The standard resolution of the product of a compactum

Houston Journal of Mathematics
c 2010 University of Houston
Volume 36, No. 3, 2010
THE STANDARD RESOLUTION OF THE PRODUCT OF A
COMPACTUM AND A POLYHEDRON CONSISTS OF ANES
FOR METRIC SPACES
ˇ C
´
SIBE MARDESI
Communicated by Yasunao Hattori
Abstract. In order to study (strong) shape properties of the Cartesian
product X × P of a compact Hausdorff space X and a (non-compact) polyhedron P , the author has introduced in 2003 a resolution (inverse limit with
some additional properties) q : X × P → Y , where Y = (Yµ , qµµ0 , M ) is an
inverse system consisting of paracompact spaces Yµ , having the homotopy
type of polyhedra. In the present paper, it is proved that the spaces Yµ are
stratifiable k-spaces and absolute neighborhood extensors for metric spaces.
This makes it possible to use the homotopy extension property, for pairs of
metric spaces (X, A), A closed in X, and maps F : (A × I) ∪ (X × 0) → Yµ , a
result needed in various arguments concerning the shape of products X × P .
1. Introduction
In studying (strong) shape properties of a topological space Z (non-compact)
an important role is played by HPol-resolutions of Z. These are resolutions
r : Z → Z, where all terms Zν of the inverse system Z = (Zν , rνν 0 , N ) belong to
the class HPol of spaces having the homotopy type of polyhedra (cf. [14], [17]).
In order to study the shape of the Cartesian product X × P of a compact Hausdorff space X and a (non-compact) polyhedron P (CW-topology), the author has
introduced in [15] an HPol-resolution q : X × P → Y , called the standard resolution of X × P (see Section 2). It was proved in [15], Theorem 4, that the terms
Yµ of the inverse system Y = (Yµ , qµµ0 , M ) are (Hausdorff) paracompact spaces
2000 Mathematics Subject Classification. 54B35, 54C55, 54C56, 54E20.
Key words and phrases. Inverse system, inverse limit, resolution, direct limit, Cartesian
product, stratifiable space, absolute neighborhood extensor (ANE), absolute neighborhood retract (ANR), shape.
887
888
ˇ C
´
SIBE MARDESI
(hence, also normal spaces). Since X × P is paracompact (hence, also a topologically complete space), it follows that the standard resolution q is an inverse
limit satisfying two additional conditions (B1) and (B2), described in Section 2
(cf [14], Theorem 6.16).
In the present paper we will improve these results by showing that the spaces
Yµ are stratifiable k-spaces (Lemmas 4 and 5). Stratifiable spaces were introduced
in 1961 by J. Ceder, under the name of M3 -spaces [6]. They were named stratifiable by C.J.R. Borges in [1]. Ceder proved that metric spaces are stratifiable,
separable stratifiable spaces are Lindel¨
of and stratifiable spaces are paracompact
and perfectly normal ([6], Theorems 2.2 and 2.5). While non-locally finite polyhedra fail to be metrizable, Ceder proved that polyhedra (even CW-complexes)
are stratifiable spaces ([6], Corollary 8.6 and Theorem 2.2).
A pair of spaces (Z, B), where B is a closed subset of Z, is said to have the
homotopy extension property (HEP) with respect to a space Y , provided every
mapping F : (B × I) ∪ (Z × 0) → Y admits an extension H : (Z × I) → Y .
Dowker’s lemma asserts that this is the case, whenever Z is normal and there is a
neighborhood U of B in Z such that F admits an extension G : (U ×I)∪(Z ×0) →
Y ([11], Lemma IV.2.1). Consequently, if C is a class of normal spaces, having the
property that Z ∈ C implies Z × I ∈ C, it follows that every pair (Z, B), where
Z ∈ C, has HEP with respect to any space Y , which is an absolute neighborhood
extensor (ANE) for the class C.
In various arguments concerning the shape of X × P , one would like to be able
to use HEP for pairs (Z, B) with respect to terms Yµ of the standard resolution
of X × P . Consequently, it is important to detect classes C of normal spaces such
that Z ∈ C implies Z × I ∈ C and Yµ is an absolute neighborhood extensor (ANE)
for the class C.
In some situations the terms Yµ of the standard resolution Y are (non-compact)
polyhedra. In general, polyhedra are not ANEs for Lindel¨of spaces [7], let alone
ANEs for paracompact spaces [2]. Therefore, one cannot take for C neither of
these two classes. On the other hand, it was proved in 1952 by J. Dugundji
[8] that polyhedra are ANEs for metric spaces, abbreviated to ANE(M) (also see
[11], Theorem III,10.4). The main result of the present paper (Theorem 1) asserts
that also the spaces Yµ are ANE(M). It follows that one can take for C the class
M of metrizable spaces and one can use HEP for metrizable pairs (Z, B), B closed
in Z, and mappings F : (B × I) ∪ (Z × 0) → Yµ (see Corollary 3). This fact plays
an important role in some forthcoming work of the author [16]. In particular, it is
used in the proof of the following assertion. Let X be a compact metric space, let
P be a separable polyhedron and let Z be a metric space. Then, for every strong
THE STANDARD RESOLUTION
889
shape morphism F : Z → X and every homotopy class of mappings [g] : Z → P ,
there exists a strong shape morphism H : Z → X × P such that S[πX ]H = F and
S[πP ]H = S[g]. Here S : H(Top)→SSh(Top) denotes the strong shape functor,
while πX : X × P → X and πP : X × P → P denote the canonical projections.
Note that the class S of stratifiable spaces is contained in the class of normal
spaces and has the property that Z ∈ S implies Z × I ∈ S, because a product
of two (even countably many) stratifiable spaces is a stratifiable space (see [6],
Theorem 2.4 or [13], Lemma 2.6, for a detailed proof). R. Cauty proved that
polyhedra (even CW -complexes) are ANEs for stratifiable spaces ([5], Theorem
2.3 and Corollary 1.5) (for CW-complexes see [4], Theorem 8). However, we
do not know if the spaces Yµ are also ANEs for stratifiable spaces and thus, at
present, in the homotopy extension theorem we cannot take for C the class of
stratifiable spaces.
2. The standard resolution of a product X × P
A resolution of a space Z consists of an inverse system Z = (Zν , rνν 0 , N ) of
spaces and a mapping r = (rν , N ) : Z → Z such hat the following conditions (B1)
and (B2) are satisfied ([17], I.6.3, Corollary 1 or [14], Theorem 6.7).
(B1) For every normal covering U of Z, there are a ν ∈ N and a normal
covering Uν of Zν such that the covering rν−1 (Uν ) refines U.
(B2) For every ν ∈ N and every normal covering Uν of Zν , there exists a ν 0 ≥ ν
such that
(1)
rνν 0 (Zν 0 ) ⊆ St(rν (Z), Uν ).
If the spaces Zν are completely regular (Tychonoff) and the space Z is topologically complete (e.g., paracompact), then resolutions are inverse limits ([14],
Theorem 6.16). Every limit r : Z → Z, where all Zν are compact Hausdorff
spaces, is a resolution ([14], Theorem 6.20). Every topological space admits a
resolution formed by polyhedra ([14], Theorem 6.22). Every compact Hausdorff
space is the limit of an inverse system of compact polyhedra ([17], I.5,Theorem
2).
