avewright/proofwiki-math huggingface dataset:
Theorem: Compact Closure of Element is Principal Ideal on Compact Subset iff Element is Compact

Let $L = \struct {S, \vee, \preceq}$ be a bounded below algebraic join semilattice.
Let $P = \struct {\map K L, \precsim}$ be an ordered subset of $L$
where $\map K L$ denotes the compact subset of $L$.
Let $x \in S$.
Then $x^{\mathrm{compact} }$ is principal ideal in $P$ {{iff}} $x$ is a compact element.

Proof:
=== Sufficient Condition ===
Assume that
:$x^{\mathrm{compact} }$ is principal ideal in $P$.
By definitions of compact subset and compact closure:
:$x^{\mathrm{compact} } \subseteq \map K L$
By definition of principal ideal:
:$\exists y \in x^{\mathrm{compact} }: y$ is upper bound for $x^{\mathrm{compact} }$ in $P$.
By definition of ordered subset:
:$y$ is upper bound for $x^{\mathrm{compact} }$ in $L$.
By definition of supremum:
:$\map {\sup_L} {x^{\mathrm{compact} } } \preceq y$
By definition of algebraic:
:$L$ satisfies the axiom of $K$-approximation.
By the axiom of $K$-approximation:
:$x \preceq y$
By definition of compact closure:
:$y \preceq x$ and $y$ is a compact element.
Thus by definition of antisymmetry:
:$x$ is a compact element.
{{qed|lemma}}
Theorem: Multiple of 6 is Semiperfect

Let $n \in \Z_{>0}$ be a multiple of $6$.
Then $n$ is semiperfect.

Proof:
Let $n = 6 k$.
Then:
:$n = 2 \times 3 k$
and so $3 k$ is a factor of $n$.
:$n = 3 \times 2 k$
and so $2 k$ is a factor of $n$.
:$n = 6 \times k$
and so $k$ is a factor of $n$.
But:
:$n = k + 2 k + 3 k$
and so is the sum of a subset of its factors
Hence the result by definition of semiperfect.
{{qed}}
Category:Semiperfect Numbers
Category:6
3askxluwp6e4d1z6jxg3a37go0jam49
Theorem: Open Projection and Closed Graph Implies Quotient is Hausdorff

Let $\RR \subseteq X \times X$ be an equivalence relation on a topological space $\struct {X, \tau}$.
Let $X / \RR$ be the quotient space.
Let $p$ denote the quotient mapping.
Let:
:$\RR$ be closed in $X \times X$
:$p$ be an open mapping.
Then $X / \RR$ is Hausdorff.

Proof:
Let $\eqclass x \RR, \eqclass y \RR \in X / \RR$ such that $\eqclass x \RR \ne \eqclass y \RR$.
Then $\tuple {x, y} \notin \RR$.
Since $\RR$ is closed, $\paren {X \times X} \setminus \RR$ is open.
Let:
:$\BB = \set { U \times V : U, V \in \tau}$
By Natural Basis of Product Topology, $\BB$ is a basis for the product topology on $X \times X$.
Then choose $U, V \in \tau$ such that $\tuple {x, y} \in U \times V \subseteq \paren {X \times X} \setminus \RR$.
Thus $x \in U$, so:
:$\eqclass x \RR = \map p x \in p \sqbrk U$
Similarly, $y \in V$, so:
:$\eqclass y \RR = \map p y \in p \sqbrk V$
Since $p$ is open, $p \sqbrk U$ and $p \sqbrk V$ are open.
{{AimForCont}} that $p \sqbrk U \cap p \sqbrk V \ne \O$.
Then there exist $z \in U$ and $w \in V$ such that $\eqclass z \RR = \eqclass w \RR$.
Then $z \RR w$, and so $\tuple {z, w} \in \RR$.
However, $\tuple {z, w} \in U \times V$, contradicting $U \times V \subseteq \paren {X \times X} \setminus \RR$.
Thus:
:$p \sqbrk U \cap p \sqbrk V = \O$
Hence the result.
{{qed}}
Theorem: Subgroup is Closed iff Quotient is Hausdorff

