nyuuzyou/wikis huggingface dataset:
Homomorphic Image of R-Module is R-Module

== Theorem ==
Let $\struct {R, +_R, \times_R}$ be a ring.
Let $\struct {G, +_G, \circ_G}_R$ be an $R$-module.
Let $\struct {H, +_H, \circ_H}_R$ be an $R$-algebraic structure.
Let $\phi: G \to H$ be a homomorphism.
Then the homomorphic image of $\phi$ is an $R$-module.
== Proof ==
Let us write $\phi \sqbrk G$ to denote the homomorphic image of $\phi$.
From Image of Group Homomorphism is Subgroup, $\phi \sqbrk G$ is a subgroup of $\struct {H, +_H}$.
For any $\map \phi g$ and $\map \phi {g'}$ in $\phi \sqbrk G$, we have:
hence $\phi \sqbrk G$ is an abelian group.
Now we can turn to showing that $\phi \sqbrk G$ is an $R$-module.
To do this, we take the module axioms in turn.
=== ===
It is to be shown that for all $\lambda \in R$ and $\map \phi g, \map \phi {g'} \in \phi \sqbrk G$:
:$\lambda \circ_H \paren {\map \phi g +_H \map \phi {g'} } = \paren {\lambda \circ_H \map \phi g} +_H \paren {\lambda \circ_H \map \phi {g'} }$
Compute, using that $\phi$ is a homomorphism repetitively:
=== ===
It is to be shown that for all $\lambda, \mu \in R$ and $\map \phi g \in \phi \sqbrk G$:
:$\paren {\lambda +_R \mu} \circ_H \map \phi g = \paren {\lambda \circ_H \map \phi g} +_H \paren {\mu \circ_H \map \phi g}$
Compute, using that $\phi$ is a homomorphism repetitively:
=== ===
It is to be shown that for all $\lambda, \mu \in R$ and $\map \phi g \in \phi \sqbrk G$:
:$\paren {\lambda \times_R \mu} \circ_H \map \phi g = \lambda \circ_H \paren {\mu \circ_H \map \phi g}$
Compute, using that $\phi$ is a homomorphism repetitively:
Having verified that $\phi \sqbrk G$ satisfies the three module axioms, we conclude it is an $R$-module.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations
Category:Module Theory
Homomorphic Image of Vector Space

== Theorem ==
Let $\struct {K, +_K, \times_K}$ be a division ring.
Let $\struct {V, +_V, \circ_V}_K$ be a $K$-vector space.
Let $\struct {W, +_W, \circ_W}_K$ be a $K$-algebraic structure.
Let $\phi: V \to W$ be a homomorphism, that is, a linear transformation.
Then the homomorphic image of $\phi$ is a $K$-vector space.
== Proof ==
Let us write $\phi \sqbrk V$ for the homomorphic image of $\phi$.
From Homomorphic Image of R-Module is R-Module, $\phi \sqbrk V$ is a $K$-module.
It thus suffices to show that $\phi \sqbrk V$ is unitary, since then it will be a $K$-vector space.
To this end, let $1_K$ be the unity of $K$.
Then for any $\map \phi {\mathbf v} \in \phi \sqbrk V$, compute:
Hence $\phi \sqbrk V$ is unitary, and so a $K$-vector space.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations
Category:Homomorphisms (Abstract Algebra)
Category:Linear Algebra
Homomorphism Preserves Subsemigroups

== Theorem ==
Let $\struct {S, \circ}$ and $\struct {T, *}$ be semigroups.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be a homomorphism.
Let $S'$ be a subsemigroup of $S$.
Then $\phi \paren {S'}$ is a subsemigroup of $T$.
== Proof ==
By Restriction of Homomorphism to Image is Epimorphism, $\phi$ is an epimorphism onto its image .
Then by Epimorphism Preserves Semigroups, it follows that the image of $S'$ is a semigroup.
The result follows.
Category:Homomorphisms (Abstract Algebra)
Category:Subsemigroups
Homomorphism from Group of Cube Roots of Unity to Itself

