hcju/proofwikiqa huggingface dataset:
{{AimForCont}} $Y$ is not a complete lattice. Then there is an $S \subseteq Y$ with no least upper bound in $Y$. Let $U = \left\{{b \in Y: b \text{ is an upper bound of } S}\right\}$. Note that $U$ may be empty. Let: : $\mathcal U = \left\{ {u^\succeq: u \in U}\right\}$ : $\mathcal L = \left\{ {s^\preceq: s \in S}\right\}$. where: : $u^\succeq$ is the upper closure of $u$ in $U$ : $s^\preceq$ is the lower closure of $s$ in $S$. Then $\mathcal U \cup \mathcal L$ covers $Y$: Let $y \in Y$. If $y$ is an upper bound for $S$, then since $S$ has no least upper bound in $Y$, there is a $y' \in U$ such that $y' \prec y$. Thus $y \in y'^\succeq$, so $\displaystyle y \in \bigcup \mathcal U$. Otherwise, there is an $s \in S$ such that $y \prec s$, so $\displaystyle y \in \bigcup \mathcal L$. $\mathcal U \cup \mathcal L$ has no finite subcover: Let $F_U$ and $F_L$ be finite subsets of $U$ and $S$, respectively. Define: :$\mathcal F_U = \left\{ {u^\succeq: u \in F_S}\right\}$ :$\mathcal F_L = \left\{ {s^\preceq: s \in F_L}\right\}$ We will show that $\mathcal F_U \cup \mathcal F_L$ does not cover $Y$. Suppose to the contrary that $\mathcal F_U \cup \mathcal F_L$ is a cover for $Y$. Since $Y$ is non-empty, $F_L$ or $F_U$ is non-empty. If $F_L$ is non-empty, it has a maximum $m$. Since $S$ has no least upper bound, it has no maximum Therefore there is an $s \in S$ such that $m \prec s$. Thus $\displaystyle s \notin \bigcup \mathcal F_L$. By the definition of upper bound, no upper bound of $S$ precedes $m$. So $\displaystyle m \notin \bigcup \mathcal F_U$. Thus $\mathcal F_L \cup \mathcal F_U$ does not cover $Y$, a contradiction. If $F_U$ is non-empty, it has a minimum $m$. Then since $S$ has no least upper bound in $Y$, there is a $u \in U$ such that $u \prec m$. Thus $\displaystyle u \notin \bigcup \mathcal F_U$. But since $u$ is an upper bound of $S$, it is not succeeded by any element of $S$. So $\displaystyle u \notin \bigcup \mathcal F_L$. Thus $\mathcal F_L \cup \mathcal F_U$ does not cover $Y$, a contradiction. So $\mathcal U \cup \mathcal L$ is an open cover of $Y$ with no finite subcover So $Y$ is not compact, which contradicts the supposition. By Proof by Contradiction, it follows that $\left({Y, \preceq \restriction_Y}\right)$ is a complete lattice. {{qed}} Category:Complete Lattices Category:Compact Subspace of Linearly Ordered Space 6bbqncbqivqa8toel2klaar7xfp2cg1
Let $X$ be a set. Let $\SS \subseteq \powerset X$ be a synthetic sub-basis on $X$. Let $\BB$ be the synthetic basis on $X$ generated by the synthetic sub-basis $\SS$. Let $\map \tau \SS$ be the topology on $X$ generated by the synthetic basis $\BB$. We now show that: :$(1): \quad$ $\SS \subseteq \map \tau \SS$ :$(2): \quad$ For any topology $\TT$ on $X$, the implication $\SS \subseteq \TT \implies \map \tau \SS \subseteq \TT$ holds. We have that: :$\SS \subseteq \BB \subseteq \map \tau \SS$ Hence by transitivity of $\subseteq$: :$\SS \subseteq \map \tau \SS$ Suppose that $\TT$ is a topology on $X$ such that $\SS \subseteq \TT$. From General Intersection Property of Topological Space: :$\BB \subseteq \TT$ By open set axiom $(1)$: :$\map \tau \SS \subseteq \TT$ It follows that $\map \tau \SS$ is the unique topology on $X$ satisfying conditions $(1)$ and $(2)$. {{qed}} Category:Sub-Bases oqm1jc8rte8ftibb6yxmzch5w8zo1v0