Let $G$ be a topological group.
Let $H \le G$ be a subgroup.
Let $G / H$ be their quotient.
{{TFAE}}
{{begin-itemize}}
{{item|(1):|$H$ is closed in $G$}}
{{item|(2):|$G / H$ is Hausdorff}}
{{end-itemize}}

Proof:
{{ProofWanted|use Group Acts by Homeomorphisms Implies Projection on Quotient Space is Open and Open Projection and Closed Graph Implies Quotient is Hausdorff}}
Category:Topological Groups
Category:Hausdorff Spaces
q2a4yzdbsj9cz236j6j2e36yywa9a9o
Theorem: Higher Homotopy Groups are Abelian

Let $T = \left({S, \tau}\right)$ be a topological space.
Let $x_0 \in S$.
Let $n \ge 2$ be a integer.
Let $\pi_n \left({T, x_0}\right)$ be the $n$th homotopy group with base point $x_0$.
Then $\pi_n \left({T, x_0}\right)$ is abelian.

Proof:
{{proof wanted}}
Category:Homotopy Theory
Category:Algebraic Topology
g4qe2zuesl24pqvto0uiq5uikf4igaz
Theorem: Multiple of Semiperfect Number is Semiperfect

Let $n \in \Z_{>0}$ be a semiperfect number.
Let $k \in \Z_{>0}$ be a (strictly) positive integer.
Then $k n$ is also a semiperfect number.

Proof:
Let $P$ be a subset of the divisors of $n$ such that the sum of the elements of $P$ equals $n$.
Let $\sigma = \ds \sum_{p \mathop \in P} p$ be the sum of the elements of $P$.
Let:
:$Q = \set {k p: p \in P}$
be the set of elements of $P$ multiplied by $k$.
We have by definition that:
:$\forall p \in P: p \divides P$
where $p \divides P$ denotes divisibility.
Then we have:
:$\forall p \in P: k p \divides k P$
Thus $Q$ constitutes a subset of the divisors of $k n$.
But:
{{begin-eqn}}
{{eqn | l = n k
      | r = k \sigma
      | c = 
}}
{{eqn | r = k \sum_{p \mathop \in P} p
      | c = 
}}
{{eqn | r = \sum_{p \mathop \in P} k p
      | c = 
}}
{{eqn | r = \sum_{q \mathop \in Q} q
      | c = 
}}
{{end-eqn}}
demonstrating that $Q$ constitutes a subset of the divisors of $k n$ equal to $k n$.
Hence the result, by definition of semiperfect number.
{{qed}}
Category:Semiperfect Numbers
tl840gr1kesrvjp3cb7ksqa5jrtj1gj
Theorem: Image of Compact Subset under Directed Suprema Preserving Closure Operator is Subset of Compact Subset

Let $L = \struct {S, \preceq}$ be an algebric lattice.
Let $c: S \to S$ be a closure operator that preserves directed suprema.
Then:
:$c \sqbrk {\map K L} \subseteq \map K {\struct {c \sqbrk S, \precsim} }$
where:
:$\map K L$ denotes the compact subset of $L$
:$c \sqbrk S$ denotes the image of $S$ under $c$
:$\mathord \precsim = \mathord \preceq \cap \paren {c \sqbrk S \times \map c S}$