== Theorem ==
Let $\struct {U_3, \times}$ denote the multiplicative group of the complex cube roots of unity.
Here, $U_3 = \set {1, \omega, \omega^2}$ where $\omega = e^{2 i \pi / 3}$.
Let $\phi: U_3 \to U_3$ be defined as:
:$\forall z \in U_3: \map \phi z = \begin{cases} 1 & : z = 1 \\ \omega^2 & : z = \omega \\ \omega & : z = \omega^2 \end{cases}$
Then $\phi$ is a group homomorphism.
== Proof ==
It is noted that
:$\paren {\omega^2}^2 = \omega$
and so $\phi$ is the square function.
By Roots of Unity under Multiplication form Cyclic Group and Cyclic Group is Abelian, $U_3$ is abelian.
Thus for all $a, b \in U_3$:
showing that $\phi$ is a group homomorphism.
== Sources ==
* : Chapter $8$: The Homomorphism Theorem: Example $8.4$
Category:Multiplicative Groups of Complex Roots of Unity
Category:Cyclic Group of Order 3
Homomorphism from Integers into Ring with Unity

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring with unity whose zero is $0_R$ and whose unity is $1_R$.
Let the characteristic of $R$ be $p$.
Let $g_a: \Z \to R$ be the mapping from the integers into $R$ defined as:
:$\forall n \in \Z:\forall a \in R: \map {g_a} n = n \cdot a$
where $\cdot$ denotes the multiple operation.
Then the following hold:
=== Multiple Function on Ring is Homomorphism ===
=== Principal Ideal of Characteristic of Ring is Subset of Kernel of Multiple Function ===
=== Multiplication Function on Ring with Unity is Zero if Characteristic is Divisor ===
=== Kernel of Non-Zero Divisor Multiple Function is Primary Ideal of Characteristic ===
=== Kernel of Non-Zero Divisor Multiple Function is Primary Ideal of Characteristic ===
=== Kernel of Multiple Function on Ring with Characteristic Zero is Trivial ===
=== Multiple Function on Ring is Zero iff Characteristic is Divisor ===
== Examples ==
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $24$. The Division Algorithm: Theorem $24.8$
Category:Ideal Theory
Category:Group Homomorphisms
Category:Integers
Category:Rings with Unity
Category:Homomorphism from Integers into Ring with Unity
Homomorphism from Integers into Ring with Unity/Examples/Characteristic 2

== Example of Use of Homomorphism from Integers into Ring with Unity ==
Let $\struct {R, +, \circ}$ be a ring with unity whose zero is $0_R$ and whose unity is $1_R$.
Let the characteristic of $R$ be $2$.
Then:
:$\forall a \in R: 2 \cdot a = 0$
or equivalently:
:$\forall a \in R: a = -a$
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $24$. The Division Algorithm
Category:Homomorphism from Integers into Ring with Unity
Homomorphism from Reals to Circle Group

== Theorem ==
Let $\struct {\R, +}$ be the additive group of real numbers.
Let $\struct {K, \times}$ be the circle group.
Let $\phi: \struct {\R, +} \to \struct {K, \times}$ be the mapping defined as:
:$\forall x \in \R: \map \phi x = e^{i x}$
Then $\phi$ is a (group) homomorphism.
=== Corollary ===
== Proof ==
Let $x, y \in \R$.
Then:
== Sources ==
* : Chapter $\text{II}$: Groups: Morphisms
Category:Circle Group
Category:Examples of Group Homomorphisms
Homomorphism from Reals to Circle Group/Corollary

