Google 翻译生成的粗略草稿，质量较低。术语依照术语表，但句子可能有误；之后会由仔细的译文取代。
7. 逆极限和直极限
1. 逆极限
令I I I ( E α ) α ∈ I (E_\alpha)_{\alpha\in I} ( E α ) α ∈ I I I I I I I ( α , β ) (\alpha,\beta) ( α , β ) α ≤ β \alpha\leq\beta α ≤ β f α β f_{\alpha\beta} f α β E α E_\alpha E α E β E_\beta E β f α β f_{\alpha\beta} f α β
( L P I ) (\mathrm{LP}_I) ( LP I ) 关系 α ≤ β ≤ γ \alpha\leq\beta\leq\gamma α ≤ β ≤ γ f α γ = f α β ∘ f β γ f_{\alpha\gamma}=f_{\alpha\beta}\circ f_{\beta\gamma} f α γ = f α β ∘ f β γ
( L P I I ) (\mathrm{LP}_{II}) ( LP I I ) 对于每个 α ∈ I \alpha\in I α ∈ I f α α f_{\alpha\alpha} f α α E α E_\alpha E α
令G = ∏ α ∈ I E α G=\prod_{\alpha\in I}E_\alpha G = ∏ α ∈ I E α ( E α ) α ∈ I (E_\alpha)_{\alpha\in I} ( E α ) α ∈ I E E E x x x G G G
(1)
pr α x = f α β ( pr β x ) \operatorname{pr}_{\alpha}x=f_{\alpha\beta}(\operatorname{pr}_{\beta}x) pr α x = f α β ( pr β x ) 对于指标( α , β ) (\alpha,\beta) ( α , β ) α ≤ β \alpha\leq\beta α ≤ β E E E ( E α ) α ∈ I (E_\alpha)_{\alpha\in I} ( E α ) α ∈ I ( f α β ) (f_{\alpha\beta}) ( f α β ) E = lim ⟵ ( E α , f α β ) E=\underset{\longleftarrow}{\lim}\,(E_\alpha,f_{\alpha\beta}) E = ⟵ lim ( E α , f α β ) E = lim ⟵ E α E=\underset{\longleftarrow}{\lim}\,E_\alpha E = ⟵ lim E α ( ( E α ) , ( f α β ) ) ((E_\alpha),(f_{\alpha\beta})) (( E α ) , ( f α β )) ( E α β , f α β ) (E_{\alpha\beta},f_{\alpha\beta}) ( E α β , f α β ) I I I pr α \operatorname{pr}_\alpha pr α E E E f α f_\alpha f α E E E E α E_\alpha E α
(2)
f α = f α β ∘ f β f_\alpha=f_{\alpha\beta}\circ f_\beta f α = f α β ∘ f β 每当α ≤ β \alpha\leq\beta α ≤ β E E E
例
假设I I I ( α , β ) (\alpha,\beta) ( α , β ) α ≤ β \alpha\leq\beta α ≤ β ( α , α ) (\alpha,\alpha) ( α , α ) α ∈ I \alpha\in I α ∈ I f α α f_{\alpha\alpha} f α α 所有 x ∈ G x\in G x ∈ G lim ⟵ E α \underset{\longleftarrow}{\lim}\,E_\alpha ⟵ lim E α ∏ α ∈ I E α \prod_{\alpha\in I}E_\alpha ∏ α ∈ I E α
被以下引用 1 被以下引用 推论 2
假设I I I E α E_\alpha E α α ∈ I \alpha\in I α ∈ I F F F α ≤ β \alpha\leq\beta α ≤ β f α β f_{\alpha\beta} f α β F F F E = lim ⟵ E α E=\underset{\longleftarrow}{\lim}\,E_\alpha E = ⟵ lim E α ∏ α ∈ I E α = F I \prod_{\alpha\in I}E_\alpha=F^I ∏ α ∈ I E α = F I Δ \Delta Δ x ∈ Δ x\in\Delta x ∈ Δ x x x E E E ( α , β ) (\alpha, \beta) ( α , β ) p r α x = p r β x \mathrm{pr}_\alpha x = \mathrm{pr}_\beta x pr α x = pr β x γ ∈ I \gamma \in I γ ∈ I α ⩽ γ \alpha \leqslant \gamma α ⩽ γ β ⩽ γ \beta \leqslant \gamma β ⩽ γ p r α x = f α γ ( p r γ x ) = p r γ x \mathrm{pr}_\alpha x = f_{\alpha\gamma}(\mathrm{pr}_\gamma x) = \mathrm{pr}_\gamma x pr α x = f α γ ( pr γ x ) = pr γ x p r β x = p r γ x \mathrm{pr}_\beta x = \mathrm{pr}_\gamma x pr β x = pr γ x
应该注意的是，即使所有 E α E_\alpha E α f α β f_{\alpha\beta} f α β E = l i m ← E α E = \varprojlim E_\alpha E = lim E α
¶ 很明显，对于( E α ) α ∈ J (E_\alpha)_{\alpha \in J} ( E α ) α ∈ J ( f α β ) (f_{\alpha\beta}) ( f α β ) α ∈ J \alpha \in J α ∈ J β ∈ J \beta \in J β ∈ J α ⩽ β \alpha \leqslant \beta α ⩽ β ( E α ) α ∈ I (E_\alpha)_{\alpha \in I} ( E α ) α ∈ I ( E α ) α ∈ J (E_\alpha)_{\alpha \in J} ( E α ) α ∈ J x ∈ E x \in E x ∈ E
g ( x ) = ( f α ( x ) ) α ∈ J (3) g(x) = (f_\alpha(x))_{\alpha \in J} \tag{3} g ( x ) = ( f α ( x ) ) α ∈ J ( 3 ) 凭借(2)属于到E'； 映射 g : E → E ′ g : E \to E' g : E → E ′ ( E α ) α ∈ J ′ (E_\alpha)_{\alpha \in J'} ( E α ) α ∈ J ′ g ′ : E ′ → E ′ ′ g' : E' \to E'' g ′ : E ′ → E ′′ g ′ ′ : E → E ′ ′ g'' : E \to E'' g ′′ : E → E ′′
g ′ ′ = g ′ ∘ g . (4) g'' = g' \circ g. \tag{4} g ′′ = g ′ ∘ g . ( 4 ) 引用了 1 · 被以下引用 1 引用了 习题 4, § 7
被以下引用 注 2
2. 映射的逆系
令* I 为有序集合，令 ( E α , f α β ) (E_\alpha, f_{\alpha\beta}) ( E α , f α β ) 为集合相对于 I的逆系，*让* E = l i m ← E α E = \varprojlim E_\alpha E = lim E α 为其逆极限，并且对于每个 α ∈ I \alpha \in I α ∈ I
被以下引用 4 被以下引用 § 7. 逆极限和直极限 , 推论 1 , 命题 3 , 命题 5
f α : E → E α f_\alpha : E \to E_\alpha f α : E → E α 为典范映射。对于每个 α ∈ I \alpha \in I α ∈ I u α u_\alpha u α E α E_\alpha E α
f α β ∘ u β = u α whenever α ⩽ β . (5) f_{\alpha\beta} \circ u_\beta = u_\alpha \qquad \textit{whenever } \alpha \leqslant \beta. \tag{5} f α β ∘ u β = u α whenever α ⩽ β . ( 5 ) 然后:
(a) 存在 F 进入 E 的唯一映射 u u u 使得
u α = f α ∘ u for all α ∈ I ; (6) u_\alpha = f_\alpha \circ u \qquad \textit{for all } \alpha \in I;\tag{6} u α = f α ∘ u for all α ∈ I ; ( 6 ) (b) 映射 u u u 是单射当且仅当，对于 F 的不同元素* y y y z z z 的每个对， 存在* α ∈ I \alpha \in I α ∈ I 使得 u α ( y ) ≠ u α ( z ) u_\alpha(y) \neq u_\alpha(z) u α ( y ) = u α ( z )
对于关系 u α = f α ∘ u u_\alpha = f_\alpha \circ u u α = f α ∘ u y ∈ F y \in F y ∈ F
p r α ( u ( y ) ) = u α ( y ) ; \mathrm{pr}_\alpha(u(y)) = u_\alpha(y); pr α ( u ( y )) = u α ( y ) ; 元素 u ( y ) ∈ ∏ α ∈ I E α u(y) \in \prod_{\alpha \in I} E_\alpha u ( y ) ∈ ∏ α ∈ I E α u ( y ) = ( u α ( y ) ) α ∈ I u(y) = (u_\alpha(y))_{\alpha \in I} u ( y ) = ( u α ( y ) ) α ∈ I u ( y ) ∈ E u(y) \in E u ( y ) ∈ E y ∈ F y \in F y ∈ F
p r α ( u ( y ) ) = f α β ( p r β ( u ( y ) ) ) \mathrm{pr}_\alpha(u(y)) = f_{\alpha\beta}(\mathrm{pr}_\beta(u(y))) pr α ( u ( y )) = f α β ( pr β ( u ( y ))) 每当α ⩽ β \alpha \leqslant \beta α ⩽ β
u α ( y ) = f α β ( u β ( y ) ) u_\alpha(y) = f_{\alpha\beta}(u_\beta(y)) u α ( y ) = f α β ( u β ( y )) 因此由 (5) 得出。 命题的第二个部分紧随定义之后。
令*( E α , f α β ) (E_\alpha, f_{\alpha\beta}) ( E α , f α β ) ( F α , g α β ) (F_\alpha, g_{\alpha\beta}) ( F α , g α β ) 让 E = l i m ← E α E = \varprojlim E_\alpha E = lim E α F = l i m ← F α F = \varprojlim F_\alpha F = lim F α f α f_\alpha f α （分别 g α g_\alpha g α 到 E α E_\alpha E α （分别为 F 到 F α F_\alpha F α α ∈ I \alpha \in I α ∈ I α ∈ I \alpha \in I α ∈ I u α u_\alpha u α F α F_\alpha F α
引用了 1 引用了 命题 1
E β ⟶ u β F β f α β ↓ ↓ g α β E α ⟶ u α F α \begin{array}{ccc} E_\beta & \overset{u_\beta}{\longrightarrow} & F_\beta \\ {\scriptstyle f_{\alpha\beta}}\big\downarrow & & \big\downarrow{\scriptstyle g_{\alpha\beta}} \\ E_\alpha & \underset{u_\alpha}{\longrightarrow} & F_\alpha \end{array} E β f α β ↓ ⏐ E α ⟶ u β u α ⟶ F β ↓ ⏐ g α β F α 可交换 [^1] 每当 α ⩽ β \alpha \leqslant \beta α ⩽ β 然后存在唯一映射 u : E → F u : E \to F u : E → F 使得对于每个 α ∈ I \alpha \in I α ∈ I 图表
E ⟶ u F f α ↓ ↓ g α E α ⟶ u α F α \begin{array}{ccc} E & \overset{u}{\longrightarrow} & F \\ {\scriptstyle f_\alpha}\big\downarrow & & \big\downarrow{\scriptstyle g_\alpha} \\ E_\alpha & \underset{u_\alpha}{\longrightarrow} & F_\alpha \end{array} E f α ↓ ⏐ E α ⟶ u u α ⟶ F ↓ ⏐ g α F α 是可交换的。
输入v α = u α ∘ f α v_\alpha = u_\alpha \circ f_\alpha v α = u α ∘ f α α ⩽ β \alpha \leqslant \beta α ⩽ β
g α β ∘ v β = g α β ∘ u β ∘ f β = u α ∘ f α β ∘ f β = u α ∘ f α = v α , g_{\alpha\beta} \circ v_\beta = g_{\alpha\beta} \circ u_\beta \circ f_\beta = u_\alpha \circ f_{\alpha\beta} \circ f_\beta = u_\alpha \circ f_\alpha = v_\alpha, g α β ∘ v β = g α β ∘ u β ∘ f β = u α ∘ f α β ∘ f β = u α ∘ f α = v α , 因此我们可以应用命题 1 到映射 v α v_\alpha v α u : E → F u:E\to F u : E → F
g α ∘ u = v α = u α ∘ f α
g_\alpha\circ u=v_\alpha=u_\alpha\circ f_\alpha
g α ∘ u = v α = u α ∘ f α 对于每个α ∈ I \alpha\in I α ∈ I
¶ 满足推论 1 的条件的映射 u α : E α → F α u_\alpha:E_\alpha\to F_\alpha u α : E α → F α ( E α , f α β ) (E_\alpha,f_{\alpha\beta}) ( E α , f α β ) ( F α , g α β ) (F_\alpha,g_{\alpha\beta}) ( F α , g α β ) 逆系。推论1中定义的映射u u u ( u α ) (u_\alpha) ( u α ) 逆极限*，并且在不存在混淆风险的情况下写作u = lim ← u α u=\underset{\leftarrow}{\lim}\,u_\alpha u = ← lim u α
令( E α , f α β ) (E_\alpha,f_{\alpha\beta}) ( E α , f α β ) ( F α , g α β ) (F_\alpha,g_{\alpha\beta}) ( F α , g α β ) ( G α , h α β ) (G_\alpha,h_{\alpha\beta}) ( G α , h α β ) I I I E = lim ← E α E=\underset{\leftarrow}{\lim}\,E_\alpha E = ← lim E α F = lim ← F α F=\underset{\leftarrow}{\lim}\,F_\alpha F = ← lim F α G = lim ← G α G=\underset{\leftarrow}{\lim}\,G_\alpha G = ← lim G α f α f_\alpha f α g α , h α g_\alpha,h_\alpha g α , h α E E E F , G F,G F , G E α E_\alpha E α F α , G α F_\alpha,G_\alpha F α , G α ( u α ) (u_\alpha) ( u α ) ( v α ) (v_\alpha) ( v α ) u α : E α → F α u_\alpha:E_\alpha\to F_\alpha u α : E α → F α v α : F α → G α v_\alpha:F_\alpha\to G_\alpha v α : F α → G α v α ∘ u α : E α → G α v_\alpha\circ u_\alpha:E_\alpha\to G_\alpha v α ∘ u α : E α → G α
被以下引用 1 被以下引用 § 7. 逆极限和直极限
(7)
lim ← ( v α ∘ u α ) = ( lim ← v α ) ∘ ( lim ← u α ) .
