nyuuzyou/wikis huggingface dataset:
Lefschetz Principle (First-Order)

== Theorem ==
Let $\phi$ be a sentence in the language $\LL_r = \set {0, 1, +, -, \cdot}$ for rings, where $0, 1$ are constants and $+, -, \cdot$ are binary functions.
:$(1): \quad \phi$ is true in every algebraically closed field of characteristic $0$.
:$(2): \quad \phi$ is true in some algebraically closed field of characteristic $0$.
:$(3): \quad \phi$ is true in algebraically closed fields of characteristic $p$ for arbitrarily large primes $p$.
:$(4): \quad \phi$ is true in algebraically closed fields of characteristic $p$ for sufficiently large primes $p$.
Note in particular that since $\C$ is an algebraically closed field of characteristic $0$, these are equivalent to $\phi$ being true in the field $\C$.
== Proof ==
=== $(1)$ iff $(2)$ ===
From Theory of Algebraically Closed Fields of Characteristic $p$ is Complete:
the theory $ACF_p$ of algebraically closed fields of characteristic $p$ is complete.
That is, all such fields satisfy the exact same $\LL_r$ sentences.
=== $(2)$ implies $(4)$ ===
Let $\phi$ be true in some such field.
Then:
:$ACF_0 \models \phi$
By Gödel's Completeness Theorem and the finiteness of proofs, it follows that there is a finite subset $\Delta$ of $ACF_0$ such that $\Delta \models \phi$.
Such a $\Delta$ can only make finitely many assertions about the character of its models.
Hence, as long as $p$ is selected sufficiently large, an algebraically closed field of characteristic $p$ will satisfy $\phi$.
=== $(4)$ implies $(3)$ ===
We have that all sufficiently large $p$ work.
Hence it follows that it is always possible to find arbitrarily large $p$ that work.
=== $(3)$ implies $(2)$ ===
We prove this by contraposition.
Suppose there is no algebraically closed field of characteristic $0$ where $\phi$ is true.
Then $\phi$ is false in algebraically closed fields of characteristic $0$
Since $ACF_0$ is complete, this means that $ACF_0 \models \neg \phi$.
Similarly to the case of $(2)$ implies $(4)$, there must then be a finite subset $\Delta$ of $ACF_0$ such that $\Delta \models \neg \phi$.
But then, for all sufficiently large $p$, we have that $\phi$ is false in the algebraically closed fields of characteristic $p$.
Hence, it cannot be true for arbitrarily large $p$.
By Rule of Transposition, $\phi$ is true in algebraically closed fields for arbitrarily large primes $p$ implies there exists an algebraically closed field of characteristic $0$ where $\phi$ is true.
Category:Algebraic Geometry
Category:Field Theory
Category:Model Theory for Predicate Logic
Category:Proofs by Contraposition
Left-Hand Differentiable Function is Left-Continuous

== Theorem ==
Let $f$ be a real function defined on an interval $I$.
Let $a$ be a point in $I$ where $f$ is left-hand differentiable.
Then $f$ is left-continuous at $a$.
== Proof ==
By hypothesis, $\map {f'_-} a$ exists.
First we note that $a$ cannot be the left-hand end point of $I$ because values in $I$ less than $a$ need to exist for $\map {f'_-} a$ to exist.
We form the following expression:
:$\ds \lim_{x \mathop \to a^-} \paren {\map f x - \map f a}$
We need to show that it is defined and to find its value.
We find:
Note that this proves that $\ds \lim_{x \mathop \to a^-} \paren {\map f x - \map f a}$ exists.
We continue by manipulating the result above:
	
which means that $f$ is left-continuous at $a$.
Category:Continuous Real Functions
Category:Differentiable Real Functions
Left-Hand and Right-Hand Differentiable Function is Continuous