== Theorem ==
Let $\struct {\R, +}$ be the additive group of real numbers.
Let $\struct {C_{\ne 0}, \times}$ be the multiplicative group of complex numbers.
Let $\phi: \struct {\R, +} \to \struct {C_{\ne 0}, \times}$ be the mapping defined as:
:$\forall x \in \R: \map \phi x = \cos x + i \sin x$
Then $\phi$ is a (group) homomorphism.
== Proof ==
By Euler's Identity, $\phi$ can also be expressed as:
:$\forall x \in \R: \map \phi x = e^{i x}$
From Homomorphism from Reals to Circle Group, $\phi$ is a homomorphism from $\struct {\R, +}$ to the circle group $\struct {K, \times}$.
From Circle Group is Infinite Abelian Group, we note that $\struct {K, \times}$ is a subgroup of the multiplicative group of complex numbers.
== Sources ==
* : Chapter $\text{II}$: Groups: Exercise $\text{F}$
Category:Circle Group
Category:Examples of Group Homomorphisms
Homomorphism of Chain Complexes induces Homomorphism of Homology

== Theorem ==
Let $A_\bullet$ and $B_\bullet$ be chain complexes of abelian groups.
Let $f: A_\bullet \to B_\bullet$ be a homomorphism.
Then for every $n$, $f$ induces a morphism $H_n \left({A_\bullet}\right) \to H_n \left({B_\bullet}\right)$ of homology groups.
== Proof ==
Let $\partial^A_\bullet$, $\partial^B_\bullet$ denote the differential on $A_\bullet$, respectively $B_\bullet$. 
First it will be demonstrated that:
:$\forall a \in \ker \left({\partial^A_n}\right) \subseteq A_n: f_n \left({a}\right) \in \ker \left({\partial^B_n}\right)$
Thus:
:$\partial^B_n f_n \left({a}\right) = f_{n - 1} \left({\partial^A_n a}\right) = f_{n - 1} \left({0}\right) = 0$
Thus there exists a map:
:$\bar f_n: \ker \left({\partial^A_n}\right) \to \ker \left({\partial^B_n}\right)$
via the restriction of $f$.
Next will be shown that:
:$\forall a \in \operatorname {Im} \left({\partial^A_{n + 1} }\right): f_n \left({a}\right) \in \operatorname {Im} \left({\partial^B_{n + 1} }\right)$
Let $a = \partial^A_{n + 1}a'$
Then:
:$\partial^B_{n + 1} \left({f_{n + 1} \left({a'}\right)}\right) = f_n \left({\partial^A_n a'}\right) = f_n \left({a}\right)$.
Let $\pi: \ker \left({\partial^B_n}\right) \to H_n \left({B_\bullet}\right)$ be the quotient map.
Let $\rho = \pi \circ \bar f_n$.
From above:
:$\operatorname{Im} \left({\partial^A_{n + 1} }\right) \subseteq \ker \left({\rho}\right)$
Thus $\rho$ factors through a map:
:$\tilde f_n: H_n \left({A_\bullet}\right) \to H_n \left({B_\bullet}\right)$
completing the proof.
Category:Homological Algebra
Homomorphism of External Direct Products

== Theorem ==
Let:
:$\struct {S_1 \times S_2, \circ}$ be the external direct product of two algebraic structures $\struct {S_1, \circ_1}$ and $\struct {S_2, \circ_2}$
:$\struct {T_1 \times T_2, *}$ be the external direct product of two algebraic structures $\struct {T_1, *_1}$ and $\struct {T_2, *_2}$
:$\phi_1$ be a homomorphism from $\struct {S_1, \circ_1}$ onto $\struct {T_1, *_1}$
:$\phi_2$ be a homomorphism from $\struct {S_2, \circ_2}$ onto $\struct {T_2, *_2}$.
Then the mapping $\phi_1 \times \phi_2: \struct {S_1 \times S_2, \circ} \to \struct {T_1 \times T_2, *}$ defined as:
:$\map {\paren {\phi_1 \times \phi_2} } {x, y} = \tuple {\map {\phi_1} x, \map {\phi_2} y}$
is a homomorphism from $\struct {S_1 \times S_2, \circ}$ to $\struct {T_1 \times T_2, *}$.
=== General Result ===
== Proof ==
Let $\tuple {x_1, x_2}, \tuple {y_1, y_2} \in S_1 \circ S_2$.
Then:
thus demonstrating the morphism property.
Category:Homomorphism of External Direct Products
Category:Homomorphisms (Abstract Algebra)
Category:External Direct Products
Homomorphism of External Direct Products/General Result