Let $h \in H$ such that $\phi_h: G \to H$ is a (group) homomorphism, where $\phi_h$ is defined as: :$\forall g \in G: \map {\phi_h} g = h$ Then from Group Homomorphism Preserves Identity: :$\map {\phi_h} {e_G} = e_H$ and so $h = e_H$. Hence the result by definition of constant mapping. It remains to prove that such a constant mapping is in fact a homomorphism. Let $x, y \in G$. {{begin-eqn}} {{eqn | l = \map {\phi_{e_H} } {x \circ y}       | r = e_H       | c = Definition of $\phi_{e_H}$ }} {{eqn | r = e_H * e_H       | c = {{Defof|Identity Element}} }} {{eqn | r = \map {\phi_{e_H} } x \circ \map {\phi_{e_H} } y       | c = by definition of $\phi_{e_H}$ }} {{end-eqn}} Thus the morphism property is demonstrated, and $\phi_{e_H}$ is seen to be a homomorphism. {{qed}}
By the definition of set union, we have: :$\displaystyle \forall i \in I: \forall S \in \mathbb S_i: S \in \bigcup_{j \mathop \in I} \mathbb S_j$ Hence, by Set is Subset of Union: General Result: :$\displaystyle \forall i \in I: \forall S \in \mathbb S_i: S \subseteq \bigcup \bigcup_{j \mathop \in I} \mathbb S_j$ By Union is Smallest Superset: General Result, it follows that: :$\displaystyle \forall i \in I: \bigcup \mathbb S_i \subseteq \bigcup \bigcup_{j \mathop \in I} \mathbb S_j$ Therefore, by Union is Smallest Superset: Family of Sets, we conclude that: :$\displaystyle \bigcup_{i \mathop \in I} \bigcup \mathbb S_i \subseteq \bigcup \bigcup_{j \mathop \in I} \mathbb S_j$ By Set is Subset of Union General Result and then Set is Subset of Union: Family of Sets, we have: :$\displaystyle \forall i \in I: \forall S \in \mathbb S_i: S \subseteq \bigcup \mathbb S_i \subseteq \bigcup_{j \mathop \in I} \bigcup \mathbb S_j$ Because $\subseteq$ is transitive, we can apply the definition set union to conclude that: :$\displaystyle \forall S \in \bigcup_{i \mathop \in I} \mathbb S_i: S \subseteq \bigcup_{j \mathop \in I} \bigcup \mathbb S_j$ Hence, by Union is Smallest Superset: General Result: :$\displaystyle \bigcup \bigcup_{i \mathop \in I} \mathbb S_i \subseteq \bigcup_{j \mathop \in I} \bigcup \mathbb S_j$ By definition of set equality, the result follows. {{qed}} Category:Union Distributes over Union 8ojqstck027fv727ed67nneo45t7n6g
=== Existence === Define: :$\mathbb E = \leftset {\TT \subseteq \powerset X: \SS \subseteq \TT}$ and $\TT$ is a topology on $\rightset X$ Since $\powerset X$ is a topology on $X$, it follows that $\mathbb E$ is non-empty. Hence, we can define: :$\displaystyle \map \tau \SS = \bigcap \mathbb E$ It follows that $\map \tau \SS$ is a topology on $X$. By Intersection is Largest Subset: General Result, it follows that $\SS \subseteq \map \tau \SS$. By Intersection is Subset: General Result, it follows that if $\SS \subseteq \TT$ and $\TT$ is a topology on $X$, then $\map \tau \SS \subseteq \TT$. {{qed|lemma}} === Uniqueness === Suppose that $\TT_1$ and $\TT_2$ are both topologies on $X$ satisfying conditions $(1)$ and $(2)$. By condition $(1)$, we have $\SS \subseteq \TT_2$; hence, we can apply condition $(2)$ to conclude that: :$\TT_1 \subseteq \TT_2$ Similarly: :$\TT_2 \subseteq \TT_1$ By definition of set equality: :$\TT_1 = \TT_2$ {{qed}}