== Theorem ==
Let $f$ be a real function defined on an interval $I$.
Let $a$ be a point in $I$ where $f$ is left- and right-hand differentiable.
Then $f$ is continuous at $a$.
== Proof ==
By Left-Hand Differentiable Function is Left-Continuous, $f$ is left-continuous at $a$.
By Right-Hand Differentiable Function is Right-Continuous, $f$ is right-continuous at $a$.
By Continuous at Point iff Left-Continuous and Right-Continuous, $f$ is continuous at $a$.
Category:Continuous Real Functions
Category:Differentiable Real Functions
Left-Truncatable Prime/Examples/357,686,312,646,216,567,629,137

== Theorem ==
The largest left-truncatable prime is $357 \, 686 \, 312 \, 646 \, 216 \, 567 \, 629 \, 137$.
== Proof ==
First it is demonstrated that $357 \, 686 \, 312 \, 646 \, 216 \, 567 \, 629 \, 137$ is indeed a left-truncatable prime:
== Sources ==
* 
* : $357,686,312,646,216,567,629,137$
* 
* : $357,686,312,646,216,567,629,137$
Category:Left-Truncatable Primes
Left-Truncatable Prime/Sequence

== Sequence ==
The sequence of left-truncatable primes begins:
:$2, 3, 5, 7, 13, 17, 23, 37, 43, 47, 53, 67, 73, 83, 97, 113, 137, 167, 173, 197, \ldots$
Category:Left-Truncatable Primes
Left-Truncated Automorphic Number is Automorphic

== Theorem ==
Let $n$ be an automorphic number, expressed in some conventional number base.
Let any number of digits be removed from the left-hand end of $n$.
Then what remains is also an automorphic number.
== Proof ==
Let $n$ be an automorphic number of $d$ digits, expressed in base $b$.
By , we have:
:$n^2 \equiv n \pmod {b^d}$
Let some digits be removed from the left-hand end of $n$, so that only $d'$ digits remain.
This only makes sense when $d' < d$.
Define this new number as $n'$.
Then we have:
:$n \equiv n' \pmod {b^{d'} }$
Thus we have:
Hence $n'$ is an automorphic number of $d'$ digits in base $b$.
== Also see ==
Similar proofs can give similar results for other similarly defined numbers, e.g. Trimorphic Numbers, Tri-Automorphic Numbers.
== Examples ==
== Sources ==
* : $1,787,109,376$
* : $1,787,109,376$
Category:Automorphic Numbers
Category:Left-Truncated Automorphic Number is Automorphic
Left-Truncated Automorphic Number is Automorphic/Examples/1,787,109,376

== Example of Left-Truncated Automorphic Number is Automorphic ==
We have that $1 \, 787 \, 109 \, 376$ is automorphic:
:$1 \, 787 \, 109 \, 376^2 = 3 \, 193 \, 759 \, 92 \, \mathbf {1 \, 787 \, 109 \, 376}$
Hence so is $109 \, 376$:
::$109 \, 376^2 = 11 \, 963 \, \mathbf {109 \, 376}$
== Sources ==
* : $1,787,109,376$
* : $1,787,109,376$
Category:Left-Truncated Automorphic Number is Automorphic
Left Cancellable Commutative Operation is Right Cancellable

== Theorem ==
Let $\struct {S, \circ}$ be an algebraic structure.
Let $\circ$ be left cancellable and also commutative.
Then $\circ$ is also right cancellable.
== Proof ==
Let $\circ$ be both left cancellable and commutative on a set $S$.
Then:
== Also see ==
* Right Cancellable Commutative Operation is Left Cancellable
Category:Commutativity
Category:Cancellability
Left Cancellable Element is Left Cancellable in Subset

== Theorem ==
Let $\struct {S, \circ}$ be an algebraic structure.
Let $\struct {T, \circ} \subseteq \struct {S, \circ}$.
Let $x \in T$ be left cancellable in $S$.
Then $x$ is also left cancellable in $T$.
== Proof ==
Let $x \in T$ be left cancellable in $S$.
That is:
:$\forall a, b \in S: x \circ a = x \circ b \implies a = b$
Therefore:
:$\forall c, d \in T: x \circ c = x \circ d \implies c = d$
Thus $x$ is left cancellable in $T$.
Category:Abstract Algebra
Left Cancellable Elements of Semigroup form Subsemigroup

