hcju/proofwikiqa huggingface dataset:
=== Sufficient condition === Let $U$ be cyclic. Let $n \ge 0$ be an integer. Let $n = p_1^{e_1} \cdots p_r^{e_r}$, be the decomposition of $n$ into distinct prime powers given by the Fundamental Theorem of Arithmetic. Then by the corollary to the Chinese remainder theorem we have an isomorphism: :$\Z / n \Z \simeq \Z / p_1 \Z \times \cdots \times \Z / p_r \Z$ By Units of Direct Product are Direct Product of Units we have: :$\paren {\Z / n \Z}^\times \simeq \paren {\Z / p_1 \Z}^\times \times \cdots \times \paren {\Z / p_r \Z}^\times$ Suppose that $r \ge 2$, and choose $i, j \in \set {1, \ldots, r}$ such that $i \ne j$. If $\paren {\Z / p_i \Z}^\times$ or $\paren {\Z / p_j \Z}^\times$ is not cyclic, then $\paren {\Z / n \Z}^\times$ cannot be cyclic. {{explain|Link to a result establishing the above fact.}} Therefore suppose that $\paren {\Z / p_i \Z}^\times$ and $\paren {\Z / p_j \Z}^\times$ are cyclic. By Order of Group of Units of Integers Modulo m these groups have orders: :$\map \phi {p_i^{e_i} }$ and: :$\map \phi { p_j^{e_j} }$ respectively, where $\phi$ is the Euler $\phi$ function. By Euler Phi Function of Integer we have: :$\map \phi {p_i^{e_i} } = p_i^{e_i - 1} \paren {p_i - 1}$ and :$\map \phi {p_j^{e_j} } = p_j^{e_j - 1} \paren {p_j - 1}$ If $p_i, p_j$ are odd, $2$ divides $p_i - 1$ and $p_j - 1$. Therefore $2$ divides $\map \phi {p_i^{e_i} }$ and $\map \phi {p_j^{e_j} }$. In particular, $\map \phi {p_i^{e_i} }$ and $\map \phi {p_j^{e_j} }$ are not coprime. Now by Group Direct Product of Cyclic Groups, $\left({\Z / n \Z}\right)^\times$ is not cyclic. Let $p_i$ or $p_j$ be even. {{WLOG}}, we can assume $p_i = 2$. Then: :$\map \phi {p_i^{e_i} } = \map \phi {2^{e_i} } = p_i^{e_i - 1} \paren {p_i - 1}$ So if $e_i \ge 2$, then $2$ divides $\map \phi {p_i^{e_i} }$ and $\map \phi {p_j^{e_j} }$. In particular $\map \phi {p_i^{e_i} }$ and $\map \phi {p_j^{e_j} }$ are not coprime. Again by Group Direct Product of Cyclic Groups, $\paren {\Z / n \Z}^\times$ is not cyclic. Thus if $\paren {\Z / n \Z}^\times$ is cyclic, then $n = 2^e \times p^\alpha$ with $e = 0$ or $e = 1$, $\alpha \ge 0$ and $p \ge 3$ prime. {{Qed|lemma}} === Necessary Condition === {{ProofWanted}} Category:Ring of Integers Modulo m kzf5rgrqazm8xul4km4k01d3vtkttqk
Define a mapping: :$\phi: \Z / N \Z \to \Z / n_1 \Z \times \cdots \times \Z / n_r \Z$ by: :$\phi \left({d \pmod N}\right) = \left({d \pmod {n_1}, \ldots, d \pmod {n_r} }\right)$ Then, by Mappings Between Residue Classes, $\phi$ is well-defined. By the definition of multiplication and addition in $\Z / k \Z$, $k \in \Z$ we have: :$\left({a \pmod k}\right) + \left({b \pmod k}\right) = \left({a + b}\right) \pmod k$ and :$\left({a \pmod k}\right) \cdot \left({b \pmod k}\right) = \left({a \cdot b}\right) \pmod k$ Thus taking $k = n_1, \ldots, n_r$ separately we see that $\phi$ is a ring homomorphism. Let: :$\left({a_1 \pmod {n_1}, \ldots, a_r \pmod {n_r} }\right) \in \Z / n_1 \Z \times \cdots \times \Z / n_r \Z$ By the Chinese Remainder Theorem there exists a unique $x \in \Z / N \Z$ such that: :$\phi \left({x}\right) = \left({a_1 \pmod {n_1}, \ldots, a_r \pmod {n_r} }\right)$ Since such an $x$ exists, $\phi$ is surjective. Since this $x$ is unique modulo $N$, it follows that $\phi$ is injective. {{Qed}}
