \implies \map f x \ne \map f {x'}$
To this end, consider a singleton $\set a$, and define:
:$\bar x: \set a \to X, \map {\bar x} a := x$
:$\bar x': \set a \to X, \map {\bar x'} a := x'$
In particular, $\bar x \ne \bar x'$, and so, $f$ being monic, we deduce:
:$f \circ \bar x \ne f \circ \bar x'$
It follows that it must be that:
Hence $f$ is injective.
== Sources ==
* : $\S 2.1$: Proposition $2.2$
Category:Injections
Category:Category of Sets
Injection implies Cardinal Inequality

== Theorem ==
Let $S$ and $T$ be sets.
Let $f: S \to T$ be an injection.
Let $\card T$ denote the cardinal number of $T$.
Let:
:$T \sim \card T$
where $\sim$ denotes set equivalence
Then:
:$\card S \le \card T$
== Proof ==
Let $f \sqbrk S$ denote the image of $S$ under $f$.
Category:Cardinals
Injection is Bijection iff Inverse is Injection

== Theorem ==
Let $\phi: S \to T$ be an injection.
Then $\phi$ is a bijection its inverse $\phi^{-1}$ is also an injection.
== Proof ==
=== Necessary Condition ===
Let $\phi$ be a bijection.
Then from Bijection iff Inverse is Bijection, its inverse $\phi^{-1}$ is also a bijection and therefore by definition an injection.
=== Sufficient Condition ===
Let $\phi$ be an injection such that $\phi^{-1}$ is also an injection.
By Cardinality of Codomain of Surjection, and the Cantor-Bernstein-Schröder Theorem, the result follows.
Category:Injections
Category:Bijections