\underset{\leftarrow}{\lim}\,(v_\alpha\circ u_\alpha)
=
\left(\underset{\leftarrow}{\lim}\,v_\alpha\right)\circ
\left(\underset{\leftarrow}{\lim}\,u_\alpha\right).
← lim ( v α ∘ u α ) = ( ← lim v α ) ∘ ( ← lim u α ) . 因为如果我们输入w α = v α ∘ u α w_\alpha=v_\alpha\circ u_\alpha w α = v α ∘ u α α ≤ β \alpha\leq\beta α ≤ β
w α ∘ f α β = v α ∘ ( u α ∘ f α β ) = v α ∘ ( g α β ∘ v β ) = ( h α β ∘ v β ) ∘ u β = h α β ∘ w β ,
w_\alpha\circ f_{\alpha\beta}
=v_\alpha\circ(u_\alpha\circ f_{\alpha\beta})
=v_\alpha\circ(g_{\alpha\beta}\circ v_\beta)
=(h_{\alpha\beta}\circ v_\beta)\circ u_\beta
=h_{\alpha\beta}\circ w_\beta,
w α ∘ f α β = v α ∘ ( u α ∘ f α β ) = v α ∘ ( g α β ∘ v β ) = ( h α β ∘ v β ) ∘ u β = h α β ∘ w β , 这表明( w α ) (w_\alpha) ( w α ) u = lim ← u α u=\underset{\leftarrow}{\lim}\,u_\alpha u = ← lim u α v = lim ← v α v=\underset{\leftarrow}{\lim}\,v_\alpha v = ← lim v α α ∈ I \alpha\in I α ∈ I h α ∘ ( v ∘ u ) = ( v α ∘ g α ) ∘ u = ( v α ∘ u α ) ∘ f α h_\alpha\circ(v\circ u)=(v_\alpha\circ g_\alpha)\circ u=(v_\alpha\circ u_\alpha)\circ f_\alpha h α ∘ ( v ∘ u ) = ( v α ∘ g α ) ∘ u = ( v α ∘ u α ) ∘ f α v ∘ u = lim ← w α v\circ u=\underset{\leftarrow}{\lim}\,w_\alpha v ∘ u = ← lim w α
¶ 令 ( E α , f α β ) (E_\alpha,f_{\alpha\beta}) ( E α , f α β ) α ∈ I \alpha\in I α ∈ I M α M_\alpha M α E α E_\alpha E α f α β ( M β ) ⊂ M α f_{\alpha\beta}(M_\beta)\subset M_\alpha f α β ( M β ) ⊂ M α α ≤ β \alpha\leq\beta α ≤ β M α M_\alpha M α E α E_\alpha E α g α β g_{\alpha\beta} g α β M β M_\beta M β M α M_\alpha M α α ≤ β \alpha\leq\beta α ≤ β f α β f_{\alpha\beta} f α β M β M_\beta M β ( M α , g α β ) (M_\alpha,g_{\alpha\beta}) ( M α , g α β )
(8)
lim ← M α = ( lim ← E α ) ∩ ∏ α ∈ I M α .
\underset{\leftarrow}{\lim}\,M_\alpha
=
\left(\underset{\leftarrow}{\lim}\,E_\alpha\right)\cap\prod_{\alpha\in I}M_\alpha.
← lim M α = ( ← lim E α ) ∩ α ∈ I ∏ M α .
令( E α , f α β ) (E_\alpha,f_{\alpha\beta}) ( E α , f α β ) ( E α ′ , f α β ′ ) (E'_\alpha,f'_{\alpha\beta}) ( E α ′ , f α β ′ ) I I I u α u_\alpha u α α ∈ I \alpha\in I α ∈ I E α E_\alpha E α E α ′ E'_\alpha E α ′ u α u_\alpha u α u = lim ← u α u=\underset{\leftarrow}{\lim}\,u_\alpha u = ← lim u α x ′ = ( x α ′ ) ∈ E ′ = lim ← E α ′ x'=(x'_\alpha)\in E'=\underset{\leftarrow}{\lim}\,E'_\alpha x ′ = ( x α ′ ) ∈ E ′ = ← lim E α ′ u ˉ α − 1 ( x α ′ ) \bar u_\alpha^{-1}(x'_\alpha) u ˉ α − 1 ( x α ′ ) E α E_\alpha E α u ˉ − 1 ( x ′ ) = lim ← u ˉ α − 1 ( x α ′ ) \bar u^{-1}(x')=\underset{\leftarrow}{\lim}\,\bar u_\alpha^{-1}(x'_\alpha) u ˉ − 1 ( x ′ ) = ← lim u ˉ α − 1 ( x α ′ )
被以下引用 2 被以下引用 命题 1, A II § 6 , 推论
对于如果 α ⩽ β \alpha \leqslant \beta α ⩽ β x β ∈ u β − 1 ( x β ′ ) x_\beta \in \overset{-1}{u_\beta}(x'_\beta) x β ∈ u β − 1 ( x β ′ )
u α ( f α β ( x β ) ) = f α β ′ ( u β ( x β ) ) = f α β ′ ( x β ′ ) = x α ′ , u_\alpha(f_{\alpha\beta}(x_\beta)) = f'_{\alpha\beta}(u_\beta(x_\beta)) = f'_{\alpha\beta}(x'_\beta) = x'_\alpha, u α ( f α β ( x β )) = f α β ′ ( u β ( x β )) = f α β ′ ( x β ′ ) = x α ′ , 第一个断言随之而来；并说 x = ( x α ) ∈ E = l i m ← E α x = (x_\alpha) \in \mathrm{E} = \varprojlim \mathrm{E}_\alpha x = ( x α ) ∈ E = lim E α u ( x ) = x ′ u(x) = x' u ( x ) = x ′ α ∈ I \alpha \in \mathrm{I} α ∈ I u α ( x α ) = x α ′ u_\alpha(x_\alpha) = x'_\alpha u α ( x α ) = x α ′
如果u α u_\alpha u α α ∈ I \alpha \in \mathrm{I} α ∈ I u u u
引用了 2 引用了 命题 2 , 习题 4, § 7
有了命题2的记号，像u α ( E α ) u_\alpha(\mathrm{E}_\alpha) u α ( E α ) E α ′ \mathrm{E}'_\alpha E α ′
(9)
u ( E ) ⊂ l i m ← u α ( E α ) ; u(\mathrm{E}) \subset \varprojlim u_\alpha(\mathrm{E}_\alpha); u ( E ) ⊂ lim u α ( E α ) ; 但这个关系的两侧*不一定相等*（习题4）。
令I \mathrm{I} I ( E α , f α β ) (\mathrm{E}_\alpha, f_{\alpha\beta}) ( E α , f α β ) I \mathrm{I} I E = l i m ← E α \mathrm{E} = \varprojlim \mathrm{E}_\alpha E = lim E α J \mathrm{J} J I \mathrm{I} I J \mathrm{J} J E ′ \mathrm{E}' E ′ ( E α , f α β ) (\mathrm{E}_\alpha, f_{\alpha\beta}) ( E α , f α β ) J \mathrm{J} J E \mathrm{E} E g g g E ′ \mathrm{E}' E ′ 公式 (3)) 是双射 。
对于每个α ∈ J \alpha \in \mathrm{J} α ∈ J f α ′ f'_\alpha f α ′ E ′ → E α \mathrm{E}' \to \mathrm{E}_\alpha E ′ → E α g g g E \mathrm{E} E E ′ \mathrm{E}' E ′ f α = f α ′ ∘ g f_\alpha = f'_\alpha \circ g f α = f α ′ ∘ g α ∈ J \alpha \in \mathrm{J} α ∈ J g g g x x x y y y E \mathrm{E} E α ∈ I \alpha \in \mathrm{I} α ∈ I f α ( x ) ≠ f α ( y ) f_\alpha(x) \neq f_\alpha(y) f α ( x ) = f α ( y ) J \mathrm{J} J I \mathrm{I} I λ ∈ J \lambda \in \mathrm{J} λ ∈ J α ⩽ λ \alpha \leqslant \lambda α ⩽ λ f α λ ( f λ ( x ) ) ≠ f α λ ( f λ ( y ) ) f_{\alpha\lambda}(f_\lambda(x)) \neq f_{\alpha\lambda}(f_\lambda(y)) f α λ ( f λ ( x )) = f α λ ( f λ ( y )) f λ ( x ) ≠ f λ ( y ) f_\lambda(x) \neq f_\lambda(y) f λ ( x ) = f λ ( y ) g g g x ′ = ( x λ ′ ) λ ∈ J x' = (x'_\lambda)_{\lambda \in \mathrm{J}} x ′ = ( x λ ′ ) λ ∈ J E ′ \mathrm{E}' E ′ α ∈ I \alpha \in \mathrm{I} α ∈ I λ ∈ J \lambda \in \mathrm{J} λ ∈ J α ⩽ λ \alpha \leqslant \lambda α ⩽ λ f α λ ( x λ ′ ) f_{\alpha\lambda}(x'_\lambda) f α λ ( x λ ′ ) λ ∈ J \lambda \in \mathrm{J} λ ∈ J α ⩽ λ \alpha \leqslant \lambda α ⩽ λ μ ∈ J \mu \in \mathrm{J} μ ∈ J α ⩽ μ \alpha \leqslant \mu α ⩽ μ ν ∈ J \nu \in \mathrm{J} ν ∈ J λ ⩽ ν \lambda \leqslant \nu λ ⩽ ν μ ⩽ ν \mu \leqslant \nu μ ⩽ ν f α λ ( x λ ′ ) = f α λ ( f λ ν ( x ν ′ ) ) = f α ν ( x ν ′ ) f_{\alpha\lambda}(x'_\lambda) = f_{\alpha\lambda}(f_{\lambda\nu}(x'_\nu)) = f_{\alpha\nu}(x'_\nu) f α λ ( x λ ′ ) = f α λ ( f λ ν ( x ν ′ )) = f α ν ( x ν ′ ) f α μ ( x μ ′ ) = f α ν ( x ν ′ ) f_{\alpha\mu}(x'_\mu) = f_{\alpha\nu}(x'_\nu) f α μ ( x μ ′ ) = f α ν ( x ν ′ ) x α x_\alpha x α λ ∈ J \lambda \in \mathrm{J} λ ∈ J f α λ ( x λ ′ ) f_{\alpha\lambda}(x'_\lambda) f α λ ( x λ ′ ) α ⩽ λ \alpha \leqslant \lambda α ⩽ λ x = ( x α ) α ∈ I x = (x_\alpha)_{\alpha \in \mathrm{I}} x = ( x α ) α ∈ I x ∈ E x \in \mathrm{E} x ∈ E α ⩽ β \alpha \leqslant \beta α ⩽ β λ ∈ J \lambda \in \mathrm{J} λ ∈ J β ⩽ λ \beta \leqslant \lambda β ⩽ λ
f α β ( x β ) = f α β ( f β λ ( x λ ′ ) ) = f α λ ( x λ ′ ) = x α . f_{\alpha\beta}(x_\beta) = f_{\alpha\beta}(f_{\beta\lambda}(x'_\lambda)) = f_{\alpha\lambda}(x'_\lambda) = x_\alpha. f α β ( x β ) = f α β ( f β λ ( x λ ′ )) = f α λ ( x λ ′ ) = x α . 最后，我们对所有λ ∈ J \lambda \in \mathrm{J} λ ∈ J x λ ′ = f λ λ ( x λ ′ ) x'_\lambda = f_{\lambda\lambda}(x'_\lambda) x λ ′ = f λλ ( x λ ′ ) λ ∈ J \lambda \in \mathrm{J} λ ∈ J x λ = x λ ′ x_\lambda = x'_\lambda x λ = x λ ′ f ( x λ ) = x λ ′ f(x_\lambda) = x'_\lambda f ( x λ ) = x λ ′ g ( x ) = x ′ g(x) = x' g ( x ) = x ′ g g g
特别是，如果I \mathrm{I} I ω \omega ω J = { ω } \mathrm{J} = \{\omega\} J = { ω } l i m ← E α \varprojlim \mathrm{E}_\alpha lim E α E ω \mathrm{E}_\omega E ω
注
引用了 1 引用了 命题 1
3. 双重逆极限
令I、L为两个预先订购的集合，以及I × L I \times L I × L ( E α λ , f α β λ μ ) (E^\lambda_\alpha, f^{\lambda\mu}_{\alpha\beta}) ( E α λ , f α β λ μ ) I × L I \times L I × L
我们有
(11) 每当α ⩽ β ⩽ γ \alpha \leqslant \beta \leqslant \gamma α ⩽ β ⩽ γ λ ⩽ μ ⩽ ν \lambda \leqslant \mu \leqslant \nu λ ⩽ μ ⩽ ν f α γ λ ν = f α β λ μ ∘ f β γ μ ν \qquad f^{\lambda\nu}_{\alpha\gamma} = f^{\lambda\mu}_{\alpha\beta} \circ f^{\mu\nu}_{\beta\gamma} f α γ λ ν = f α β λ μ ∘ f β γ μν
让E或lim E α λ \lim \mathrm{E}^{\lambda}_{\alpha} lim E α λ
对于每个λ ∈ L \lambda \in \mathrm{L} λ ∈ L g α β λ = f α β λ λ : E β λ → E α λ g^{\lambda}_{\alpha\beta} = f^{\lambda\lambda}_{\alpha\beta} : \mathrm{E}^{\lambda}_{\beta} \to \mathrm{E}^{\lambda}_{\alpha} g α β λ = f α β λλ : E β λ → E α λ
(12) g α γ λ = g α β λ ∘ g β γ λ \qquad g^{\lambda}_{\alpha\gamma} = g^{\lambda}_{\alpha\beta} \circ g^{\lambda}_{\beta\gamma} g α γ λ = g α β λ ∘ g β γ λ α ⩽ β ⩽ γ \alpha \leqslant \beta \leqslant \gamma α ⩽ β ⩽ γ