=== $(1)$ implies $(2)$ === By definition, $\mathcal C$ covers $X$ {{iff}} $X \subseteq \displaystyle \bigcup \mathcal C$. By Union of Subsets is Subset, we have that: :$\displaystyle \bigcup \mathcal C \subseteq X$ since $\mathcal C \subseteq \mathcal P \left({X}\right)$. By definition of set equality, it follows that $X = \displaystyle \bigcup \mathcal C$. {{qed|lemma}} === $(2)$ implies $(3)$ === Since $\mathcal C \subseteq \mathcal C$, we can take $\mathcal S = \mathcal C$. Hence $(3)$ is immediate from $(2)$. {{qed|lemma}} === $(3)$ implies $(1)$ === By assumption, $X = \displaystyle \bigcup \mathcal S$ for some $\mathcal S \subseteq \mathcal C$. By Union is Increasing, we have that: :$\displaystyle \bigcup \mathcal S \subseteq \bigcup \mathcal C$ Hence, $X \subseteq \displaystyle \bigcup \mathcal C$, i.e. $\mathcal C$ is a cover for $X$. {{qed}} Category:Covers i6k1g0bz2xv6x1acrm9np06vcdovjap
By definition, a subgroup is normal in $G$ {{iff}}: :$\forall g \in G: g \circ N = N \circ g$ First note that: :$(1): \quad \paren {\forall g \in G: g \circ N \circ g^{-1} = N} \iff \paren {\forall g \in G: g^{-1} \circ N \circ g = N}$ which is shown by, for example, setting $h := g^{-1}$ and substituting. === Necessary Condition === Suppose that $N$ is normal in $G$. Then: {{begin-eqn}} {{eqn | lo= \forall g \in G:       | l = g \circ N       | r = N \circ g       | c = {{Defof|Normal Subgroup}} of $G$ }} {{eqn | ll= \leadsto       | l = \paren {g \circ N} \circ g^{-1}       | r = \paren {N \circ g} \circ g^{-1}       | c = {{Defof|Subset Product}} }} {{eqn | ll= \leadsto       | l = g \circ N \circ g^{-1}       | r = N \circ \paren {g \circ g^{-1} }       | c = Subset Product within Semigroup is Associative: Corollary }} {{eqn | ll= \leadsto       | l = g \circ N \circ g^{-1}       | r = N \circ e       | c = {{Defof|Inverse Element}} }} {{eqn | ll= \leadsto       | l = g \circ N \circ g^{-1}       | r = N       | c = Coset by Identity }} {{end-eqn}} Similarly: {{begin-eqn}} {{eqn | lo= \forall g \in G:       | l = N \circ g       | r = g \circ N       | c = {{Defof|Normal Subgroup}} }} {{eqn | ll= \leadsto       | r = g^{-1} \circ \paren {N \circ g}       | l = g^{-1} \circ \paren {g \circ N}       | c = {{Defof|Subset Product}} }} {{eqn | ll= \leadsto       | l = g^{-1} \circ N \circ g       | r = \paren {g^{-1} \circ g} \circ N       | c = Subset Product within Semigroup is Associative: Corollary }} {{eqn | ll= \leadsto       | l = g^{-1} \circ N \circ g       | r = e \circ N       | c = {{Defof|Inverse Element}} }} {{eqn | ll= \leadsto       | l = g^{-1} \circ N \circ g       | r = N       | c = Coset by Identity }} {{end-eqn}} {{qed|lemma}} === Sufficient Condition === Let $N$ be a subgroup of $G$ such that: : $\forall g \in G: g \circ N \circ g^{-1} = N$ and so from $(1)$ above: : $\forall g \in G: g^{-1} \circ N \circ g = N$ Then: {{begin-eqn}} {{eqn | lo= \forall g \in G:       | l = g \circ N \circ g^{-1}       | r = N       | c =  }} {{eqn | ll= \leadsto       | l = \paren {g \circ N \circ g^{-1} } \circ g       | r = N \circ g       | c = {{Defof|Subset Product}} }} {{eqn | ll= \leadsto       | l = \paren {g \circ N} \circ \paren {g^{-1} \circ g}       | r = N \circ g       | c = Subset Product within Semigroup is Associative: Corollary }} {{eqn | ll= \leadsto       | l = \paren {g \circ N} \circ e       | r = N \circ g       | c = {{Defof|Inverse Element}} }} {{eqn | ll= \leadsto       | l = g \circ N       | r = N \circ g       | c = Coset by Identity }} {{end-eqn}} {{qed}}
