On approximate inverse systems and resolutions by
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Having this in mind, in 1981 the author introduced resolutions [3] (see also [6]), which can be viewed as special cases of inverse limits and behave in the non-compact case just as inverse limits behave in the compact case. A resolution of a space X consists of an inverse system of spaces and mappings X = (X a , p aa0
An approximate inverse system was defined in [7] (see also [5]) as a collection X = (X a , U a , p aa0
(A1) (p aa1
(A2) For any a ∈ A and any U ∈ Cov(X a ), there exists an a 0 ≥ a such that (p aa1
(A3) For any a ∈ A and any U ∈ Cov(X a ), there exists an a 0 ≥ a such that U a00
Definition 1. An approximate inverse system is a collection X = (X a , p aa0
(AS) For any a ∈ A and any U ∈ Cov(X a ), there exists an a 0 ≥ a such that (p aa00
(R1) For any polyhedron P , open covering V of P and mapping f : X → P , there exists an a ∈ A such that for any a 0 ≥ a there exists a mapping g : X a0
(R2) For any polyhedron P and open covering V of P , there exists a V 0 ∈ Cov(P ) such that, for any a ∈ A and mappings g, g 0 : X a → P which satisfy (gp a , g 0 p a ) ≺ V 0 , there exists an a 0 ≥ a such that (gp aa00
Theorem 1. Let X = (X a , p aa0
(ii) Y b = X s(b) , q bb0
we define the spaces Y b and the mappings q bb0
We will also define, for each pair c i ∈ C i , c i+1 ∈ C i+1 , c i < c i+1 , a covering V ci
(1) (p s(b)s(b1
b, b 1 ∈ B n , a 0 ∈ A, b ≤ b 1 , s(b 1 ) ≤ a 0 , (2) (p s(ci
i + 1 ≤ n, a 0 , a 00 ∈ A, s(c i+1 ) ≤ a 0 ≤ a 00 , (3) U b00
Finally, we require that for any c i ∈ C i , and any covering V ci
(4) c i < c i+1 , V ci
(5) c n+1 = (c n , c 0 n , V cn
where c n , c 0 n are different elements of C n and V cn
(7) V cn
(8) (p s(b)a0
(9) (p s(cn
Now define s(c n+1 ) = a. Formulas (8) and (9) show that (1) and (2) remain valid also for n + 1. Since each b 00 ∈ B n has only finitely many predecessors, it is possible to choose a sufficiently fine covering U b00
It is now easy to verify that X is a gauged approximate system. Indeed, condition (A1) follows from (1) by putting a 0 = s(b 2 ), because b ≤ b 1 ≤ b 2 implies s(b 1 ) ≤ s(b 2 ) = a 0 , p s(b)s(b1
c i+1 and V ci
P r o o f. We must show that g satisfies condition (AS). Consider any b ∈ B and U ∈ Cov(Y b ). Put a = s(b) and note that Y b = X s(b) . Since f has property (AS), there exists an a 0 ≥ a such that (f a , p aa00
P r o o f. Note that t is not assumed to be increasing, which makes the proof more difficult. We must verify condition (AS) for the collection of mappings (f a ). Consider any a ∈ A and U ∈ Cov(X a ). Choose a star- refinement V of U . Applying (A2) to p, we obtain an a 0 ≥ a such that (1) (p aa1
(2) (f a , p aa00
Indeed, let a 00 ≥ a 0 . Put b = t(a), b 00 = t(a 00 ), V 00 = (p aa00
(g b , q bb∗
(g b00
Note that s(b) = st(a) = a and put a ∗ = s(b ∗ ). Then q bb∗
(5) (f a , p aa∗
Now note that s(b 00 ) = st(a 00 ) = a 00 and therefore q b00
f a00
(7) (p aa00
(8) (p aa00
(13) (g b , q bb0
Now note that q ts(b)b0
(R2) ∗ For any polyhedron P and open covering V of P , there exists a V 0 ∈ Cov(P ) such that, for any a ∈ A and mappings g, g 0 : X a → P which satisfy (gp a , g 0 p a ) ≺ V 0 , there exists an a 0 ≥ a such that (gp aa0
P r o o f o f T h e o r e m 7. First assume that p is a resolution. We must show that q satisfies conditions (R1) and (R2). Let P be a polyhedron, V ∈ Cov(P ) and let f : X → P be a mapping. Since p satisfies (R1), there exists an a ∈ A such that, for any a 0 ≥ a, there is a mapping g : X a0
(1) (gp s(b)a00
(2) (gq bb00
In order to verify condition (R2) ∗ for p, consider again a polyhedron P and a covering V ∈ Cov(P ). Choose V 0 ∈ Cov(P ) by applying (R2) ∗ to q. We claim that V 0 satisfies (R2) ∗ also for p. Indeed, let a ∈ A and let g, g 0 : X a → P be mappings such that (gp a , g 0 p a ) ≺ V 0 . Choose b ∈ B so that s(b) = a. Then, by the choice of V 0 , there exists a b 0 ≥ b such that (gq bb0
(B1) For any U ∈ Cov(X), there exists an a ∈ A such that for any a 0 ≥ a there exists a W ∈ Cov(X a0
(B2) For any a ∈ A and W ∈ Cov(X a ), there exists an a 0 ≥ a such that p aa00
(B2) ∗ For any a ∈ A and W ∈ Cov(X a ), there exists an a 0 ≥ a such that p aa0
P r o o f o f L e m m a 2. That (B1) implies (B1) ∗ and (B2) implies (B2) ∗ is obvious. Now assume that p = (p a ) has property (B1) ∗ . Then for a given U ∈ Cov(X), there exist an a ∗ ∈ A and a V ∈ Cov(X a∗
≺ U . Therefore, if we put W = (p a∗
(p aa1
(p aa1
Put W ∗ = (p aa∗
(3) p a∗
(4) p aa∗
p aa0
p aa∗
Since p a∗
p aa∗
p aa00
p aa00
(6) p aa00
Since s : B → A is onto, there exists a b ∗ ∈ B such that s(b ∗ ) = a 0 . Now choose for b 0 any element of B such that b 0 ≥ b, b ∗ . If b 00 ≥ b 0 , we see that a 00 = s(b 00 ) ≥ a 0 and thus q bb00
(7) q bb00
Now assume that q has properties (B1) and (B2) of [7]. By Lemma 2, it suffices to prove that p has properties (B1) ∗ and (B2) ∗ . If U ∈ Cov(X), (B1) ∗ for q (which is equivalent to (B1) of [7] by [7, Remark 2.9]) yields a b ∈ B and a V ∈ Cov(Y b ) which satisfy (5). However, this yields the anal- ogous relation for a = s(b), which is property (B1) ∗ for p. Now let a ∈ A, V ∈ Cov(X a ). Choose b ∈ B so that s(b) = a. Then (B2) ∗ for q (which is equivalent to (B2) of [7] by [7, Remark 2.10]) yields a b 0 ≥ b such that q bb0
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