这样( E α λ , g α β λ ) (\mathrm{E}^{\lambda}_{\alpha}, g^{\lambda}_{\alpha\beta}) ( E α λ , g α β λ ) F λ = lim E α λ \mathrm{F}^{\lambda} = \lim \mathrm{E}^{\lambda}_{\alpha} F λ = lim E α λ λ ⩽ μ \lambda \leqslant \mu λ ⩽ μ h α λ μ = f α α λ μ : E α μ → E α λ h^{\lambda\mu}_{\alpha} = f^{\lambda\mu}_{\alpha\alpha} : \mathrm{E}^{\mu}_{\alpha} \to \mathrm{E}^{\lambda}_{\alpha} h α λ μ = f α α λ μ : E α μ → E α λ h λ μ = lim h α λ μ : F μ → F λ h^{\lambda\mu} = \lim h^{\lambda\mu}_{\alpha} : \mathrm{F}^{\mu} \to \mathrm{F}^{\lambda} h λ μ = lim h α λ μ : F μ → F λ λ ⩽ μ ⩽ ν \lambda \leqslant \mu \leqslant \nu λ ⩽ μ ⩽ ν
(13)
h λ ν = h λ μ ∘ h μ ν h^{\lambda\nu} = h^{\lambda\mu} \circ h^{\mu\nu} h λ ν = h λ μ ∘ h μν （no. 2，命题2，推论2）；因此，相对于L，( F λ , h λ μ ) (\mathrm{F}^{\lambda}, h^{\lambda\mu}) ( F λ , h λ μ ) F = lim F λ \mathrm{F} = \lim \mathrm{F}^{\lambda} F = lim F λ F → E \mathrm{F} \to \mathrm{E} F → E ∏ λ ∈ L F λ \prod_{\lambda \in \mathrm{L}} \mathrm{F}^{\lambda} ∏ λ ∈ L F λ F λ \mathrm{F}^{\lambda} F λ ∏ α ∈ I E α λ \prod_{\alpha \in \mathrm{I}} \mathrm{E}^{\lambda}_{\alpha} ∏ α ∈ I E α λ ∏ ( α , λ ) ∈ I × L E α λ = G \prod_{(\alpha, \lambda) \in \mathrm{I} \times \mathrm{L}} \mathrm{E}^{\lambda}_{\alpha} = \mathrm{G} ∏ ( α , λ ) ∈ I × L E α λ = G z ∈ G z \in \mathrm{G} z ∈ G p r λ ( z ) \mathrm{pr}^{\lambda}(z) pr λ ( z ) ∏ α ∈ I E α λ \prod_{\alpha \in \mathrm{I}} \mathrm{E}^{\lambda}_{\alpha} ∏ α ∈ I E α λ ( p r α λ ( z ) ) α ∈ I (\mathrm{pr}^{\lambda}_{\alpha}(z))_{\alpha \in \mathrm{I}} ( pr α λ ( z ) ) α ∈ I z ∈ F z \in \mathrm{F} z ∈ F
(14) p r λ ( z ) = h λ μ ( p r μ ( z ) ) \qquad \mathrm{pr}^{\lambda}(z) = h^{\lambda\mu}(\mathrm{pr}^{\mu}(z)) pr λ ( z ) = h λ μ ( pr μ ( z )) λ ⩽ μ \lambda \leqslant \mu λ ⩽ μ
和 p r λ ( z ) ∈ F λ \mathrm{pr}^{\lambda}(z) \in \mathrm{F}^{\lambda} pr λ ( z ) ∈ F λ λ ∈ L \lambda \in \mathrm{L} λ ∈ L α ⩽ β \alpha \leqslant \beta α ⩽ β
(15)
p r α λ ( z ) = f α β λ λ ( p r β λ ( z ) ) . \mathrm{pr}^{\lambda}_{\alpha}(z) = f^{\lambda\lambda}_{\alpha\beta}(\mathrm{pr}^{\lambda}_{\beta}(z)). pr α λ ( z ) = f α β λλ ( pr β λ ( z )) . 但是h λ μ ( p r μ ( z ) ) = ( f α α μ λ ( p r α μ ( z ) ) α ∈ I h^{\lambda\mu}(\mathrm{pr}^{\mu}(z)) = (f^{\mu\lambda}_{\alpha\alpha}(\mathrm{pr}^{\mu}_{\alpha}(z))_{\alpha \in \mathrm{I}} h λ μ ( pr μ ( z )) = ( f α α μ λ ( pr α μ ( z ) ) α ∈ I α ⩽ β \alpha \leqslant \beta α ⩽ β λ ⩽ μ \lambda \leqslant \mu λ ⩽ μ
p r α λ ( z ) = f α α λ μ ( f α β μ μ ( p r β μ ( z ) ) ) = f α β λ μ ( p r β μ ( z ) ) , \mathrm{pr}^{\lambda}_{\alpha}(z) = f^{\lambda\mu}_{\alpha\alpha}(f^{\mu\mu}_{\alpha\beta}(\mathrm{pr}^{\mu}_{\beta}(z))) = f^{\lambda\mu}_{\alpha\beta}(\mathrm{pr}^{\mu}_{\beta}(z)), pr α λ ( z ) = f α α λ μ ( f α β μμ ( pr β μ ( z ))) = f α β λ μ ( pr β μ ( z )) , 这意味着z ∈ E z \in \mathrm{E} z ∈ E
如果* ( E α λ , f α β λ μ ) (\mathrm{E}^\lambda_\alpha, f^{\lambda\mu}_{\alpha\beta}) ( E α λ , f α β λ μ ) 是集合的逆系相对于乘积 I × L \mathrm{I} \times \mathrm{L} I × L
被以下引用 2 被以下引用 推论 1 , 推论 2
( 16 ) l i m ← α , λ E α λ = l i m ← λ ( l i m ← α E α λ ) . (16) \qquad \varprojlim_{\alpha,\lambda} \mathrm{E}^\lambda_\alpha = \varprojlim_\lambda \; (\varprojlim_\alpha \mathrm{E}^\lambda_\alpha). ( 16 ) α , λ lim E α λ = λ lim ( α lim E α λ ) .
设*( E ′ α λ , f ′ α β μ λ ) (\mathrm{{E}'}^\lambda_\alpha, {f'}^{\mu\lambda}_{\alpha\beta}) ( E ′ α λ , f ′ α β μ λ ) I × L \mathrm{I} \times \mathrm{L} I × L ( α , λ ) ∈ I × L (\alpha, \lambda) \in \mathrm{I} \times \mathrm{L} ( α , λ ) ∈ I × L u α λ u^\lambda_\alpha u α λ E α λ \mathrm{E}^\lambda_\alpha E α λ E ′ α λ \mathrm{{E}'}^\lambda_\alpha E ′ α λ u α λ u^\lambda_\alpha u α λ
引用了 1 引用了 命题 4
( 17 ) l i m ← α , λ u α λ = l i m ← λ ( l i m ← α u α λ ) . (17) \qquad \varprojlim_{\alpha,\lambda} u^\lambda_\alpha = \varprojlim_\lambda \; (\varprojlim_\alpha u^\lambda_\alpha). ( 17 ) α , λ lim u α λ = λ lim ( α lim u α λ ) . 验证与命题4类似。
设*( E α λ , f α β λ ) λ ∈ L (\mathrm{E}^\lambda_\alpha, f^\lambda_{\alpha\beta})_{\lambda \in \mathrm{L}} ( E α λ , f α β λ ) λ ∈ L 如果 ∏ λ ∈ L f α β λ \prod_{\lambda \in \mathrm{L}} f^\lambda_{\alpha\beta} ∏ λ ∈ L f α β λ 表示扩张至乘积 (章 II, § 5, no. 7, 定义 2) 属于映射 ( f α β λ ) λ ∈ L (f^\lambda_{\alpha\beta})_{\lambda \in \mathrm{L}} ( f α β λ ) λ ∈ L ( ∏ λ ∈ L E α λ , ∏ λ ∈ L f α β λ ) \left( \prod_{\lambda \in \mathrm{L}} \mathrm{E}^\lambda_\alpha, \prod_{\lambda \in \mathrm{L}} f^\lambda_{\alpha\beta} \right) ( ∏ λ ∈ L E α λ , ∏ λ ∈ L f α β λ ) 是集合相对于 I的逆系，*并且（直到规范的双射）我们有
引用了 3 · 被以下引用 1 引用了 定义 2, II § 5 , 例 1 , 命题 4
被以下引用 引理 3, LIE IX Appendix 1
( 18 ) l i m ← α ∏ λ ∈ L E α λ = ∏ λ ∈ L ( l i m ← α E α λ ) . (18) \qquad \varprojlim_\alpha \prod_{\lambda \in \mathrm{L}} \mathrm{E}^\lambda_\alpha = \prod_{\lambda \in \mathrm{L}} (\varprojlim_\alpha \mathrm{E}^\lambda_\alpha). ( 18 ) α lim λ ∈ L ∏ E α λ = λ ∈ L ∏ ( α lim E α λ ) . 考虑双逆系( E α λ , g α β λ μ ) (\mathrm{E}^\lambda_\alpha, g^{\lambda\mu}_{\alpha\beta}) ( E α λ , g α β λ μ ) I × L \mathrm{I} \times \mathrm{L} I × L
4. 条件逆极限非空
在这个小节中，我们将给两个最常用的足够的条件，使逆极限非空（另见习题5）。
设*( E α , f α β ) (\mathrm{E}_\alpha, f_{\alpha\beta}) ( E α , f α β ) f α β f_{\alpha\beta} f α β 然后，如果 E = l i m ← E α \mathrm{E} = \varprojlim \mathrm{E}_\alpha E = lim E α f α : E → E α f_\alpha : \mathrm{E} \to \mathrm{E}_\alpha f α : E → E α 对于每个 α ∈ I \alpha \in \mathrm{I} α ∈ I 都是满射（并且，更不必说， E 不为空，前提是 E α \mathrm{E}_\alpha E α
引用了 1 引用了 命题 1
令( α n ) (\alpha_n) ( α n ) i < n i < n i < n β n ⩾ α n \beta_n \geqslant \alpha_n β n ⩾ α n β 0 = α 0 , β n ⩾ β i \beta_0 = \alpha_0, \beta_n \geqslant \beta_i β 0 = α 0 , β n ⩾ β i ( β n ) (\beta_n) ( β n ) ( β n ) (\beta_n) ( β n ) α ⩽ β n \alpha \leqslant \beta_n α ⩽ β n f α = f α β n ∘ f β n f_\alpha = f_{\alpha\beta_n} \circ f_{\beta_n} f α = f α β n ∘ f β n I = N \mathrm{I} = \mathbf{N} I = N f 0 f_0 f 0 x 0 ∈ E 0 x_0 \in \mathrm{E}_0 x 0 ∈ E 0 x n ∈ E n x_n \in \mathrm{E}_n x n ∈ E n n ⩾ 1 n \geqslant 1 n ⩾ 1 f − 1 n − 1 , n ( x n − 1 ) \overset{-1}{f}_{n-1,n}(x_{n-1}) f − 1 n − 1 , n ( x n − 1 ) n − m n - m n − m x m = f m n ( x n ) x_m = f_{mn}(x_n) x m = f mn ( x n ) m ⩽ n m \leqslant n m ⩽ n x = ( x n ) x = (x_n) x = ( x n )
第二个标准涉及逆系( E α , f α β ) (\mathrm{E}_\alpha, f_{\alpha\beta}) ( E α , f α β ) α ∈ I \alpha \in \mathrm{I} α ∈ I S α \mathfrak{S}_\alpha S α E α \mathrm{E}_\alpha E α
(i) 集合属于至S α \mathfrak{S}_\alpha S α S α \mathfrak{S}_\alpha S α
特别是（通过考虑空族的交）E α ∈ S α \mathrm{E}_\alpha \in \mathfrak{S}_\alpha E α ∈ S α
(ii) 如果子集F ⊂ S α \mathfrak{F} \subset \mathfrak{S}_\alpha F ⊂ S α F \mathfrak{F} F ⋂ M ∈ F M \bigcap\limits_{\mathbf{M} \in \mathfrak{F}} \mathrm{M} M ∈ F ⋂ M
鉴于(i)，显然(ii)等价于以下条件：
(ii)′ 如果 G ⊂ S α \mathfrak{G} \subset \mathfrak{S}_\alpha G ⊂ S α ⋂ M ∈ G M \bigcap\limits_{\mathbf{M} \in \mathfrak{G}} \mathrm{M} M ∈ G ⋂ M
(iii) 对于指标α \alpha α β \beta β α ⩽ β \alpha \leqslant \beta α ⩽ β x α ∈ E α x_\alpha \in \mathrm{E}_\alpha x α ∈ E α f − 1 α β ( x α ) ∈ S β \overset{-1}{f}_{\alpha\beta}(x_\alpha) \in \mathfrak{S}_\beta f − 1 α β ( x α ) ∈ S β
(iv) 对于指标α \alpha α β \beta β α ⩽ β \alpha \leqslant \beta α ⩽ β M β ∈ S β \mathrm{M}_\beta \in \mathfrak{S}_\beta M β ∈ S β f α β ( M β ) ∈ S α f_{\alpha\beta}(\mathrm{M}_\beta) \in \mathfrak{S}_\alpha f α β ( M β ) ∈ S α
对于每个α ∈ I \alpha \in \mathrm{I} α ∈ I E = l i m ← E α \mathrm{E} = \varprojlim \mathrm{E}_\alpha E = lim E α f α : E → E α f_\alpha : \mathrm{E} \to \mathrm{E}_\alpha f α : E → E α
(a) 对于每个 α ∈ I \alpha \in \mathrm{I} α ∈ I
