nyuuzyou/wikis huggingface dataset:
Group Homomorphism/Examples/Pointwise Addition on Continuous Real Functions on Closed Unit Interval/Example 10/Kernel

== Example of Group Homomorphism ==
Let $J \subseteq \R$ denote the closed unit interval $\closedint 0 1$.
Let $\map {\mathscr C} J$ denote the set of all continuous real functions from $J$ to $\R$.
Let $G = \struct {\map {\mathscr C} J, +}$ denote the group formed on $\map {\mathscr C} J$ by pointwise addition.
Let $\struct {\R, +}$ denote the additive group of real numbers.
Let $I_J$ denote the identity mapping on $J$:
:$\forall x \in J: \map {I_J} x = x$
Let $\phi: \struct {\map {\mathscr C} J, +} \to \struct {\R, +}$ be the homomorphism defined as:
:$\forall f \in \map {\mathscr C} J: \map \phi f = -\map f 0 + \ds \int_{-2}^0 \map f {e^x} \rd x$
The kernel of $\phi$ is given by:
:$\map \ker \phi = I_J - f_m$
where:
:$f_m: \R \to \R$ denotes the constant mapping on $\R$
:$m = 1 - e^{-2}$
:$I_J$ denotes the identity mapping on $J$.
== Proof ==
From Group Homomorphism: Example 10, we have that $\phi$ is indeed a homomorphism.
For all $c \in \R$, let $f_c: \R \to \R$ be the constant mapping:
:$\forall x \in \R: \map {f_c} x = c$
First we show that:
:$\forall c \in \R: \map \phi {f_c} = c$
Let $c \in \R$ be arbitrary.
We have:
Then we show that there exists a unique $m \in \R$ such that:
:$\map \phi {I_J - f_m} = 0$
where in this case:
:$m = 1 - e^{-2}$
We have:
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Exercise $13.18 \ \text {(j)}$
Category:Examples of Group Homomorphisms
Group Homomorphism/Examples/Pointwise Addition on Continuous Real Functions on Closed Unit Interval/Example 2

== Example of Group Homomorphism ==
Let $J \subseteq \R$ denote the closed unit interval $\closedint 0 1$.
Let $\map {\mathscr C} J$ denote the set of all continuous real functions from $J$ to $\R$.
Let $G = \struct {\map {\mathscr C} J, +}$ be the group formed on $\map {\mathscr C} J$ by pointwise addition.
Let $\struct {\R, +}$ denote the additive group of real numbers.
From Pointwise Addition on Continuous Real Functions on Closed Unit Interval forms Group we have that $G$ is indeed a group.
Let $\phi: \struct {\map {\mathscr C} J, +} \to \struct {\R, +}$ be the mapping defined as:
:$\forall f \in \map {\mathscr C} J: \map \phi f = \size {\map f 0}$
where $\size {\, \cdot \,}$ denotes the absolute value function.
Then $\phi$ is not a homomorphism.
== Proof ==
Let $f, g \in \map {\mathscr C} J$ be such that:
By Constant Real Function is Continuous, both $f$ and $g$ are continuous on $J$.
We have:
However:
Thus:
:$\map \phi {f + g} \ne \map \phi f + \map \phi g$
and so $\phi$ is not a homomorphism by definition.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Exercise $13.18 \ \text {(b)}$
Category:Examples of Group Homomorphisms
Group Homomorphism/Examples/Pointwise Addition on Continuous Real Functions on Closed Unit Interval/Example 3