The standard resolution q : X × P → Y of the Cartesian product X × P of
a compact Hausdorff space X and a polyhedron P is defined as follows. One
chooses an inverse system of compact polyhedra X = (Xλ , pλλ0 , Λ) with inverse
limit p = (pλ , Λ) : X × P → X and one chooses a triangulation K of P . To
define the indexing set M in the inverse system Y = (Yµ , pµµ0 , M ) one orders the
simplicial complex K by putting ζ ≤ σ, whenever ζ ∈ K is a face of the simplex
ˇ C
´
SIBE MARDESI
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σ ∈ K. Then M is the set of all increasing functions µ : K → Λ. An ordering ≤
on M is defined by putting µ ≤ µ0 , whenever µ(σ) ≤ µ0 (σ), for every σ ∈ K. To
define the spaces Yµ , µ ∈ M , one first associates with every σ ∈ K the Cartesian
product Xµ(σ) × σ. Then one considers the coproduct (disjoint sum)
a
(2)
Y˜µ =
(Xµ(σ) × σ).
σ∈K
By definition, Yµ is the quotient space
Yµ = Y˜µ /∼µ ,
(3)
where ∼µ denotes the equivalence relation generated by putting (x1 , t1 ) ∼µ (x2 , t2 ),
for points (xk , tk ) ∈ Xµ(σk ) × σk ⊆ Y˜µ , k = 1, 2, whenever σ1 ≤ σ2 , x1 =
pµ(σ1 )µ(σ2 ) (x2 ) and t2 = iσ1 σ2 (t1 ), where iσ1 σ2 : σ1 → σ2 denotes the inclusion
mapping (we will usually omit iσ1 σ2 and just write t1 = t2 ). The corresponding
quotient mapping is denoted by φµ : Y˜µ → Yµ .
In order to define the bonding mappings qµµ0 : Yµ0 → Yµ , µ ≤ µ0 , one first
defines mappings q˜µµ0 : Y˜µ0 → Y˜µ , by putting
a
(4)
q˜µµ0 =
(pµ(σ)µ0 (σ) × 1σ ).
σ∈K
Then qµµ0 : Yµ0 → Yµ is the only mapping for which
(5)
qµµ0 φµ0 = φµ q˜µµ0 .
To define the mappings qµ : X × P → Yµ , for µ ∈ M , one associates with every
simplex σ ∈ K the mapping pµ(σ) × 1 : X × σ → Xµ(σ) × σ. Putting
a
(6)
Y˜ =
(X × σ),
σ∈K
one defines mappings q˜µ : Y˜ → Y˜µ by the formula
a
(7)
q˜µ =
(pµ(σ) × 1).
σ∈K
One also considers the mapping φ : Y˜ → Y , defined by the requirement that
φ|(X × σ) is the inclusion mapping X × σ ,→ X × P . There exists a unique
mapping qµ : Y → Y˜µ such that
(8)
φµ q˜µ = qµ φ.
0
Since qµ = qµµ0 qµ0 , for µ ≤ µ , the mappings qµ : Y → Yµ , µ ∈ M , form a mapping
q = (qµ , M ) : Y → Y , which is the desired resolution q of X × P .
The following lemma gives an explicit description of the ∼µ -equivalence.
THE STANDARD RESOLUTION
891
Lemma 1. Points (xk , tk ) ∈ Xµ(σk ) × σk ⊆ Y˜µ , k = 1, 2, are ∼µ -equivalent if and
only if there exists a simplex τ ≤ σ1 , σ2 such that pµ(τ )µ(σ1 ) (x1 ) = pµ(τ )µ(σ2 ) (x2 )
and t1 = t2 ∈ τ .
Proof. Let us denote temporarily the relation described in the lemma by
≈µ and let us show that it is an equivalence relation. Only transitivity requires verification. Let (xk , tk ) ∈ Xµ(σk ) × σk ⊆ Y˜µ , k = 1, 2, 3, and let
(x1 , t1 ) ≈µ (x2 , t2 ) and (x2 , t2 ) ≈µ (x3 , t3 ). Then there exist simplices τ 0 , τ 00 ∈ K
such that τ 0 ≤ σ1 , σ2 , pµ(τ 0 )µ(σ1 ) (x1 ) = pµ(τ 0 )µ(σ2 ) (x2 ), t1 = t2 ∈ τ 0 and
τ 00 ≤ σ2 , σ3 , pµ(τ 00 )µ(σ2 ) (x2 ) = pµ(τ 00 )µ(σ3 ) (x3 ), t2 = t3 ∈ τ 00 . Consider the
point t = t1 = t2 = t3 and note that t belongs to the simplex τ = τ 0 ∩ τ 00 .
Since, τ ≤ τ 0 , τ 00 , one concludes that τ ≤ σ1 , σ3 . Moreover, pµ(τ )µ(σ1 ) (x1 ) =
pµ(τ )µ(τ 0 ) pµ(τ 0 )µ(σ1 ) (x1 ) = pµ(τ )µ(τ 0 ) pµ(τ 0 )µ(σ2 ) (x2 ) = pµ(τ )µ(σ2 ) (x2 ). Analogously,
pµ(τ )µ(σ2 ) (x2 ) = pµ(τ )µ(σ3 ) (x3 ) and thus, pµ(τ )µ(σ1 ) (x1 ) = pµ(τ )µ(σ3 ) (x3 ), which
shows that indeed, (x1 , t1 ) ≈µ (x3 , t3 ).
Let (xk , tk ) ∈ Xµ(σk ) × σk ⊆ Y˜µ , k = 1, 2. Let us first show that (x1 , t1 ) ∼µ
(x2 , t2 ) implies (x1 , t1 ) ≈µ (x2 , t2 ). It suffices to consider the case when σ1 ≤
σ2 , x1 = pµ(σ1 )µ(σ2 ) (x2 ) and t1 = t2 . Clearly, in this case (x1 , t1 ) ≈µ (x2 , t2 ).
Conversely, if (x1 , t1 ) ≈µ (x2 , t2 ), then there exists a simplex τ ≤ σ1 , σ2 such that
pµ(τ )µ(σ1 ) (x1 ) = pµ(τ )µ(σ2 ) (x2 ) and t1 = t2 ∈ τ . Putting x = pµ(τ )µ(σ1 ) (x1 ) and
t = t1 , we see that (x, t) ∈ Xµ(τ ) × τ ∈ Y˜µ and (x, t) ∼µ (x1 , t1 ). Analogously,
since x = pµ(τ )µ(σ2 ) (x2 ) and t = t2 , we see that (x, t) ∼µ (x2 , t2 ). Consequently,
(x1 , t1 ) ∼µ (x2 , t2 ).
Let K (n) denote the n-th skeleton of K and let
a
(9)
Y˜µ(n) =
(Xµ(σ) × σ).
σ∈K (n)
(n)
(n)
(n)
Note that Y˜µ is an open and closed subset of Y˜ . Put Yµ = φµ (Y˜µ ). Clearly,
(n+1)
(n)
(n+1)
(n)
(n)
.
, it follows that Yµ ⊆ Yµ
Yµ ⊆ φµ (Y˜µ ) = Yµ . Since Y˜µ ⊆ Y˜µ
(n)
Lemma 2. Yµ is a closed subset of Yµ , n ∈ N. The space Yµ is the colimit of
(1)
(0)
the direct sequence Yµ ,→ Yµ ,→ . . ..
Proof. Let us first show that


(10)
(n)
˜ (n)
φ−1
µ (Yµ ) = Yµ
[
[

(Xµ(σ) × σ (n) ) ,
σ∈K\K (n)
where σ (n) denotes the n-skeleton of σ. Consider a point (x, t) ∈ Xµ(σ) × σ ⊆ Y˜µ .
First assume that (x, t) belongs to the left side of (10). If dim σ ≤ n, i.e., σ ∈ K (n) ,
ˇ C
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SIBE MARDESI
892
(n)
(n)
then Xµ(σ) ×σ ⊆ Y˜µ and thus, (x, t) ∈ Y˜µ also belongs to the right side of (10).