( 19 ) f α ( E ) = ⋂ β ⩾ α f α β ( E β ) . (19) \qquad f_\alpha(\mathrm{E}) = \bigcap_{\beta \geqslant \alpha} f_{\alpha\beta}(\mathrm{E}_\beta). ( 19 ) f α ( E ) = β ⩾ α ⋂ f α β ( E β ) . (b) 如果对于每个 α ∈ I \alpha \in \mathrm{I} α ∈ I E α \mathrm{E}_\alpha E α
令Σ \Sigma Σ A = ( A α ) α ∈ I \mathfrak{A} = (\mathrm{A}_\alpha)_{\alpha \in \mathrm{I}} A = ( A α ) α ∈ I
( 20 ) A α ≠ ∅ and A α ∈ S α for all α ∈ I ; (20) \qquad \mathrm{A}_\alpha \neq \emptyset \quad \textit{and} \quad \mathrm{A}_\alpha \in \mathfrak{S}_\alpha \quad \textit{for all} \quad \alpha \in \mathrm{I}; ( 20 ) A α = ∅ and A α ∈ S α for all α ∈ I ; ( 21 ) f α β ( A β ) ⊂ A α whenever α ⩽ β . (21) \qquad f_{\alpha\beta}(\mathrm{A}_\beta) \subset \mathrm{A}_\alpha \quad \textit{whenever} \quad \alpha \leqslant \beta. ( 21 ) f α β ( A β ) ⊂ A α whenever α ⩽ β . 如果A = ( A α ) \mathfrak{A} = (A_\alpha) A = ( A α ) A ′ = ( A α ′ ) \mathfrak{A}' = (A'_\alpha) A ′ = ( A α ′ ) Σ \Sigma Σ A ⩽ A ′ \mathfrak{A} \leqslant \mathfrak{A}' A ⩽ A ′ α \alpha α A α ⊃ A α ′ A_\alpha \supset A'_\alpha A α ⊃ A α ′ Σ \Sigma Σ
(1) 我们首先证明有序的集合 Σ \Sigma Σ λ → A λ = ( A α λ ) α ∈ I \lambda \to \mathfrak{A}^\lambda = (A^\lambda_\alpha)_{\alpha \in I} λ → A λ = ( A α λ ) α ∈ I Σ \Sigma Σ α ∈ I \alpha \in I α ∈ I B α = ⋂ λ ∈ L A α λ B_\alpha = \bigcap_{\lambda \in L} A^\lambda_\alpha B α = ⋂ λ ∈ L A α λ B = ( B α ) α ∈ I \mathfrak{B} = (B_\alpha)_{\alpha \in I} B = ( B α ) α ∈ I Σ \Sigma Σ B \mathfrak{B} B A λ \mathfrak{A}^\lambda A λ
(2) 设 A = ( A α ) \mathfrak{A} = (A_\alpha) A = ( A α ) Σ \Sigma Σ α ⩽ β \alpha \leqslant \beta α ⩽ β A α = f α β ( A β ) A_\alpha = f_{\alpha\beta}(A_\beta) A α = f α β ( A β ) A α ′ = ⋂ β ⩾ α f α β ( A β ) A'_\alpha = \bigcap_{\beta \geqslant \alpha} f_{\alpha\beta}(A_\beta) A α ′ = ⋂ β ⩾ α f α β ( A β ) α ∈ I \alpha \in I α ∈ I A ′ = ( A α ′ ) \mathfrak{A}' = (A'_\alpha) A ′ = ( A α ′ ) Σ \Sigma Σ α ⩽ β ⩽ γ \alpha \leqslant \beta \leqslant \gamma α ⩽ β ⩽ γ f α γ ( A γ ) = f α β ( f β γ ( A γ ) ) ⊂ f α β ( A β ) f_{\alpha\gamma}(A_\gamma) = f_{\alpha\beta}(f_{\beta\gamma}(A_\gamma)) \subset f_{\alpha\beta}(A_\beta) f α γ ( A γ ) = f α β ( f β γ ( A γ )) ⊂ f α β ( A β ) f α β ( A β ) ∈ S α f_{\alpha\beta}(A_\beta) \in \mathfrak{S}_\alpha f α β ( A β ) ∈ S α f α β ( A β ) ≠ ∅ f_{\alpha\beta}(A_\beta) \neq \emptyset f α β ( A β ) = ∅ A ′ \mathfrak{A}' A ′ α ⩽ β \alpha \leqslant \beta α ⩽ β
f α β ( A β ′ ) ⊂ ⋂ γ ⩾ β f α β ( f β γ ( A γ ) ) = ⋂ γ ⩾ β f α γ ( A γ ) ; f_{\alpha\beta}(A'_\beta) \subset \bigcap_{\gamma \geqslant \beta} f_{\alpha\beta}(f_{\beta\gamma}(A_\gamma)) = \bigcap_{\gamma \geqslant \beta} f_{\alpha\gamma}(A_\gamma); f α β ( A β ′ ) ⊂ γ ⩾ β ⋂ f α β ( f β γ ( A γ )) = γ ⩾ β ⋂ f α γ ( A γ ) ; 对于每个δ ⩾ α \delta \geqslant \alpha δ ⩾ α γ ∈ I \gamma \in I γ ∈ I γ ⩾ δ \gamma \geqslant \delta γ ⩾ δ γ ⩾ β \gamma \geqslant \beta γ ⩾ β f α γ ( A γ ) ⊂ f α δ ( A δ ) f_{\alpha\gamma}(A_\gamma) \subset f_{\alpha\delta}(A_\delta) f α γ ( A γ ) ⊂ f α δ ( A δ )
⋂ γ ⩾ β f α γ ( A γ ) = ⋂ δ ⩾ α f α δ ( A δ ) = A α ′ . \bigcap_{\gamma \geqslant \beta} f_{\alpha\gamma}(A_\gamma) = \bigcap_{\delta \geqslant \alpha} f_{\alpha\delta}(A_\delta) = A'_\alpha. γ ⩾ β ⋂ f α γ ( A γ ) = δ ⩾ α ⋂ f α δ ( A δ ) = A α ′ . 因而 A ′ \mathfrak{A}' A ′ Σ \Sigma Σ A α ′ ⊂ A α A'_\alpha \subset A_\alpha A α ′ ⊂ A α α \alpha α Σ \Sigma Σ A \mathfrak{A} A A ′ = A \mathfrak{A}' = \mathfrak{A} A ′ = A
(3) 接下来我们将确定，如果A = ( A α ) \mathfrak{A} = (A_\alpha) A = ( A α ) Σ \Sigma Σ A α A_\alpha A α x α ∈ A α x_\alpha \in A_\alpha x α ∈ A α β ⩾ α \beta \geqslant \alpha β ⩾ α B β = A β ∩ f − 1 α β ( x α ) B_\beta = A_\beta \cap \overset{-1}{f}_{\alpha\beta}(x_\alpha) B β = A β ∩ f − 1 α β ( x α ) β \beta β ⩾ α \geqslant \alpha ⩾ α B β = A β B_\beta = A_\beta B β = A β B = ( B β ) \mathfrak{B} = (B_\beta) B = ( B β ) Σ \Sigma Σ β \beta β ⩾ α \geqslant \alpha ⩾ α β ⩽ γ \beta \leqslant \gamma β ⩽ γ f β γ ( B γ ) ⊂ f β γ ( A γ ) ⊂ A β = B β f_{\beta\gamma}(B_\gamma) \subset f_{\beta\gamma}(A_\gamma) \subset A_\beta = B_\beta f β γ ( B γ ) ⊂ f β γ ( A γ ) ⊂ A β = B β α ⩽ β ⩽ γ \alpha \leqslant \beta \leqslant \gamma α ⩽ β ⩽ γ
f − 1 α γ ( x α ) = f − 1 β γ ( f − 1 α β ( x α ) ) , \overset{-1}{f}_{\alpha\gamma}(x_\alpha) = \overset{-1}{f}_{\beta\gamma}(\overset{-1}{f}_{\alpha\beta}(x_\alpha)), f − 1 α γ ( x α ) = f − 1 β γ ( f − 1 α β ( x α )) , 我们有
f β γ ( f − 1 α γ ( x α ) ) ⊂ f − 1 α β ( x α ) , f_{\beta\gamma}(\overset{-1}{f}_{\alpha\gamma}(x_\alpha)) \subset \overset{-1}{f}_{\alpha\beta}(x_\alpha), f β γ ( f − 1 α γ ( x α )) ⊂ f − 1 α β ( x α ) , 从f β γ ( A γ ) ⊂ A β f_{\beta\gamma}(A_\gamma) \subset A_\beta f β γ ( A γ ) ⊂ A β f β γ ( B γ ) ⊂ B β f_{\beta\gamma}(B_\gamma) \subset B_\beta f β γ ( B γ ) ⊂ B β B \mathfrak{B} B A α = f α β ( A β ) A_\alpha = f_{\alpha\beta}(A_\beta) A α = f α β ( A β ) α ⩽ β \alpha \leqslant \beta α ⩽ β B β ≠ ∅ \mathrm{B}_\beta \neq \emptyset B β = ∅ β ∈ I \beta \in \mathrm{I} β ∈ I β ∈ I \beta \in \mathrm{I} β ∈ I B β ∈ S β \mathrm{B}_\beta \in \mathfrak{S}_\beta B β ∈ S β B ∈ Σ \mathfrak{B} \in \Sigma B ∈ Σ B β ⊂ A β \mathrm{B}_\beta \subset \mathrm{A}_\beta B β ⊂ A β β ∈ I \beta \in \mathrm{I} β ∈ I A \mathfrak{A} A B β = A β \mathrm{B}_\beta = \mathrm{A}_\beta B β = A β β \beta β A α = { x α } \mathrm{A}_\alpha = \{x_\alpha\} A α = { x α }
(4) 我们现在可以证明定理 1。让我们从(a)开始。我们有
f α ( E ) ⊂ ⋂ β ⩾ α f α β ( E β ) . f_\alpha(\mathrm{E}) \subset \bigcap_{\beta \geqslant \alpha} f_{\alpha\beta}(\mathrm{E}_\beta). f α ( E ) ⊂ β ⩾ α ⋂ f α β ( E β ) . 相反，让
x α ∈ ⋂ β ⩾ α f α β ( E β ) , x_\alpha \in \bigcap_{\beta \geqslant \alpha} f_{\alpha\beta}(\mathrm{E}_\beta), x α ∈ β ⩾ α ⋂ f α β ( E β ) , 并把
B β = f − 1 α β ( x α ) \mathrm{B}_\beta = \overset{-1}{f}_{\alpha\beta}(x_\alpha) B β = f − 1 α β ( x α ) 如果β ⩾ α \beta \geqslant \alpha β ⩾ α B β = E β \mathrm{B}_\beta = \mathrm{E}_\beta B β = E β x α x_\alpha x α B β \mathrm{B}_\beta B β β ∈ I \beta \in \mathrm{I} β ∈ I B β ∈ S β \mathrm{B}_\beta \in \mathfrak{S}_\beta B β ∈ S β β ⩽ γ \beta \leqslant \gamma β ⩽ γ f β γ ( B γ ) ⊂ B β f_{\beta\gamma}(\mathrm{B}_\gamma) \subset \mathrm{B}_\beta f β γ ( B γ ) ⊂ B β B = ( B β ) ∈ Σ \mathfrak{B} = (\mathrm{B}_\beta) \in \Sigma B = ( B β ) ∈ Σ A = ( A α ) \mathfrak{A} = (\mathrm{A}_\alpha) A = ( A α ) Σ \Sigma Σ A ⩾ B \mathfrak{A} \geqslant \mathfrak{B} A ⩾ B A \mathfrak{A} A β ∈ I \beta \in \mathrm{I} β ∈ I A β \mathrm{A}_\beta A β { y β } \{y_\beta\} { y β } y = ( y β ) y = (y_\beta) y = ( y β ) f α ( y ) = y α = x α f_\alpha(y) = y_\alpha = x_\alpha f α ( y ) = y α = x α
最后，(a) 暗示(b)。我们不妨假设I不为空（否则没有什么可证明的）。 假设 E α \mathrm{E}_\alpha E α f α β ( E β ) ≠ ∅ f_{\alpha\beta}(\mathrm{E}_\beta) \neq \emptyset f α β ( E β ) = ∅ β ⩾ α \beta \geqslant \alpha β ⩾ α f α β ( E β ) f_{\alpha\beta}(\mathrm{E}_\beta) f α β ( E β ) α \alpha α β ⩾ α \beta \geqslant \alpha β ⩾ α E α \mathrm{E}_\alpha E α S α \mathfrak{S}_\alpha S α ′ ' ′
⋂ β ⩾ α f α β ( E β ) ≠ ∅ . \bigcap_{\beta \geqslant \alpha} f_{\alpha\beta}(\mathrm{E}_\beta) \neq \emptyset. β ⩾ α ⋂ f α β ( E β ) = ∅. 因此 (a) 为 f α ( E ) ≠ ∅ f_\alpha(\mathrm{E}) \neq \emptyset f α ( E ) = ∅ 更不必说 E ≠ ∅ \mathrm{E} \neq \emptyset E = ∅
如果E α E_\alpha E α S α \mathfrak{S}_\alpha S α E α E_\alpha E α E α E_\alpha E α f α β f_{\alpha\beta} f α β S α \mathfrak{S}_\alpha S α E α E_\alpha E α
引用了 2 引用了 定理 1 , TG I § 9. 紧凑空间和局部紧凑空间
\* 设 A 为环和单位元素，对于每个 α ∈ I \alpha \in I α ∈ I T α T_\alpha T α 阿廷的 左A-模。令E α E_\alpha E α T α T_\alpha T α T α T_\alpha T α E α E_\alpha E α T α T_\alpha T α β ⩾ α \beta \geqslant \alpha β ⩾ α f α β : E β → E α f_{\alpha\beta} : E_\beta \to E_\alpha f α β : E β → E α S α \mathfrak{S}_\alpha S α E α E_\alpha E α T α T_\alpha T α M ∈ F M \in \mathfrak{F} M ∈ F ⋂ M ∈ F M \bigcap_{M \in \mathfrak{F}} M ⋂ M ∈ F M f α β f_{\alpha\beta} f α β