We proceed by induction over $n \ge 1$. === Basis for the induction === Suppose that $n = 1$. Then $A \sqbrk {x_1, \ldots, x_n} = A \sqbrk {x_1}$ is the ring of polynomial forms in a single indeterminate $x_1$. Thus by Hilbert's Basis Theorem, $A \sqbrk {x_1}$ is Noetherian. === Induction step === Suppose that $A \sqbrk {x_1, \ldots, x_{n - 1} }$ is Noetherian. We wish to show that $A \sqbrk {x_1, \ldots, x_n}$ is Noetherian. By Adjunction of Indeterminates to Ring is Associative: :$A \sqbrk {x_1, \ldots, x_n} = A \sqbrk {x_1, \ldots, x_{n - 1} } \sqbrk {x_n}$ is the polynomial ring in a single indeterminate $x_n$ over the Noetherian ring $A \sqbrk {x_1, \ldots, x_{n - 1} }$. Thus by Hilbert's Basis Theorem, A \sqbrk {x_1, \ldots, x_n}$ is Noetherian. {{qed}} Category:Commutative Algebra r2olgak94ntrzs0mz2t19yop119et2v
Assume that none of the $Y_k$ are empty. If not, then any empty ones can simply be removed. === Uniqueness === {{AimForCont}} that $Y_j, Y_k \in \mathcal U$ with $j \ne k$. Then since $Y_1, \dots, Y_n$ are a pairwise disjoint: : $Y_j \cap Y_k = \varnothing$ But by the definition of an ultrafilter, $\mathcal U$ has the finite intersection property. It follows from this contradiction that no such $Y_j, Y_k \in \mathcal U$ with $j \ne k$ exists. {{qed|lemma}} === Existence === Aiming for a contradiction, suppose that $Y_1, \dots, Y_n \notin \mathcal U$. Since the $Y_k$ are pairwise disjoint: : $\displaystyle Y_i^c = \bigcup \left\{{Y_k: k \ne i}\right\}$ Then by the definition of an ultrafilter: : $\displaystyle \forall i: Y_i^c = \bigcup \left\{{Y_k: k \ne i}\right\} \in \mathcal U$ But: : $\displaystyle \bigcap_{i \mathop = 1}^n \bigcup \left\{{Y_k: k \ne i}\right\} = \varnothing$ contradicting the fact that $\mathcal U$ has the finite intersection property. Thus $Y_k \in \mathcal U$ for some $k$. {{qed}}
Let $A \in \mathcal A$. Let: :$\mathcal B = \left\{ {B: B \subseteq X \text{ and } A \cup B \in \mathcal A}\right\}$ It is to be shown that $\mathcal B$ has finite character: First suppose that $B \in \mathcal B$ and $F$ is a finite subset of $B$. Then since $B \in \mathcal B$, $B \subseteq X$ and $A \cup B \in \mathcal A$. We wish to show that $F \in \mathcal B$. Since $F \subseteq B \subseteq X$: :$F \subseteq X$ It remains to be shown that: :$A \cup F \in \mathcal A$. $A \cup F \subseteq A \cup B$. Let $G$ be a finite subset of $A \cup F$. Then $G$ is a finite subset of $A \cup B$. Since $A \cup B \in \mathcal A$ and $\mathcal A$ has finite character, $G \in \mathcal A$. Thus every finite subset of $A \cup F$ is in $\mathcal A$. Since $\mathcal A$ has finite character, $A \cup F \in \mathcal A$. Thus $F \in \mathcal B$. Suppose instead that $B \subseteq X$ and every finite