== Example of Group Homomorphism ==
Let $J \subseteq \R$ denote the closed unit interval $\closedint 0 1$.
Let $\map {\mathscr C} J$ denote the set of all continuous real functions from $J$ to $\R$.
Let $G = \struct {\map {\mathscr C} J, +}$ be the group formed on $\map {\mathscr C} J$ by pointwise addition.
Let $\struct {\R, +}$ denote the additive group of real numbers.
From Pointwise Addition on Continuous Real Functions on Closed Unit Interval forms Group we have that $G$ is indeed a group.
Let $\phi: \struct {\map {\mathscr C} J, +} \to \struct {\R, +}$ be the mapping defined as:
:$\forall f \in \map {\mathscr C} J: \map \phi f = \ds \int_0^1 \map f x \rd x$
Then $\phi$ is a homomorphism.
=== Kernel ===
== Proof ==
Let $f, g \in \map {\mathscr C} J$ be arbitrary.
As both $f$ and $g$ are continuous real functions on $J$, they are integrable on $\closedint 0 1$.
We have:
Thus $\phi$ is a homomorphism by definition.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Exercise $13.18 \ \text {(c)}$
Category:Examples of Group Homomorphisms
Group Homomorphism/Examples/Pointwise Addition on Continuous Real Functions on Closed Unit Interval/Example 3/Kernel

== Example of Group Homomorphism ==
Let $J \subseteq \R$ denote the closed unit interval $\closedint 0 1$.
Let $\map {\mathscr C} J$ denote the set of all continuous real functions from $J$ to $\R$.
Let $G = \struct {\map {\mathscr C} J, +}$ denote the group formed on $\map {\mathscr C} J$ by pointwise addition.
Let $\struct {\R, +}$ denote the additive group of real numbers.
Let $I_J$ denote the identity mapping on $J$:
:$\forall x \in J: \map {I_J} x = x$
Let $\phi: \struct {\map {\mathscr C} J, +} \to \struct {\R, +}$ be the homomorphism defined as:
:$\forall f \in \map {\mathscr C} J: \map \phi f = \ds \int_0^1 \map f x \rd x$
The kernel of $\phi$ is given by:
:$\map \ker \phi = I_J - f_m$
where:
:$f_m: \R \to \R$ denotes the constant mapping on $\R$
:$m = \dfrac 1 2$
:$I_J$ denotes the identity mapping on $J$.
== Proof ==
From Group Homomorphism: Example 3, we have that $\phi$ is indeed a homomorphism.
For all $c \in \R$, let $f_c: \R \to \R$ be the constant mapping:
:$\forall x \in \R: \map {f_c} x = c$
First we show that:
:$\forall c \in \R: \map \phi {f_c} = c$
Let $c \in \R$ be arbitrary.
We have:
Then we show that there exists a unique $m \in \R$ such that:
:$\map \phi {I_J - f_m} = 0$
where in this case:
:$m = \dfrac 1 2$
We have:
Hence the result by definition of kernel.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Exercise $13.18 \ \text {(c)}$
Category:Examples of Group Homomorphisms
Group Homomorphism/Examples/Pointwise Addition on Continuous Real Functions on Closed Unit Interval/Example 4

== Example of Group Homomorphism ==
Let $J \subseteq \R$ denote the closed unit interval $\closedint 0 1$.
Let $\map {\mathscr C} J$ denote the set of all continuous real functions from $J$ to $\R$.
Let $G = \struct {\map {\mathscr C} J, +}$ be the group formed on $\map {\mathscr C} J$ by pointwise addition.
Let $\struct {\R, +}$ denote the additive group of real numbers.
From Pointwise Addition on Continuous Real Functions on Closed Unit Interval forms Group we have that $G$ is indeed a group.
Let $\phi: \struct {\map {\mathscr C} J, +} \to \struct {\R, +}$ be the mapping defined as:
:$\forall f \in \map {\mathscr C} J: \map \phi f = \ds \dfrac \pi 3 \int_0^1 \map f x \cos \dfrac {\pi x} 6 \rd x$
Then $\phi$ is a homomorphism.
=== Kernel ===
== Proof ==
Let $f, g \in \map {\mathscr C} J$ be arbitrary.
As both $f$ and $g$ are continuous real functions on $J$, they are integrable on $\closedint 0 1$.
We have:
Thus $\phi$ is a homomorphism by definition.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Exercise $13.18 \ \text {(d)}$
Category:Examples of Group Homomorphisms
Group Homomorphism/Examples/Pointwise Addition on Continuous Real Functions on Closed Unit Interval/Example 4/Kernel