5. 直极限
令I为（*右*）*定向*预排序的集合，并让( E α ) α ∈ I (E_\alpha)_{\alpha \in I} ( E α ) α ∈ I ( α , β ) (\alpha, \beta) ( α , β ) α ⩽ β \alpha \leqslant \beta α ⩽ β f β α f_{\beta\alpha} f β α E α E_\alpha E α E β E_\beta E β f β α f_{\beta\alpha} f β α
(LII _\mathrm{I} I 关系 α ⩽ β ⩽ γ \alpha \leqslant \beta \leqslant \gamma α ⩽ β ⩽ γ 暗示 f γ α = f γ β ∘ f β α f_{\gamma\alpha} = f_{\gamma\beta} \circ f_{\beta\alpha} f γ α = f γ β ∘ f β α I I _\mathrm{II} II 对于每个 α ∈ I , f α α \alpha \in I, f_{\alpha\alpha} α ∈ I , f α α 是 E α E_\alpha E α
Let G be the set which is the sum of the family of sets ( E α ) α ∈ I (E_\alpha)_{\alpha \in I} ( E α ) α ∈ I E α E_\alpha E α x ∈ G x \in G x ∈ G λ ( x ) \lambda(x) λ ( x ) α ∈ I \alpha \in I α ∈ I x ∈ E α x \in E_\alpha x ∈ E α R { x , y } R\{x, y\} R { x , y } x x x y y y γ ∈ I \gamma \in I γ ∈ I γ ⩾ α = λ ( x ) \gamma \geqslant \alpha = \lambda(x) γ ⩾ α = λ ( x ) γ ⩾ β = λ ( y ) \gamma \geqslant \beta = \lambda(y) γ ⩾ β = λ ( y ) f γ α ( x ) = f γ β ( y ) f_{\gamma\alpha}(x) = f_{\gamma\beta}(y) f γ α ( x ) = f γ β ( y ) x ∈ E α , y ∈ E β , z ∈ E γ x \in E_\alpha, y \in E_\beta, z \in E_\gamma x ∈ E α , y ∈ E β , z ∈ E γ λ ∈ I \lambda \in I λ ∈ I λ ⩾ α \lambda \geqslant \alpha λ ⩾ α λ ⩾ β \lambda \geqslant \beta λ ⩾ β f λ α ( x ) = f λ β ( y ) f_{\lambda\alpha}(x) = f_{\lambda\beta}(y) f λ α ( x ) = f λ β ( y ) μ ∈ I \mu \in I μ ∈ I μ ⩾ β \mu \geqslant \beta μ ⩾ β μ ⩾ γ \mu \geqslant \gamma μ ⩾ γ f μ β ( y ) = f μ γ ( z ) f_{\mu\beta}(y) = f_{\mu\gamma}(z) f μ β ( y ) = f μ γ ( z ) ν ∈ I \nu \in I ν ∈ I ν ⩾ λ \nu \geqslant \lambda ν ⩾ λ ν ⩾ μ \nu \geqslant \mu ν ⩾ μ
到 ( L I I ) (\mathrm{LI_I}) ( L I I )
f ν α ( x ) = f ν λ ( f λ α ( x ) ) = f ν λ ( f λ β ( y ) ) = f ν β ( y ) f_{\nu\alpha}(x) = f_{\nu\lambda}(f_{\lambda\alpha}(x)) = f_{\nu\lambda}(f_{\lambda\beta}(y)) = f_{\nu\beta}(y) f ν α ( x ) = f ν λ ( f λ α ( x )) = f ν λ ( f λ β ( y )) = f ν β ( y ) = f ν μ ( f μ β ( y ) ) = f ν μ ( f μ γ ( z ) ) = f ν γ ( z ) , = f_{\nu\mu}(f_{\mu\beta}(y)) = f_{\nu\mu}(f_{\mu\gamma}(z)) = f_{\nu\gamma}(z), = f ν μ ( f μ β ( y )) = f ν μ ( f μ γ ( z )) = f ν γ ( z ) , 这就建立了我们的断言。 商集 E = G / R \mathrm{E} = \mathrm{G/R} E = G/R ( f β α ) (f_{\beta\alpha}) ( f β α ) 直极限家族 ( E α ) α ∈ I (\mathrm{E}_\alpha)_{\alpha \in \mathrm{I}} ( E α ) α ∈ I ，并写作 E = l i m → ( E α , f β α ) \mathrm{E} = \varinjlim (\mathrm{E}_\alpha, f_{\beta\alpha}) E = lim ( E α , f β α ) E = l i m → E α \mathrm{E} = \varinjlim \mathrm{E}_\alpha E = lim E α ( ( E α ) , ( f β α ) ) ((\mathrm{E}_\alpha), (f_{\beta\alpha})) (( E α ) , ( f β α )) ( E α , f β α ) (\mathrm{E}_\alpha, f_{\beta\alpha}) ( E α , f β α ) 的*直接系统，相对于有向集合I。
¶ 显然，只要 E α \mathrm{E}_\alpha E α f f f f α f_\alpha f α E α \mathrm{E}_\alpha E α E = G / R \mathrm{E} = \mathrm{G/R} E = G/R f α f_\alpha f α E α \mathrm{E}_\alpha E α α ⩽ β \alpha \leqslant \beta α ⩽ β
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f β ∘ f β α = f α f_\beta \circ f_{\beta\alpha} = f_\alpha f β ∘ f β α = f α 因为对于每个x ∈ E α x \in \mathrm{E}_\alpha x ∈ E α f β β ( f β α ( x ) ) = f β α ( x ) f_{\beta\beta}(f_{\beta\alpha}(x)) = f_{\beta\alpha}(x) f β β ( f β α ( x )) = f β α ( x ) ( L I I ) (\mathrm{LI_I}) ( L I I ) x ∈ E α x \in \mathrm{E}_\alpha x ∈ E α f β α ( x ) ∈ E β f_{\beta\alpha}(x) \in \mathrm{E}_\beta f β α ( x ) ∈ E β
例
设A，B为两个集合，并令( V α ) α ∈ I (\mathrm{V}_\alpha)_{\alpha \in \mathrm{I}} ( V α ) α ∈ I α ⩽ β \alpha \leqslant \beta α ⩽ β V β ⊂ V α \mathrm{V}_\beta \subset \mathrm{V}_\alpha V β ⊂ V α E α \mathrm{E}_\alpha E α V α \mathrm{V}_\alpha V α ( α , β ) (\alpha, \beta) ( α , β ) α ⩽ β \alpha \leqslant \beta α ⩽ β f β α f_{\beta\alpha} f β α E α \mathrm{E}_\alpha E α E β \mathrm{E}_\beta E β u ∈ E α u \in \mathrm{E}_\alpha u ∈ E α 限制 到 V β \mathrm{V}_\beta V β ( L I I ) (\mathrm{LI_I}) ( L I I ) ( L I I I ) (\mathrm{LI_{II}}) ( L I II ) E = l i m → E α \mathrm{E} = \varinjlim \mathrm{E}_\alpha E = lim E α V α \mathrm{V}_\alpha V α ( V α ) (\mathrm{V}_\alpha) ( V α ) 邻域 （*一般拓扑*、章 I, §6, no. 10）。 \
引用了 1 引用了 TG I § 6. 过滤器
假设对于每个 α ∈ I \alpha \in \mathrm{I} α ∈ I E α \mathrm{E}_\alpha E α α ⩽ β \alpha \leqslant \beta α ⩽ β f β α f_{\beta\alpha} f β α l i m → E α \varinjlim \mathrm{E}_\alpha lim E α l i m → E α \varinjlim \mathrm{E}_\alpha lim E α ( E α ) (\mathrm{E}_\alpha) ( E α ) ( G α ) (\mathrm{G}_\alpha) ( G α ) α ∈ I \alpha \in \mathrm{I} α ∈ I h α : F → G α h_\alpha : \mathrm{F} \to \mathrm{G}_\alpha h α : F → G α ( P y ) y ∈ F (\mathrm{P}_y)_{y \in \mathrm{F}} ( P y ) y ∈ F P y \mathrm{P}_y P y h α ( y ) h_\alpha(y) h α ( y ) α \alpha α y → P y y \to \mathrm{P}_y y → P y l i m → E α \varinjlim \mathrm{E}_\alpha lim E α
被以下引用 1 被以下引用 注
设*( E α , f β α ) (\mathrm{E}_\alpha, f_{\beta\alpha}) ( E α , f β α ) E = l i m → E α \mathrm{E} = \varinjlim \mathrm{E}_\alpha E = lim E α α ∈ I \alpha \in \mathrm{I} α ∈ I f α : E α → E f_\alpha \colon \mathrm{E}_\alpha \to \mathrm{E} f α : E α → E
被以下引用 11 被以下引用 推论, A II § 6 , 命题 5, A II § 6 , 命题 7, A II § 6 , 推论 4, A II § 6 , 命题 26, A II § 7 , § 7. 逆极限和直极限 , 推论 , 命题 10 , 命题 6 , 命题 7 , 命题 8
(i) 令 ( x ( i ) ) 1 ⩽ i ⩽ n (x^{(i)})_{1 \leqslant i \leqslant n} ( x ( i ) ) 1 ⩽ i ⩽ n E 的元素的有限系统。 则存在 α ∈ I \alpha \in I α ∈ I 和 E α E_\alpha E α 的元素的有限系统 ( x α ( i ) ) 1 ⩽ i ⩽ n (x_\alpha^{(i)})_{1 \leqslant i \leqslant n} ( x α ( i ) ) 1 ⩽ i ⩽ n 这样 x ( i ) = f α ( x α ( i ) ) x^{(i)} = f_\alpha(x_\alpha^{(i)}) x ( i ) = f α ( x α ( i ) ) 1 ⩽ i ⩽ n 1 \leqslant i \leqslant n 1 ⩽ i ⩽ n
(ii) 设 ( y α ( i ) ) 1 ⩽ i ⩽ n (y_\alpha^{(i)})_{1 \leqslant i \leqslant n} ( y α ( i ) ) 1 ⩽ i ⩽ n 为某些 E α E_\alpha E α 如果 f α ( y α ( i ) ) = f α ( y α ( j ) ) f_\alpha(y_\alpha^{(i)}) = f_\alpha(y_\alpha^{(j)}) f α ( y α ( i ) ) = f α ( y α ( j ) ) ( i , j ) (i, j) ( i , j ) β ⩾ α \beta \geqslant \alpha β ⩾ α f β α ( y α ( i ) ) = f β α ( y α ( j ) ) f_{\beta\alpha}(y_\alpha^{(i)}) = f_{\beta\alpha}(y_\alpha^{(j)}) f β α ( y α ( i ) ) = f β α ( y α ( j ) ) ( i , j ) (i, j) ( i , j )
(i) 通过 E 的定义，每个 i i i β i ∈ I \beta_i \in I β i ∈ I z β i ∈ E β i z_{\beta_i} \in E_{\beta_i} z β i ∈ E β i x ( i ) = f β i ( z β i ) x^{(i)} = f_{\beta_i}(z_{\beta_i}) x ( i ) = f β i ( z β i ) α \alpha α α ⩾ β i \alpha \geqslant \beta_i α ⩾ β i 1 ⩽ i ⩽ n 1 \leqslant i \leqslant n 1 ⩽ i ⩽ n x α ( i ) = f α β i ( z β i ) x_\alpha^{(i)} = f_{\alpha\beta_i}(z_{\beta_i}) x α ( i ) = f α β i ( z β i )
(ii) 通过 E 的定义，对于每个对( i , j ) (i, j) ( i , j ) γ i j ∈ I \gamma_{ij} \in I γ ij ∈ I γ i j ⩾ α \gamma_{ij} \geqslant \alpha γ ij ⩾ α f γ i j α ( y α ( i ) ) = f γ i j α ( y α ( j ) ) f_{\gamma_{ij}\alpha}(y_\alpha^{(i)}) = f_{\gamma_{ij}\alpha}(y_\alpha^{(j)}) f γ ij α ( y α ( i ) ) = f γ ij α ( y α ( j ) ) β \beta β β ⩾ γ i j \beta \geqslant \gamma_{ij} β ⩾ γ ij ( i , j ) (i, j) ( i , j ) f β α = f β γ i j ∘ f γ i j α f_{\beta\alpha} = f_{\beta\gamma_{ij}} \circ f_{\gamma_{ij}\alpha} f β α = f β γ ij ∘ f γ ij α
6. 直接系统 / 映射
设*I*为有向集合，让*( E α , f β α ) (E_\alpha, f_{\beta\alpha}) ( E α , f β α ) E = l i m → E α E = \varinjlim E_\alpha E = lim E α α ∈ I \alpha \in I α ∈ I
引用了 3 · 被以下引用 8 引用了 命题 8, II § 4 , II § 6. 等价关系 , 引理 1
被以下引用 命题 2, A I § 10 , § 7. 逆极限和直极限 , 注 , 注 , 命题 10 , 推论 1 , 命题 7 , 命题 8
f α : E α → E f_\alpha : E_\alpha \to E f α : E α → E 为典范映射。对于每个 α ∈ I \alpha \in I α ∈ I u α u_\alpha u α 成为 E α E_\alpha E α 进入集合 F 这样
