OpenLLM-France/wikimedia huggingface dataset:
# Base change/123/471/025 and 024/661/35-2/Point with coordinates 254/General/Exercise

We consider the families of vectors
    ${\mathfrak {v}}={\begin{pmatrix}1\\2\\3\end{pmatrix}},\,{\begin{pmatrix}4\\7\\1\end{pmatrix}},\,{\begin{pmatrix}0\\2\\5\end{pmatrix}}\,\,{\text{ and }}\,\,{\mathfrak {u}}={\begin{pmatrix}0\\2\\4\end{pmatrix}},\,{\begin{pmatrix}6\\6\\1\end{pmatrix}},\,{\begin{pmatrix}3\\5\\-2\end{pmatrix}}$
in $\mathbb {R} ^{3}$.
a) Show that ${\mathfrak {v}}$ and ${\mathfrak {u}}$ are both a basis of $\mathbb {R} ^{3}$.
b) Let $P\in \mathbb {R} ^{3}$ denote the point that has the coordinates $(2,5,4)$ with respect to the basis ${\mathfrak {v}}$. What are the coordinates of this point with respect to the basis ${\mathfrak {u}}$?
c) Determine the transformation matrix that describes the change of basis from ${\mathfrak {v}}$ to ${\mathfrak {u}}$.
# Cyclic group/Canonical representation/Example

The subgroups of the integers are of the form $\mathbb {Z} n$ with $n\geq 0$, due to fact. The factor group are denoted by
    $\mathbb {Z} /(n)$
("$\mathbb {Z}$ modulo $n$“). For $n=0$, this is just $\mathbb {Z}$ itself; for $n=1$, this is the trivial group. In general, the equivalence relation on $\mathbb {Z}$ defined by the subgroup $\mathbb {Z} n$ is given in the way that $a$ and $b$ are equivalent if and only if their difference $a-b$ belongs to $\mathbb {Z} n$, that is, if it is a multiple of $n$. Therefore, ($n\geq 1$), every integer number is equivalent to exactly one of the $n$ numbers
    $0,1,2,\ldots ,n-1$
(or, as we also say, congruent modulo $n$), namely to the remainder upon division through $n$. These remainders form a system of representatives for the factor group, and contains $n$ elements. The fact that the quotient mapping
    $\mathbb {Z} \longrightarrow \mathbb {Z} /(n),a\longmapsto [a]=a\!\!\!\mod n,$
is a homomorphism might be expressed by saying that the remainder of a sum of two integers depends only on their remainders, not on the numbers themselves. As an image of the cyclic group $\mathbb {Z}$, the group $\mathbb {Z} /(n)$ is also cyclic; $1$ (but also $-1$) is always a generator.
# Subgroup/Normal subgroup/Definition

Let $G$ be a group, and let $H\subseteq G$ denote a subgroup. $H$ is called a normal subgroup if
    $xH=Hx\,$
holds for all $x\in G$, that is, if every left coset of $x$ coincides with the right coset of $x$.
# Normal subgroup/Characterization/Fact

Let $G$ be a group, and let $H\subseteq G$ be a subgroup.
* $H$ is a normal subgroup of $G$.
* We have $xhx^{-1}\in H$ for all $x\in G$ and $h\in H$.
* $H$ is invariant under every inner automorphism of $G$.
# Normal subgroup/Characterization/Fact/Proof

(1) means for given $h\in H$ that we can write $xh={\tilde {h}}x$ with some ${\tilde {h}}\in H$. Multiplication by $x^{-1}$ from the right yields $xhx^{-1}={\tilde {h}}\in H$; therefore, $(2)$ holds. Reading this argument backwards gives the implication $(2)\Rightarrow (1)$. Moreover, $(2)$ is an explicit reformulation of $(3)$.
# Permutation group S3/Subgroups and normal subgroup/Example