== Theorem ==
Let $\struct {S, \circ}$ be a semigroup.
Let $C_\lambda$ be the set of left cancellable elements of $\struct {S, \circ}$.
Then $\struct {C_\lambda, \circ}$ is a subsemigroup of $\struct {S, \circ}$.
== Proof ==
Let $C_\lambda$ be the set of left cancellable elements of $\struct {S, \circ}$:
:$C_\lambda = \set {x \in S: \forall a, b \in S: x \circ a = x \circ b \implies a = b}$
Let $x, y \in C_\lambda$.
Then:
Thus $\struct {C_\lambda, \circ}$ is closed.
Therefore by the Subsemigroup Closure Test $\struct {C_\lambda, \circ}$ is a subsemigroup of $\struct {S, \circ}$.
Category:Subsemigroups
Category:Cancellability
Left Cancellable iff Left Regular Representation Injective

== Theorem ==
Let $\struct {S, \circ}$ be an algebraic structure.
Then $a \in S$ is left cancellable the left regular representation $\map {\lambda_a} x$ is injective.
== Proof ==
Suppose $a \in S$ is left cancellable.
Then:
:$\forall x, y \in S: a \circ x = a \circ y \implies x = y$
From the definition of the left regular representation:
:$\map {\lambda_a} x = a \circ x$
Thus:
:$\forall x, y \in S: \map {\lambda_a} x = \map {\lambda_a} y \implies x = y$
and so the left regular representation is injective.
Suppose $\map {\lambda_a} x$ is injective.
Then:
:$\forall x, y \in S: \map {\lambda_a} x = \map {\lambda_a} y \implies x = y$
From the definition of the left regular representation:
:$\map {\lambda_a} x = a \circ x$
Thus:
:$\forall x, y \in S: a \circ x = a \circ y \implies x = y$
and so $a$ is left cancellable.
== Also see ==
* Right Cancellable iff Right Regular Representation Injective
* Cancellable iff Regular Representations Injective
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Exercise $7.6$
Category:Regular Representations
Category:Cancellability
Left Composition of Compact Linear Transformation with Bounded Linear Transformation is Compact

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $\struct {X, \norm {\, \cdot \,}_X}$, $\struct {Y, \norm {\, \cdot \,}_Y}$ and $\struct {Y, \norm {\, \cdot \,}_Y}$ be normed vector spaces over $\GF$.
Let $T : X \to Y$ be a compact linear transformation.
Let $S : Y \to Z$ be a bounded linear transformation.
Then $S T : X \to Z$ is a compact linear transformation.
== Proof ==
Let $\sequence {x_n}_{n \mathop \in \N}$ be a bounded sequence.
Since $T$ is compact, there exists a subsequence $\sequence {x_{n_j} }_{j \mathop \in \N}$ such that: 
:$T x_{n_j} \to y$
as $n \to \infty$ for some $y \in Y$. 
Since $S$ is bounded, it is continuous from Continuity of Linear Transformations, and so:
:$S T x_{n_j} \to S y$
So for each bounded sequence $\sequence {x_n}_{n \mathop \in \N}$, there exists a subsequence $\sequence {x_{n_j} }_{j \mathop \in \N}$ such that $\sequence {S T x_n}_{n \mathop \in \N}$ converges. 
Category:Compact Linear Transformations
Left Congruence Class Modulo Subgroup is Left Coset