== Example of Group Homomorphism ==
Let $J \subseteq \R$ denote the closed unit interval $\closedint 0 1$.
Let $\map {\mathscr C} J$ denote the set of all continuous real functions from $J$ to $\R$.
Let $G = \struct {\map {\mathscr C} J, +}$ denote the group formed on $\map {\mathscr C} J$ by pointwise addition.
Let $\struct {\R, +}$ denote the additive group of real numbers.
Let $I_J$ denote the identity mapping on $J$:
:$\forall x \in J: \map {I_J} x = x$
Let $\phi: \struct {\map {\mathscr C} J, +} \to \struct {\R, +}$ be the homomorphism defined as:
:$\forall f \in \map {\mathscr C} J: \map \phi f = \ds \dfrac \pi 3 \int_0^1 \map f x \cos \dfrac {\pi x} 6 \rd x$
The kernel of $\phi$ is given by:
:$\map \ker \phi = I_J - f_m$
where:
:$f_m: \R \to \R$ denotes the constant mapping on $\R$
:$m = \dfrac {6 \paren {\sqrt 3 - 2} } \pi$
:$I_J$ denotes the identity mapping on $J$.
== Proof ==
From Group Homomorphism: Example 4, we have that $\phi$ is indeed a homomorphism.
For all $c \in \R$, let $f_c: \R \to \R$ be the constant mapping:
:$\forall x \in \R: \map {f_c} x = c$
First we show that:
:$\forall c \in \R: \map \phi {f_c} = c$
Let $c \in \R$ be arbitrary.
We have:
Then we show that there exists a unique $m \in \R$ such that:
:$\map \phi {I_J - f_m} = 0$
where in this case:
:$m = \dfrac {6 \paren {\sqrt 3 - 2} } \pi$
We have:
Hence the result by definition of kernel.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Exercise $13.18 \ \text {(d)}$
Category:Examples of Group Homomorphisms
Group Homomorphism/Examples/Pointwise Addition on Continuous Real Functions on Closed Unit Interval/Example 5

== Example of Group Homomorphism ==
Let $J \subseteq \R$ denote the closed unit interval $\closedint 0 1$.
Let $\map {\mathscr C} J$ denote the set of all continuous real functions from $J$ to $\R$.
Let $G = \struct {\map {\mathscr C} J, +}$ be the group formed on $\map {\mathscr C} J$ by pointwise addition.
Let $\struct {\R, +}$ denote the additive group of real numbers.
From Pointwise Addition on Continuous Real Functions on Closed Unit Interval forms Group we have that $G$ is indeed a group.
Let $\phi: \struct {\map {\mathscr C} J, +} \to \struct {\R, +}$ be the mapping defined as:
:$\forall f \in \map {\mathscr C} J: \map \phi f = \ds \int_0^1 \map \cos {\dfrac {\pi \map f x} 6} \rd x$
Then $\phi$ is not a homomorphism.
== Proof ==
Let $f, g \in \map {\mathscr C} J$ be arbitrary.
As both $f$ and $g$ are continuous real functions on $J$, both $\map \cos {\dfrac {\pi \map f x} 6}$ and $\map \cos {\dfrac {\pi \map g x} 6}$ are integrable on $\closedint 0 1$.
We have:
However:
Thus:
:$\map \phi {f + g} \ne \map \phi f + \map \phi g$
and so $\phi$ is not a homomorphism by definition.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Exercise $13.18 \ \text {(e)}$
Category:Examples of Group Homomorphisms
Group Homomorphism/Examples/Pointwise Addition on Continuous Real Functions on Closed Unit Interval/Example 6