subset of $B$ is an element of $\mathcal B$. We wish to show that $B \in \mathcal B$. In order to do this, we must show that $A \cup B \in \mathcal A$. Let $F$ be a finite subset of $A \cup B$. $\mathcal A$ has finite character, $F \cap A \in \mathcal A$. Since $F \cap B$ is a finite subset of $B$, $F \cap B \in \mathcal B$ by assumption. Then by the definition of $\mathcal B$: :$\left({F \cap B}\right) \cup A \in \mathcal A$ But $F$ is a finite subset of $\left({F \cap B}\right) \cup A \in \mathcal A$. Since $\mathcal A$ has finite character, $F \in \mathcal A$. As this holds for all finite subsets of $A \cup B$ and $\mathcal A$ has finite character: :$A \cup B \in \mathcal A$ {{qed|lemma}} Let $B \in \mathcal B$ and $x \in X$. Then: :$B \cup A \in \mathcal A$ so either $B \cup A \cup \left\{ {x}\right\}$ or $B \cup A \cup \left\{ {x'}\right\}$ is in $\mathcal A$. But then $B \cup \left\{ {x}\right\}$ or $B \cup \left\{ {x'}\right\}$ is in $\mathcal B$ by the definition of $\mathcal B$. Thus $\mathcal B$ satisfies the premises of the weak form of the Restricted Tukey-Teichmüller Theorem. Thus there is a $B \in \mathcal B$ such that for all $x \in X$, either $x \in B$ or $x' \in B$. Let $C = A \cup B$. Then since $B \subseteq C$, if $x \in X$ then either $x \in C$ or $x' \in C$. But since $B \in \mathcal B$: :$C = A \cup B \in \mathcal A$ {{qed}} {{namedfor|John Wilder Tukey|name2 = Oswald Teichmüller|cat = Tukey|cat2 = Teichmüller}}
{{AimForCont}} that neither $J \vee z$ nor $J \vee \neg z$ has the finite join property. Then there are $x_1, \dots, x_n, y_1, \dots, y_m \in J$ such that $x_1 \vee \dots \vee x_n \vee z = y_1 \vee \dots \vee y_m \vee \neg z = \top$. Let $q = x_1 \vee \dots \vee x_n \vee y_1 \vee \dots \vee y_m$. Then $q \vee z = q \vee \neg z = \top$. Thus: {{begin-eqn}} {{eqn | l = \top       | r = \paren {q \vee z} \wedge \paren {q \vee \neg z}       | c =  }} {{eqn | r = \paren {q \wedge q} \vee \paren {q \wedge \neg z} \vee \paren {q \wedge z} \vee \paren {z \wedge \neg z}       | c =  }} {{eqn | r = q \wedge \paren {q \vee \neg z \vee z \vee \top}       | c =  }} {{eqn | r = q       | c =  }} {{end-eqn}} This contradicts the fact that $J$ has the finite join property. {{qed}} Category:Lattice Theory jec44gmhujcir8cxdixrvr7hbiq6deh
=== $(1)$ implies $(2)$ === Let $\mathcal F$ have finite character in the sense of Definition:Finite Character/Mappings. That is, suppose that for each partial mapping $f \subseteq S \times T$: :$f \in \mathcal F$ {{iff}} for each finite subset $K$ of the domain of $f$, the restriction of $f$ to $K$ is in $\mathcal F$. Let $q \subseteq S \times T$. First suppose that $q \in \mathcal F$. Let $r$ be a finite subset of $q$. Then $r$ is a partial mapping from its domain into $T$. Since $r$ is finite, its domain, $\operatorname{Dom} \left({r}\right)$, is a finite subset of the domain of $q$. Since $\mathcal F$ is of finite character as a set of mappings, $q \restriction \operatorname{Dom} \left({r}\right)$ is in $\mathcal F$. But $q \restriction \operatorname{Dom} \left({r}\right) = r$. Therefore $r \in \mathcal F$. {{explain}} Suppose instead that for each finite subset $r$ of $q$, $r \in \mathcal F$. Then for each finite subset $r$ of $q$, $r$ is a partial mapping from $S$ to $T$. Let $x \in \operatorname{Dom} \left({q}\right)$. Let $y_1, y_2 \in T$. Suppose that $\left({x, y_1}\right), \left({x, y_2}\right) \in q$. Then $\left\{ {\left({x, y_1}\right), \left({x, y_2}\right)}\right\}$ is a finite subset of $q$. Therefore it must be a partial mapping. Thus $y_1 = y_2$. As this holds for all such $x, y_1, y_2$, $q$ is a partial mapping. Let $K$ be a finite subset of $\operatorname{Dom} \left({q}\right)$. Then the restriction of $q$ to $K$ is a finite subset of $q$. Thus $q \restriction K \in \mathcal F$. As this holds for all finite subsets $K$ of $\operatorname{Dom} \left({q}\right)$, $q \in \mathcal F$. So we see that $\mathcal F$ has finite character in the sense of Definition:Finite Character. {{qed|lemma}} === $(2)$ implies $(1)$ === Let $\mathcal F$ have finite character in the sense of Definition:Finite Character. {{finish}} Category:Set Theory cxsnryj9jgj9zr84ewrzzk5ycwdwlvn
Let $\map {\operatorname {Fin} } X$ be the set of finite subsets of $X$. For each $S \in \map {\operatorname {Fin} } X$, let $\Gamma_S = \set {f \in M: S \subseteq \Dom f}$. By $(2)$, $\Gamma_S$ is non-empty for each $S \in \map {\operatorname {Fin} } X$. For each $S \in \map {\operatorname {Fin} } X$, $\Gamma_S \cap \Gamma_T = \Gamma_{S \cup T}$: If $f \in \Gamma_{S \cup T}$, then $S \cup T \subseteq \Dom f$. Then $S \subseteq \Dom f$ and $T \subseteq \Dom f$, so $f \in \Gamma_S$ and $f \in \Gamma_T$. Thus $f \in \Gamma_S \cap \Gamma_T$. If $f \in \Gamma_S \cap \Gamma_T$, then $f \in \Gamma_S$ and $f \in \Gamma_T$. Thus $S \subseteq \Dom f$ and $T \subseteq \Dom f$. Thus $S \cup T \subseteq \Dom f$. So $f \in \Gamma_{S \cup T}$. Thus $\set {\Gamma_S: S \in \map {\operatorname {Fin} } X}$ has the finite intersection property. Then by the corollary to the Ultrafilter Lemma, there is an ultrafilter $\mathcal U$ on on $M$ which includes $\set {\Gamma_S: S \in \map {\operatorname {Fin} } X}$ as a subset. We will now construct a mapping $\phi: X \to Y$. Fix $x \in X$. Note that by the definitions of $\Phi$ and $\Gamma$: :$\map \Phi x = \set {\map f x: f \in \Gamma_{\set x} }$ For each $y \in \map \Phi x$, let $K_y = \set {f \in \Gamma_{\set x} : \map f x = y}$. Then $\set {K_y : y \in \map \Phi x}$ is a partition of $\Gamma_{\set x}$. {{explain}} Because $\set x \in \map {\operatorname {Fin} } X$: :$\Gamma_{\set x} \in \mathcal U$ By $(1)$, $\map \Phi x$ is finite. Thus by Component of Finite Union in Ultrafilter, there is exactly one $y \in \map \Phi x$ such that $K_y \in \mathcal U$. We define $\map \phi x$ to be that $y$. By the definition of $K_y$, for each $x \in X$ we have: :$\set {f \in \Gamma_{\set x}: \map f x = \map \phi x} \in \mathcal U$ We must show that $\phi \in \mathcal M$. Let $S$ be a finite subset of $X$. Let $\displaystyle \Psi = \bigcap_{x \mathop \in S} \set {f \in \Gamma_{\set x}: \map f x = \map \phi x}$. Since: :$\set {f \in \Gamma_{\set x}: \map f x = \map \phi x} \in \mathcal U$ for each $x \in X$ :$S$ is finite :$\mathcal U$ is a filter it follows that: :$\Psi \in \mathcal U$ Thus $\Psi \ne \O$. Choose any $f \in \Psi$. Then $\map f x \in \Gamma_{\set x}$ for each $x \in S$. So $f \in M$ and $S$ is a finite subset of the domain of $f$. Since $M$ has finite character, $f \restriction S \in M$. Furthermore, by the definition of $\Psi$: :$\map f x = \map \phi x$ for each $x \in S$. Thus: :$\phi \restriction S = f \restriction S \in \phi$ So we see that $\phi \restriction S \in M$ for each finite subset $S$ of $X$. Since $M$ has finite character, $\phi \in M$. {{qed}} {{BPI|Ultrafilter Lemma}} {{Namedfor|Robert H. Cowen|name2 = Erwin Engeler|cat = Cowen|cat2 = Engeler}}
Let $\mathcal I$ be the set of intersections of non-empty finite subsets of $\mathcal A$. Let $\mathcal F = \left\{ {T \in \mathcal P \left({S}\right) : \exists B \in \mathcal I: B \subseteq T}\right\}$. Note that $\mathcal A \subseteq \mathcal I \subseteq \mathcal F$. $\mathcal F$ is a filter on $S$: Because $\mathcal A$ has the finite intersection property: : $\varnothing \notin \mathcal I$ Because each element of $\mathcal F$ is a superset of some element of $\mathcal I$: : $\varnothing \notin \mathcal F$ Let $P \in \mathcal F$ and $P \subseteq Q \subseteq S$. Then: :$\exists B \in \mathcal I: B \subseteq P$ and so: : $B \subseteq Q \subseteq S$ Thus: :$Q \in \mathcal F$ Let $P \in \mathcal F$ and $Q \in \mathcal F$. Then: :$\exists B, C \in \mathcal I: B \subseteq P$ and $C \subseteq Q$ Then: :$B \cap C \in \mathcal I$ and: : $B \cap C \subseteq P \cap Q \subseteq S$ Thus: : $P \cap Q \in \mathcal F$ Thus, by definition, $\mathcal F$ is a filter on $S$. By the Ultrafilter Lemma, there exists an ultrafilter $\mathcal U$ on $S$ such that: : $\mathcal F \subseteq \mathcal U$ Because $\mathcal A \subseteq \mathcal F$: :$\mathcal A \subseteq \mathcal U$ {{qed}} Category:Set Theory Category:Filter Theory ns8cq9ep3o3s706dakzxbrobkbo66tr
{{ProofWanted|Use Strict Ordering can be Expanded to Compare Additional Pair.}} Category:Set Theory Category:Order Theory 74m8k8i9rhy8socwg7laaok2ie828p2
{{WLOG}}, suppose that $\phi \notin T$. Suppose that $T \cup \left\{ {\phi}\right\}$ is not finitely satisfiable. Then by definition there must be a finite subset $K$ of $T \cup \left\{ {\phi}\right\}$ which is not satisfiable. Since $T$ is finitely satisfiable, $\phi \in K$. Therefore $\Delta = K \setminus \left\{ {\phi}\right\}$ is a finite subset of $T$ such that $\Delta \models \neg \phi$. {{explain|$\models$}} Let $\Sigma$ be a finite subset of $T$. Since $\Delta \cup \Sigma$ is a finite subset of $T$, it is satisfiable. But $\Delta \models \neg \phi$. Hence $\left\{ {\neg \phi}\right\} \cup \Sigma$ is satisfiable. But any finite subset of $T \cup \left\{ {\neg \phi}\right\}$ is either in $T$ or of the form $\Sigma \cup \left\{{\neg \phi}\right\}$. This demonstrates that $T \cup \left\{ {\neg \phi}\right\}$ is finitely satisfiable.  {{qed}} {{LEM||3}} Category:Model Theory d6cgnnwob2jy4johg3o5xfioii1lc8t