We consider the permutation group $G=S_{3}$ for a set with three elements, that is, $S_{3}$ consists of all bijective mappings of the set $\{1,2,3\}$ to itself. The trivial group $\{\operatorname {id} \}$ and the whole group are normal subgroups. The subset $H=\{\operatorname {id} \,,\varphi \}$, where $\varphi$ is the element that swops $1$ and $2$ and fixes $3$, is a subgroup. However, it is not a normal subgroup. To show this, let $\psi$ denote the bijection that fixes $1$ and swops $2$ and $3$. The inverse of $\psi$ is $\psi$ itself. The conjugation $\psi \varphi \psi ^{-1}=\psi \varphi \psi$ is the mapping that sends $1$ to $3$, $2$ to $2$, and $3$ to $1$. This bijection does not belong to $H$.
# Group homomorphism/Kernel/Normal subgroup/Fact

Let $G$ and $H$ be groups, and let
    $\varphi \colon G\longrightarrow H$
be a group homomorphism.
kernel $\operatorname {kern} \varphi$ is a normal subgroup
$G$
# Automotive Technology/Suspension and Steering

This page is written for those taking the ASE A4 test

## Steering

Always disable the Air Bag system before any change to the steering column.
Steering Columns should be locked into place to preserve timing and prevent overextending the clock-spring.
Replace Bushings when they no longer hold the part in position. Replace Bearings when they have any visible play, or have a turning torque outside of specification.
Adjustments can affect front wheel toe

### Diagnosis

Bump steer can be caused by
* Loose idler arm
* Loose rack mountings
* Loose or damaged mounting
Steering wheel free play can be caused by
* Worn couplings
* worn u joint
* Loose mounting
Steering problems are sometimes subframe alignment problems

#### Manual Steering Gears

Also known as worm gears
Noises, binding, looseness, hard steering point to bearing and lubrication problems.

#### Other parts

Air bag Steering wheel Clock spring
Worm gear Idler arm Rack and pinion
Tie rods

## Power Steering Units

Power steering uses hydraulic pressure to apply force to the worm gear or rack

### Diagnosis

Weeping, or a slight fluid leak requires repair or replacement
* inspect power steering fluid
* Remove and replace power steering pump belt, tensioner, pulley
* diagnose power steering pump problems
* Remove and replace power steering pump
* power steering pressure testio
* Remove and replace power steering hydraulics
* Steering gear worm bearing preload and sector lash
* Remove and replace steering gear seals and gaskets
* Remove and replace rack and pinion unit
* Remove and replace rack and pinion steering gear bellows/boots.
* flush, fill, bleed power steering hydraulics
* Diagnose, Remove and replace variable-assist steering systems

#### Rack and Pinion Steering

    can provide assistance by applying pressure to one side

## Steering Linkage

* Inspect and adjust steering linkage geometry

#### Pitman arm

Pitman arm is only bent in collisions, and wears on other parts. These are pressed on, and are difficult to separate from the splines.

#### Center link

If bent, will affect front wheel toe

#### Idler arm

* Remove and Replace tie rods
* Remove and Replace steering linkage damper
Remove and Replace

## Frame
### Frame
### Springs

You should understand spring frequency, even if you never measure the frequency of the spring. Most spring are made of a chrome-vanadium alloy. Very tough stuff.

#### Constant-velocity joint

    Usable in torque transfer

### Suspension
#### Short/Long Arm
## ASE Task List
## Glossary
### Bead

    The lip of the tire that seals the rim flange

### Belt

    Material under the tread

### Center section AKA Spider

    Center of a steel wheel

### Unsprung Weight

    Weight below the spring in a vehicle

### Uniform Tire Quality Grading System (UTQG)

    Tires sold in the USA are rated in Tread wear, Traction, and Temperature Resistance as per NHTSA.