== Theorem ==
Let $G$ be a group, and let $H \le G$ be a subgroup.
Let $\RR^l_H$ be the equivalence defined as left congruence modulo $H$.
The equivalence class $\eqclass g {\RR^l_H}$ of an element $g \in G$ is the left coset $g H$.
This is known as the left congruence class of $g \bmod H$.
== Proof ==
Let $x \in \eqclass g {\RR^l_H}$.
Then:
Now let $x \in g H$.
Then:
Thus:
:$\eqclass g {\RR^l_H} = g H$
that is, the equivalence class $\eqclass g {\RR^l_H}$ of an element $g \in G$ equals the left coset $g H$.
== Also see ==
* Right Congruence Class Modulo Subgroup is Right Coset
* Left Coset Space forms Partition
* Element of Group is in Unique Left Coset of Subgroup
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 11$: Quotient Structures: Theorem $11.1$
* : Chapter $2$: Subgroups and Cosets: $\S 37$
* : $\S 42.3$ Another approach to cosets
* : Chapter $5$: Cosets and Lagrange's Theorem: Proposition $5.4$
Category:Cosets
Left Congruence Modulo Subgroup is Equivalence Relation

== Theorem ==
Let $G$ be a group, and let $H$ be a subgroup of $G$.
Let $x, y \in G$.
Let $x \equiv^l y \pmod H$ denote the relation that $x$ is left congruent modulo $H$ to $y$.
Then the relation $\equiv^l$ is an equivalence relation.
== Proof 1==
Let $G$ be a group whose identity is $e$.
Let $H$ be a subgroup of $G$.
For clarity of expression, we will use the notation:
:$\tuple {x, y} \in \RR^l_H$
for:
:$x \equiv^l y \pmod H$
From the definition of left congruence modulo a subgroup, we have:
:$\RR^l_H = \set {\tuple {x, y} \in G \times G: x^{-1} y \in H}$
We show that $\RR^l_H$ is an equivalence:
=== Reflexive ===
We have that $H$ is a subgroup of $G$.
From Identity of Subgroup:
:$e \in H$
Hence:
:$\forall x \in G: x^{-1} x = e \in H \implies \tuple {x, x} \in \RR^l_H$
and so $\RR^l_H$ is reflexive.
=== Symmetric ===
But then:
:$\tuple {x^{-1} y}^{-1} = y^{-1} x \implies \tuple {y, x} \in \RR^l_H$
Thus $\RR^l_H$ is symmetric.
=== Transitive ===
Thus $\RR^l_H$ is transitive.
So $\RR^l_H$ is an equivalence relation.
== Proof 2==
Let $G$ be a group, and let $H$ be a subgroup of $G$.
Then $G$ is a monoid.
From Condition for Cosets of Subgroup of Monoid to be Partition, $\RR^l_H$ is a relation induced by partition.
The result follows from Relation Induced by Partition is Equivalence.
== Also see ==
* Definition:Left Coset
* Definition:Left Coset Space
* Right Congruence Modulo Subgroup is Equivalence Relation
== Sources ==
* : $\S 6.1$. The quotient sets of a subgroup: Lemma $\text {(ii)}$
* : Chapter $\text {II}$: New Structures from Old: $\S 11$: Quotient Structures: Theorem $11.1$
* : $\S 1.9$: Subgroups: Theorem $11$
* : Chapter $2$: Subgroups and Cosets: $\S 37$
* : $\S 42.1$ Another approach to cosets
* : Chapter $5$: Cosets and Lagrange's Theorem: Proposition $5.3$
Category:Congruence Modulo Subgroup
Category:Examples of Equivalence Relations
Left Coset Equals Subgroup iff Element in Subgroup

== Theorem ==
Let $G$ be a group whose identity is $e$.
Let $H$ be a subgroup of $G$.
Let $x \in G$.
Let $x H$ denote the left coset of $H$ by $x$.
Then:
:$x H = H \iff x \in H$
== Proof ==
== Also see ==
* Right Coset Equals Subgroup iff Element in Subgroup
== Sources ==
* : $\S 6.1$. The quotient sets of a subgroup: Example $112$
* : $\S 42.6 \ \text {(3L)}$ Another approach to cosets
* : Chapter $5$: Cosets and Lagrange's Theorem: Exercise $2$
Category:Cosets
Left Coset Space forms Partition

== Theorem ==
Let $G$ be a group, and let $H \le G$ be a subgroup.
The left coset space of $H$ forms a partition of its group $G$, and hence:
 