=== Sufficient condition === Let $\struct {F, +, \cdot}$ be a field of order $q$. By Characteristic of Galois Field is Prime, the characteristic of $F$ is a prime number $p$. By Field of Prime Characteristic has Unique Prime Subfield the prime subfield of $F$ is $\F_p := \Z / p \Z$. By Vector Space on Field Extension is Vector Space, $F$ is an $\F_p$-vector space. Since $F$ is finite, $F$ has a finite basis over $\F_p$. By Same Dimensional Vector Spaces are Isomorphic, this means that with $k$ equal to the dimension of $F$ there is an isomorphism of vector spaces: :$F \simeq \F_p^k$ Finally by the definition of the product of cardinals: :$\card F = \card {\F_p}^k = p^k$ So the order of $F$ is a prime power. {{qed|lemma}} === Necessary condition === {{ProofWanted|need splitting fields to complete}} Category:Galois Fields ngc5i1ix4oowgkypqf7oklmvvqqayo2
Let $\mathcal A$ be the set of relations $A$ on $S$ with the property that the transitive closure $A^+$ of $A$ is a strict ordering of $S$. For each $\left({x, y}\right) \in S \times S$, let $\left({x, y}\right)' = \left({y, x}\right)$. Let $A \in \mathcal A$. Let $\left({x, y}\right) \in S \times S$. Let $\left({x, y}\right) \in A^+$. Then: :$\left({A \cup \left\{ {\left({x, y}\right)}\right\} }\right)^+ = A^+$ so: : $A \cup \left\{ {\left({x, y}\right)}\right\} \in \mathcal A$ Let $\left({y, x}\right) \in A^+$. Then: :$\left({ A \cup \left\{ {\left({y, x}\right)}\right\} }\right)^+ = A^+$ so: :$A \cup \left\{ {\left({x, y}\right)'}\right\} = A \cup \left\{ {\left({y, x}\right)}\right\} \in \mathcal A$ Otherwise, $x$ and $y$ are non-comparable by $A^+$. So by Strict Ordering can be Expanded to Compare Additional Pair: :$A \cup \left\{ {\left({x, y}\right)}\right\} \in \mathcal A$ Thus it has been shown that for each $A \in \mathcal A$ and each $\left({x, y}\right) \in S$, either: :$A \cup \left\{ {\left({x, y}\right)}\right\} \in \mathcal A$ or: :$A \cup \left\{ {\left({x, y}\right)'}\right\} \in \mathcal A$ {{qed|lemma}} Let $A \subseteq S \times S$. Let $A \in \mathcal A$. Let $F$ be a finite subset of $A$. Since $A^+$ is a strict ordering, it is asymmetric. Since Transitive Closure is Closure Operator, $F^+ \subseteq A^+$. So $F^+$ is also asymmetric. Since $F^+$ is also transitive, it is a strict ordering. So $F \in \mathcal A$. Suppose instead that each finite subset of $A$ is in $\mathcal A$. We must show that $A^+$ is antireflexive. Suppose for the sake of contradiction that for some $x \in S$, $\left({x, x}\right) \in A^+$. Then by the definition of transitive closure, there are elements $y_0, \dots, y_n$ such that $x = y_0 = y_n$ and: :$\left({y_0, y_1}\right), \left({y_1, y_2}\right), \dotsc, \left({y_{n - 1}, y_n}\right) \in A$ Let $F = \left\{ {\left({y_0, y_1}\right), \left({y_1, y_2}\right), \dotsc, \left({y_{n - 1}, y_n}\right)}\right\}$. Then $F$ is a finite subset of $A$. But $\left({x, x}\right) \in F^+$, contradicting the fact that $F \in \mathcal A$. Thus we see that $A^+$ is antireflexive, and thus a strict ordering of $S$. Therefore, $\mathcal A$ has finite character. {{qed|lemma}} Note that ${\prec} = {\prec^+}$, so ${\prec} \in \mathcal A$. By the Restricted Tukey-Teichmüller Theorem (Strong Form), there exists an $R \in \mathcal A$ such that: : ${\prec} \subseteq R$ : For each $\left({m, n}\right) \in S \times S$, either $\left({m, n}\right) \in R$ or $\left({n, m}\right) = \left({m, n}\right)' \in R$. Then: : $R^+$ is a strict total ordering of $S$. : $\forall a, b \in S: a \prec b \implies a < b$ {{qed}}
Let $\Gamma$ be a branch of a finite rooted tree $T$. Aiming for a contradiction, suppose $\Gamma$ were an infinite branch of $T$. By definition $\Gamma$ contains an infinite number of nodes. {{explain|Technically, from the definition currently posted, it doesn't - that still needs to be demonstrated.}} From Subset of Finite Set is Finite, it follows that $\Gamma$ has a finite number of nodes. From this contradiction it follows that $\Gamma$ can not be an infinite branch of $T$. Hence the result. {{qed}}
First suppose that $\phi_2$ and $\phi_4$ are surjective and $\phi_5$ is injective. Let $n_3 \in N_3$ be any element. We want to find $x \in M_3$ such that $\map {\phi_3} x = n_3$. Let $n_4 = \map {\beta_3} {n_3} \in N_4$. Since $\phi_4$ is surjective, there exists $m_4 \in M_4$ such that $\map {\phi_4} {m_4} = n_4$. Since the rows are exact, we have that: :$\map {\beta_4} {n_4} = \map {\beta_4 \circ \beta_3} {n_3} = 0$ That is: :$\map {\beta_4 \circ \phi_4} {m_4} = 0$ Since the diagram is commutative, this implies that $\map {\phi_5 \circ \alpha_4} {m_4} = 0$. Since $\phi_5$ is injective, this means that $\map {\alpha_4} {m_4} = 0$. That is: :$m_4 \in \ker \alpha_4$ Since the rows are exact, this means that $m_4 \in \Img {\alpha_3}$, say: :$m_4 = \map {\alpha_3} {m_3}$ Thus we have: :$\map {\beta_3} {n_3} = n_4 = \map {\phi_4 \circ \alpha_3} {m_3} = \map {\beta_3 \circ \phi_3} {m_3}$ Using the fact that $\beta_3$ is a homomorphism this means that: :$n_3 - \map {\phi_3} {m_3} \in \ker \beta_3$ By exactness of the rows we therefore have $n_3 - \map {\phi_3} {m_3} = \map {\beta_2} {n_2}$ for some $n_2 \in N_2$. By hypothesis $\phi_2$ is surjective, so there exists $m_2 \in M_2$ such that: :$n_3 - \map {\phi_3} {m_3} = \map {\beta_2 \circ \phi_2} {m_2}$ Since the diagram is commutative, we have therefore: :$n_3 - \map {\phi_3} {m_3} = \map {\phi_3 \circ \alpha_2} {m_2}$ Finally by the homomorphism property this shows that: :$n_3 = \map {\phi_3} {m_3 + \map {\alpha_2} {m_2} }$ as required. Now suppose that $\phi_2$ and $\phi_4$ are injective and $\phi_1$ is surjective. Let $m_3 \in M_3$ such that $\map {\phi_3} {m_3} = 0$. We want to show that $m_3 = 0$. First of all $\map {\phi_3} {m_3} = 0$ implies that $\map {\beta_3 \circ \phi_3} {m_3} = 0$. Since the diagram is commutative, this implies