From Subgroup is Superset of Conjugate iff Normal, $N$ is normal in $G$ {{iff}}: :$\forall g \in G: N \supseteq g \circ N \circ g^{-1}$ :$\forall g \in G: N \supseteq g^{-1} \circ N \circ g$ From Subgroup is Subset of Conjugate iff Normal, $N$ is normal in $G$ {{iff}}: :$\forall g \in G: N \subseteq g \circ N \circ g^{-1}$ :$\forall g \in G: N \subseteq g^{-1} \circ N \circ g$ The result follows by definition of set equality. {{qed}}
By definition, a subgroup is normal in $G$ {{iff}}: :$\forall g \in G: g \circ N = N \circ g$ === Necessary Condition === Suppose that $g \circ N = N \circ g$, by definition 1 of normality in $G$. Let $n \in N$. Then: {{begin-eqn}} {{eqn | l = g \circ n       | o = \in       | r = N \circ g       | c = {{Defof|Coset}} }} {{eqn | ll= \leadstoandfrom       | lo= \exists n_1 \in N:       | l = g \circ n       | r = n_1 \circ g       | c = {{Defof|Coset}} }} {{eqn | ll= \leadstoandfrom       | l = g \circ n \circ g^{-1}       | r = n_1 \circ g \circ g^{-1}       | c =  }} {{eqn | r = n_1 \circ e       | c = {{Defof|Inverse Element}} }} {{eqn | r = n_1       | c = {{Defof|Identity Element}} }} {{eqn | ll= \leadstoandfrom       | l = g \circ n \circ g^{-1}       | o = \in       | r = N       | c = Definition of $n_1$ }} {{end-eqn}} {{qed|lemma}} === Sufficient Condition === Suppose that: : $\forall g \in G: \paren {n \in N \iff g \circ n \circ g^{-1} \in N}$ Let $g \circ n \circ g^{-1} \in N$. {{begin-eqn}} {{eqn | lo= \exists n_1 \in N:       | l = g \circ n \circ g^{-1}       | r = n_1       | c =  }} {{eqn | ll= \leadsto       | l = g \circ n       | r = n_1 \circ g       | c = Group Axioms }} {{eqn | ll= \leadsto       | l = g \circ n       | o = \in       | r = N \circ g       | c = {{Defof|Coset}} }} {{eqn | ll= \leadsto       | l = g \circ N       | o = \subseteq       | r = N \circ g       | c = {{Defof|Subset}} }} {{end-eqn}} Similarly: :$n \circ g \in N \circ g \implies N \circ G = g \circ N$ {{begin-eqn}} {{eqn | lo= \exists n_1 \in N:       | l = g \circ n \circ g^{-1}       | r = n_2       | c =  }} {{eqn | ll= \leadsto       | l = n \circ g^{-1}       | r = g^{-1} \circ n_2       | c = Group Axioms }} {{eqn | ll= \leadsto       | l = n \circ g^{-1}       | o = \in       | r = g^{-1} \circ N       | c = {{Defof|Coset}} }} {{eqn | ll= \leadsto       | l = N \circ g^{-1}       | o = \subseteq       | r = g^{-1} \circ N       | c = {{Defof|Subset}} }} {{end-eqn}} As $g$ is arbitrary, then so is $g^{-1}$. Thus: :$N \circ g \subseteq g \circ N$ By definition of set equality: : $g \circ N = N \circ g$
=== Necessary Condition === Let $N$ be a normal subgroup of $G$ by Definition 1. Then the equality of the coset spaces follows directly from definition of normal subgroup and coset. {{qed|lemma}} === Sufficient Condition === Suppose that every right coset of $N$ in $G$ is a left coset of $N$ in $G$. Let $g \in G$. Since every right coset of $N$ in $G$ is a left coset, there exists an $h \in G$ such that $N \circ g = h \circ N$. By Element of Group is in its own Coset: :$g \in N \circ g = h \circ N$ From Element in Left Coset iff Product with Inverse in Subgroup: : $g^{-1} \circ h \in N$ Then: {{begin-eqn}} {{eqn | l = N \circ g       | r = \paren {g \circ g^{-1} } \circ \paren {h \circ N}       | c = {{Defof|Inverse Element}} and Coset by Identity }} {{eqn | r = g \circ \paren {\paren { g^{-1} \circ h } \circ N}       | c = Subset Product within Semigroup is Associative: Corollary }} {{eqn | r = g \circ N       | c = $g^{-1} \circ h \in N$ and Left Coset Equals Subgroup iff Element in Subgroup }} {{end-eqn}} Since this holds for all $g \in G$, $N$ is normal in $G$ (by definition 1). {{Qed}}