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u β ∘ f β α = u α whenever α ⩽ β . u_\beta \circ f_{\beta\alpha} = u_\alpha \qquad \textit{whenever } \alpha \leqslant \beta. u β ∘ f β α = u α whenever α ⩽ β . 然后：
(a) 存在唯一映射 u u u 进入 F 这样
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u α = u ∘ f α for all α ∈ I . u_\alpha = u \circ f_\alpha \qquad \textit{for all } \alpha \in I. u α = u ∘ f α for all α ∈ I . (b) u u u 是满射当且仅当 F 是并 of 集合 u α ( E α ) u_\alpha(E_\alpha) u α ( E α )
(c) u u u 是单射当且仅当对于每个 α ∈ I \alpha \in I α ∈ I 关系 x ∈ E α x \in E_\alpha x ∈ E α y ∈ E α y \in E_\alpha y ∈ E α u α ( x ) = u α ( y ) u_\alpha(x) = u_\alpha(y) u α ( x ) = u α ( y ) 意味着存在 β ⩾ α \beta \geqslant \alpha β ⩾ α 使得 f β α ( x ) = f β α ( y ) f_{\beta\alpha}(x) = f_{\beta\alpha}(y) f β α ( x ) = f β α ( y )
(a) 对于no. 5的记号，令v v v α ∈ I \alpha \in I α ∈ I E α E_\alpha E α u α u_\alpha u α v v v E = G / R E = G/R E = G / R u u u v = u ∘ f v = u \circ f v = u ∘ f loc.cit. )。
(b) 由于 E 是 f α ( E α ) f_\alpha(E_\alpha) f α ( E α ) F = ⋃ α ∈ I u α ( E α ) F = \bigcup_{\alpha \in I} u_\alpha(E_\alpha) F = ⋃ α ∈ I u α ( E α ) u u u
(c) 通过no. 5的引理 1，E \mathbf{E} E f α ( x ) f_\alpha(x) f α ( x ) f α ( y ) f_\alpha(y) f α ( y ) x ∈ E α x \in \mathrm{E}_\alpha x ∈ E α y ∈ E α y \in \mathrm{E}_\alpha y ∈ E α α ∈ I \alpha \in \mathrm{I} α ∈ I f α ( x ) = f α ( y ) f_\alpha(x) = f_\alpha(y) f α ( x ) = f α ( y ) β ⩾ α \beta \geqslant \alpha β ⩾ α f β α ( x ) = f β α ( y ) f_{\beta\alpha}(x) = f_{\beta\alpha}(y) f β α ( x ) = f β α ( y ) u α ( x ) = u ( f α ( x ) ) u_\alpha(x) = u(f_\alpha(x)) u α ( x ) = u ( f α ( x )) u α ( y ) = u ( f α ( y ) ) u_\alpha(y) = u(f_\alpha(y)) u α ( y ) = u ( f α ( y ))
¶ 如果映射 u u u ( E α ) (\mathrm{E}_\alpha) ( E α )
令*( E α , f β α ) (\mathrm{E}_\alpha, f_{\beta\alpha}) ( E α , f β α ) ( F α , g β α ) (\mathrm{F}_\alpha, g_{\beta\alpha}) ( F α , g β α ) 让 E = l i m → E α \mathrm{E} = \varinjlim \mathrm{E}_\alpha E = lim E α F = l i m → F α \mathrm{F} = \varinjlim \mathrm{F}_\alpha F = lim F α α ∈ I \alpha \in \mathrm{I} α ∈ I 让 f α f_\alpha f α g α g_\alpha g α E α \mathrm{E}_\alpha E α F α \mathrm{F}_\alpha F α 对于每个 α ∈ I \alpha \in \mathrm{I} α ∈ I 让 u α u_\alpha u α 成为 E α \mathrm{E}_\alpha E α 进入 F α \mathrm{F}_\alpha F α 这样，每当 α ⩽ β \alpha \leqslant \beta α ⩽ β
引用了 1 引用了 命题 6
E α ⟶ u α F α f β α ↓ ↓ g β α E β ⟶ u β F β \begin{array}{ccc} \mathrm{E}_\alpha & \overset{u_\alpha}{\longrightarrow} & \mathrm{F}_\alpha \\ {\scriptstyle f_{\beta\alpha}}\big\downarrow & & \big\downarrow{\scriptstyle g_{\beta\alpha}} \\ \mathrm{E}_\beta & \underset{u_\beta}{\longrightarrow} & \mathrm{F}_\beta \end{array} E α f β α ↓ ⏐ E β ⟶ u α u β ⟶ F α ↓ ⏐ g β α F β * 是可交换的。那么存在唯一映射* u : E → F u : \mathrm{E} \rightarrow \mathrm{F} u : E → F 这样，对于每个 α ∈ I \alpha \in \mathrm{I} α ∈ I
E α ⟶ u α F α f α ↓ ↓ g α E ⟶ u F \begin{array}{ccc} \mathrm{E}_\alpha & \overset{u_\alpha}{\longrightarrow} & \mathrm{F}_\alpha \\ {\scriptstyle f_\alpha}\big\downarrow & & \big\downarrow{\scriptstyle g_\alpha} \\ \mathrm{E} & \underset{u}{\longrightarrow} & \mathrm{F} \end{array} E α f α ↓ ⏐ E ⟶ u α u ⟶ F α ↓ ⏐ g α F 是可交换的。
把v α = g α ∘ u α v_\alpha = g_\alpha \circ u_\alpha v α = g α ∘ u α α ⩽ β \alpha \leqslant \beta α ⩽ β
v β ∘ f β α = g β ∘ u β ∘ f β α = g β ∘ g β α ∘ u α = g α ∘ u α = v α . v_\beta \circ f_{\beta\alpha} = g_\beta \circ u_\beta \circ f_{\beta\alpha} = g_\beta \circ g_{\beta\alpha} \circ u_\alpha = g_\alpha \circ u_\alpha = v_\alpha. v β ∘ f β α = g β ∘ u β ∘ f β α = g β ∘ g β α ∘ u α = g α ∘ u α = v α . 因此，我们可以应用命题 6 到映射 v α v_\alpha v α u : E → F u : \mathrm{E} \to \mathrm{F} u : E → F
u ∘ f α = v α = g α ∘ u α u \circ f_\alpha = v_\alpha = g_\alpha \circ u_\alpha u ∘ f α = v α = g α ∘ u α 对于所有α ∈ I \alpha \in \mathrm{I} α ∈ I
¶ 满足推论 1 的条件的映射 u α : E α → F α u_\alpha : \mathrm{E}_\alpha \to \mathrm{F}_\alpha u α : E α → F α ( E α , f β α ) (\mathrm{E}_\alpha, f_{\beta\alpha}) ( E α , f β α ) 直接系统到 ( F α , g β α ) (\mathrm{F}_\alpha, g_{\beta\alpha}) ( F α , g β α ) ( u α ) (u_\alpha) ( u α ) 直极限*，并且在不存在歧义的情况下写作u = l i m → u α u = \varinjlim u_\alpha u = lim u α
令*( E α , f β α ) (\mathrm{E}_\alpha, f_{\beta\alpha}) ( E α , f β α ) ( F α , g β α ) (\mathrm{F}_\alpha, g_{\beta\alpha}) ( F α , g β α ) ( G α , h β α ) (\mathrm{G}_\alpha, h_{\beta\alpha}) ( G α , h β α ) 令 E = l i m → E α \mathrm{E} = \varinjlim \mathrm{E}_\alpha E = lim E α F = l i m → F α \mathrm{F} = \varinjlim \mathrm{F}_\alpha F = lim F α G = l i m → G α \mathrm{G} = \varinjlim \mathrm{G}_\alpha G = lim G α f α f_\alpha f α （分别 g α g_\alpha g α h α h_\alpha h α E α \mathrm{E}_\alpha E α （分别 F α \mathrm{F}_\alpha F α G α \mathrm{G}_\alpha G α （分别 F、G*）。如果*( u α ) (u_\alpha) ( u α ) ( v α ) (v_\alpha) ( v α ) u α : E α → F α u_\alpha : \mathrm{E}_\alpha \to \mathrm{F}_\alpha u α : E α → F α v α : F α → G α v_\alpha : \mathrm{F}_\alpha \to \mathrm{G}_\alpha v α : F α → G α v α ∘ u α : E α → G α v_\alpha \circ u_\alpha : \mathrm{E}_\alpha \to \mathrm{G}_\alpha v α ∘ u α : E α → G α
被以下引用 1 被以下引用 § 7. 逆极限和直极限
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l i m → ( v α ∘ u α ) = ( l i m → v α ) ∘ ( l i m → u α ) . \varinjlim (v_\alpha \circ u_\alpha) = (\varinjlim v_\alpha) \circ (\varinjlim u_\alpha). lim ( v α ∘ u α ) = ( lim v α ) ∘ ( lim u α ) . 因为如果我们输入w α = v α ∘ u α w_\alpha = v_\alpha \circ u_\alpha w α = v α ∘ u α α ⩽ β \alpha \leqslant \beta α ⩽ β
h β α ∘ w α = ( h β α ∘ v α ) ∘ u α = ( v β ∘ g β α ) ∘ u α = v β ∘ ( u β ∘ f β α ) = w β ∘ f β α , h_{\beta\alpha} \circ w_\alpha = (h_{\beta\alpha} \circ v_\alpha) \circ u_\alpha = (v_\beta \circ g_{\beta\alpha}) \circ u_\alpha = v_\beta \circ (u_\beta \circ f_{\beta\alpha}) = w_\beta \circ f_{\beta\alpha}, h β α ∘ w α = ( h β α ∘ v α ) ∘ u α = ( v β ∘ g β α ) ∘ u α = v β ∘ ( u β ∘ f β α ) = w β ∘ f β α , 这表明 ( w α ) (w_\alpha) ( w α ) u = l i m → u α u = \varinjlim u_\alpha u = lim u α v = l i m → v α v = \varinjlim v_\alpha v = lim v α α ∈ I \alpha \in \mathrm{I} α ∈ I
( v ∘ u ) ∘ f α = v ∘ ( g α ∘ u α ) = h α ∘ ( v α ∘ u α ) , (v \circ u) \circ f_\alpha = v \circ (g_\alpha \circ u_\alpha) = h_\alpha \circ (v_\alpha \circ u_\alpha), ( v ∘ u ) ∘ f α = v ∘ ( g α ∘ u α ) = h α ∘ ( v α ∘ u α ) , 凭借直极限的唯一性，我们有v ∘ u = l i m → w α v \circ u = \varinjlim w_\alpha v ∘ u = lim w α
令*( E α , f β α ) (\mathrm{E}_\alpha, f_{\beta\alpha}) ( E α , f β α ) ( E α ′ , f β α ′ ) (\mathrm{E}'_\alpha, f'_{\beta\alpha}) ( E α ′ , f β α ′ ) α ∈ I \alpha \in \mathrm{I} α ∈ I u α u_\alpha u α E α \mathrm{E}_\alpha E α E α ′ \mathrm{E}'_\alpha E α ′ u α u_\alpha u α u = l i m → u α u = \varinjlim u_\alpha u = lim u α 如果每个 u α u_\alpha u α 是单射 （或*满射*）*则* u u u 是单射 （或*满射*）。
引用了 2 · 被以下引用 1 引用了 引理 1 , 命题 6
被以下引用 推论
令E = l i m → E α \mathrm{E} = \varinjlim \mathrm{E}_\alpha E = lim E α E ′ = l i m → E α ′ \mathrm{E}' = \varinjlim \mathrm{E}'_\alpha E ′ = lim E α ′ f α : E α → E f_\alpha : \mathrm{E}_\alpha \to \mathrm{E} f α : E α → E f α ′ : E α ′ → E ′ f'_\alpha : \mathrm{E}'_\alpha \to \mathrm{E}' f α ′ : E α ′ → E ′ u α u_\alpha u α u u u x ∈ E α x \in \mathrm{E}_\alpha x ∈ E α y ∈ E α y \in \mathrm{E}_\alpha y ∈ E α f α ′ ( u α ( x ) ) = f α ′ ( u α ( y ) ) f'_\alpha(u_\alpha(x)) = f'_\alpha(u_\alpha(y)) f α ′ ( u α ( x )) = f α ′ ( u α ( y )) β ⩾ α \beta \geqslant \alpha β ⩾ α f β α ( x ) = f β α ( y ) f_{\beta\alpha}(x) = f_{\beta\alpha}(y) f β α ( x ) = f β α ( y ) β ⩾ α \beta \geqslant \alpha β ⩾ α