### Wear bars

    raised features located at the bottom of the tread grooves that indicate the tire has reached its wear limit
Automotive Diagnostics
# Group homomorphism/Kernel/Normal subgroup/Fact/Proof

By fact, we know that the kernel is a subgroup. We use fact. Hence, let $x\in G$ be arbitrary, and $h\in \operatorname {kern} \varphi$. Then
    $\varphi {\left(xhx^{-1}\right)}=\varphi (x)\varphi (h)\varphi {\left(x^{-1}\right)}=\varphi (x)e_{H}\varphi {\left(x^{-1}\right)}=\varphi (x)\varphi (x)^{-1}=e_{H}\,,$
therefore, $xhx^{-1}$ belongs to the kernel.
# Group/Normal subgroup/Residue class group/Fact

Normal subgroup/Situation Let $G/H$ be the set of all cosets (the quotient set), and let
    $q\colon G\longrightarrow G/H,g\longmapsto [g],$
denote the canonical projection.
$G/H$
$q$
# Group/Normal subgroup/Residue class group/Fact/Proof

Since the canonical projection shall be a group homomorphism, the operation must fulfill
    $[x][y]=[xy]\,.$
We have to show that this rule gives a well-defined operation on $G/H$, that is, is independent of the choice of representatives. Hence, we have to show for $[x]=[x']$ and $[y]=[y']$ that $[xy]=[x'y']$ holds. Due to the condition, we can write $x'=xh$ and $hy'={\tilde {h}}y=yh'$ with $h,{\tilde {h}},h'\in H$. Therefore,
    $x'y'=(xh)y'=x(hy')=x(yh')=xyh'\,.$
This means $[xy]=[x'y']$. From this, the group property, the homomorphism property of the projection and the uniqueness follows.
# Residue class group/Representative/Definition

Let $G$ be a group, and let $H\subseteq G$ be a normal subgroup. The quotient set
    $G/H,$
endowed with the (according to fact) uniquely determined group structure, is called the factor group of $G$ modulo $H$. The elements $[g]\in G/H$ are called residue classes. For a residue class $[g]$, every element $g'\in G$ with $[g']=[g]$ is called a representative of $[g]$.
# Electronics/Resistance

The electrical resistance of an object is a measure of its opposition to the flow of electric current. The reciprocal/opposite electrical property is Conductance. Resistance is measured in ohms, symbolized by the Greek letter omega (Ω), named after Georg Simon Ohm.
There is a material science behind what makes a good electrical conductor (things like aluminum and copper, which are good at transmitting electricity), and resistors (things like rubber, wood and plastics, which do not like to transmit electricity). There are items like heating elements and items like Electronics/Resistors that have a fixed resistance value. Resistance is used to ensure the current and voltage after the item causing the resistance is at a specified value. If the resistor is the only path of electrical current flow (series circuit), there will be a voltage drop after the resistance item. If there are multiple path of electrical current flow (parallel circuit), there will be a current drop where the parallel path meets.

### Ohms Law:

With some of the other electrical properties, the resistance of an object can be calculated.
$Ohms=Volts^{2}/Watts$
$Ohms=Watts/Amps^{2}$
$Ohms=Volts/Amps$
# Group theory/Homomorphism theorems/Examples/Section
## Theorem
### Proof

$\Box$

## Example

The mapping constructed in this theorem is called induced mapping or induced homomorphism, and the theorem is called the theorem about the induced homomorphism.

## Corollary
### Proof

$\Box$

## Theorem
### Proof

$\Box$
This statement is often briefly expressed by saying:
image $=$ preimage modulo kernel.
# Group homomorphism/Homomorphism theorem/Surjective and kernel/Fact

Let $G,Q$ and $H$ be groups, and let $\varphi \colon G\rightarrow H$ be a group homomorphism and $\psi \colon G\rightarrow Q$ a surjective group homomorphism. Suppose that
    $\operatorname {kern} \psi \subseteq \operatorname {kern} \varphi \,.$
ist.
    ${\tilde {\varphi }}\colon Q\longrightarrow H$
such that $\varphi ={\tilde {\varphi }}\circ \psi$
Put differently, the diagram
    ${\begin{matrix}G&{\stackrel {\varphi }{\longrightarrow }}&H&\\\!\!\!\!\!\psi \downarrow &\nearrow {\tilde {\varphi }}\!\!\!\!\!&\\Q&&&&\!\!\!\!\!\!\!\!\\\end{matrix}}$
commutes.
# Group homomorphism/Homomorphism theorem/Surjective and kernel/Fact/Proof