(n)
(n)
Now assume that dim σ > n, i.e., σ ∈ K\K (n) . Since φµ (x, t) ∈ Yµ = φµ (Y˜µ ),
there exist a simplex σ1 ∈ K (n) and a point (x1 , t1 ) ∈ Xµ(σ1 ) × σ1 such that
φµ (x, t) = φµ (x1 , t1 ), i.e., (x, t) ∼µ (x1 , t1 ). By Lemma 1, there exists a simplex
τ ∈ K such that τ ≤ σ, σ1 , pµ(τ )µ(σ) (x) = pµ(τ )µ(σ1 ) (x1 ) and t = t1 ∈ τ . Since
dim σ1 ≤ n and τ ≤ σ1 , it follows that dim τ ≤ n and thus t ∈ τ ⊆ σ (n) .
Consequently, (x, t) ∈ Xµ(σ) × σ (n) and we see that in this case too, (x, t) belongs
to the right side of (10).
To prove the opposite inclusion, first assume that (x, t) belongs to the first
(n)
(n)
summand of the right side of (10), i.e., (x, t) ∈ Y˜µ . Then φµ (x, t) ∈ φµ (Y˜µ ) =
(n)
(n)
Yµ and thus, (x, t) ∈ φ−1
µ (Yµ ), i.e., (x, t) also belongs to the left side of (10).
Now assume that (x, t) ∈ Xµ(σ) × σ (n) , where dim σ > n. Then t belongs to a face
σ1 of σ of dimension dim σ1 ≤ n. Consequently, (pµ(σ1 )µ(σ) (x), t) ∈ Xµ(σ1 ) × σ1 ⊆
(n)
(n)
Y˜µ and thus, φµ (pµ(σ )µ(σ) (x), t) ∈ Yµ . Moreover, (pµ(σ )µ(σ) (x), t) ∼µ (x, t),
1
1
(n)
i.e., φµ (pµ(σ1 )µ(σ) (x), t) = φµ (x, t). It follows that φµ (x, t) ∈ Yµ , which implies
(n)
that (x, t) ∈ φ−1
µ (Yµ ), i.e., in this case too, (x, t) belongs to the left side of (10).
This completes the proof of formula (10).
Now recall that φµ : Y˜µ → Yµ is a quotient mapping. Therefore, to prove
(n)
(n)
that Yµ is a closed subset of Yµ , it suffices to prove that φ−1
µ (Yµ ) is a closed
subset of Y˜µ . Clearly, Xµ(σ) × σ (n) is a closed subset of Xµ(σ) × σ and thus,
S
`
(n)
) is a closed subset in the coproduct dim σ>n (Xµ(σ) × σ),
dim σ>n (Xµ(σ) × σ
(n)
hence also in Y˜µ . Since Y˜µ is also a closed subset of Y˜µ , we see that the right
(n)
˜
side of (10), hence also its left side φ−1
µ (Yµ ), is a closed subset of Yµ .
Let us now show that
(11)
Yµ =
[
Yµ(n) .
n∈N
If y ∈ Yµ , then y = φµ (x, t), where (x, t) ∈ Xµ(σ) × σ ⊆ Y˜µ . If dim σ = n, then
(n)
(n)
(n)
Xµ(σ) × σ ⊆ Y˜µ and thus, y = φµ (x, t) ∈ φµ (Y˜µ ) = Yµ , which shows that y
belongs to the right side of (11). The converse implication is obvious.
It remains to show that the topology of Yµ is generated by the topology of the
(n)
(n)
(n)
sets Yµ , i.e., if U ⊆ Yµ is a set such U ∩ Yµ is open in Yµ , for every n ∈ N,
(n) ˜ (n)
(n)
(n)
(n)
then U is open in Yµ . Denote by φµ : Yµ → Yµ the restriction φµ = φµ |Y˜µ .
Let us first show that, for an arbitrary subset U ⊆ Yµ , one has
(12)
˜ (n) = (φµ(n) )−1 (U ∩ Yµ(n) ).
φ−1
µ (U ) ∩ Yµ
THE STANDARD RESOLUTION
893
Indeed, if (x, t) belongs to the left side of (12), then φµ (x, t) ∈ U and since also
(n)
(n)
(n)
(x, t) ∈ Y˜µ , one has φµ (x, t) = φµ (x, t) and thus, φµ (x, t) ∈ U . It follows that
(n)
(n)
(n)
(n)
(x, t) ∈ (φµ )−1 (U ). Moreover, φµ (x, t) = φµ (x, t) ∈ φµ (Y˜µ ) = Yµ and thus,
(n) −1
(n)
(n) −1
(n)
(n) −1
(n) −1
(n)
(x, t) ∈ (φµ ) (Yµ ). Since (φµ ) (U ∩ Yµ ) = (φµ ) (U ) ∩ (φµ ) (Yµ ),
(x, t) is contained in the right side of (12). Conversely, if (x, t) belongs to the
(n)
(n)
right side of (12), then φµ (x, t) = φµ (x, t) and thus, φµ (x, t) ∈ U ∩ Yµ ⊆ U ,
(n)
(n)
˜
˜
hence also (x, t) ∈ φ−1
µ (U ). Moreover, (x, t) ∈ Yµ , because Yµ is the domain
(n)
(n)
of φµ . It follows that (x, t) ∈ φ−1 (U ) ∩ Y˜µ , i.e., it belongs to the left side of
µ
(12). This completes the proof of formula (12).
(n)
(n)
Now assume that U ⊆ Yµ is a set such that U ∩ Yµ is open in Yµ , for
˜
every n ∈ N. If we show that φ−1
µ (U ) is open in Yµ , the desired conclusion that
U is open in Yµ will follow from the fact that φµ is a quotient mapping. Since
(n)
(n)
(n)
(n)
(n)
(n)
φµ : Y˜µ → Yµ is a mapping, the set (φµ )−1 (U ∩ Yµ ) is open in Y˜µ .
(n)
(n)
˜
Therefore, (12) implies that the set φ−1
is open in Y˜µ . However,
µ (U ) ∩ Yµ
(n)
(n)
Y˜µ is open in Y˜µ and thus, φ−1 (U ) ∩ Y˜µ is open also in Y˜µ . Consequently,
µ
−1
˜
˜ (n)
˜ (n)
the union ∪n∈N (φ−1
µ (U ) ∩ Yµ ) = φµ (U ) ∩ (∪n∈N Yµ ) is open in Yµ . Since
(n)
−1
˜
˜
˜
∪n∈N (Yµ ) = Yµ and φµ (U ) ⊆ Yµ , we see that the latter union equals φ−1
µ (U ),
˜
which is thus, an open set in Yµ , as desired.
(n)
(n)
(n)
We will now describe the structure of the spaces Yµ . Let Z˜µ and A˜µ be
(n)
subsets of Y˜µ , defined as follows.
a
(13)
Z˜µ(n) =
(Xµ(σ) × σ).
dim σ=n
(14)
A˜(n)
µ =
a
(Xµ(σ) × ∂σ).
dim σ=n
(n−1)
(n)
(n)
(n)
(n)
be a mapping
Note that A˜µ is a closed subset of Z˜µ . Let fµ : A˜µ → Yµ
(n)
defined as follows. If dim σ = n and τ < σ is a proper face of σ, then fµ |(Xµ(σ) ×
τ ) is given by
(15)
fµ(n) (x, t) = φµ (pµ(τ )µ(σ) (x), t).