== Theorem ==
Let $n \in \N_{>0}$.
Let:
be external direct products of algebraic structures.
Let $\Phi_n: \struct {\SS_n, \circledcirc_n} \to \struct {\TT_n, \circledast_n}$ be the mapping defined as:
:$\Phi_n: \tuple {s_1, \ldots, s_n} := \begin{cases}
\map {\phi_1} {s_1} & : n = 1 \\
\tuple {\map {\phi_1} {s_1}, \map {\phi_2} {s_2} } & : n = 2 \\
\tuple {\map {\Phi_n} {s_1, \ldots, s_{n - 1} }, \map {\phi_n} {s_n} } & : n > 2 \\
\end{cases}$
That is:
:$\Phi_n: \tuple {s_1, \ldots, s_n} := \tuple {\map {\phi_1} {s_1}, \map {\phi_2} {s_2}, \ldots, \map {\phi_n} {s_n} }$
Let $\phi_k: \struct {S_k, \circ_k} \to \struct {T_k, \ast_k}$ be a homomorphism for each $k \in \set {1, 2, \ldots, n}$.
Then $\Phi_n$ is a homomorphism from $\struct {\SS_n, \circledcirc_n}$ to $\struct {\TT_n, \circledast_n}$.
== Proof ==
Proof by induction:
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$\Phi_n: \struct {\SS_n, \circledcirc_n} \to \struct {\TT_n, \circledast_n}$ is a homomorphism.
=== Basis for the Induction ===
$\map P 1$ is true, as this just says:
:$\Phi_1 = \phi_1: \struct {S_1, \circ_1} \to \struct {T_1, \ast_1}$ is a homomorphism
which holds .
$\map P 2$ is the case:
:$\Phi_2 = \phi_1 \times \phi_2: \struct {S_1 \times S_2, \circledcirc_2} \to \struct {T_1 \times T_2, \circledast_2}$ is a homomorphism
which has been proved in Homomorphism of External Direct Products.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P m$ is true, where $m \ge 2$, then it logically follows that $\map P {m + 1}$ is true.
So this is our induction hypothesis:
:$\Phi_m: \struct {\SS_m, \circledcirc_m} \to \struct {\TT_m, \circledast_m}$ is a homomorphism.
Then we need to show:
:$\Phi_{m + 1}: \struct {\SS_{m + 1}, \circledcirc_{m + 1} } \to \struct {\TT_{m + 1}, \circledast_{m + 1} }$ is a homomorphism.
=== Induction Step ===
This is our induction step:
Let $\tuple {s_1, \ldots, s_{m + 1} }$ and $\tuple {t_1, \ldots, t_{m + 1} }$ be arbitrary elements of $\SS_{m + 1}$.
So $\map P m \implies \map P {m + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \N_{>0}: \Phi_n: \struct {\SS_n, \circledcirc_n} \to \struct {\TT_n, \circledast_n}$ is a homomorphism.
Category:Homomorphism of External Direct Products
Homomorphism of Generated Group