Let $S_n$ act on the set $S$. Let $e$ be the identity of $S_n$, by definition the identity mapping $I_S$ on $S$. As $S$ has at least three elements, three can be arbitrary selected and called $a$, $b$ and $c$. Let $\rho$ be a transposition of $S_n$, transposing elements $a$ and $b$. $\rho$ can be described in cycle notation as $\paren {a \ b}$. From Transposition is Self-Inverse it follows that $\set {e, \rho}$ is a subgroup of $S_n$. Let $\pi$ be the permutation on $S$ described in cycle notation as $\paren {a \ b \ c}$. By inspection it is found that $\pi^{-1} = \paren {a \ c \ b}$. Then we have: {{begin-eqn}} {{eqn | l = \pi^{-1} \rho \pi       | r = \paren {a \ c \ b} \paren {a \ b} \paren {a \ b \ c}       | c =  }} {{eqn | r = \paren {a \ c}       | c = by evaluation }} {{eqn | o = \notin       | r = \set {e, \rho}       | c =  }} {{end-eqn}} So, by definition, $\set {e, \rho}$ is not a normal subgroup. {{qed}}
We have the implicit mappings $i_1: S \to S \sqcup T$ and $i_2: T \to S \sqcup T$ defined by: :$i_1 \left({s}\right) = \left({s, 1}\right)$ :$i_2 \left({t}\right) = \left({t, 2}\right)$ Now given a set $V$ and mappings $f: S \to V$ and $g: T \to V$, there is to be a unique $\left[{f, g}\right]: S \sqcup T \to V$ such that: :$f = \left[{f, g}\right] \circ i_1$ :$g = \left[{f, g}\right] \circ i_2$ Define $\left[{f, g}\right]$ by: :$\left[{f, g}\right] \left({x, \delta}\right) := \begin{cases}  f \left({x}\right) & \text{if $\delta = 1$} \\  g \left({x}\right) & \text{if $\delta = 2$}\end{cases}$ Now it is immediate that $\left[{f, g}\right]$ so defined satisfies the two conditions above. Furthermore, these conditions fix $\left[{f, g}\right]$ uniquely, since every $\left({x, \delta}\right) \in S \sqcup T$ has $\delta = 1$ or $\delta = 2$. Hence the result, by definition of coproduct. {{qed}}
We have: * Inclusion Mapping on Subgroup is Homomorphism * Inclusion Mapping is Injection The result follows by definition of (group) monomorphism. {{qed}}
=== Necessary Condition === Let $\phi: G \to H$ be an isomorphism. Then by definition $\phi$ is a bijection. From Bijection iff Inverse is Bijection it follows that: :$\exists \phi^{-1}: \struct {H, *} \to \struct {G, \circ}$ such that $\phi^{-1}$ is also a bijection. Thus: {{begin-eqn}} {{eqn | ll= \forall g \in G, h \in H:       | o =        | r = \map \phi g = h \iff \map {\phi^{-1} } h = g       | c = Inverse Element of Bijection }} {{eqn | o = \leadsto       | r = \map \phi {g_1 \circ g_2} = h_1 * h_2       | c = {{Defof|Morphism Property}} }} {{eqn | o = \leadsto       | r = \map {\phi^{-1} } {h_1 * h_2} = g_1 \circ g_2 = \map {\phi^{-1} } {h_1} \circ \map {\phi^{-1} } {h_2}       | c = Inverse Element of Bijection }} {{end-eqn}} So $\phi^{-1}: \struct {H, *} \to \struct {G, \circ}$ is a homomorphism. $\phi^{-1}$ is also (from above) a bijection. Thus, by definition, $\phi^{-1}$ is an isomorphism. === Sufficient Condition === Let $\phi^{-1}: \struct {H, *} \to \struct {G, \circ}$ be an isomorphism. Applying the same result as above in reverse, we have that $\paren {\phi^{-1} }^{-1}: \struct {G, \circ} \to \struct {H, *}$ is also an isomorphism. But by Inverse of Inverse of Bijection: : $\paren {\phi^{-1} }^{-1} = \phi$ and hence the result. {{qed}}