(n−1)
Note that, for (x, t) ∈ Xµ(σ) × τ , (pµ(τ )µ(σ) (x), t) ∈ Xµ(τ ) × τ ⊆ Y˜µ
and thus,
(n−1)
˜
(pµ(τ )µ(σ) (x), t) is a well-defined point of Yµ
. Therefore, φµ (pµ(τ )µ(σ) (x), t)
(n−1)
(n)
. To see that the mappings fµ |(Xµ(σ) × τ ),
(n)
(n)
(n−1)
τ < σ, determine a well-defined mapping fµ : A˜µ → Yµ
, we need to
(n)
(n)
show that the mappings fµ |(Xµ(σ) × τ1 ) and fµ |(Xµ(σ) × τ2 ) coincide on
is a well-defined point of Yµ
ˇ C
´
SIBE MARDESI
894
Xµ(σ) × ζ, where ζ = τ1 ∩ τ2 . Indeed, to (x, t) ∈ Xµ(σ) × ζ, the first of these
mappings assigns the value φµ (pµ(τ1 )µ(σ) (x), t) and the second one assigns the
value φµ (pµ(τ2 )µ(σ) (x), t). However, since ζ ≤ τ1 , τ2 , one has (pµ(τ1 )µ(σ) (x), t) ∼µ
(pµ(ζ)µ(τ1 ) pµ(τ1 )µ(σ) (x), t) = (pµ(ζ)µ(σ) (x), t). Analogously, (pµ(τ2 )µ(σ) (x), t) ∼µ
(pµ(ζ)µ(σ) (x), t) and thus, φµ (pµ(τ1 )µ(σ) (x), t) = φµ (pµ(τ2 )µ(σ) (x), t), as desired.
(n)
Lemma 3. For every n ∈ N, the space Yµ
coincides with the adjunction space
Yµ(n) = Z˜µ(n) tf (n) Yµ(n−1) .
(16)
µ
Proof. Recall that the adjunction space Z tf Y of a pair of spaces (Z, A), A
closed in Z, and a mapping f : A → Y is the quotient space Z t Y / ∼, where ∼ is
the equivalence relation associated with the decomposition of ZtY , which consists
of the following sets. The singletons {z} and {y}, where z ∈ Z\A and y ∈ Y \f (A)
and the sets {y} ∪ f −1 (y), where y ∈ f (A). We will denote by ψ : Z t Y → Z tf Y
the corresponding quotient mapping. It is well known that ψ|Y : Y → Z tf Y
is a closed embedding of Y into Z tf Y and ψ|(Z\A) is an embedding of Z\A
into Z tf Y . To simplify notations, we will identify points ψ(y) and y, for y ∈ Y .
Then, for z ∈ A, one has ψ(z) = ψf (z) = f (z), because f (z) ∈ Y . In our case
(n−1)
(n)
(n−1)
(n)
(n)
and
→ Z˜µ tf (n) Yµ
we consider the quotient mappings ψµ : Z˜µ t Yµ
µ
(n)
(n−1)
(n)
(n−1)
(n)
(n−1)
(n−1) ˜ (n)
= Y˜µ and
. Note that Z˜µ t Y˜µ
→ Z˜µ t Yµ
: Zµ t Y˜µ
1 t φµ
(n−1)
(n)
(n)
(n−1)
(n)
(n)
˜
˜
.
) : Yµ → Zµ t (n) Yµ
consider the quotient mapping χµ = ψµ (1 t φµ
fµ
(n)
(n)
We will show that the decomposition of Y˜µ induced by χµ coincides with the
(n)
(n)
(n)
(n)
decomposition of Y˜µ induced by φµ : Y˜µ → Yµ . It is then clear that the
corresponding quotient spaces coincide, i.e., (16) holds. Note that it suffices
(n)
(n)
(n)
(n)
to show that, for points y1 , y2 ∈ Y˜µ , φµ (y1 ) = φµ (y2 ) implies χµ (y1 ) =
(n)
(n)
χµ (y2 ), because in that case, every equivalence class, determined by φµ , is
(n)
contained in an equivalence class, determined by χµ , which implies that the
corresponding classes coincide.
Assume that yi = (xi , ti ) ∈ Xµ(σi ) × σi , where dim σi ≤ n, i = 1, 2, and
(n)
(n)
φµ (y1 ) = φµ (y2 ), i.e., (y1 ) ∼µ (y2 ). Then, by Lemma 1, there is a simplex
ζ ≤ σ1 , σ2 such that
(17)
pµ(ζ)µ(σ1 ) (x1 ) = pµ(ζ)µ(σ2 ) (x2 ), t1 = t2 ∈ ζ.
(n)
(n)
For symmetry reasons, it suffices to prove that χµ (y1 ) = χµ (y2 ) in the following
(n)
(n)
(n)
(n)
(n−1)
four cases: (i) y1 ∈ Z˜µ \A˜µ , (ii) y1 , y2 ∈ A˜µ , (iii) y1 ∈ A˜µ and y2 ∈ Y˜µ
,
(n−1)
˜
(iv) y1 , y2 ∈ Yµ
.
THE STANDARD RESOLUTION
895
(n)
(n)
Case (i). Since y1 = (x1 , t1 ) ∈ Z˜µ \A˜µ , it follows that t1 ∈ Int (σ1 ). Since
also t1 ∈ ζ ≤ σ1 , ζ cannot be a proper face of σ1 and thus, ζ = σ1 . Now
ζ ≤ σ2 implies σ1 ≤ σ2 . However, σ1 cannot be a proper face of σ2 , because
this would imply that dim σ1 < dim σ2 ≤ n, but dim σ1 = n. Consequently,
ζ = σ1 = σ2 . Now (17) shows that x1 = x2 and thus, y1 = y2 , which implies
(n)
(n)
χµ (y1 ) = χµ (y2 ).
Case (ii). In this case dim σi = n and there exist proper faces τi < σ1
(n)
(n−1)
such that ti ∈ τi , i = 1, 2. Since yi ∈ A˜µ , we see that (1 t φµ
)(yi ) = yi
(n)
(n)
(n)
and thus, χµ (yi ) = ψµ (yi ), i = 1, 2. Since yi = (xi , ti ) ∈ A˜µ , one has
(n)
(n)
(n−1)
ψµ (yi ) = fµ (xi , ti ) = φµ
(pµ(τi )µ(σi ) (xi ), ti ), i = 1, 2. Now note that
(n−1)
φµ
(n−1)
(pµ(τi )µ(σi ) (xi ), ti ) = φµ
(pµ(ζ)µ(τi ) pµ(τi )µ(σi ) (xi ), ti ), i = 1, 2.
(n)
(n−1)
pµ(ζ)µ(τi ) pµ(τi )µ(σi ) = pµ(ζ)µ(σi ) , we see that ψµ (yi ) = φµ
(n)
Since
(pµ(ζ)µ(σi ) (xi ), ti ),
(n)
i = 1, 2. Consequently, by (17), χµ (y1 ) = χµ (y2 ).
Case (iii). In this case dim σ1 = n, dim σ2 ≤ n − 1 and there exist a
proper face τ1 < σ1 and a face τ2 ≤ σ2 such that ti ∈ τi , i = 1, 2. More(n−1)
(n−1)
(n−1)
(y2 ). Therefore,
)(y2 ) = φµ
)(y1 ) = y1 and (1 t φµ
over, (1 t φµ
(n)
(n−1)
(n−1) (n)
(n)
(n)
(pµ(τ1 )µ(σ1 ) (x1 ), t1 ) and χµ (y2 ) =
fµ (y1 ) = φµ
χµ (y1 ) = ψµ (y1 ) = φµ
(n−1)
(n−1)
(n) (n−1)
(y) was identified with y, for points
(y2 ), because ψµ
(y2 ) = φµ
ψµ φµ
(n−1)
(n−1)
(n−1)
(x2 , t2 ).