== Theorem ==
Let $\struct {G, \circ}$ and $\struct {H, \circ}$ be groups.
Let $\phi: G \to H$ and $\psi: G \to H$ be homomorphisms. 
Let $\gen S = G$ be the group generated by $S$.
Let:
:$\forall x \in S: \map \phi x = \map \psi x$
Then:
:$\phi = \psi$
== Proof ==
Let $H = \set {x \in G: \map \phi x = \map \psi x}$.
From Elements of Group with Equal Images under Homomorphisms form Subgroup, $H$ is a subgroup of $G$.
But from the definition of the group generated by $S$, the smallest subgroup that contains $S$ is $G$ itself.
Thus:
:$G = \set {x \in G: \map \phi x = \map \psi x}$
and so:
:$\phi = \psi$
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Theorem $14.8$
Category:Group Homomorphisms
Category:Generators of Groups
Homomorphism of Power of Group Element

== Theorem ==
Let $\struct {G, \circ}$ and $\struct {H, \ast}$ be groups.
Let $\phi: S \to T$ be a group homomorphism.
Then:
:$\forall n \in \Z: \forall g \in G: \map \phi {g^n} = \paren {\map \phi g}^n$
== Proof ==
The result for $n \in \N_{>0}$ follows directly from General Morphism Property for Semigroups.
For $n = 0$, we use Homomorphism with Cancellable Codomain Preserves Identity.
For $n < 0$, we use Homomorphism with Identity Preserves Inverses, along with Index Laws for Monoids: Negative Index.
== Sources ==
* : $\S 7.2$. Some lemmas on homomorphisms: Example $133$
* : $\S 47.5$ Homomorphisms and their elementary properties
* : Chapter $8$: The Homomorphism Theorem: Exercise $4$
Category:Group Homomorphisms
Homomorphism of Powers

== Theorem ==
Let $\struct {T_1, \odot}$ and $\struct {T_2, \oplus}$ be semigroups.
Let $\phi: \struct {T_1, \odot} \to \struct {T_2, \oplus}$ be a (semigroup) homomorphism.
=== Naturally Ordered Semigroup ===
=== Natural Numbers ===
=== Integers ===
Category:Homomorphism of Powers
Category:Homomorphisms (Abstract Algebra)
Homomorphism of Powers/Integers

== Theorem ==
Let $\struct {T_1, \odot}$ and $\struct {T_2, \oplus}$ be monoids.
Let $\phi: \struct {T_1, \odot} \to \struct {T_2, \oplus}$ be a (semigroup) homomorphism.
Let $a$ be an invertible element of $T_1$.
Let $n \in \Z$.
Let $\odot^n$ and $\oplus^n$ be as defined as in Index Laws for Monoids.
Then:
:$\forall n \in \Z: \map \phi {\map {\odot^n} a} = \map {\oplus^n} {\map \phi a}$
== Proof ==
By Homomorphism of Powers: Natural Numbers, we need show this only for negative $n$, that is:
:$\forall n \in \N^*: \map \phi {\map {\odot^{-n} } a} = \map {\oplus^{-n} } {\map \phi a}$
But by Homomorphism with Identity Preserves Inverses:
:$\map \phi {a^{-1} } = \paren {\map \phi a}^{-1}$
Hence by Homomorphism of Powers: Natural Numbers:
:$\map {\oplus^{-n} } {\map \phi a} = \map {\oplus^n} {\map \phi {a^{-1} } } = \map \phi {\map {\odot^n} {a^{-1} } } = \map \phi {\map {\odot^{-n} } a}$
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $20$. The Integers: Theorem $20.13$
Category:Homomorphism of Powers
Category:Integers
Homomorphism of Powers/Natural Numbers

== Theorem ==
Let $\struct {T_1, \odot}$ and $\struct {T_2, \oplus}$ be semigroups.
Let $\phi: \struct {T_1, \odot} \to \struct {T_2, \oplus}$ be a (semigroup) homomorphism.
Let $n \in \N$.
Let $\odot^n$ and $\oplus^n$ be the $n$th powers of $\odot$ and $\oplus$, respectively.
Then:
:$\forall a \in T_1: \forall n \in \N: \map \phi {\map {\odot^n} a} = \map {\oplus^n} {\map \phi a}$
== Proof ==
Consider the natural numbers $\N$ defined as a naturally ordered semigroup.
Then the result follows from Homomorphism of Powers: Naturally Ordered Semigroup.
Category:Homomorphism of Powers
Category:Natural Numbers
Homomorphism of Powers/Naturally Ordered Semigroup