== Example of Group Homomorphism ==
Let $J \subseteq \R$ denote the closed unit interval $\closedint 0 1$.
Let $\map {\mathscr C} J$ denote the set of all continuous real functions from $J$ to $\R$.
Let $G = \struct {\map {\mathscr C} J, +}$ be the group formed on $\map {\mathscr C} J$ by pointwise addition.
Let $\struct {\R, +}$ denote the additive group of real numbers.
From Pointwise Addition on Continuous Real Functions on Closed Unit Interval forms Group we have that $G$ is indeed a group.
Let $\phi: \struct {\map {\mathscr C} J, +} \to \struct {\R, +}$ be the mapping defined as:
:$\forall f \in \map {\mathscr C} J: \map \phi f = \ds \int_0^1 \map f {\cos \dfrac {\pi x} 6} \rd x$
Then $\phi$ is a homomorphism.
=== Kernel ===
== Proof ==
Let $f, g \in \map {\mathscr C} J$ be arbitrary.
As both $f$ and $g$ are continuous real functions on $J$, they are integrable on $\closedint 0 1$.
We have:
Thus $\phi$ is a homomorphism by definition.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Exercise $13.18 \ \text {(f)}$
Category:Examples of Group Homomorphisms
Group Homomorphism/Examples/Pointwise Addition on Continuous Real Functions on Closed Unit Interval/Example 6/Kernel

== Example of Group Homomorphism ==
Let $J \subseteq \R$ denote the closed unit interval $\closedint 0 1$.
Let $\map {\mathscr C} J$ denote the set of all continuous real functions from $J$ to $\R$.
Let $G = \struct {\map {\mathscr C} J, +}$ denote the group formed on $\map {\mathscr C} J$ by pointwise addition.
Let $\struct {\R, +}$ denote the additive group of real numbers.
Let $I_J$ denote the identity mapping on $J$:
:$\forall x \in J: \map {I_J} x = x$
Let $\phi: \struct {\map {\mathscr C} J, +} \to \struct {\R, +}$ be the homomorphism defined as:
:$\forall f \in \map {\mathscr C} J: \map \phi f = \ds \int_0^1 \map f {\cos \dfrac {\pi x} 6} \rd x$
The kernel of $\phi$ is given by:
:$\map \ker \phi = I_J - f_m$
where:
:$f_m: \R \to \R$ denotes the constant mapping on $\R$
:$m = \dfrac 3 \pi$
:$I_J$ denotes the identity mapping on $J$.
== Proof ==
From Group Homomorphism: Example 6, we have that $\phi$ is indeed a homomorphism.
For all $c \in \R$, let $f_c: \R \to \R$ be the constant mapping:
:$\forall x \in \R: \map {f_c} x = c$
First we show that:
:$\forall c \in \R: \map \phi {f_c} = c$
Let $c \in \R$ be arbitrary.
We have:
Then we show that there exists a unique $m \in \R$ such that:
:$\map \phi {I_J - f_m} = 0$
where in this case:
:$m = \dfrac 3 \pi$
We have:
Hence the result by definition of kernel.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Exercise $13.18 \ \text {(f)}$
Category:Examples of Group Homomorphisms
Group Homomorphism/Examples/Pointwise Addition on Continuous Real Functions on Closed Unit Interval/Example 7

== Example of Group Homomorphism ==
Let $J \subseteq \R$ denote the closed unit interval $\closedint 0 1$.
Let $\map {\mathscr C} J$ denote the set of all continuous real functions from $J$ to $\R$.
Let $G = \struct {\map {\mathscr C} J, +}$ be the group formed on $\map {\mathscr C} J$ by pointwise addition.
Let $\struct {\R, +}$ denote the additive group of real numbers.
From Pointwise Addition on Continuous Real Functions on Closed Unit Interval forms Group we have that $G$ is indeed a group.
Let $\phi: \struct {\map {\mathscr C} J, +} \to \struct {\R, +}$ be the mapping defined as:
:$\forall f \in \map {\mathscr C} J: \map \phi f = \ds \int_0^1 \int_0^1 \map f x \map f y \rd y \rd x$
Then $\phi$ is not a homomorphism.
== Proof ==
Let $f, g \in \map {\mathscr C} J$ be arbitrary.
As both $f$ and $g$ are continuous real functions on $J$, they are integrable on $\closedint 0 1$.
We have:
Thus $\phi$ is not a homomorphism by definition.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Exercise $13.18 \ \text {(g)}$
Category:Examples of Group Homomorphisms
Group Homomorphism/Examples/Pointwise Addition on Continuous Real Functions on Closed Unit Interval/Example 8