We show firstly the uniqueness. For every Element $u\in Q$, there exists some $g\in G$ with $\psi (g)=u$. The commutativity of the diagram ensures that
    ${\tilde {\varphi }}(u)=\varphi (g)\,$
holds. This means that there exists at most one ${\tilde {\varphi }}$.
We have to show that this condition yields a well-defined mapping. Hence, let $g,g'\in G$ be two preimages of $u$. Then
    $\psi {\left(g'g^{-1}\right)}=uu^{-1}=e_{Q}\,;$
therefore, we have $g'g^{-1}\in \operatorname {kern} \psi \subseteq \operatorname {kern} \varphi$. Hence, $\varphi (g)=\varphi (g')$, and the mapping is well-defined. Let $u,v\in Q$ be given, and let $g,h\in G$ be preimages. Then $gh$ is a preimage of $uv$. Therefore,
    ${\tilde {\varphi }}(uv)=\varphi (gh)=\varphi (g)\varphi (h)={\tilde {\varphi }}(u){\tilde {\varphi }}(v)\,.$
This means that ${\tilde {\varphi }}$ is a group homomorphism.
# Group homomorphism/Homomorphism theorem/Residue class groups of Z/Example

We consider the surjective group homomorphisms
    $\varphi \colon \mathbb {Z} \longrightarrow \mathbb {Z} /(4)$
and
    $\psi \colon \mathbb {Z} \longrightarrow \mathbb {Z} /(12).$
We have
    $\operatorname {kern} \psi =\mathbb {Z} \cdot 12\subseteq \mathbb {Z} \cdot 4=\operatorname {kern} \varphi \,.$
Due to fact, there exists a uniquely determined group homomorphism
    ${\tilde {\varphi }}\colon \mathbb {Z} /(12)\longrightarrow \mathbb {Z} /(4),$
which is compatible with the remainder mappings. The morphism ${\tilde {\varphi }}$ sends the remainder of a number after division by $12$ to the remainder after division by $4$. In particular, the theorem implies that the second remainder does only depend on the first remainder, not on the number itself.
If, to the contrary, we consider
    $\varphi \colon \mathbb {Z} \longrightarrow \mathbb {Z} /(5)$
and
    $\psi \colon \mathbb {Z} \longrightarrow \mathbb {Z} /(12),$
then
    $\operatorname {kern} \psi =\mathbb {Z} \cdot 12\not \subseteq \mathbb {Z} \cdot 5=\operatorname {kern} \varphi \,,$
and there does not exists a natural mapping
    $\mathbb {Z} /(12)\longrightarrow \mathbb {Z} /(5).$
For example, the numbers $1,13,25,37,49$ have modulo $12$ the remainder $1$ but modulo $5$ their remainders are $1,3,0,2,4$.
# Group homomorphism/Surjective and residue class group/Fact

Let $G$ and $H$ be groups, and let
    $\varphi \colon G\longrightarrow H$
be a surjective group homomorphism.
isomorphism
    ${\tilde {\varphi }}\colon G/\operatorname {kern} \varphi \longrightarrow H.$
# Group homomorphism/Surjective and residue class group/Fact/Proof