(pµ(τ1 )µ(σ1 ) (x1 ), t1 ) = φµ
. It remains to show that φµ
y ∈ Yµ
(n−1)
As in (ii), φµ
(n−1)
(pµ(τ1 )µ(σ1 ) (x1 ), t1 ) = φµ
(n−1)
(x2 , t2 )
φµ
(pµ(ζ)µ(σ1 ) (x1 ), t1 ). On the other
(n−1)
(pµ(ζ)µ(σ2 ) (x2 ), t2 ).
φµ
Now (17) shows that the two
=
hand,
expressions do coincide.
(n−1)
(n−1)
(n−1)
(yi ) =
(yi ) and thus, χµ
)(yi ) = φµ
Case (iv). In this case (1 ∪ φµ
(n)
(n)
(n−1)
(n−1) (n−1)
(yi ) = φµ (yi ) and since φµ (y1 ) =
(yi ), i = 1, 2. However, φµ
φµ
ψµ
(n−1)
(n−1)
(n)
(y1 ).
(y1 ) = φµ
φµ (y1 ), it follows that also φµ
Remark 1. Lemmas 2 and 3 were already established in [15]. However, we believe
that the new proofs given here are more transparent.
(n)
Lemma 4. The spaces Yµ , n ≥ 0, and Yµ are stratifiable.
(n)
Proof. Let us first show that the spaces Yµ , n ∈ N, are stratifiable. By (13),
(n)
Z˜µ is a coproduct of metrizable spaces. Therefore, it is metrizable, hence also
(0)
(0)
(0)
stratifiable. In particular, Yµ = Y˜µ = Z˜µ is stratifiable. Assume that we
(n−1)
, n ≥ 1, is stratifiable. Borges proved that the
have already proved that Yµ
`
adjunction space Z f Y , where A ⊆ Z is closed and f : A → Y is a mapping,
is stratifiable, whenever Z and Y are stratifiable ([1], Theorem 6.2). Therefore,
(n)
(16) shows that also Yµ is stratifiable. It follows from a more general result
ˇ C
´
SIBE MARDESI
896
of Borges ([1], Theorem 7.2) that the colimit of an increasing sequence of closed
(n)
stratifiable spaces is a stratifiable space. Since the spaces Yµ are stratifiable,
Lemma 2 implies that also Yµ is stratifiable.
Remark 2. In general, if Z and Y are Hausdorff spaces, the adjunction space
Z tf Y need not be Hausdorff ([11], I, Proposition 2.2). However, if the spaces Z
and Y are (Hausdorff ) normal spaces, then so is Z tf Y ([10], Lemma 3.3).
Lemma 5. The spaces Yµ are k-spaces.
Proof. Recall that k-spaces, or compactly generated spaces, are spaces Z having
the property that a set U ⊆ Z is open in Z, provided U ∩ C is open in C, for
every compact subset C ⊆ Z. It is well known that compact Hausdorff spaces,
metric spaces and polyhedra are k-spaces. Moreover, the product of a locally
compact space and a k-space is a k-space ([9], 3.3.27). A coproduct of k-spaces
is a k-space ([9], 3.3.26). Since Y˜µ is a coproduct of metrizable spaces, it is a
k-space. It is also known that a quotient space of a k-space is a k-space, provided
it is a Hausdorff space ([9], Theorem 3.3.23). Since Yµ is a quotient space of Y˜µ
and Yµ is a Hausdorff space, it is also a k-space.
(n−1)
3. The inclusion Yµ
(n)
,→ Yµ
is a closed cofibration
Recall that an inclusion mapping A ,→ Z is a closed cofibration if A is a closed
subset of Z and the homotopy extension property holds for all topological spaces
Y , i.e., every mapping F : (A × I) ∪ (Z × 0) → Y extends to Z × I. It is known
([20], Theorem 2) that A ,→ Z is a closed cofibration if and only if the following
two conditions hold.
(i) A is a strong neighborhood deformation retract of Z, i.e., there exist an
open neighborhood U of A in Z and a homotopy H : U × I → Z such that, for
z ∈ U , H(z, 0) = z and H(z, 1) ∈ A, and for z ∈ A and t ∈ I, H(z, t) = z.
(ii) There is a mapping χ : Z → I such that χ−1 (0) = A and χ|(Z\U ) = 1.
(n−1)
Lemma 6. The inclusion mapping Yµ
(n)
,→ Yµ
is a closed cofibration.
(n)
(n−1)
(n)
Proof. We will first define an open neighborhood Uµ of Yµ
in Yµ , as
required by condition (i). Let bσ denote the barycenter of the simplex σ. Let
˜µ(n) ⊆ Y˜µ(n) denote the set
B
a
˜ (n) =
(18)
B
(Xµ(σ) × bσ ).
µ
dim σ=n
THE STANDARD RESOLUTION
897
˜µ(n) is a closed subset of Y˜µ(n) and
Clearly, B
a
a (n−1)
˜ (n) = Y˜ (n) \B
˜ (n) =
(19)
U
(Xµ(σ) × (σ\bσ ))
Y˜µ
µ
µ
µ
dim σ=n
(n)
is an open subset of Y˜µ . Note that bσ ∈ Int (σ). Therefore, by the argu(n)
ment given in the proof of Lemma 3, case (i), if (x, bσ ) ∈ Xµ(σ) × bσ ⊆ Y˜µ ,
(n)
(x0 , t0 ) ∈ Xµ(σ0 ) × σ 0 ⊆ Y˜µ and φµ (x, bσ ) = φµ (x0 , t0 ), then (x, bσ ) = (x0 , t0 ).
(n)
˜µ(n) ), we obtain a subset Bµ(n) ⊆ Yµ(n) such
Consequently, putting Bµ = φµ (B
(n) −1
(n)
(n)
(n)
(n)
(n)
˜µ . Since φ(n)
that (φµ ) (Bµ ) = B
= φµ |Y˜µ : Y˜µ → Yµ is a quoµ
˜µ(n) is closed in Y˜µ(n) , it follows that Bµ(n) is closed in Yµ(n) .
tient mapping and B
(n)
(n)
(n)
(n)
(n)
We now put Uµ = Yµ \Bµ . Clearly, Uµ is an open subset of Yµ and
(n)
(n)
(n)
−1
˜µ(n) . Since Y˜µ(n−1) ∩ B
˜µ(n) = ∅ and (φ(n)
˜µ(n) , it
(φµ )−1 (Uµ ) = U
(Bµ ) = B
µ )
(n−1)
(n)
(n)
(n−1)
follows that Yµ
∩ Bµ = ∅, i.e., Uµ is an open neighborhood of Yµ
in
(n)
Yµ .
(n−1)
(n)
(n)
. Every n-simplex can
We will now define a retraction ρµ : Uµ → Yµ
be viewed as the join bσ ∗ ∂σ, i.e., the cone with base ∂σ and vertex bσ . Let
ρ˜σ : (σ\bσ ) → ∂σ be the natural projection from bσ to the base ∂σ. Clearly, ρ˜σ is
a retraction and thus, also 1 × ρ˜σ : Xµ(σ) × (σ\bσ ) → Xµ(σ) × ∂σ is a retraction.