== Example of Group Homomorphism ==
Let $J \subseteq \R$ denote the closed unit interval $\closedint 0 1$.
Let $\map {\mathscr C} J$ denote the set of all continuous real functions from $J$ to $\R$.
Let $G = \struct {\map {\mathscr C} J, +}$ be the group formed on $\map {\mathscr C} J$ by pointwise addition.
Let $\struct {\R, +}$ denote the additive group of real numbers.
From Pointwise Addition on Continuous Real Functions on Closed Unit Interval forms Group we have that $G$ is indeed a group.
Let $\phi: \struct {\map {\mathscr C} J, +} \to \struct {\R, +}$ be the mapping defined as:
:$\forall f \in \map {\mathscr C} J: \map \phi f = \ds \int_0^1 \int_0^1 \map f {x y} \rd y \rd x$
Then $\phi$ is a homomorphism.
=== Kernel ===
== Proof ==
Let $f, g \in \map {\mathscr C} J$ be arbitrary.
As both $f$ and $g$ are continuous real functions on $J$, they are integrable on $\closedint 0 1$.
We have:
Thus $\phi$ is a homomorphism by definition.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Exercise $13.18 \ \text {(h)}$
Category:Examples of Group Homomorphisms
Group Homomorphism/Examples/Pointwise Addition on Continuous Real Functions on Closed Unit Interval/Example 8/Kernel

== Example of Group Homomorphism ==
Let $J \subseteq \R$ denote the closed unit interval $\closedint 0 1$.
Let $\map {\mathscr C} J$ denote the set of all continuous real functions from $J$ to $\R$.
Let $G = \struct {\map {\mathscr C} J, +}$ denote the group formed on $\map {\mathscr C} J$ by pointwise addition.
Let $\struct {\R, +}$ denote the additive group of real numbers.
Let $I_J$ denote the identity mapping on $J$:
:$\forall x \in J: \map {I_J} x = x$
Let $\phi: \struct {\map {\mathscr C} J, +} \to \struct {\R, +}$ be the homomorphism defined as:
:$\forall f \in \map {\mathscr C} J: \map \phi f = \ds \int_0^1 \int_0^1 \map f {x y} \rd y \rd x$
The kernel of $\phi$ is given by:
:$\map \ker \phi = I_J - f_m$
where:
:$f_m: \R \to \R$ denotes the constant mapping on $\R$
:$m = \dfrac 1 4$
:$I_J$ denotes the identity mapping on $J$.
== Proof ==
From Group Homomorphism: Example 8, we have that $\phi$ is indeed a homomorphism.
For all $c \in \R$, let $f_c: \R \to \R$ be the constant mapping:
:$\forall x \in \R: \map {f_c} x = c$
First we show that:
:$\forall c \in \R: \map \phi {f_c} = c$
Let $c \in \R$ be arbitrary.
We have:
Then we show that there exists a unique $m \in \R$ such that:
:$\map \phi {I_J - f_m} = 0$
where in this case:
:$m = \dfrac 1 4$
We have:
Hence the result by definition of kernel.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Exercise $13.18 \ \text {(h)}$
Category:Examples of Group Homomorphisms
Group Homomorphism/Examples/Pointwise Addition on Continuous Real Functions on Closed Unit Interval/Example 9