Proof:
Let $x \in c \sqbrk {\map K L}$.
By definition of image of set:
:$\exists y \in \map K L: x = \map c y$
and
:$x \in c \sqbrk S$
By definition of compact subset:
:$y$ is compact in $L$.
By definition of compact element:
:$y \ll y$
where $\ll$ denotes the way below relation.
Define $P = \struct {c \sqbrk S, \precsim}$ as an ordered subset of $L$.
We will prove that:
:for every directed subset $D$ of $c \sqbrk S$: $x \precsim \sup_P D \implies \exists d \in D: d \precsim x$
Let $D$ be a directed subset of $c \sqbrk S$.
By definition of ordered subset:
:$D$ is directed in $L$.
By definition of algebraic ordered set:
:$L$ is up-complete.
By definition of up-complete:
:$D$ admits a supremum in $L$.
{{begin-eqn}}
{{eqn | l = \map c { {\sup}_L D}
      | r = \map { {\sup}_L} {c \sqbrk D}
      | c = {{Defof|Mapping Preserves Supremum|subdef = Directed}}
}}
{{eqn | r = {\sup}_L D
      | c = {{Defof|Closure Operator}}
}}
{{end-eqn}}
By definition of image of set:
:$\sup_L D \in c \sqbrk S$
By Supremum in Ordered Subset:
:$\sup_L D = \sup_P D$
By definition of ordered subset:
:$x \preceq \sup_L D$
By definition of closure operator/inflationary:
:$y \preceq x$
By definition of transitivity:
:$y \preceq \sup_L D$
By definition of way below relation:
:$\exists d \in D: y \preceq d$
By definition of subset:
:$d \in c \sqbrk S$
By definition of closure operator/idempotent:
:$d = \map c d$
By definition of closure operator/increasing:
:$x = \map c y \preceq \map c d = d$
Thus by definition of ordered subset:
:$\exists d \in D: x \precsim d$
{{qed|lemma}}
By definition of way below relation:
:$x \ll_P x$
By definition:
:$x$ is a compact element in $P$.
Thus by definition of compact subset:
:$x \in \map K P$
{{qed}}
Theorem: Integers such that all Coprime and Less are Prime

The following positive integers have the property that all positive integers less than and coprime to it, excluding $1$, are prime:
:$1, 2, 3, 4, 6, 8, 12, 18, 24, 30$
{{OEIS|A048597}}
There are no other positive integers with this property.

Proof:
Let $S_n$ denote the set of all positive integers less than and coprime to $n$, excluding $1$.
Let $\map P n$ denote the propositional function:
:All positive integers less than and coprime to $n$, excluding $1$, are prime.
We establish that $\map P n = \T$ for all the positive integers given:
{{begin-eqn}}
{{eqn | l = S_1
      | r = \O
      | c = trivially
}}
{{eqn | l = S_2
      | r = \O
      | c = trivially
}}
{{eqn | l = S_3
      | r = \set 2
      | c = which is prime
}}
{{eqn | l = S_4
      | r = \set 3
      | c = which is prime
}}
{{eqn | l = S_6
      | r = \set 5
      | c = which is prime
}}
{{eqn | l = S_8
      | r = \set {3, 5, 7}
      | c = all prime
}}
{{eqn | l = S_{12}
      | r = \set {5, 7, 11}
      | c = all prime
}}
{{eqn | l = S_{18}
      | r = \set {5, 7, 11, 13, 17}
      | c = all prime
}}
{{eqn | l = S_{24}
      | r = \set {5, 7, 11, 13, 17, 19, 23}
      | c = all prime
}}
{{eqn | l = S_{30}
      | r = \set {7, 11, 13, 17, 19, 23, 29}
      | c = all prime
}}
{{end-eqn}}
From Schatunowsky's Theorem:
:$30$ is the greatest positive integer $n$ such that $\map P n$ is true
We note that for all primes $p$ greater than $3$, $p - 1$ is composite, and so $\map P p = \F$.
The remaining composite numbers less than $30$ are investigated:
{{begin-eqn}}
{{eqn | l = S_9
      | r = \set {2, 4, 5, 7, 8}
      | c = of which $2, 4, 8$ are composite
}}
{{eqn | l = S_{10}
      | r = \set {3, 7, 9}
      | c = of which $9$ is composite,
}}
{{eqn | l = S_{14}
      | r = \set {3, 5, 9, 11, 13}
      | c = of which $9$ is composite
}}
{{eqn | l = S_{15}
      | r = \set {2, 4, 7, 8, 11, 13, 14}
      | c = of which $4, 8, 14$ are composite
}}
{{eqn | l = S_{16}
      | r = \set {3, 5, 7, 9, 11, 13, 15}
      | c = of which $9, 15$ are composite
}}
{{eqn | l = S_{20}
      | r = \set {3, 7, 9, 11, 13, 17, 19}
      | c = of which $9$ is composite
}}
{{eqn | l = S_{21}
      | r = \set {2, 4, 5, 8, 10, 11, 13, 16, 17, 19, 20}
      | c = of which $4, 8, 10, 16, 20$ are composite
}}
{{eqn | l = S_{22}
      | r = \set {3, 5, 7, 9, 13, 15, 17, 19, 21}
      | c = of which $9, 15, 21$ are composite
}}
{{eqn | l = S_{25}
      | r = \set {2, 3, 4, 6, 7, 8, 9, 11, 12, 13, 14, 16, 17, 18, 19, 21, 22, 23, 24}
      | c = of which $4, 6, 8, 9, 12, 14, 16, 18, 21, 22, 24$ are composite
}}
{{eqn | l = S_{26}
      | r = \set {3, 5, 7, 9, 11, 15, 17, 19, 21, 23, 25}
      | c = of which $9, 15, 21, 25$ are composite
}}
{{eqn | l = S_{27}
      | r = \set {2, 4, 5, 7, 8, 10, 11, 13, 14, 16, 17, 19, 20, 22, 23, 25, 26}
      | c = of which $4, 8, 10, 14, 16, 20, 22, 25, 26$ are composite
}}
{{eqn | l = S_{28}
      | r = \set {3, 5, 9, 11, 13, 15, 17, 19, 23, 25, 27}
      | c = of which $9, 15, 25, 27$ are composite
}}
{{end-eqn}}
That exhausts the list.
Hence the result.
{{qed}}
Theorem: Pythagorean Triangles whose Area equal their Perimeter