== Proof ==
Follows directly from:
:Left Congruence Modulo Subgroup is Equivalence Relation
:Relation Partitions Set iff Equivalence
== Also see ==
* Right Coset Space forms Partition
== Sources ==
* : $\S 6.1$. The quotient sets of a subgroup: Theorem
* : Chapter $\text {II}$: New Structures from Old: $\S 11$: Quotient Structures: Theorem $11.1$
* : Chapter $\text{II}$: Groups: Subgroups
* : Chapter $2$: Subgroups and Cosets: $\S 37$
* : $0$: Some Conventions and some Basic Facts
* : $\S 42.4$ Another approach to cosets
* : Chapter $5$: Cosets and Lagrange's Theorem: Corollary $5.5$
Category:Coset Space forms Partition
Left Coset by Identity

== Theorem ==
Let $G$ be a group whose identity is $e$.
Let $H$ be a subgroup of $G$.
Then:
: $e H = H$
where $e H$ is the left coset of $H$ by $e$.
== Proof ==
We have:
So $e H = H$.
== Also see ==
This is consistent with the definition of the concept of coset by means of the subset product:
: $e H = \set e H$
* Right Coset by Identity
== Sources ==
* : $\S 6.1$. The quotient sets of a subgroup: Example $112$
* : $\S 1.9$: Example $30$
* : Chapter $5$: Cosets and Lagrange's Theorem: Definition $5.1$: Remark $1$
Category:Coset by Identity
Left Coset of Stabilizer in Group of Transformations

== Theorem ==
Let $S$ be a non-empty set.
Let $G$ be a group of permutations of $S$.
Let $t \in G$.
Let $G_t$ be the set defined as:
:$G_t = \set {g \in G: \map g t = t}$
Then each left coset of $G_t$ in $G$ consists of the elements of $G$ that map $t$ to some element of $S$.
== Proof ==
Let $x \in G$.
Consider the left coset $x G_t$.
Let $\map x t = s$.
Then:
== Sources ==
* : Chapter $7$: Cosets and Lagrange's Theorem: Exercise $15$
Category:Cosets
Left Cosets are Equal iff Element in Other Left Coset

== Theorem ==
Let $G$ be a group.
Let $H$ be a subgroup of $G$.
Let $x, y \in G$.
Let $x H$ denote the left coset of $H$ by $x$.
Then:
:$x H = y H \iff x \in y H$
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Right Cosets are Equal iff Element in Other Right Coset
Category:Cosets
Category:Left Cosets are Equal iff Element in Other Left Coset
Left Cosets are Equal iff Element in Other Left Coset/Proof 2

== Theorem ==
== Proof ==
Let $x \in y H$.
Then $x$ is of the form $y h_1$ for some $h_1 \in H$.
Thus every element of the form $x h_2 \in x H$ is of the form $y h_1 h_2$ for some $h_2 \in H$.
But:
:$h_1 h_2 \in H$
and so:
:$x h_2 \in y H$
So by definition of subset:
:$x H \subseteq y H$
Let $x \in y H$ again.
Then $x$ is of the form $y h$ for some $h \in H$.
But then:
:$y = x h^{-1} \in x H$
Thus every element of the form $y h_2 \in y H$ is of the form $x h^{-1} h_2 \in x H$.
Thus by definition of subset:
:$y H \subseteq x H$
By definition of set equality:
:$x H = y H$
Let $x H = y H$.
Then $x h_1 = y h_2$ for some $h_1, h_2 \in H$.
Hence:
:$x = y h_2 h^{-1} \in y H$
The result follows.
== Sources ==
* : Chapter $\text{II}$: Groups: Subgroups
Category:Left Cosets are Equal iff Element in Other Left Coset
Left Cosets are Equal iff Product with Inverse in Subgroup