== Example of Group Homomorphism ==
Let $J \subseteq \R$ denote the closed unit interval $\closedint 0 1$.
Let $\map {\mathscr C} J$ denote the set of all continuous real functions from $J$ to $\R$.
Let $G = \struct {\map {\mathscr C} J, +}$ be the group formed on $\map {\mathscr C} J$ by pointwise addition.
Let $\struct {\R, +}$ denote the additive group of real numbers.
From Pointwise Addition on Continuous Real Functions on Closed Unit Interval forms Group we have that $G$ is indeed a group.
Let $\phi: \struct {\map {\mathscr C} J, +} \to \struct {\R, +}$ be the mapping defined as:
:$\forall f \in \map {\mathscr C} J: \map \phi f = \ds 2 \int_0^1 \int_0^x \map f y \rd y \rd x$
Then $\phi$ is a homomorphism.
=== Kernel ===
== Proof ==
Let $f, g \in \map {\mathscr C} J$ be arbitrary.
As both $f$ and $g$ are continuous real functions on $J$, they are integrable on $\closedint 0 1$.
We have:
Thus $\phi$ is a homomorphism by definition.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Exercise $13.18 \ \text {(i)}$
Category:Examples of Group Homomorphisms
Group Homomorphism/Examples/Pointwise Addition on Continuous Real Functions on Closed Unit Interval/Example 9/Kernel

== Example of Group Homomorphism ==
Let $J \subseteq \R$ denote the closed unit interval $\closedint 0 1$.
Let $\map {\mathscr C} J$ denote the set of all continuous real functions from $J$ to $\R$.
Let $G = \struct {\map {\mathscr C} J, +}$ denote the group formed on $\map {\mathscr C} J$ by pointwise addition.
Let $\struct {\R, +}$ denote the additive group of real numbers.
Let $I_J$ denote the identity mapping on $J$:
:$\forall x \in J: \map {I_J} x = x$
Let $\phi: \struct {\map {\mathscr C} J, +} \to \struct {\R, +}$ be the homomorphism defined as:
:$\forall f \in \map {\mathscr C} J: \map \phi f = \ds 2 \int_0^1 \int_0^x \map f y \rd y \rd x$
The kernel of $\phi$ is given by:
:$\map \ker \phi = I_J - f_m$
where:
:$f_m: \R \to \R$ denotes the constant mapping on $\R$
:$m = \dfrac 1 3$
:$I_J$ denotes the identity mapping on $J$.
== Proof ==
From Group Homomorphism: Example 9, we have that $\phi$ is indeed a homomorphism.
For all $c \in \R$, let $f_c: \R \to \R$ be the constant mapping:
:$\forall x \in \R: \map {f_c} x = c$
First we show that:
:$\forall c \in \R: \map \phi {f_c} = c$
Let $c \in \R$ be arbitrary.
We have:
Then we show that there exists a unique $m \in \R$ such that:
:$\map \phi {I_J - f_m} = 0$
where in this case:
:$m = \dfrac 1 3$
We have:
Hence the result by definition of kernel.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Exercise $13.18 \ \text {(i)}$
Category:Examples of Group Homomorphisms
Group Homomorphism/Examples/Square Function

== Example of Group Homomorphism ==
Let $\struct {\R_{\ne 0}, \times}$ be the group formed from the non-zero real numbers under multiplication.
Let $\struct {\R_{>0}, \times}$ be the group formed from the (strictly) positive real numbers under multiplication.
Let $f: \R_{\ne 0} \to \R_{>0}$ be the mapping defined as:
:$\forall x \in \R_{\ge 0}: \map f x = x^2$
Then $f$ is a group homomorphism.
== Proof ==
First we note that:
:from Non-Zero Real Numbers under Multiplication form Group, $\struct {\R_{\ne 0}, \times}$ is a group
:from Strictly Positive Real Numbers under Multiplication form Uncountable Abelian Group, $\struct {\R_{>0}, \times}$ is a group
Then we note that:
:$\forall x \in \R_{\ge 0}: x^2 \in \R_{>0}$
demonstrating that $f$ is defined for all $x \in \R_{\ge 0}$.
Finally we check the morphism property:
Hence the result.
== Sources ==
* 
* 
Category:Examples of Group Homomorphisms
Group Homomorphism Preserves Identity