There exist exactly $2$ Pythagorean triples which define a Pythagorean triangle whose area equals its perimeter:
:$(1): \quad \tuple {6, 8, 10}$, leading to an area and perimeter of $24$
:$(2): \quad \tuple {5, 12, 13}$, leading to an area and perimeter of $30$.

Proof:
From Area of Right Triangle, the area $\AA$ is:
:$\AA = \dfrac {a b} 2$
where $a$ and $b$ are the legs.
$(1): \quad$ The area of the $\tuple {6, 8, 10}$ triangle is $\dfrac {6 \times 8} 2 = 24$.
Its perimeter equals $6 + 8 + 10 = 24$.
$(2): \quad$ The area of the $\tuple {5, 12, 13}$ triangle is $\dfrac {5 \times 12} 2 = 30$.
Its perimeter equals $5 + 12 + 13 = 30$.
It remains to prove that these are the only ones.
Let $a, b, c$ be the lengths of the sides of a Pythagorean triangle $T$.
Thus $a, b, c$ form a Pythagorean triple.
By definition of Pythagorean triple, $a, b, c$ are in the form:
:$2 m n, m^2 - n^2, m^2 + n^2$
We have that $m^2 + n^2$ is always the hypotenuse.
Thus the area of $T$ is given by:
:$\AA = m n \paren {m^2 - n^2}$
The perimeter of $T$ is given by:
:$\PP = m^2 - n^2 + 2 m n + m^2 + n^2 = 2 m^2 + 2 m n$
We need to find all $m$ and $n$ such that $\PP = \AA$. 
Thus:
{{begin-eqn}}
{{eqn | l = 2 m^2 + 2 m n
      | r = m n \paren {m^2 - n^2}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 2 m \paren {m + n}
      | r = n \paren {m + n} \paren {m - n}
      | c = Difference of Two Squares
}}
{{eqn | ll= \leadsto
      | l = n \paren {m - n}
      | r = 2
      | c = 
}}
{{end-eqn}}
As $m$ and $n$ are both (strictly) positive integers, it follows immediately that either:
:$n = 1$
:$m - n = 2$
and so:
:$m = 3, n = 1$
leading to the triangle:
: $a = 6, b = 8, c = 10$
or:
:$n = 2$
:$m - n = 1$
and so:
:$m = 3, n = 2$
leading to the triangle:
: $a = 12, b = 5, c = 13$
and the result follows.
{{qed}}
Theorem: Separable Extension is Contained in Galois Extension

Let $E/F$ be a separable finite field extension.
Then there exists a finite field extension $L/E$ such that $L/F$ is Galois.

Proof:
{{ProofWanted}}
Category:Galois Theory
lh89mtvcdjlsiei1e74gdkpyl4lil3k
Theorem: Galois Extension is Galois over Intermediate Field

Let $L / F$ be a Galois Extension.
Let $E$ be an intermediate field.
Then $L / E$ is Galois.