We apply fact to $Q=G/\operatorname {kern} \varphi$ and the canonical projection $q\colon G\rightarrow G/\operatorname {kern} \varphi$. This induces a group homomorphism
    ${\tilde {\varphi }}\colon G/\operatorname {kern} \varphi \longrightarrow H$
fulfilling $\varphi ={\tilde {\varphi }}\circ q$, which is surjective. Let $[x]\in G/\operatorname {kern} \varphi$ and $[x]\in \operatorname {kern} {\tilde {\varphi }}$. Then
    ${\tilde {\varphi }}([x])=\varphi (x)=e_{H}\,.$
Therefore, $x\in \operatorname {kern} \varphi$. Hence, $[x]=e_{Q}$, that is, the kernel of ${\tilde {\varphi }}$ is trivial, and so, due to fact, ${\tilde {\varphi }}$ is also injective.
# Cyclic group/Z mod n/Example

Let $G$ be a cyclic group with a generator $g$. We consider the corresponding group homomorphism
    $\varphi \colon \mathbb {Z} \longrightarrow G,n\longmapsto g^{n},$
in the sense of fact. Since we have a generator, this mapping is surjective. The kernel of this mapping is determined the order of $g$, which we denote by $k$ (or it is $0$, in case the order is $\infty$). Due to fact, there exists a canonical isomorphism
    ${\tilde {\varphi }}\colon \mathbb {Z} /(k)\longrightarrow G.$
In particular, there exists, up to isomorphism, for every $k$, exactly one cyclic group, namely $\mathbb {Z} /(k)$.
# R mod Z2 pi/Isomorphism theorem/Example

The group homomorphism
    $\mathbb {R} \longrightarrow S^{1},t\longmapsto {\begin{pmatrix}\cos t\\\sin t\end{pmatrix}},$
is surjective. Due to the periodicity
of the trigonometric functions, the kernel equals $\mathbb {Z} 2\pi$. By the isomorphism theorem, there exists a canonical isomorphism
    $\mathbb {R} /\mathbb {Z} 2\pi \cong S^{1}\,.$
# C/Cx/Exponential mapping/Isomorphism theorem/Example

The complex exponential function
    $(\mathbb {C} ,0,+)\longrightarrow (\mathbb {C} ^{\times },1,\cdot ),z\longmapsto \exp z,$
is a surjective group homomorphism. The kernel is $\mathbb {Z} 2\pi {\mathrm {i} }$. Due to the isomorphism theorem, there exists a canonical isomorphism
    $\mathbb {C} /\mathbb {Z} 2\pi {\mathrm {i} }\cong \mathbb {C} ^{\times }\,.$
# General and special linear group/Determinant/Isomorphism theorem/Example

The determinant
    $\det \colon \operatorname {GL} _{n}\!{\left(K\right)}\longrightarrow K^{\times },M\longmapsto \det M,$
is a surjective group homomorphism; its kernel is by definition the special linear group $\operatorname {SL} _{n}\!{\left(K\right)}$. Due to the isomorphism theorem, there exists a canonical isomorphism
    $\operatorname {GL} _{n}\!{\left(K\right)}/\operatorname {SL} _{n}\!{\left(K\right)}\cong K^{\times }\,.$
# Group homomorphism/Factorization/Fact

Let $G$ and $H$ be groups, and let
    $\varphi \colon G\longrightarrow H$
be a group homomorphism.
    $G{\stackrel {q}{\longrightarrow }}G/\operatorname {kern} \varphi {\stackrel {\theta }{\longrightarrow }}\operatorname {Im} \varphi {\stackrel {\iota }{\hookrightarrow }}H,$
where $q$ is the canonical projection, $\theta$ is a group isomorphism, and $\iota$ is the canonical inclusion of the
# Group homomorphism/Factorization/Fact/Proof

This follows from fact, applied to the image group $U=\operatorname {Im} \varphi \subseteq H$.
# Commutative ring/Coset for ideal/Definition

Commutative ring/Ideal/Situation For $a\in R$, the subset
    $a+I={\left\{a+f\mid f\in I\right\}}\,$
is called the coset of $a$ to the ideal $I$. Every subset of this form is called a coset to $I$.
# Commutative ring/Residue class ring/Definition