A specific instance of Inverse of Algebraic Structure Isomorphism is Isomorphism. {{qed}}
It will suffice to define partial mappings such that the axiom for a magma of sets crystallises into the axioms for a topology. Let $X$ be any set, and let $\mathcal P \left({X}\right)$ be its power set. Define: :$\phi_1: \mathcal P \left({X}\right) \to \mathcal P \left({X}\right): \phi_1 \left({S}\right) := X$ :$\phi_2: \mathcal P \left({X}\right)^2 \to \mathcal P \left({X}\right): \phi_2 \left({S, T}\right) := S \cap T$ For each index set $I$, define: :$\phi_I: \mathcal P \left({X}\right)^I \to \mathcal P \left({X}\right): \phi_I \left({\left({S_i}\right)_{i \in I}}\right) := \displaystyle \bigcup_{i \mathop \in I} S_i$ It is blatantly obvious that these partial mappings capture the axioms for a topology. {{finish}} {{qed}} Category:Topology Category:Magmas of Sets t0nurgf55mollmgw3yf2eobuitc6d7m
So as to alleviate possible confusion over notation, let the composite of $\phi$ and $\psi$ be denoted $\psi \bullet \phi$ instead of the more usual $\psi \circ \phi$. Then what we are trying to prove is denoted: :$\paren {\psi \bullet \phi}: \struct {G_1, \circ} \to \struct {G_3, \oplus}$ is a homomorphism. To prove the above is the case, we need to demonstrate that the morphism property is held by $\circ$ under $\psi \bullet \phi$. We take two elements $x, y \in G_1$, and put them through the following wringer: {{begin-eqn}} {{eqn | l = \map {\paren {\psi \bullet \phi} } {x \circ y}       | r = \map \psi {\map \phi {x \circ y} }       | c = {{Defof|Composition of Mappings}} }} {{eqn | r = \map \psi {\map \phi x * \map \phi y}       | c = {{Defof|Morphism Property}}: applied to $\circ$ under $\phi$ }} {{eqn | r = \map \psi {\map \phi x} \oplus \map \psi {\map \phi y}       | c = {{Defof|Morphism Property}}: applied to $*$ under $\psi$ }} {{eqn | r = \map {\paren {\psi \bullet \phi} } x \oplus \map {\paren {\psi \bullet \phi} } y       | c = {{Defof|Composition of Mappings}} }} {{end-eqn}} Disentangling the confusing and tortuous expressions above, we (eventually) see that this shows that the morphism property is indeed held by $\circ$ under $\psi \bullet \phi$. Thus $\paren {\psi \bullet \phi}: \struct {G_1, \circ} \to \struct {G_3, \oplus}$ is indeed a homomorphism. {{qed}}
A specific instance of Composite of Homomorphisms on Algebraic Structure is Homomorphism. {{qed}}
A monomorphism is a homomorphism which is also an injection. From Composite of Group Homomorphisms is Homomorphism, $\psi \circ \phi$ is a homomorphism. From Composite of Injections is Injection, $\psi \circ \phi$ is an injection. {{qed}}
A group epimorphism is a group homomorphism which is also a surection. From Composite of Group Homomorphisms is Homomorphism, $\psi \circ \phi$ is a group homomorphism. From Composite of Surjections is Surjection, $\psi \circ \phi$ is a surection. {{qed}}
A group isomorphism is a group homomorphism which is also a bijection. From Composite of Group Homomorphisms is Homomorphism, $\psi \circ \phi$ is a group homomorphism. From Composite of Bijections is Bijection, $\psi \circ \phi$ is a bijection. {{qed}}