f β α ′ ( u α ( x ) ) = f β α ′ ( u α ( y ) ) , i.e., u β ( f β α ( x ) ) = u β ( f β α ( y ) ) , f'_{\beta\alpha}(u_\alpha(x)) = f'_{\beta\alpha}(u_\alpha(y)), \qquad \text{i.e.,} \qquad u_\beta(f_{\beta\alpha}(x)) = u_\beta(f_{\beta\alpha}(y)), f β α ′ ( u α ( x )) = f β α ′ ( u α ( y )) , i.e., u β ( f β α ( x )) = u β ( f β α ( y )) , 因此 f β α ( x ) = f β α ( y ) f_{\beta\alpha}(x) = f_{\beta\alpha}(y) f β α ( x ) = f β α ( y ) u β u_\beta u β
现在假设 u α u_\alpha u α
E ′ = ⋃ α f α ′ ( E α ′ ) = ⋃ α f α ′ ( u α ( E α ) ) = ⋃ α u ( f α ( E α ) ) = u ( ⋃ α f α ( E α ) ) = u ( E ) . \mathrm{E}' = \bigcup_\alpha f'_\alpha(\mathrm{E}'_\alpha) = \bigcup_\alpha f'_\alpha(u_\alpha(\mathrm{E}_\alpha)) = \bigcup_\alpha u(f_\alpha(\mathrm{E}_\alpha)) = u\Big(\bigcup_\alpha f_\alpha(\mathrm{E}_\alpha)\Big) = u(\mathrm{E}). E ′ = α ⋃ f α ′ ( E α ′ ) = α ⋃ f α ′ ( u α ( E α )) = α ⋃ u ( f α ( E α )) = u ( α ⋃ f α ( E α ) ) = u ( E ) . 当记号为命题7时，对于每个α ∈ I \alpha \in \mathrm{I} α ∈ I M α \mathrm{M}_\alpha M α E α \mathrm{E}_\alpha E α α ⩽ β \alpha \leqslant \beta α ⩽ β f β α ( M α ) ∈ M β f_{\beta\alpha}(\mathrm{M}_\alpha) \in \mathrm{M}_\beta f β α ( M α ) ∈ M β ( M α ) α ∈ I (\mathrm{M}_\alpha)_{\alpha \in \mathrm{I}} ( M α ) α ∈ I E α \mathrm{E}_\alpha E α 直接系统。设g β α g_{\beta\alpha} g β α α ⩽ β \alpha \leqslant \beta α ⩽ β M α \mathrm{M}_\alpha M α M β \mathrm{M}_\beta M β f β α f_{\beta\alpha} f β α M α \mathrm{M}_\alpha M α ( M α , g β α ) (\mathrm{M}_\alpha, g_{\beta\alpha}) ( M α , g β α ) j α : M α → E α j_\alpha : \mathrm{M}_\alpha \to \mathrm{E}_\alpha j α : M α → E α j = l i m → j α j = \varinjlim j_\alpha j = lim j α 认同*M = l i m → M α \mathrm{M} = \varinjlim \mathrm{M}_\alpha M = lim M α
设*( E α , f β α ) (\mathrm{E}_\alpha, f_{\beta\alpha}) ( E α , f β α ) ( E α ′ , f β α ′ ) (\mathrm{E}'_\alpha, f'_{\beta\alpha}) ( E α ′ , f β α ′ ) ( u α ) (u_\alpha) ( u α ) u α : E α → E α ′ u_\alpha : \mathrm{E}_\alpha \to \mathrm{E}'_\alpha u α : E α → E α ′ u = l i m → u α u = \varinjlim u_\alpha u = lim u α
(i) 令 ( M α ) (\mathrm{M}_\alpha) ( M α ) E α \mathrm{E}_\alpha E α 那么 ( u α ( M α ) ) (u_\alpha(\mathrm{M}_\alpha)) ( u α ( M α )) E α ′ \mathrm{E}'_\alpha E α ′
引用了 2 · 被以下引用 1 引用了 命题 7 , 引理 1
被以下引用 命题 3, A II § 6
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l i m → u α ( M α ) = u ( l i m → M α ) . \varinjlim u_\alpha(\mathrm{M}_\alpha) = u\,(\varinjlim \mathrm{M}_\alpha). lim u α ( M α ) = u ( lim M α ) . (ii) 让 ( a α ′ ) α ∈ I (a'_\alpha)_{\alpha \in \mathbf{I}} ( a α ′ ) α ∈ I 成为一个家庭，使得 a α ′ ∈ E α ′ a'_\alpha \in \mathrm{E}'_\alpha a α ′ ∈ E α ′ 对于每个 α ∈ I \alpha \in \mathrm{I} α ∈ I 和 f β α ′ ( a α ′ ) = a β ′ f'_{\beta\alpha}(a'_\alpha) = a'_\beta f β α ′ ( a α ′ ) = a β ′ 每当 α ⩽ β \alpha \leqslant \beta α ⩽ β 然后 E α \mathrm{E}_\alpha E α u − 1 α ( a α ′ ) \overset{-1}{u}_\alpha(a'_\alpha) u − 1 α ( a α ′ ) 形式直接系统， 我们有*
(27)
l i m → u − 1 α ( a α ′ ) = u − 1 ( a ′ ) , \varinjlim \overset{-1}{u}_\alpha(a'_\alpha) = \overset{-1}{u}(a'), lim u − 1 α ( a α ′ ) = u − 1 ( a ′ ) , 其中 a ′ a' a ′ l i m → E α ′ \varinjlim \mathrm{E}'_\alpha lim E α ′ α ∈ I \alpha \in \mathrm{I} α ∈ I a α ′ a'_\alpha a α ′
(i) 显然，E α ′ \mathrm{E}'_\alpha E α ′ u α ( M α ) u_\alpha(\mathrm{M}_\alpha) u α ( M α ) u α ( M α ) = v α ( M α ) u_\alpha(\mathrm{M}_\alpha) = v_\alpha(\mathrm{M}_\alpha) u α ( M α ) = v α ( M α ) v α v_\alpha v α M α \mathrm{M}_\alpha M α u α ( M α ) u_\alpha(\mathrm{M}_\alpha) u α ( M α ) u α u_\alpha u α M α \mathrm{M}_\alpha M α v α v_\alpha v α
(ii) 设 N α = u − 1 α ( a α ′ ) \mathrm{N}_\alpha = \overset{-1}{u}_\alpha(a'_\alpha) N α = u − 1 α ( a α ′ ) α ⩽ β \alpha \leqslant \beta α ⩽ β x α ∈ N α x_\alpha \in \mathrm{N}_\alpha x α ∈ N α
u β ( f β α ( x α ) ) = f β α ′ ( u α ( x α ) = f β α ′ ( a α ′ ) = a β ′ ; u_\beta(f_{\beta\alpha}(x_\alpha)) = f'_{\beta\alpha}(u_\alpha(x_\alpha) = f'_{\beta\alpha}(a'_\alpha) = a'_\beta; u β ( f β α ( x α )) = f β α ′ ( u α ( x α ) = f β α ′ ( a α ′ ) = a β ′ ; 因而f β α ( x α ) ∈ N β f_{\beta\alpha}(x_\alpha) \in \mathrm{N}_\beta f β α ( x α ) ∈ N β N α \mathrm{N}_\alpha N α E α \mathrm{E}_\alpha E α x ∈ l i m → N α x \in \varinjlim \mathrm{N}_\alpha x ∈ lim N α α ∈ I \alpha \in \mathrm{I} α ∈ I x α ∈ N α x_\alpha \in \mathrm{N}_\alpha x α ∈ N α x = f α ( x α ) x = f_\alpha(x_\alpha) x = f α ( x α ) u ( x ) = u ( f α ( x α ) ) = f α ′ ( u α ( x α ) ) = f α ′ ( a α ′ ) = a ′ u(x) = u(f_\alpha(x_\alpha)) = f'_\alpha(u_\alpha(x_\alpha)) = f'_\alpha(a'_\alpha) = a' u ( x ) = u ( f α ( x α )) = f α ′ ( u α ( x α )) = f α ′ ( a α ′ ) = a ′ x ∈ u − 1 ( a ′ ) x \in \overset{-1}{u}(a') x ∈ u − 1 ( a ′ ) x = f α ( x α ) x = f_\alpha(x_\alpha) x = f α ( x α ) α ∈ I \alpha \in \mathrm{I} α ∈ I x α ∈ E α x_\alpha \in \mathrm{E}_\alpha x α ∈ E α a ′ = u ( f α ( x α ) ) = f α ′ ( u α ( x α ) ) = f α ′ ( a α ′ ) a' = u(f_\alpha(x_\alpha)) = f'_\alpha(u_\alpha(x_\alpha)) = f'_\alpha(a'_\alpha) a ′ = u ( f α ( x α )) = f α ′ ( u α ( x α )) = f α ′ ( a α ′ ) β ⩾ α \beta \geqslant \alpha β ⩾ α f β α ′ ( u α ′ ( x α ) ) = f β α ′ ( a α ′ ) = a β ′ f'_{\beta\alpha}(u'_\alpha(x_\alpha)) = f'_{\beta\alpha}(a'_\alpha) = a'_\beta f β α ′ ( u α ′ ( x α )) = f β α ′ ( a α ′ ) = a β ′ u α ( f β α ( x α ) ) = a β ′ u_\alpha(f_{\beta\alpha}(x_\alpha)) = a'_\beta u α ( f β α ( x α )) = a β ′ f β α ( x α ) ∈ N β f_{\beta\alpha}(x_\alpha) \in \mathrm{N}_\beta f β α ( x α ) ∈ N β x = f β ( f β α ( x α ) ) x = f_\beta(f_{\beta\alpha}(x_\alpha)) x = f β ( f β α ( x α )) x ∈ l i m → N α x \in \varinjlim \mathrm{N}_\alpha x ∈ lim N α
设*I*为有向集合，让*( E α , f β α ) (\mathrm{E}_\alpha, f_{\beta\alpha}) ( E α , f β α ) E = l i m → E α \mathrm{E} = \varinjlim \mathrm{E}_\alpha E = lim E α 为 I 的共尾子集，并令 E′ 为从 ( E α , f β α ) (\mathrm{E}_\alpha, f_{\beta\alpha}) ( E α , f β α ) 通过限制指标集合至 J 获得的集合直接系统的直极限。 则典范映射 g g g 的 E′ 到 E *是双射的。
引用了 3 引用了 III § 1. 序关系。 有序集 , 引理 1 , 命题 6
J 必然是有向集合 (§ 1, no. 10)。我们将使用命题 6 的判据来证明g g g g g g α ∈ J \alpha \in \mathrm{J} α ∈ J
g ( E α ) = f α ( E α ) . g(\mathrm{E}_\alpha) = f_\alpha(\mathrm{E}_\alpha). g ( E α ) = f α ( E α ) . 现在，对于每个β ∈ I \beta \in I β ∈ I γ ∈ J \gamma \in J γ ∈ J β ⩽ γ \beta \leqslant \gamma β ⩽ γ g ( E γ ) ⊃ g ( f γ β ( E β ) ) = f β ( E β ) g(E_\gamma) \supset g(f_{\gamma\beta}(E_\beta)) = f_\beta(E_\beta) g ( E γ ) ⊃ g ( f γ β ( E β )) = f β ( E β ) g ( E α ) g(E_\alpha) g ( E α ) α \alpha α
7. 双直极限。 乘积直极限
令I、L为两个有向集合，并令I × L I \times L I × L ( E α λ , f β α μ λ ) (E^\lambda_\alpha, f^{\mu\lambda}_{\beta\alpha}) ( E α λ , f β α μ λ ) I × L I \times L I × L
( 29 ) f γ α ν λ = f γ β μ ν ∘ f β α μ λ (29) \qquad f^{\nu\lambda}_{\gamma\alpha} = f^{\mu\nu}_{\gamma\beta} \circ f^{\mu\lambda}_{\beta\alpha} ( 29 ) f γ α ν λ = f γ β μν ∘ f β α μ λ 每当α ⩽ β ⩽ γ \alpha \leqslant \beta \leqslant \gamma α ⩽ β ⩽ γ λ ⩽ μ ⩽ ν \lambda \leqslant \mu \leqslant \nu λ ⩽ μ ⩽ ν
让E或l i m → α , λ E α λ \varinjlim_{\alpha,\lambda} E^\lambda_\alpha lim α , λ E α λ λ ∈ L \lambda \in L λ ∈ L g β α λ = f β α λ λ : E α λ → E β λ g^\lambda_{\beta\alpha} = f^{\lambda\lambda}_{\beta\alpha} : E^\lambda_\alpha \to E^\lambda_\beta g β α λ = f β α λλ : E α λ → E β λ
( 30 ) g γ α λ = g γ β λ ∘ g β α λ whenever α ⩽ β ⩽ γ ; (30) \qquad g^\lambda_{\gamma\alpha} = g^\lambda_{\gamma\beta} \circ g^\lambda_{\beta\alpha} \qquad \text{whenever } \alpha \leqslant \beta \leqslant \gamma; ( 30 ) g γ α λ = g γ β λ ∘ g β α λ whenever α ⩽ β ⩽ γ ; 在其他字中，( E α λ , g β α λ ) (E^\lambda_\alpha, g^\lambda_{\beta\alpha}) ( E α λ , g β α λ ) F λ = l i m → α E α λ F^\lambda = \varinjlim_\alpha E^\lambda_\alpha F λ = lim α E α λ λ ⩽ μ \lambda \leqslant \mu λ ⩽ μ h α μ λ = f α α λ μ : E α λ → E α μ h^{\mu\lambda}_\alpha = f^{\lambda\mu}_{\alpha\alpha} : E^\lambda_\alpha \to E^\mu_\alpha h α μ λ = f α α λ μ : E α λ → E α μ h μ λ = l i m → α h α μ λ : F λ → F μ h^{\mu\lambda} = \varinjlim_\alpha h^{\mu\lambda}_\alpha : F^\lambda \to F^\mu h μ λ = lim α h α μ λ : F λ → F μ λ ⩽ μ ⩽ ν \lambda \leqslant \mu \leqslant \nu λ ⩽ μ ⩽ ν