Let $R$ be a commutative ring, and let $I\subseteq R$ be an ideal in $R$. Then the residue class ring $R/I$ (R modulo I) is a commutative ring that is determined by the following data.
* As a set, $R/I$ is the set of all cosets to $I$.
* An addition of cosets is defined by
    $(a+I)+(b+I):=(a+b+I)\,.$
* A multiplication of cosets is defined by
    $(a+I)\cdot (b+I):=(a\cdot b+I)\,.$
* ${\bar {0}}=0+I=I$ is the neutral element of the addition (the zero class).
* ${\bar {1}}=1+I$ is the neutral element of the multiplication (the unit class).
# Group theory/Isomorphism theorem for residue class groups/Fact

Let $G$ be a group, and let $N\subseteq G$ be a normal subgroup with residue class group $Q=G/N$. Let $H\subseteq G$ be another normal subgroup in $G$ with $N\subseteq H$.
image ${\overline {H}}$ of $H$ in $Q$ is a normal subgroup, and we have a canonical isomorphism
    $G/H\cong Q/{\overline {H}}\,.$
# Group theory/Isomorphism theorem for residue class groups/Fact/Proof

For the first statement, see exercise. Therefore, the residue class group $Q/{\overline {H}}$ is well-defined. We consider the composition
    $p\circ q:G\longrightarrow Q\longrightarrow Q/{\overline {H}}.$
Because of
    ${\begin{aligned}\operatorname {kern} {\left(p\circ q\right)}&={\left\{x\in G\mid (p\circ q)(x)=e\right\}}\\&={\left\{x\in G\mid q(x)\in \operatorname {kern} p\right\}}\\&={\left\{x\in G\mid q(x)\in {\overline {H}}\right\}}\\&=H,\end{aligned}}$
we have $\operatorname {kern} {\left(p\circ q\right)}=H$. Hence, according to fact, we get the canonical isomorphism
    $G/H\longrightarrow Q/{\overline {H}}.$
# Ring theory/Ring homomorphism/Definition

Let $R$ and $S$ be rings. A mapping
    $\varphi \colon R\longrightarrow S$
is called a ring homomorphism, if the following properties hold:
* $\varphi (a+b)=\varphi (a)+\varphi (b)$.
* $\varphi (1)=1$.
* $\varphi (a\cdot b)=\varphi (a)\cdot \varphi (b)$.
# Commutative ring/Ring homomorphism/Kernel/Ideal/Fact

Let $R$ and $S$ be commutative rings, and let
    $\varphi \colon R\longrightarrow S$
be a ring homomorphism.
kernel
    $\operatorname {kern} \varphi ={\left\{f\in R\mid \varphi (f)=0\right\}}\,$
is an ideal
$R$
# Linear algebra (Osnabrück 2024-2025)/Commutative ring/Ring homomorphism/Kernel/Ideal/Fact/Factreferencenumber

Lemma 47.11
# Residue class rings of Z/Ring homomorphism/Fact

Let $d\geq 0$ be a natural number.
$\mathbb {Z} /(d)$
residue class mapping
    $\mathbb {Z} \longrightarrow \mathbb {Z} /(d),a\longmapsto {\overline {a}}\,,$
is a
$\mathbb {Z} /(d)$ is a commutative ring with $d$ elements (for $d\geq 1$).
# Residue class rings of Z/Field/Integer/Prime number/Fact

Let $n\geq 1$ be a natural number, and let $\mathbb {Z} /(n)$ be the corresponding residue class ring.
* $\mathbb {Z} /(n)$ is a field.
* $\mathbb {Z} /(n)$ is a domain.
* $n$ is a prime number.
# Sets/Relation/Definition

Let $X$ and $Y$ denote sets. A relation between $X$ and $Y$ is a subset $R\subseteq X\times Y$.
# Set/Order relation/Definition

A relation $\preccurlyeq$ on a set $I$ is called an order relation, if the following three conditions are fulfilled.
* $i\preccurlyeq i$ for all $i\in I$.
* From $i\preccurlyeq j$ and $j\preccurlyeq k$ we can deduce $i\preccurlyeq k$.
* From $i\preccurlyeq j$ and $j\preccurlyeq i$ we can deduce $i=j$.
# Set/Equivalence relation/Definition