== Theorem ==
Let $G$ be a group.
Let $H$ be a subgroup of $G$.
Let $x, y \in G$.
Let $x H$ denote the left coset of $H$ by $x$.
Then:
:$x H = y H \iff x^{-1} y \in H$
== Proof ==
== Also see ==
* Right Cosets are Equal iff Product with Inverse in Subgroup
== Sources ==
* : $\S 6.2$. Mappings of quotient sets: $\text{(ii)}$
* : Chapter $\text {II}$: New Structures from Old: $\S 11$: Quotient Structures: Theorem $11.1$
* : Chapter $\text{II}$: Groups: Subgroups
* : $0$: Some Conventions and some Basic Facts
* : $\S 42.6 \ \text {(2L)}$ Another approach to cosets
* : Chapter $7$: Cosets and Lagrange's Theorem: Exercise $7 \ \text{(iv)}$
Category:Cosets
Left Distributive Law for Natural Numbers

== Theorem ==
The operation of multiplication is left distributive over addition on the set of natural numbers $\N$:
:$\forall x, y, n \in \N_{> 0}: n \times \paren {x + y} = \paren {n \times x} + \paren {n \times y}$
== Proof ==
Using the axiomatization:
Let us cast the proposition in the form:
:$\forall a, b, n \in \N_{> 0}: a \times \paren {b + n} = \paren {a \times b} + \paren {a \times n}$
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$\forall a, b \in \N_{> 0}: a \times \paren {b + n} = \paren {a \times b} + \paren {a \times n}$
=== Basis for the Induction ===
$\map P 1$ is the case:
and so $\map P 1$ holds.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:$\forall a, b \in \N_{> 0}: a \times \paren {b + k} = \paren {a \times b} + \paren {a \times k}$
Then we need to show:
:$\forall a, b \in \N_{> 0}: a \times \paren {b + \paren {k + 1} } = \paren {a \times b} + \paren {a \times \paren {k + 1} }$
=== Induction Step ===
This is our induction step:
The result follows by the Principle of Mathematical Induction.
== Also see ==
* Right Distributive Law for Natural Numbers
== Sources ==
* : $\S 2.1$: Theorem $2.4$
Category:Natural Numbers/1-Based
Left Distributive and Commutative implies Distributive

== Theorem ==
Let $\struct {S, \circ, *}$ be an algebraic structure.
Let the operation $\circ$ be left distributive over the operation $*$.
Let $\circ$ be commutative.
Then $\circ$ is distributive over $*$.
== Proof ==
Let $a, b, c \in S$.
Then
So $\circ$ is right distributive over $*$.
Since $\circ$ is both left distributive and right distributive over $*$, it is distributive over $*$.
Category:Commutativity
Category:Distributive Operations
Left Ideal is Left Module over Ring

== Theorem ==
Let $\struct {R, +, \times}$ be a ring.
Let $J \subseteq R$ be a left ideal of $R$.
Let $\circ : R \times J \to J$ be the restriction of $\times$ to $R \times J$.
Then $\struct {J, +, \circ}$ is a left module over $\struct {R, +, \times}$.
=== Corollary ===
== Proof ==
By definition of a left ideal then $\circ$ is well-defined.
=== ===
Follows directly from 
=== ===
Follows directly from 
=== ===
Follows directly from 
== Also see ==
* Right Ideal is Right Module over Ring
== Sources ==
* : Chapter $4$: Rings and Modules: $\S 4.1$: The Definitions Recalled
Category:Module Theory
Category:Ideal Theory
Category:Left Ideal is Left Module over Ring
Left Ideal is Left Module over Ring/Ring is Left Module over Ring

== Theorem ==
Let $\struct {R, +, \times}$ be a ring.
Then $\struct {R, +, \times}$ is a left module over $\struct {R, +, \times}$.
== Proof ==
From Ring is Ideal of Itself, $R$ is a left ideal.
From Left Ideal is Left Module over Ring, $\struct {R, +, \times}$ is a left module over $\struct {R, +, \times}$.
== Also see ==
* Ring is Right Module over Ring
== Sources ==
* : Chapter $4$: Rings and Modules: $\S 4.1$: The Definitions Recalled
Category:Left Ideal is Left Module over Ring
Left Identity Element is Idempotent