(n)
Now note that Xµ(σ) × ∂σ ⊆ A˜µ . Recall that in Section 2, we have defined a
(n−1)
(n)
(n−1)
(n) ˜(n)
. Therefore, fµ (1 × ρ˜σ ) : Xµ(σ) × (σ\bσ ) → Yµ
mapping fµ : Aµ → Yµ
(n)
is a well-defined mapping, which has the property that fµ (1× ρ˜σ )|(Xµ(σ) ×∂σ) =
(n)
(n)
fµ |(Xµ(σ) × ∂σ). These mappings define a mapping ρ˜µ of the first summand of
(n−1)
(n−1)
of (19),
. We extend this mapping to the second summand Y˜µ
(19) to Yµ
(n−1)
(n) ˜ (n)
(n−1)
(n) ˜ (n−1)
.
and thus, obtain a mapping ρ˜µ : Uµ → Yµ
= φµ
by putting ρ˜µ |Yµ
(n)
(n)
(n−1)
such that
Let us show that there is a unique mapping ρµ : Uµ → Yµ
(n)
(n) ˜ (n) ˜ (n)
(n) (n) ˜ (n)
is
a
quotient
→
U
:
U
|
U
).
Indeed,
since
φ
= ρµ (φµ |U
µ
µ
µ
µ
µ
˜µ(n)
mapping, it suffices to verify that φµ (x1 , t1 ) = φµ (x2 , t2 ), for (xi , ti ) ∈ U
(n)
(n)
and (xi , ti ) ∈ Xµ(σi ) × σi , i = 1, 2, implies ρ˜µ (x1 , t1 ) = ρ˜µ (x2 , t2 ). As
in the proof of Lemma 3, we distinguish four cases. In case (i), (x1 , t1 ) ∈
Xµ(σ1 ) × (σ\∂σ1 ), dim σ1 = n and the assertion holds, because, as seen above,
(x1 , t1 ) = (x2 , t2 ). In case (ii), (xi , ti ) ∈ Xµ(σi ) × ∂σi , dim σi = n, i = 1, 2,
(n)
ρ˜µ
(n)
(n)
(n)
and then, as seen above, ψµ (x1 , t1 ) = ψµ (x2 , t2 ). However, ψµ (xi , ti ) =
(n)
(n)
(n)
(n)
fµ (xi , ti ) and ρ˜µ (xi , ti ) = fµ (1 × ρ˜σ )(xi , ti ) = fµ (xi , ti ), i = 1, 2 and
(n)
(n)
thus, indeed, ρ˜µ (x1 , t1 ) = ρ˜µ (x2 , t2 ). In case (iii), (x1 , t1 ) ∈ Xµ(σ1 ) × ∂σ1 ,
(n)
dim σ1 = n, and (x2 , t2 ) ∈ Xµ(σ2 ) × ∂σ2 , dim σ2 = n − 1. Therefore, ρ˜µ (x1 , t1 ) =
ˇ C
´
SIBE MARDESI
898
(n)
(n)
(n)
fµ (1 × ρ˜σ )(x1 , t1 ) = fµ (x1 , t1 ). We saw above that fµ (x1 , t1 ) = (x2 , t2 ) and
(n)
(n)
(n−1)
thus, ρ˜µ (x1 , t1 ) = (x2 , t2 ). On the other hand, ρ˜µ (x2 , t2 ) = φµ
(x2 , t2 ) =
(n)
(n)
(x2 , t2 ). Consequently, again ρ˜µ (x1 , t1 ) = ρ˜µ (x2 , t2 ). In case (iv), (xi , ti ) ∈
(n−1)
(n)
(n−1)
(n)
Y˜µ
, i = 1, 2. Therefore, ρ˜µ (xi , ti ) = φµ
(xi , ti ) = φµ (xi , ti ) and thus,
(n)
(n)
(n)
(n)
φµ (x1 , t1 ) = φµ (x2 , t2 ) implies ρ˜µ (x1 , t1 ) = ρ˜µ (x2 , t2 ). Moreover, by defini(n)
(n−1)
(n−1)
(n)
(n−1)
(n−1)
(n)
(n−1)
tion, ρ˜µ |Y˜µ
= φµ
and also φµ |Y˜µ
= φµ
. Therefore, ρµ |Yµ
(n−1)
(n)
(n)
(n−1)
is the identity mapping on Yµ
and thus, ρµ : Uµ → Yµ
is indeed a
retraction.
(n)
(n)
(n)
Similarly, we can define a homotopy Rµ : Uµ ×I → Yµ , which connects the
(n)
(n)
(n)
(n−1)
inclusion mapping Uµ ,→ Yµ and the retraction ρµ and is fixed on Yµ
.
˜ σ : Xµ(σ) ×
For (x, t) ∈ Xµ(σ) × (σ\bσ ), where dim σ = n, we define a homotopy R
(n)
˜ σ (x, t, u) = (x, (1 − u)t + u˜
(σ\bσ ) × I → Xµ(σ) × σ ⊆ Y˜µ , by putting R
ρσ (t))).
Note that t belongs to the segment [˜
ρσ (t), bσ ] and thus, (1−u)t+u˜
ρσ (t)) is a wellσ
σ
˜
defined point of the segment [˜
ρ (t), t] ⊆ σ. Also note that R (x, t, 0) = (x, t) and
˜ σ (x, t, 1) = (x, ρ˜σ (t)). Moreover, if t ∈ ∂σ, then ρ˜σ (t) = t and thus, R
˜ σ (x, t, u) =
R
˜ σ is a homotopy rel (∂σ ×I). The mappings R
˜ σ define a
(x, t), which shows that R
(n−1)
(n)
(n) ˜ (n)
(n)
˜
˜
˜
× I,
mapping Rµ : (Yµ \Bµ ) × I → Yµ . We extend this mapping to Yµ
(n−1)
(n−1)
(n)
˜
˜
× I. We thus obtain a
(y), for (y, u) ∈ Yµ
by putting Rµ (y, u) = φµ
˜µ(n) × I → Yµ(n) . An argument, similar to the above argument
˜ µ(n) : U
mapping R
(n)
(n)
(n)
(n)
(n)
for ρ˜µ and ρµ , shows that there is a unique mapping Rµ : Uµ × I → Yµ
(n)
˜ µ(n) = Rµ(n) (φ(n)
such that R
µ × 1). It is readily seen that Rµ (y, 0) = y and
(n−1)
(n)
(n)
(n)
(n)
× I.
Rµ (y, 1) = ρµ (y), for y ∈ Uµ , and Rµ (y, u) = y, for (y, u) ∈ Yµ
This completes the verification of condition (i).
To verify condition (ii) recall that stratifiable spaces are perfectly normal, i.e.,
they are normal and every closed subset is a Gδ -set. Moreover, if Z is a perfectly
normal space and A, B are closed disjoint subsets, then there exists a mapping
χ : Z → I such that A = χ−1 (0) and B = χ−1 (1) ([9], 1.5.19).
4. The spaces Yµ are ANEs for metrizable spaces
The main result of the present paper is the following theorem.
Theorem 1. The spaces Yµ are ANEs for metrizable spaces.
We will first prove the following lemma.
(n)
Lemma 7. The spaces Yµ
are ANEs for metrizable spaces.
The proof of this lemma is based on a theorem of D.M. Hyman ([12], Theorem
2.2), here stated as Proposition 1. To state that theorem, we need a definition.
THE STANDARD RESOLUTION
899
Definition 1. Let (Z, B) be a pair of spaces, where B is closed in Z. An open covering W of Z\B is called a semi-canonical covering for the pair (Z, B), provided
it has the following property: Every point b ∈ B and every open neighborhood U
of B in Z admit an open neighborhood V of b in Z such that, for every W ∈ W,
W ∩ V 6= ∅ implies W ⊆ U . A pair (Z, B) is called semi-canonical if it admits a
semi-canonical covering of Z\B.