== Theorem ==
Let $\struct {G, \circ}$ and $\struct {H, *}$ be groups.
Let $\phi: \struct {G, \circ} \to \struct {H, *}$ be a group homomorphism.
Let:
:$e_G$ be the identity of $G$
:$e_H$ be the identity of $H$.
Then:
:$\map \phi {e_G} = e_H$
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Sources ==
* : Chapter $2$: Group Homomorphism and Isomorphism: $\S 60 \alpha$
Category:Group Homomorphisms
Category:Group Homomorphism Preserves Identity
Group Homomorphism Preserves Identity/Proof 3

== Theorem ==
== Proof ==
From Group Homomorphism of Product with Inverse, we have:
:$\forall x, y \in G: \map \phi {x \circ y^{-1} } = \map \phi x * \paren {\map \phi y}^{-1}$
Putting $x = y$ we have:
== Sources ==
* : Chapter $8$: The Homomorphism Theorem: Corollary $8.7 \ \text{(i)}$
Category:Group Homomorphism Preserves Identity
Group Homomorphism Preserves Inverses

== Theorem ==
Let $\struct {G, \circ}$ and $\struct {H, *}$ be groups.
Let $\phi: \struct {G, \circ} \to\struct {H, *}$ be a group homomorphism.
Let:
:$e_G$ be the identity of $G$
:$e_H$ be the identity of $H$
Then:
:$\forall x \in G: \map \phi {x^{-1} } = \paren {\map \phi x}^{-1}$
Hence the notation $\map \phi x^{-1}$ can be used unambiguously.
This can be illustrated using the following commutative diagram:
$\quad\quad\begin{xy} \xymatrix@L+2mu@+1em{
G \ar[r]^*{\phi}
\ar[d]_*{\iota_G}
&
H \ar[d]^*{\iota_H}
\\
G \ar[r]_*{\phi} &
H
}\end{xy}$
where $\iota_G$ and $\iota_H$ are the inversion mappings on $G$ and $H$ respectively.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Sources ==
* : Chapter $\text{II}$: Groups: Morphisms
* : Chapter $2$: Group Homomorphism and Isomorphism: $\S 60 \alpha$
Category:Group Homomorphisms
Category:Group Homomorphism Preserves Inverses
Group Homomorphism Preserves Inverses/Proof 1

== Theorem ==
== Proof ==
Let $x \in G$.
Then:
So, by definition, $\map \phi {x^{-1} }$ is the right inverse of $\map \phi x$.
Similarly:
So, again by definition, $\map \phi {x^{-1} }$ is the left inverse of $\map \phi x$.
Finally, as $\map \phi {x^{-1} }$ is both:
:a left inverse of $\map \phi x$
and:
:a right inverse of $\map \phi x$
it is by definition an inverse.
From Inverse in Group is Unique, $\map \phi {x^{-1} }$ is the only such element.
Hence the result.
== Sources ==
* : Chapter $\text{II}$: Groups: Morphisms
* : $\S 47.2$ Homomorphisms and their elementary properties
* : $\S 3.2$: Groups; the axioms
Category:Group Homomorphism Preserves Inverses
Group Homomorphism Preserves Inverses/Proof 3

== Theorem ==
== Proof ==
From Group Homomorphism of Product with Inverse, we have:
:$\forall x, y \in G: \map \phi {x \circ y^{-1} } = \map \phi x * \paren {\map \phi y}^{-1}$
Putting $x = e_G$ and $y = x$ we have:
== Sources ==
* : Chapter $8$: The Homomorphism Theorem: Corollary $8.7 \ \text{(ii)}$
Category:Group Homomorphism Preserves Inverses