== Theorem ==
Let $\struct {T_1, \odot}$ and $\struct {T_2, \oplus}$ be semigroups.
Let $\phi: \struct {T_1, \odot} \to \struct {T_2, \oplus}$ be a (semigroup) homomorphism.
Let $\struct {S, \circ, \preceq}$ be a naturally ordered semigroup.
For a given $a \in T_1$, let $\map {\odot^n} a$ be the $n$th power of $a$ in $T_1$.
For a given $a \in T_2$, let $\map {\oplus^n} a$ be the $n$th power of $a$ in $T_2$.
Then:
:$\forall a \in T_1: \forall n \in \struct {S^*, \circ, \preceq}: \map \phi {\map {\odot^n} a} = \map {\oplus^n} {\map \phi a}$
where $S^* = S \setminus \set 0$.
== Proof ==
The proof proceeds by the Principle of Mathematical Induction for a Naturally Ordered Semigroup.
Let $A := \set {n \in S^*: \forall a \in T_1: \map \phi {\map {\odot^n} a} = \map {\oplus^n} {\map \phi a} }$
That is, $A$ is defined as the set of all $n$ such that:
:$\forall a \in T_1 \map \phi {\map {\odot^n} a} = \map {\oplus^n} {\map \phi a}$
=== Basis for the Induction ===
We have that:
So $1 \in A$.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $k \in A$ where $k \ge 1$, then it logically follows that $k \circ 1 \in A$.
So this is our induction hypothesis:
:$\forall a \in T_1: \map \phi {\map {\odot^k} a} = \map {\oplus^k} {\map \phi a}$
Then we need to show:
:$\forall a \in T_1: \map \phi {\map {\odot^{k \circ 1} } a} = \map {\oplus^{k \circ 1} } {\map \phi a}$
=== Induction Step ===
This is our induction step:
So $k \in A \implies k \circ 1 \in A$ and the result follows by the Principle of Mathematical Induction:
:$\forall n \in \struct {S^*, \circ, \preceq}: \map \phi {\map {\odot^n} a} = \map {\oplus^n} {\map \phi a}$
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Theorem $16.14$
* : Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Exercise $16.28 \ \text {(a)}$
Category:Homomorphism of Powers
Category:Naturally Ordered Semigroup
Homomorphism of Ring Subtraction

== Theorem ==
Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a ring homomorphism.
Then:
:$\forall a, b \in R_1: \map \phi {a -_1 b} = \map \phi a -_2 \map \phi b$
where $a -_1 b$ denotes subtraction of $b$ from $a$.
== Proof ==
== Sources ==
* : Chapter $5$: Rings: $\S 24$. Homomorphisms: Theorem $44 \ \text{(iii)}$
Category:Ring Homomorphisms
Homomorphism on Induced Structure to Commutative Semigroup

== Theorem ==
Let $\struct {S, \circ}$ be an algebraic structure.
Let $\struct {T, \oplus}$ be a commutative semigroup.
Let $T^S$ be the set of all mappings from $S$ to $T$.
Let $f$ and $g$ be homomorphisms from $S$ into $T$.
Let $f \oplus' g$ be the pointwise operation on $T^S$ induced by $\oplus$.
Then $f \oplus' g$ is a homomorphism from $\struct {S, \circ}$ into $\struct {T, \oplus}$.
== Proof ==
Let $\struct {T, \oplus}$ be a commutative semigroup.
Let $x, y \in S$.
Then:
Notice that for this to work, $\oplus$ needs to be both associative and commutative.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Theorem $13.7: \ 1^\circ$
Category:Homomorphisms (Abstract Algebra)
Category:Pointwise Operations
Category:Commutative Semigroups
Homomorphism to Group Preserves Identity