An equivalence relation on a set $M$ is a relation $R\subseteq M\times M$ fulfilling the following three properties (for arbitrary $x,y,z\in M$).
* $x\sim x$ (reflexive).
* $x\sim y$ implies $y\sim x$ (symmetric).
* From $x\sim y$ and $y\sim z$ we can deduce $x\sim z$ (transitive).
Here, $x\sim y$ means that the pair $(x,y)$ belongs to $R$.
# Equivalence relation/Equality/Example

A basic example of an equivalence relation is the equality on an arbitrary set $M$. Under equality, every element is only equivalent with itself.
# Equivalence relation/Equivalence classes/Section

An equivalence relation $R\subseteq M\times M$ on a set $M$ may also be considered as a partition of the set $M$. In order to see this, the concept of an equivalence class is helpful.

## Definition

Equivalence relation/Equivalence class/Definition
Hence, $[x]$ is the subset of all elements of $M$ that are equivalent with $x$. Every element $y\in [x]$ is called a representative of the equivalence class $[x]$.

## Definition

Equivalence relation/System of representatives/Definition
# Equivalence relation/Quotient set/Definition

Let $R\subseteq M\times M$ denote an equivalence relation. The set
    $M/R:={\left\{[x]\mid x\in M\right\}}\,$
is called the quotient set of $R$.
# Equivalence relation/Canonical projection/Definition

Let $R\subseteq M\times M$ denote an equivalence relation, and let $M/R$ be the quotient set. The mapping
    $q_{R}\colon M\longrightarrow M/R,x\longmapsto [x],$
is called the canonical projection of $R$.
# Equivalence classes/Partition/Quotient set/Properties/Fact/Proof

* Let $x$ and $y$ be equivalent, and $u\in [x]$. Then $x\sim u$, and, by transitivity, also $y\sim u$, thus $u\in [y]$. Therefore, the equivalence classes coincide. The implication from the middle to the right is clear, because, due to $x\sim x$, the equivalence classes are not empty. Suppose now that $[x]\cap [y]\neq \emptyset$, and let $z$ denote an element in the intersection. Then $x\sim z$ and $y\sim z$, and, by transitivity, $x\sim y$.
* Because of the reflexivity, we have $x\in [x]$; therefore, $M=\bigcup _{[x]\in M/\sim }[x]$. This union is disjoint by part (1).
* The surjectivity is clear because of the definition of the quotient set, and since $x$ is sent to the class $[x]$.
* We have
    $q^{-1}([x])={\left\{y\in M\mid q(y)=[x]\right\}}={\left\{y\in M\mid [y]=[x]\right\}}={\left\{y\in M\mid y\sim x\right\}}=[x]\,.$
# Plane/Parallel lines/Equivalence classes/Example

On the set of all lines in the plane, we can consider the property of being parallel as an equivalence relation. A line is parallel to itself, the relation is obviously symmetric, and if $G_{1}$ and $G_{2}$ are parallel, and $G_{2}$ and $G_{3}$ are parallel, then also $G_{1}$ and $G_{3}$ are parallel. The equivalence class of a line $G$ consists of all lines that are parallel to $G$; these lines form a parallel linenschar. We fix a point $M$ in the plane. Then there exists, for every line $G$, a parallel line $G'$ running through $M$. Therefore, every equivalence class can be uniquely represented by a line through the point $M$. The set of lines through $M$ form a system of representatives for this equivalence relation.
# Felidae

A Felid Is A Animal That Meows.
| Cats (Felidae) |||||
| Subfamily |
| Felinae | A ocelot In 2023 | Rohrkate | Coming Soon! |
| Pantherinae | Neofelis nebulosa | Panthera leo | Uncia uncia |
| Acinonychinae | A Cheetah In A Africa Spot |