For brevity, let us write $\displaystyle \sum$ for $\displaystyle \sum_{i \mathop = 1}^n$. Then: {{begin-eqn}} {{eqn | l =  \sum \left({x_i - \overline x}\right)       | r = x_1 - \overline x + x_2 - \overline x + \cdots + x_n - \overline x       | c = definition of $\sum$ }} {{eqn | r = x_1 - \sum \frac {x_i} n + x_2 - \sum \frac {x_i} n + \cdots + x_n - \sum \frac {x_i} n       | c = definition of $\overline x$ }} {{eqn | r = \left({x_1 + x_2 + \cdots + x_n}\right) - n \left({\sum \frac {x_i} n}\right) }} {{eqn | r = \sum x_i - \sum x_i }} {{eqn | r = 0 }} {{end-eqn}} {{qed}}
=== Necessary Condition === Suppose $\psi$ exists as defined. Then: :$K = \map {\phi^{-1} } {e_H} = \map {q^{-1} } {\map {\psi^{-1} } {e_H} }$ That is: $\map {\psi^{-1} } {e_H}$ is the kernel of $\psi$. So by Kernel is Normal Subgroup of Domain, $\map {\psi^{-1} } {e_H}$ is a normal subgroup of $G / N$. This corresponds via $q^{-1}$ to $K$, which must then be a normal subgroup of $G$ which contains $N$. {{qed|lemma}} === Sufficient Condition === Suppose $N \subseteq K$. Let $\psi: G / N \to H$ be defined as: :$\forall g N \in G / N: \map \psi {g N} = \map \phi g \in H$ We have that: :$n \in g N \implies n^{-1} g \in N \subseteq K$ So: :$\map \phi {n^{-1} g} = \map \phi n^{-1} \map \phi g = e_H$ and so $\map \phi n = \map \phi g$. Thus $\psi$ is well-defined. As $N \subseteq K$ , the Quotient Theorem for Group Homomorphisms has that $\Img \phi$ is isomorphic to $G / K$. We have that: :$\Img \phi = \Img {\psi \circ q} = \Img \psi$ and: :$\map \ker \psi = K / N = \set {k n: k \in K}$ Thus: :$\Img \psi$ is isomorphic to $G / K$ and also: :$\Img \psi$ is isomorphic to the quotient of the domain $G / N$ of $\psi$ modulo the kernel $K / N$ of $\psi$. {{qed}}
A specific instance of External Direct Product Identity, where the algebraic structures in question are groups. {{qed}}
Let $e_G$ be the identity for $\struct {G, \circ_1}$ Let $e_H$ be the identity for $\struct {H, \circ_2}$. Then: {{begin-eqn}} {{eqn | l = \tuple {g, h} \circ \tuple {g^{-1}, h^{-1} }       | r = \tuple {g \circ_1 g^{-1}, h \circ_2 h^{-1} }       | c =  }} {{eqn | r = \tuple {e_G, e_H}       | c =  }} {{eqn | r = \tuple {g^{-1} \circ_1 g, h^{-1} \circ_2 h}       | c =  }} {{eqn | r = \tuple {g^{-1}, h^{-1} } \circ \tuple {g, h}       | c =  }} {{end-eqn}} So the inverse of $\tuple {g, h}$ is $\tuple {g^{-1}, h^{-1} }$. {{qed}}
A specific instance of External Direct Product Inverses, where the algebraic structures in question are groups. {{qed}}
{{begin-eqn}} {{eqn | l=\left({\left({g_1, h_1}\right) \circ \left({g_2, h_2}\right)}\right) \circ \left({g_3, h_3}\right)       | r=\left({\left({g_1 \circ_1 g_2}\right) \circ_1 g_3, \left({h_1 \circ_2 h_2}\right) \circ_2 h_3}\right)       | c= }} {{eqn | r=\left({g_1 \circ_1 g_2 \circ_1 g_3, h_1 \circ_2 h_2 \circ_2 h_3}\right)       | c= }} {{eqn | r=\left({g_1 \circ_1 \left({g_2 \circ_1 g_3}\right), h_1 \circ_2 \left({h_2 \circ_2 h_3}\right)}\right)       | c= }} {{eqn | r=\left({g_1, h_1}\right) \circ \left({\left({g_2, h_2}\right) \circ \left({g_3, h_3}\right)}\right)       | c= }} {{end-eqn}} Thus $\circ$ is seen to be associative in $\left({G \times H, \circ}\right)$. {{qed}}
By definition of group, both $\circ_1$ and $\circ_2$ are associative operations. The result follows from External Direct Product Associativity, where the algebraic structures in question are groups. {{qed}}