== Theorem ==
Let $\struct {S, \circ}$ be an algebraic structure.
Let $e_L \in S$ be a left identity with respect to $\circ$.
Then $e_L$ is idempotent under $\circ$.
== Proof ==
By the definition of a left identity:
:$\forall x \in S: e_L \circ x = x$
Thus in particular:
:$e_L \circ e_L = e_L$
Therefore $e_L$ is idempotent under $\circ$.
== Also see ==
* Right Identity Element is Idempotent
* Identity Element is Idempotent
Category:Identities are Idempotent
Left Identity in Semigroup may not be Unique

== Theorem ==
Let $\struct {S, \circ}$ be a semigroup.
Let $e_L$ be a left identity of $\struct {S, \circ}$.
Then it is not necessarily the case that $e_L$ is unique.
== Proof ==
;Proof by Counterexample
Let $\struct {S, \gets}$ be an algebraic structure in which the operation $\to$ is the right operation.
From Structure under Right Operation is Semigroup, $\struct {S, \to}$ is a semigroup.
From Element under Right Operation is Left Identity, every element of $\struct {S, \to}$ is a left identity.
The result follows.
== Sources ==
* : Chapter $4$. Groups: Exercise $4$
Category:Identity Elements
Category:Semigroups
Left Identity while exists Left Inverse for All is Identity

== Theorem ==
Let $\struct {S, \circ}$ be a semigroup with a left identity $e_L$ such that:
:$\forall x \in S: \exists x_L: x_L \circ x = e_L$
That is, every element of $S$ has a left inverse with respect to the left identity.
Then $e_L$ is also a right identity, that is, is an identity.
== Proof ==
From Left Inverse for All is Right Inverse we have that:
:$x \circ x_L = e_L$
Then:
So $e_L$ behaves as a right identity as well as a left identity.
That is, by definition, $e_L$ is an identity element.
== Also see ==
* Right Identity while exists Right Inverse for All is Identity
== Sources ==
* : $\S 1.4$
* : Chapter $5$: Semigroups: Exercise $5$
Category:Semigroups
Left Inverse Mapping/Examples/Inclusion of Reals in Complex Plane

== Example of Right Inverse Mapping ==
Let $i_\R: \R \to \C$ be the inclusion mapping of the real numbers into the complex plane:
:$\forall x \in \R: \map {i_\R} z = x + 0 i$
From Inclusion Mapping is Injection, $i_\R$ is an injection.
Hence it has a left inverse $g: \C \to \R$ which, for example, can be defined as:
:$\forall z \in \C: \map g z = \map \Re z$
This left inverse is not unique.
For example, the mapping $h: \R \to \C$ defined as:
:$\forall z \in \C: \map g z = \map \Re z + \map \Im z$
is also a left inverse.
== Sources ==
* : Chapter $\text{I}$: Sets and Functions: Exercise $\text{Q}$
Category:Examples of Left Inverse Mappings
Category:Inclusion Mappings
Left Inverse Mapping is Surjection

== Theorem ==
Let $f: S \to T$ be an injection.
Let $g: T \to S$ be a left inverse of $f$.
Then $g$ is a surjection.
== Proof ==
Let $f: S \to T$ be an injection.
Then from Injection iff Left Inverse there exists at least one left inverse $g: T \to S$ of $f$ such that $g \circ f = I_S$.
$I_S$ is a surjection.
Thus $g \circ f$ is a surjection.
By Surjection if Composite is Surjection it follows that $g$ is also a surjection.
== Sources ==
* : $\S 10$: Inverses and Composites: Exercise $\text{(i)}$
* : Chapter $\text{I}$: Sets and Functions: Composition of Functions
* : Chapter $1$: Theory of Sets: $\S 8$: Composition of Functions and Diagrams: Exercise $3$
Category:Left Inverse Mappings
Category:Surjections