Proof:
{{ProofWanted}}
Category:Galois Theory
s3zuqy3xgnauel2aduazzo1ksnwjq25
Theorem: Frobenius Endomorphism on Field is Injective

Let $p$ be a prime number.
Let $F$ be a field of characteristic $p$.
Then the Frobenius endomorphism $\Frob: F \to F$ is injective.

Proof:
We have:
:$\map \Frob 1 = 1$
By Ring Homomorphism from Field is Monomorphism or Zero Homomorphism, $\Frob$ is injective.
{{qed}}
Theorem: Galois Field is Perfect

Let $\GF$ be a Galois field.
Then $\GF$ is perfect.

Proof:
By Characteristic of Galois Field is Prime, $\Char \GF$ is a prime number, say $p$.
By Frobenius Endomorphism on Field is Injective, $\Frob$ is injective.
By Injection from Finite Set to Itself is Surjection, $\Frob$ is bijective.
By Bijective Ring Homomorphism is Isomorphism, $\Frob$ is an automorphism.
{{qed}}
Category:Galois Fields
Category:Perfect Fields
pf6ghzyfw6taydho9xlsl66zwf7tcxw
Theorem: Algebraically Closed Field is Perfect

Let $F$ be an algebraically closed field.
Then $F$ is perfect.

Proof:
Let $E / F$ be any algebraic extension.
Since $F$ is an algebraically closed field, $E = F$.
By Field is Separable over itself, $E$ is separable over $F$.
Hence $F$ is perfect.
{{qed}}
Category:Field Extensions
Category:Perfect Fields
oe9rzy5e66rnq45gbo0dhdf23qfa3hz
Theorem: Area of Smallest Rectangle accommodating Re-Entrant Knight's Tour

The area of the smallest rectangular chessboard on which a re-entrant knight's tour is possible is $30$ squares.
This can be configured either as a $5 \times 6$ chessboard or a $3 \times 10$ chessboard.

Proof:
{{ProofWanted|Haven't even started the definitions yet for chess problems}}
Theorem: Area of Smallest Square accommodating Re-Entrant Knight's Tour

The area of the smallest square chessboard on which a re-entrant knight's tour is possible is $6 \times 6 = 36$ squares.

Proof:
{{ProofWanted|Haven't even started the definitions yet for chess problems}}
Theorem: Subgroup of Index Least Prime Divisor is Normal

Let $G$ be a finite group of order $n>1$.
Let $p$ be the least prime divisor of $n$.
Let $H$ be a subgroup of $G$ such that:
:$\index G H = p$
where $\index G H$ denotes the index of $H$ in $G$.
Then $H$ is normal.

Proof:
Let $G / H$ denote the left coset space of $G$ modulo $H$.
Consider the group action $G \curvearrowright G / H$, defined as:
:$\forall g \in G, \forall x H \in G / H: g * \paren {x H} := \paren {g x} H$
{{explain|Expand group action as defined on {{ProofWiki}} to explain the notation $G \curvearrowright G / H$}}
Let $\ds K = \bigcap_{x H \mathop \in G / H} \Stab {x H}$ be the kernel of the group action.
{{explain|The above definition of kernel needs to be added to {{ProofWiki}} }}
By Kernel of Group Action is Normal Subgroup $K$ is a normal subgroup of $G$.
In particular:
:$K \le \Stab H = H$
{{explain|Presumably the notation $\le$ means "is a subgroup of", but it would be wise to state this explicitly}}
{{explain|Prove that $\Stab H {{=}} H$. This is probably proved in Stabilizer of Coset Action on Set of Subgroups, but that is assuming that the group action defined above is that coset action. Because the language the group action uses here is different from how it is defined on {{ProofWiki}}, this is not clear and not at all obvious.}}
We now show that $K = H$.
First we note that the index of $H$ in $G$ is the order of the coset space $G / H$:
:$\index G H = \order {G / H}$
Since $\order {G / H} = p$, we may consider the map $\rho: G \to S_p$ induced by the permutation representation of this group action.
By the First Isomorphism Theorem for Groups:
:$G / K \cong \Img \rho$
{{explain|Has it been established that $\rho$ is in fact a group homomorphism?}}
By Lagrange's Theorem (Group Theory), we get that $\order {G / H}$ divides $\order G$ and $\order {S_p} = p!$.
{{explain|Why does $\order {S_p} {{=}} p!$ follow from Lagrange's Theorem (Group Theory)?}}
Since $p$ is prime, there does not exist a prime $q > p$ such that $q$ divides $p!$.
{{explain|We don't actually have a page proving the (ultimately trivial) fact that $q > n \implies q \nmid n!$ when $q$ is prime. Probably best implemented as a corollary to Divisors of Factorial.}}
But since:
:$p$ is the smallest prime divisor of $n$
:$\order {G / K}$ must divide $\order G$
this implies that $\order {G / K} = p^r$ for some $r \in \N$.
But the largest power of $p$ which divides $p!$ is $p$.
So, since $K \le H \ne G$:
:$\order {G / K} = p$
{{explain|We state that $H \ne G$ but is it therefore taken for granted that $H$ is a ''proper'' subgroup of $G$? Is it necessary for this proof to work that $H \ne G$?}}
Thus:
:$\order {G / H} = \order {G / K}$
so by Order of Quotient Group:
:$\order H = \order K$
But since $K \le H$ and they have the same order:
:$K = H$
as desired.
Thus $H$ is normal.
{{qed}}
Theorem: Alternating Group is Simple except on 4 Letters