that $\map {\phi_4 \circ \alpha_3} {m_3} = 0$. Since $\phi_4$ is injective we have: :$\map {\alpha_3} {m_3} = 0$ Since the rows are exact, we therefore have $m_3 = \map {\alpha_2} {m_2}$ for some $m_2 \in M_2$. Let $n_2 = \map {\phi_2} {m_2} \in N_2$. By commutativity of the diagram, we have: :$\map {\beta_2} {n_2} = \map {\phi_3 \circ \alpha_2} {m_2} = \map {\phi_3} {m_3} = 0$ By exactness of the rows it follows that there exists $n_1 \in N_1$ such that $n_2 = \map {\beta_1} {n_1}$. Since $\phi_1$ is assumed surjective this means that $n_1 = \map {\phi_1} {m_1}$ for some $m_1 \in M_1$. Now: :$\map {\phi_2} {m_2} = \map {\beta_1 \circ \phi_1} {m_1} = \map {\phi_2 \circ \alpha_1} {m_1}$ Thus by the homomorphism property: :$m_2 - \map {\alpha_1} {m_1} \in \ker \phi_2$ Since $\phi_2$ is assumed injective this means that $m_2 = \map {\alpha_1} {m_1}$. Finally, since the rows are exact we have: :$m_3 = \map {\alpha_2} {m_2} = \map {\alpha_2 \circ \alpha_1} {m_1} = 0$ as required. {{qed}} Category:Homological Algebra Category:Named Theorems hnd32cu6olnyjt3rnytt792wbeksdzq
=== $(2)$ implies $(1)$ === Suppose that for some finite $F \subseteq I$, $\card F > \card {Y_F}$. Then $\card F \not \le \card {Y_F}$. By contrapositive of Injection implies Cardinal Inequality, there can be no injection from $F$ to $Y_F$. Thus there can be no injection from $I$ to $Y$ satisfying the requirements. {{qed|lemma}} === $(1)$ implies $(2)$ === Let $\map \Phi x$ be the set of all elements of $y$ in the expression of $x$. {{explain|Explain what that means: what $x$ is and what the "expression of $x$" is.}} Let $\FF$ be the set of all partial functions $g$ from $I$ to $Y$ such that: : $g$ is one-to-one. : If $\tuple {k, y} \in g$, then $y \in S_k$. Then for any $k \in I$, $\set {\map g x: g \in \FF \land k \in \map {\operatorname {Dom}} g}$ is finite. $\FF$ has finite character: Let $g \in \FF$. Let $F$ be a finite subset of $I$. Then $g \restriction F$ is obviously in $\FF$. Suppose instead that $g$ is a partial function from $I$ to $Y$. Let the restriction of $g$ to each finite subset of $\map {\operatorname{Dom}} g$ be in $\FF$. Then $g$ is one-to-one, as follows: Let $p, q \in \map {\operatorname{Dom} } g$ with $p \ne q$. Then $\set {p, q}$ is a finite subset of $\map {\operatorname{Dom}} g$. Thus: :$g \restriction \set {p, q} \in \FF$ Thus: :$\map g p \ne \map g q$ Let $\tuple {k, y} \in g$. Then: :$\tuple {k, y} \in g \restriction \set k$ So: :$y \in S_k$ Thus: :$g \in \FF$ So we see that $\FF$ has finite character. Let $F$ be a finite subset of $I$. By the finite case, there is an element of $\FF$ with domain $F$. By the Cowen-Engeler Lemma, $\FF$ has an element $\phi$ whose domain is $I$. That is, $\phi$ is a mapping from $I$ to $Y$. But a one-to-one mapping is an injection, so $\phi$ is an injection from $X$ to $Y$. By the definition of $\FF$: :$\forall k \in I: \map \phi k \in S_k$ {{qed}} {{BPI|Cowen-Engeler Lemma}} {{Namedfor|Philip Hall}} Category:Set Theory rxne73qad7gti7230do5ig64yugp9cr