Proposition 1. (Hyman’s theorem) Let (Z, B) be a semi-canonical pair and let
B be a strong neighborhood deformation retract of Z. If both spaces B and Z\B
are ANEs for metrizable spaces, then also Z is an ANE for metrizable spaces.
(0)
Proof of Lemma 7. The proof is by induction on n. If n = 0, then Yµ =
(0)
(0)
Y˜µ = Z˜µ is a polyhedron and thus, it is an ANE(M). Let us now assume that
(n−1)
(n)
Yµ
is an ANE(M). We will conclude that also Yµ is an ANE(M), by apply(n)
(n−1)
(n)
(n−1)
ing Hyman’s theorem to the pair (Yµ , Yµ
). To verify that Yµ \Yµ
is an
(n)
(n)
(n−1)
(n)
is homeomorphic to Z˜µ \A˜µ .
ANE(M), note that (16) shows that Yµ \Yµ
(n)
However, this is an open subset of Z˜µ , which is an ANE(M), because it is a
polyhedron and the property ANE(M) is inherited by open subsets. By property
(n−1)
is a strong neighborhood deformation retract
(i) in the proof of Lemma 6, Yµ
(n−1)
(n)
(n)
) is a semi-canonical
of Yµ . Consequently, it remains to show that (Yµ , Yµ
pair.
(n)
(n)
(n)
First note that (Z˜µ , A˜µ ) is a semi-canonical pair, indeed A˜µ is closed in
(n)
(n)
Z˜µ and Z˜µ is a metrizable space, because it is a coproduct of metrizable spaces.
However, it is a well-known elementary fact that metrizable pairs (Z, B), B closed
in Z, are semi-canonical ([12], Lemma 3.1). By Lemma 3.3 of [12], if (Z, A) is
(n−1)
(n)
(n−1)
)
, Yµ
semi-canonical, then so is (Z tf Y, Y ). It follows that (Z˜µ tf (n) Yµ
µ
(n)
(n−1)
is semi-canonical. However, by (16), this is just (Yµ , Yµ
).
Remark 3. In the 1972 paper [3] by R. Cauty one finds in Theorem 3 the statement that the adjunction space Z tf Y , induced by a mapping f : A → Y , is an
ANR for stratifiable spaces, provided the spaces Z, A and Y are ANRs for stratifiable spaces. Since stratifiable spaces are ANRs for stratifiable spaces if and only if
they are ANEs for stratifiable spaces ([1], Corollary 6.3), one readily comes to the
(n)
conclusion that the spaces Yµ are ANEs for stratifiable spaces. Unfortunately,
the proof of Theorem 3 in [3] is not correct, because it uses a lemma, which asserts
that stratifiable pairs are always semi-canonical. Counter-examples to this lemma
were constructed in 1977 by S. San-ou [19]. In 1978 A. Okuyama and Y. Yasui
ˇ C
´
SIBE MARDESI
900
showed that for T1 spaces X the pair (X × I, X × 0) is a semi-canonical pair if
and only if X is metrizable [18].
A class of spaces C is said to be weakly hereditary provided X ∈ C implies that
every closed subset A ⊆ X also belongs to C. In the proof of Theorem 1 we will
use the following theorem.
Theorem 2. Let Y1 ⊆ Y2 ⊆ . . . be a direct sequence of normal spaces with colimit
Y such that each Yi is a strong neighborhood deformation retract of Yi+1 . Let C
be a weakly hereditary class of normal spaces such that Z ∈ C implies Z × I ∈ C.
If all Yi are ANEs for the class C, then also Y is an ANE for the class C.
The proof of Theorem 2 uses the following lemma.
Lemma 8. Let C be a weakly hereditary class of normal spaces such that Z ∈ C
implies Z × I ∈ C. Let (X, A) be a pair of spaces, where A is a closed subset of
X and let f : A → Y be a mapping to a space Y . Let Y1 ⊆ Y2 be closed subsets of
Y and let Y1 be a strong neighborhood deformation retract of Y2 . If X ∈ C, Y2 is
normal and the spaces Y1 and Y2 are ANEs for the class C, then there is an open
set W1 in X and there is a mapping h1 : Cl (W1 ) ∪ A → Y such that
(20)
f −1 (Y1 ) ⊆ W1 ,
(21)
h1 (Cl (W1 )) ⊆ Y2 ,
(22)
h1 |A = f.
.
Proof. By the assumptions, there exists an open neighborhood U10 of Y1 in Y2 ,
which admits a strong deformation retraction to Y1 . Since Y2 is normal, there
exists an open neighborhood U1 of Y1 in Y2 such that Y1 ⊆ U1 ⊆ Cl (U1 ) ⊆ U10 .
Consequently, there is a strong deformation retraction of Cl (U1 ) to Y1 , i.e., there
is a homotopy G : Cl (U1 ) × I → Y2 such that, for y ∈ Cl (U1 ), G(y, 0) = y and
G(y, 1) ∈ Y1 , and for y ∈ Y1 and t ∈ I, G(y, t) = y. Note that the mapping
r : Cl (U1 ) → Y1 , given by r(y) = G(y, 1), is a retraction. Also note that V1 =
f −1 (U1 ) is an open set in A, which contains f −1 (Y1 ) and f (Cl (V1 )) ⊆ Cl f (V1 ) ⊆
Cl (U1 ). It follows that, rf |(Cl (V1 )) : Cl (V1 ) → Y1 is a well-defined mapping.
Note that Cl (V1 ) ∈ C, because it is a closed subset of the space X ∈ C. Since
Y1 is an ANE for the class C, the mapping rf |(Cl (V1 )) extends to a mapping
g : W → Y1 , where W is an open neighborhood of Cl (V1 ) in X.
THE STANDARD RESOLUTION
901
Since X is normal, there exists an open set W 0 in X such that Cl (V1 ) ⊆ W 0 ⊆
Cl (W 0 ) ⊆ W . We now consider the set (Cl (V1 )) × I) ∪ (Cl (W 0 ) × 1) and we
define a mapping H of that set to Y2 by putting H|(Cl (V1 ) × I) = G(f × 1)
and H(y, 1) = g(y), for y ∈ Cl (W 0 ). Clearly, H is well defined and continuous,
because y ∈ Cl (V1 ) implies G(f × 1)(y, 1) = rf (y) = g(y). Since Cl (W 0 ) is a
closed subset of X ∈ C, it follows that Cl (W 0 ) ∈ C and thus also Cl (W 0 ) × I ∈
C. By assumption, Y2 is an ANE for the class C. Consequently, the homotopy
extension theorem (see the Introduction) enables us to extend H to a mapping
H : Cl (W 0 )×I → Y2 . Since V1 = f −1 (U1 ) is open in A, there is a set W 00 , which is
open in X and has the property that W 00 ∩ A = V1 . Note that V1 ⊆ Cl (V1 ) ⊆ W 0
and V1 ⊆ W 00 and thus V1 ⊆ W 0 ∩ W 00 . Since f −1 (Y1 ) ⊆ f −1 (U1 ) = V1 , it follows
that f −1 (Y1 ) ⊆ W 0 ∩ W 00 . By normality of X, there exists an open set W1 such
that f −1 (Y1 ) ⊆ W1 ⊆ Cl (W1 ) ⊆ W 0 ∩ W 00 . Note that (20) is satisfied. Clearly,
Cl (W1 ) ∩ A ⊆ W 00 ∩ A = V1 = f −1 (U1 ) ⊆ f −1 (Y2 ). Finally, W1 ⊆ W 0 implies
that Cl (W1 ) ⊆ Cl (W 0 ). Therefore, putting h(x) = H(x, 0), for x ∈ Cl (W1 ), we
obtain a mapping h : Cl (W1 ) → Y2 , which has the property that
(23)
h|(Cl (W1 ) ∩ A) = f |(Cl (W1 ) ∩ A).