Let $n$ be an integer such that $n \ne 4$.
Then the $n$th alternating group $A_n$ is simple.

Proof:
Recall that a group is simple if its normal subgroups are itself and the trivial subgroup.
Let $n < 4$.
$A_2$ is the trivial group and hence simple.
$A_3$ is the cyclic group of order $3$, hence a prime group.
By Prime Group is Simple, $A_3$ is simple.
{{qed|lemma}}
We note that $A_4$ is a special case.
From Normality of Subgroups of Alternating Group on 4 Letters, $A_4$ has a proper normal subgroup $K_4$, and so is not simple.
{{qed|lemma}}
It remains to investigate the case where $n \ge 5$.
Let $N$ be a non-trivial normal subgroup of $A_n$.
It is to be shown that $N = A_n$.
Theorem: Universal Property of Quotient Ring

Let $R, S$ be commutative rings.
Let $I \trianglelefteq R$ be an ideal of $R$.
Let $\pi : R \to R / I$ be the quotient epimorphism.
Let $f: R \to S$ be a ring homomorphism with $\map f I = \set 0$.
Then there exists a unique ring homomorphism $\overline f: R / I \to S$ such that $f = \overline f \circ \pi$.
:<nowiki>$\xymatrix {
R \ar[d]_\pi \ar[r]^{\forall f} & S \\
R/I \ar[ru]_{\exists ! \bar f} }$</nowiki>

Proof:
{{MissingLinks|justification for the statements and steps made in the chain of reasoning}} 
Define $\overline f: R / I \to S$ by:
:$\forall r \in R: \map {\overline f} {r + I} = \map f r$
Since $f$ is a ring homomorphism, $f$ is well-defined.
Suppose for some $r_1, r_2 \in R$ that: 
:$r_1 + I = r_2 + I$
Since $I$ is an ideal, it contains the zero $0_R$ of $R$.
Then for some $i \in I$, we have that: 
{{begin-eqn}}
{{eqn | l = r_1 + 0_R
      | r = r_1
      | c = {{Defof|Ring Zero}}
}}
{{eqn | r = r_2 + i
      | c = {{Defof|Set Equality}}
}}
{{end-eqn}}
Then: 
{{begin-eqn}}
{{eqn | l = \map {\overline f} {r_1 + I}
      | r = \map f {r_1}
      | c = Definition of $\overline f$
}}
{{eqn | r = \map f {r_2 + i}
      | c = Since $r_1=r_2+i$
}}
{{eqn | r = \map f {r_2} + \map f i
      | c = {{Defof|Ring Homomorphism}}
}}
{{eqn | r = \map f {r_2} + 0_S
      | c = Since $f(I)=\{0\}$
}}
{{eqn | r = \map f {r_2}
      | c = {{Defof|Ring Zero}}
}}
{{eqn | r = \map {\overline f} {r_2 + I}
      | c = Definition of $\overline f$
}}
{{end-eqn}}
so $\overline f$ is seen to be well-defined.
Next, for all $r_1, r_2 \in R$:
{{begin-eqn}}
{{eqn | l = \map {\overline f} {r_1 + I + r_2 + I}
      | r = \map {\overline f} {r_1 + r_2 + I}
      | c = Quotient Ring Addition is Well-Defined
}}
{{eqn | r = \map f {r_1 + r_2}
      | c = Definition of $\overline f$
}}
{{eqn | r = \map f {r_1} + \map f {r_2}
      | c = {{Defof|Ring Homomorphism}}
}}
{{eqn | r = \map {\overline f} {r_1 + I} + \map {\overline f} {r_2 + I}
      | c = Definition of $\overline f$
}}
{{end-eqn}}
Therefore $\overline f$ is a homomorphism.
It remains to demonstrate uniqueness.
Suppose there were another homomorphism $g: R / I \to S$ with $f = g \circ \pi$.
Then we would require that, for all $r\in R$:
{{begin-eqn}}
{{eqn | l = \map g {r + I}
      | r = \map f r
      | c = {{Defof|Commutative Diagram}}
}}
{{eqn | r = \map {\overline f} {r + I}
      | c = Definition of $\overline f$
}}
{{end-eqn}}
That is:
:$g = \overline f$
and so $\overline f$ is unique.
{{qed}}
Theorem: Universal Property of Quotient Space