== Theorem ==
Let $\struct {S, \circ}$ be an algebraic structure.
Let $\struct {T, *}$ be a group.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be a homomorphism.
Let $\struct {S, \circ}$ have an identity $e_S$.
Then:
:$\map \phi {e_S} = e_T$
== Proof ==
By hypothesis, $\struct {T, *}$ is a group.
By the Cancellation Laws, all elements of $T$ are cancellable.
Thus Homomorphism with Cancellable Codomain Preserves Identity can be applied.
== Sources ==
* : $\S 7.2$. Some lemmas on homomorphisms: Lemma $\text{(ii)}$
* : Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Theorem $12.3: \ 3^\circ$
* : $\S 1.10$: Theorem $22$
Category:Homomorphisms (Abstract Algebra)
Category:Group Theory
Homomorphism to Group Preserves Inverses

== Theorem ==
Let $\struct {S, \circ}$ be an algebraic structure.
Let $\struct {T, *}$ be a group.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be a homomorphism.
Let $\struct {S, \circ}$ have an identity $e_S$.
Let $x^{-1}$ be an inverse of $x$ for $\circ$.
Then $\map \phi {x^{-1} }$ is an inverse of $\map \phi x$ for $*$.
== Proof ==
By hypothesis, $\struct {T, *}$ is a group.
By , $\struct {T, *}$ has an identity.
Thus Homomorphism with Identity Preserves Inverses can be applied.
== Sources ==
* : $\S 7.2$. Some lemmas on homomorphisms: Lemma $\text{(iii)}$
* : Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Theorem $12.3: \ 3^\circ$
* : $\S 1.10$: Theorem $22$
Category:Homomorphisms (Abstract Algebra)
Category:Group Theory
Homomorphism with Cancellable Codomain Preserves Identity

== Theorem ==
Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be a homomorphism.
Let $\struct {S, \circ}$ have an identity $e_S$.
Let $\struct {T, *}$ have an identity $e_T$.
Let every element of $\struct {T, *}$ be cancellable.
Then $\map \phi {e_S}$ is the identity $e_T$.
== Proof ==
Let $\struct {S, \circ}$ be an algebraic structure in which $\circ$ has an identity $e_S$.
Let $\struct {T, *}$ be an algebraic structure in which $*$ has an identity $e_T$.
Let $\struct {T, *}$ be such that every element is cancellable.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be a homomorphism.
Every element of $\struct {T, *}$ is cancellable.
Suppose there is an idempotent element in $\struct {T, *}$
So from Identity is only Idempotent Cancellable Element, it must be the identity $e_T$.
Thus:
So $\map \phi {e_S}$ is idempotent in $\struct {T, *}$ and the result follows.
== Also see ==
* Epimorphism Preserves Identity
* Homomorphism with Identity Preserves Inverses
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Theorem $12.3: \ 1^\circ$
Category:Homomorphisms (Abstract Algebra)
Category:Identity Elements
Homomorphism with Identity Preserves Inverses

== Theorem ==
Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be a homomorphism.
Let $\struct {S, \circ}$ have an identity $e_S$.
Let $\struct {T, *}$ also have an identity $e_T = \map \phi {e_S}$.
If $x^{-1}$ is an inverse of $x$ for $\circ$, then $\map \phi {x^{-1} }$ is an inverse 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 $e_S$.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be a homomorphism.
Let $\struct {T, *}$ be an algebraic structure in which $*$ has an identity $e_T = \map \phi {e_S}$.
Let $x^{-1}$ be an inverse of $x$ for $\circ$.
Then by existence of identity $e_S$:
:$\forall x \in S: x \circ x^{-1}, x^{-1} \circ x \in \Dom \phi$
Hence:
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Theorem $12.3: \ 2^\circ$
Category:Homomorphisms (Abstract Algebra)
Category:Inverse Elements