Indeed, if x ∈ Cl (W1 ) ∩ A ⊆ Cl (W1 ), then h(x) = H(x, 0). Since Cl (W1 ) ∩
A ⊆ V1 ⊆ Cl (V1 ), we see that H(x, 0) = G(f (x), 0). Moreover, since V1 =
f −1 (U1 ), it follows that f (x) ∈ U1 ⊆ Cl (U1 ) and thus, G(f (x), 0) = f (x). Finally,
we define the mapping h1 : (Cl (W1 ) ∪ A) → Y , by taking for h1 |Cl (W1 ) the
composition of h : Cl (W1 ) → Y2 with the inclusion Y2 ⊆ Y and by putting h1 |A =
f . Formula (23) shows that h1 is well defined. Clearly, conditions (21) and (22)
are fulfilled.
Proof of Theorem 2. Let (X, A) be a pair of spaces such that X ∈ C and A
is a closed subset of X. We must prove that, for any mapping f : A → Y , there
exist an open neighborhood W of A in X and a mapping h : W → Y such that
h|A = f . In order to define W and h we will construct, by induction on i ≥ 0, a
sequence of open sets Wi ⊆ X and a sequence of mappings hi : (Cl (Wi )∪A) → Y ,
i ≥ 0, such that
(24)
Wi ⊆ Wi+1 ,
(25)
hi+1 |Cl (Wi ) = hi |Cl (Wi ),
(26)
hi |A = f,
(27)
h−1
i (Yi+1 ) ⊆ Wi+1 ,
902
(28)
ˇ C
´
SIBE MARDESI
hi+1 (Cl (Wi+1 )) ⊆ Yi+2 .
We begin the induction by putting W0 = ∅ and h0 = f : A → Y . We then
apply Lemma 8 to obtain W1 and h1 : (Cl (W1 ) ∪ A) → Y . Now assume that
we have already defined W0 , . . . , Wi−1 and h0 , . . . , hi−1 in such a way that the
equalities (24)–(28) hold, whenever both sides are already defined. To obtain
Wi+1 and hi+1 we apply again Lemma 8, this time to X 0 = X, A0 = Cl (Wi ) ∪ A,
Y 0 = Y , f 0 = hi , Y10 = Yi+1 and Y20 = Yi+2 . We obtain an open set W10 ⊆ X 0 and
−1
a mapping h01 : Cl (W10 ∪ A0 ) → Y 0 such that f 0 (Y10 ) ⊆ W10 , h01 (Cl (W10 )) ⊆ Y20
0
0
0
0
and h1 |A = f . Putting Wi+1 = W1 and hi+1 = h01 , we see that Wi+1 is
an open set in X and hi+1 : (Cl (Wi+1 ) ∪ Cl (Wi ) ∪ A) → Y is a mapping such
that h−1
i (Yi+1 ) ⊆ Wi+1 , hi+1 (Cl (Wi+1 )) ⊆ Yi+2 and hi+1 |(Cl (Wi ) ∪ A) = hi .
Consequently, conditions (25), (27) and (28) are fulfilled. Moreover, by (28)
for i − 1, one has Wi ⊆ Cl (Wi ) ⊆ h−1
i (Yi+1 ) ⊆ Wi+1 ⊆ Cl (Wi+1 ) and thus,
Cl (Wi+1 ) ∪ Cl (Wi ) = Cl (Wi+1 ), which shows that the domain of the mapping
hi+1 is Cl (Wi+1 ∪ A), as desired. Moreover, Wi ⊆ Wi+1 and thus, also (24)
holds. Finally, hi+1 |(Cl (Wi ) ∪ A) = hi implies hi+1 |A = hi |A. However, by (26),
hi |A = f and thus, also hi+1 |A = f , which completes the proof of the induction
step. Note that x ∈ f −1 (Yi ) ⊆ A and (26) imply hi−1 (x) = f (x) ∈ Yi and thus,
by (27), x ∈ h−1
i−1 (Yi ) ⊆ Wi . This shows that
(29)
f −1 (Yi ) ⊆ Wi .
To complete the proof of Theorem 2, put W = ∪i Wi . Clearly, W is an open
set in X. Moreover, by (29), A = f −1 (Y ) = ∪i f −1 (Yi ) ⊆ ∪i Wi = W , which
shows that W is an open neighborhood of A in X. By (25), hi+1 |Wi = hi |Wi and
thus, there is a well-defined mapping h : W → Y such that h|Wi = hi |Wi . Since
hi |Wi is continuous, so is h|Wi . However, since the sets Wi are open subsets of
W , it follows that h is a mapping. Finally, h|(Wi ∩ A) = hi |(Wi ∩ A) and by
(26), hi |(Wi ∩ A) = f |(Wi ∩ A). Since A ⊆ W , it follows that ∪i (Wi ∩ A) =
(∪i Wi ) ∩ A = W ∩ A = A and thus, h|(Wi ∩ A) = f |(Wi ∩ A) implies h|A = f . Taking for C the class M of metrizable spaces, Theorem 2 yields the following
corollary.
Corollary 1. Let Y1 ⊆ Y2 ⊆ . . . be a direct sequence of metrizable spaces with
colimit Y such that each Yi is a strong neighborhood deformation retract of Yi+1 .
If all Yi are ANEs for the class M of metrizable spaces, then also Y is an ANE
for the class M.
THE STANDARD RESOLUTION
903
Taking for C the class S of stratifiable spaces, Theorem 2 yields the following
corollary.
Corollary 2. Let Y1 ⊆ Y2 ⊆ . . . be a direct sequence of stratifiable spaces with
colimit Y such that each Yi is a strong neighborhood deformation retract of Yi+1 .
If all Yi are ANRs for the class S of stratifiable spaces, then also Y is an ANR
for the class S.
Proof. By Corollary 6.3 of [1], a stratifiable space is an ANR for stratifiable
spaces if and only if it is an ANR for stratifiable spaces. Therefore, the spaces Yi
are ANE’s for stratifiable spaces. Applying Theorem 2 to the class S of stratifiabe
spaces, one concludes that Y is an ANE for stratifiable spaces. However, being
the colimit of a direct sequence of stratifiable spaces, Y is stratifiable [1], Theorem
7.2. Consequently, Y is an ANR for stratifiable spaces.
Proof of Theorem 1. By Lemma 2, Yµ is the colimit of the direct sequence
(0)
(1)
(n)
(n)
Yµ ,→ Yµ ,→ . . . of closed subsets Yµ of Yµ . By Lemma 4, the spaces Yµ are
(n)
stratifiable, hence they are normal spaces. By Lemma 6 (see condition (i)), Yµ
(n+1)
. Moreover, by Lemma 7,
is a strong neighborhood deformation retract of Yµ
(n)
the spaces Yµ are ANEs for metrizable spaces. Applying Theorem 2 for C = M,
one obtains the desired conclusion that Yµ ∈ANE(M).
Corollary 3. If (X, A) is a pair of spaces, where X is metrizable and A is closed
in X, then every mapping F : (A × I) ∪ (X × 0) → Yµ extends to a mapping
H : X × I → Yµ .
The proof follows from Theorem 1 and the comments about HEP made in the
Introduction.
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Received March 10, 2008
Revised version received July 17, 2008
Second revision received July 25, 2008
Department of Mathematics, University of Zagreb, P.O.Box 335, 10 002 Zagreb,
Croatia
E-mail address: smardes@math.hr