Let $X$ and $Y$ be topological spaces.
Let $\sim$ be an equivalence relation on $X$.
Let $\pi : X \to X / {\sim}$ be the quotient mapping.
Let $f : X \to Y$ be continuous and $\sim$-invariant.
Then there exists a unique continuous map $\overline f : X / {\sim} \to Y$ such that $f = \overline f \circ \pi$.

Proof:
{{proofread}}
Theorem: Universal Property of Quotient Group

Let $G$ and $H$ be groups.
Let $N \trianglelefteq G$ be an normal subgroup.
Let $\pi: G \to G / N$ be the quotient epimorphism.
Let $f: G \to H$ be a group homomorphism with $N \subseteq \ker f$.
Then there exists a unique group homomorphism $\overline f: G / N \to H$ such that $f = \overline f \circ \pi$.
$\xymatrix{
G \ar[d]^\pi \ar[r]^{\forall f} & H \\
G/N \ar[ru]_{\exists ! \bar f} }$

Proof:
=== Existence ===
{{proofread}}
Let $\sim$ denote (left) congruence modulo $N$.
From Congruence Modulo Subgroup is Equivalence Relation, $\sim$ is an equivalence relation on $X$.
For all $g \in G$, let $\eqclass g \sim$ denote the equivalence class of $g$ under $\sim$.
Note that Group Homomorphism is Invariant under Congruence Modulo Kernel.
In particular, since $N \subseteq \ker f$, we have that $f$ is $\sim$-invariant.
Hence, by Universal Property of Quotient Set, there exists a unique such mapping $\overline f: G / \sim{} \to H$, which is given by:
:$\forall g \in G: \map {\overline f} {\eqclass g \sim} = \map f g$
From Universal Property of Quotient Set, $f$ is well-defined.
By definition of (left) coset space, $G / N = G / \sim$. 
By Left Cosets are Equal iff Product with Inverse in Subgroup:
:for all $g \in G$ the equivalence class $\eqclass g \sim$ is the (left) coset of $N$ containing $g$, which is $g N$.
Thus $\overline f$ is given by:
:$\forall g \in G: \map {\overline f} {g N} = \map f g$
We have that $\overline f$ is a group homomorphism.
Indeed, for all $x, y \in G$ we have:
{{begin-eqn}}
{{eqn | l = \map {\overline f} {\paren {x N} \paren {y N} }
      | r = \map {\overline f} {\paren {x y} N}
      | c = {{Defof|Quotient Group}}
}}
{{eqn | r = \map f {x y}
      | c = Definition of $\overline f$
}}
{{eqn | r = \map f x \map f y
      | c = {{Defof|Group Homomorphism}}
}}
{{eqn | r = \map {\overline f} {x N} \map {\overline f} {y N}
      | c = Definition of $\overline f$
}}
{{end-eqn}}
Thus we have shown the existence of such a group homomorphism.
{{qed|lemma}}