( 31 ) h ν λ = h ν μ ∘ h μ λ (31) \qquad h^{\nu\lambda} = h^{\nu\mu} \circ h^{\mu\lambda} ( 31 ) h ν λ = h ν μ ∘ h μ λ （no. 6，命题 6，推论 2），因此 ( F λ , h μ λ ) (F^\lambda, h^{\mu\lambda}) ( F λ , h μ λ ) F = l i m → F λ F = \varinjlim F^\lambda F = lim F λ E → F E \to F E → F g α λ g^\lambda_\alpha g α λ E α λ → F λ E^\lambda_\alpha \to F^\lambda E α λ → F λ h λ h^\lambda h λ F λ → F F^\lambda \to F F λ → F u α λ = h λ ∘ g α λ u^\lambda_\alpha = h^\lambda \circ g^\lambda_\alpha u α λ = h λ ∘ g α λ α ⩽ β \alpha \leqslant \beta α ⩽ β λ ⩽ μ \lambda \leqslant \mu λ ⩽ μ
u β μ ∘ f β α μ λ = h μ ∘ g β μ ∘ f β α μ λ = h μ ∘ g β μ ∘ f β α μ μ ∘ f α α μ λ = h μ ∘ g α μ ∘ f α α μ λ = h μ ∘ h μ λ ∘ g α λ = h λ ∘ g α λ = u α λ \begin{aligned} u^\mu_\beta \circ f^{\mu\lambda}_{\beta\alpha} &= h^\mu \circ g^\mu_\beta \circ f^{\mu\lambda}_{\beta\alpha} = h^\mu \circ g^\mu_\beta \circ f^{\mu\mu}_{\beta\alpha} \circ f^{\mu\lambda}_{\alpha\alpha} = h^\mu \circ g^\mu_\alpha \circ f^{\mu\lambda}_{\alpha\alpha} \\ &= h^\mu \circ h^{\mu\lambda} \circ g^\lambda_\alpha = h^\lambda \circ g^\lambda_\alpha = u^\lambda_\alpha \end{aligned} u β μ ∘ f β α μ λ = h μ ∘ g β μ ∘ f β α μ λ = h μ ∘ g β μ ∘ f β α μμ ∘ f α α μ λ = h μ ∘ g α μ ∘ f α α μ λ = h μ ∘ h μ λ ∘ g α λ = h λ ∘ g α λ = u α λ 来自 (29) 和 h μ λ h^{\mu\lambda} h μ λ I × L I \times L I × L u α λ u^\lambda_\alpha u α λ u = l i m → α , λ u α λ : E → F u = \varinjlim_{\alpha,\lambda} u^\lambda_\alpha : E \to F u = lim α , λ u α λ : E → F u u u h λ ( F λ ) h^\lambda(F^\lambda) h λ ( F λ ) F λ F^\lambda F λ g α λ ( E α λ ) g_\alpha^\lambda(\mathrm{E}_\alpha^\lambda) g α λ ( E α λ )
h λ ( g α λ ( E α λ ) ) = u α λ ( E α λ ) . h^\lambda(g_\alpha^\lambda(\mathrm{E}_\alpha^\lambda)) = u_\alpha^\lambda(\mathrm{E}_\alpha^\lambda). h λ ( g α λ ( E α λ )) = u α λ ( E α λ ) . 接下来，令x x x y y y E α λ \mathrm{E}_\alpha^\lambda E α λ u α λ ( x ) = u α λ ( y ) u_\alpha^\lambda(x) = u_\alpha^\lambda(y) u α λ ( x ) = u α λ ( y ) h λ ( g α λ ( x ) ) = h λ ( g α λ ( y ) ) h^\lambda(g_\alpha^\lambda(x)) = h^\lambda(g_\alpha^\lambda(y)) h λ ( g α λ ( x )) = h λ ( g α λ ( y )) μ ⩾ λ \mu \geqslant \lambda μ ⩾ λ h μ λ ( g α λ ( x ) ) = h μ λ ( g α λ ( y ) ) h^{\mu\lambda}(g_\alpha^\lambda(x)) = h^{\mu\lambda}(g_\alpha^\lambda(y)) h μ λ ( g α λ ( x )) = h μ λ ( g α λ ( y )) g α μ ( f α α μ λ ( x ) ) = g α μ ( f α α μ λ ( y ) ) g_\alpha^\mu(f_{\alpha\alpha}^{\mu\lambda}(x)) = g_\alpha^\mu(f_{\alpha\alpha}^{\mu\lambda}(y)) g α μ ( f α α μ λ ( x )) = g α μ ( f α α μ λ ( y )) β ⩾ α \beta \geqslant \alpha β ⩾ α g β α μ ( f α α μ λ ( x ) ) = g β α μ ( f α α μ λ ( y ) ) g_{\beta\alpha}^\mu(f_{\alpha\alpha}^{\mu\lambda}(x)) = g_{\beta\alpha}^\mu(f_{\alpha\alpha}^{\mu\lambda}(y)) g β α μ ( f α α μ λ ( x )) = g β α μ ( f α α μ λ ( y )) f β α μ λ ( x ) = f β α μ λ ( y ) f_{\beta\alpha}^{\mu\lambda}(x) = f_{\beta\alpha}^{\mu\lambda}(y) f β α μ λ ( x ) = f β α μ λ ( y ) u u u
如果* ( E α λ , f β α μ λ ) (\mathrm{E}_\alpha^\lambda, f_{\beta\alpha}^{\mu\lambda}) ( E α λ , f β α μ λ ) 是集合的直接系统相对于乘积 I × L \mathrm{I} \times \mathrm{L} I × L
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l i m → α , λ E α λ = l i m → λ ( l i m → α E α λ ) . \varinjlim_{\alpha,\ \lambda} \mathrm{E}_\alpha^\lambda = \varinjlim_{\lambda}\ (\varinjlim_{\alpha} \mathrm{E}_\alpha^\lambda). α , λ lim E α λ = λ lim ( α lim E α λ ) .
令*( E ′ α λ , f ′ β α μ λ ) (\mathrm{{E}'}^\lambda_\alpha, {f'}^{\mu\lambda}_{\beta\alpha}) ( E ′ α λ , f ′ β α μ λ ) I × L \mathrm{I} \times \mathrm{L} I × L ( α , λ ) ∈ I × L (\alpha, \lambda) \in \mathrm{I} \times \mathrm{L} ( α , λ ) ∈ I × L u α λ u_\alpha^\lambda u α λ E α λ \mathrm{E}_\alpha^\lambda E α λ E ′ α λ \mathrm{{E}'}^\lambda_\alpha E ′ α λ u α λ u_\alpha^\lambda u α λ
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l i m → α , λ u α λ = l i m → λ ( l i m → α u α λ ) . \varinjlim_{\alpha,\ \lambda} u_\alpha^\lambda = \varinjlim_{\lambda}\ (\varinjlim_{\alpha} u_\alpha^\lambda). α , λ lim u α λ = λ lim ( α lim u α λ ) . 我们将验证留给读者。
令*( E α , f β α ) (\mathrm{E}_\alpha, f_{\beta\alpha}) ( E α , f β α ) ( E α ′ , f β α ′ ) (\mathrm{E}'_\alpha, f'_{\beta\alpha}) ( E α ′ , f β α ′ ) 让 E = l i m → E α \mathrm{E} = \varinjlim \mathrm{E}_\alpha E = lim E α E ′ = l i m → E α ′ \mathrm{E}' = \varinjlim \mathrm{E}'_\alpha E ′ = lim E α ′ f α : E α → E , f α ′ : E α ′ → E ′ f_\alpha : \mathrm{E}_\alpha \to \mathrm{E}, f'_\alpha : \mathrm{E}'_\alpha \to \mathrm{E}' f α : E α → E , f α ′ : E α ′ → E ′ 表示典范映射，对于每个 α ∈ I \alpha \in \mathrm{I} α ∈ I 那么 ( E α × E α ′ , f β α × f β α ′ ) (\mathrm{E}_\alpha \times \mathrm{E}'_\alpha, f_{\beta\alpha} \times f'_{\beta\alpha}) ( E α × E α ′ , f β α × f β α ′ ) ( f α × f α ′ ) (f_\alpha \times f'_\alpha) ( f α × f α ′ ) l i m → ( f α × f α ′ ) \varinjlim (f_\alpha \times f'_\alpha) lim ( f α × f α ′ )
引用了 2 · 被以下引用 1 引用了 引理 1 , 命题 6
被以下引用 命题 7, A II § 6
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l i m → ( E α × E α ′ ) → ( l i m → E α ) × ( l i m → E α ′ ) . \varinjlim (\mathrm{E}_\alpha \times \mathrm{E}'_\alpha) \to (\varinjlim \mathrm{E}_\alpha) \times (\varinjlim \mathrm{E}'_\alpha). lim ( E α × E α ′ ) → ( lim E α ) × ( lim E α ′ ) . 命题的前两个断言立即得到验证。为了证明 g = l i m → ( f α × f α ′ ) g = \varinjlim (f_\alpha \times f'_\alpha) g = lim ( f α × f α ′ ) E × E ′ \mathrm{E} \times \mathrm{E}' E × E ′ f α ( E α ) × f α ′ ( E α ′ ) f_\alpha(\mathrm{E}_\alpha) \times f'_\alpha(\mathrm{E}'_\alpha) f α ( E α ) × f α ′ ( E α ′ ) g g g ( x , x ′ ) (x, x') ( x , x ′ ) ( y , y ′ ) (y, y') ( y , y ′ ) E α × E α ′ \mathrm{E}_\alpha \times \mathrm{E}'_\alpha E α × E α ′ f α ( x ) = f α ( y ) f_\alpha(x) = f_\alpha(y) f α ( x ) = f α ( y ) f α ′ ( x ′ ) = f α ′ ( y ′ ) f'_\alpha(x') = f'_\alpha(y') f α ′ ( x ′ ) = f α ′ ( y ′ ) β \beta β γ \gamma γ β ⩾ α \beta \geqslant \alpha β ⩾ α γ ⩾ α \gamma \geqslant \alpha γ ⩾ α f β α ( x ) = f β α ( y ) f_{\beta\alpha}(x) = f_{\beta\alpha}(y) f β α ( x ) = f β α ( y ) f γ α ′ ( x ′ ) = f γ α ′ ( y ′ ) f'_{\gamma\alpha}(x') = f'_{\gamma\alpha}(y') f γ α ′ ( x ′ ) = f γ α ′ ( y ′ ) δ ∈ I \delta \in \mathrm{I} δ ∈ I δ ⩾ β \delta \geqslant \beta δ ⩾ β δ ⩾ γ \delta \geqslant \gamma δ ⩾ γ f δ α ( x ) = f δ α ( y ) f_{\delta\alpha}(x) = f_{\delta\alpha}(y) f δ α ( x ) = f δ α ( y ) f δ α ′ ( x ′ ) = f δ α ′ ( y ′ ) f'_{\delta\alpha}(x') = f'_{\delta\alpha}(y') f δ α ′ ( x ′ ) = f δ α ′ ( y ′ )
双射 g g g
设*( F α , g β α ) (\mathrm{F}_\alpha, g_{\beta\alpha}) ( F α , g β α ) ( F α ′ , g β α ′ ) (\mathrm{F}'_\alpha, g'_{\beta\alpha}) ( F α ′ , g β α ′ ) α ∈ I \alpha \in \mathrm{I} α ∈ I u α : E α → F α u_\alpha : \mathrm{E}_\alpha \to \mathrm{F}_\alpha u α : E α → F α u α ′ : E α ′ → F α ′ u'_\alpha : \mathrm{E}'_\alpha \to \mathrm{F}'_\alpha u α ′ : E α ′ → F α ′ ( u α ) (u_\alpha) ( u α ) ( u α ′ ) (u'_\alpha) ( u α ′ ) 是映射中的两个直接系统。那么 ( u α × u α ′ ) (u_\alpha \times u'_\alpha) ( u α × u α ′ )
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l i m → ( u α × u α ′ ) = ( l i m → u α ) × ( l i m → u α ′ ) . \varinjlim (u_\alpha \times u'_\alpha) = (\varinjlim u_\alpha) \times (\varinjlim u'_\alpha). lim ( u α × u α ′ ) = ( lim u α ) × ( lim u α ′ ) . 我们将验证留给读者。
习题
参见习题为 § 7 。
[^1]: This means that u α ∘ f α β = g α β ∘ u β u_\alpha \circ f_{\alpha\beta} = g_{\alpha\beta} \circ u_\beta u α ∘ f α β = g α β ∘ u β