general, for a (topological) polyhedron P, w(P) = densP (cf. 3.2.16(4)). A triangulationof a polyhedron (or a topological polyhedron)P is a simplicial complexK such that|K| =P (or |K| ≈P). Then, it is also said that P is triangulated by K or K triangulates P. According to Theorem 3.2.10, every (topological) polyhedron has a triangulation.
3.3 Product Complexes and Homotopy Extension 153 finite. Then, for each neighborhoodW of (x, y) in|K×cL|, we can inductively choose open neighborhoodsUn ofxin |Kn|andVn ofyin |Ln| so that
clUn−1×clVn−1⊂Un×Vn⊂clUn×clVn⊂W∩ |Kn×cLn|, where clUn and clVn are compact. Observe that U =
n∈NUn and V =
n∈NVn are open neighborhoods ofxandy in|K|and|L|, respectively, and U×V ⊂W. Therefore,W is a neighborhood of (x, y) in|K| × |L|.
Case (2): As noted above, it can be assumed that L is finite. Let W be a neighborhood of (x, y) in |K×cL|. For eachD ∈ L, W ∩({x} ×D) is a neighborhood of (x, y) in{x}×DbecauseW∩(cK(x)×D) is also incK(x)×D.
Thus,W∩({x} × |L|) is a neighborhood of (x, y) in{x} × |L|. We can choose a neighborhood V ofy in|L|so that{x} ×clV ⊂W, where clV is compact because so is |L|. By induction onn ∈ ω, for each n-cellC ∈K[x], we can choose an open neighborhood UC of x in C so that clUC ×clV ⊂ W and UC ∩D = UD for each D ∈ F(∂C)[x]. In fact,
D∈F(∂C)[x]clUD ⊂ C is compact becauseF(∂C)[x] is finite. Then, we can find an open setUC′ inC such that
D∈F(∂C)[x]
clUD×clV ⊂UC′ ×clV ⊂clUC′ ×clV ⊂W.
Therefore,UC=UC′ \
D∈F(∂C)[x](D\UD) is the desired neighborhood. Now, letU =
C∈K[x]UC. Then,U is a neighborhood ofxin st(x, K) andU×V ⊂ W. Since st(x, K) is a neighborhood ofxin|K|,U is also a neighborhood of xin|K|. Hence,W is a neighborhood of (x, y) in |K| × |L|. ⊓⊔
We denoteI =F(I) (={0,1,I}), which is the cell complex with|I|=I.
It follows from Theorem 3.3.1 that |K| ×I = |K×cI| as spaces for every cell complex K. Due to the following proposition, the conditions given by Theorem 3.3.1 are essential.
3.3.2 Proposition There exist1-dimensional cell complexesK and Lwith cardK(0)= 2ℵ0 andcardL(0)=ℵ0such thatKis not locally countable,Lis not locally finite, and |K×cL| =|K| × |L|as spaces.
Proof. We defineK andLin the linear spacesRNN andRN as follows:
K={0,ea,0,ea |a∈NN} and L={0,ei,0,ei |i∈N},
whereea∈RNN andei ∈RNare the unit vectors (i.e.,eγ(γ) = 1 andeγ(γ′) = 0 forγ′=γ). For eacha∈NNandi∈N, let
ya,i=a(i)−1(ea,ei)∈ 0,ea × 0,ei ⊂RNN×RN.
Then, Y = {ya,i | a ∈NN, i ∈ N} is closed in|K×cL|, where it should be noted that (0,0)∈Y.
To see thatY is not closed in|K| × |L|, we show that (0,0)∈cl|K|×|L|Y. Let U be a neighborhood of 0 in |K|and V a neighborhood of 0 in |L|. For eacha∈NN, chooseεa>0 so that [0, εa]ea⊂U. For eachi∈N, chooseδi>0 so that [0, δi]ei⊂V. Then, we havea0∈NNsuch thata0(i)−1<min{δi, i−1} for eachi∈N. Choose i0∈Nso thati−01< εa0. Since a0(i0)−1< i−01 < εa0
and a0(i0)−1 < δi0, it follows that ya0,i0 ∈ U ×V, hence (U ×V)∩Y =∅. Thus, we have (0,0)∈cl|K|×|L|Y. ⊓⊔
For simplicial complexes K and L, the product complex K ×c L is not simplicial but does have a simplicial subdivision with the same vertices by Theorem 3.2.10. From the proof of 3.2.10, such a simplicial subdivision of K×cLcan be obtained by giving an order on (K×cL)(0) =K(0)×L(0) so thatK×cLis an ordered complex, that is, the set of vertices of each cell has the maximum. IfKandLare ordered simplicial complexes, we can define an order onK(0)×L(0) as follows:
(u, v)(u′, v′) if uu′ and vv′.
By this order,K×cLis an ordered cell complex. ByK×sL, we denote the simplicial subdivision of K×cL defined by using this order and call it the product simplicial complex of K and L. In fact, K×sL can be written as follows:
K×sL=
(u1, v1), . . . ,(uk, vk)∃σ∈K,∃τ∈L such that u1· · ·uk∈σ(0), v1· · ·vk∈τ(0)
. To prove that the above is the simplicial subdivision ofK×cLobtained by the procedure in the proof of Theorem 3.2.10, it suffices to show that the simplicial subdivision of then-skeleton (K×cL)(n)defined by this procedure can be written as follows:
¨
(u1, v1), . . . ,(uk, vk)
¬
¬∃σ×τ ∈(K×cL)(n) such that u1 · · · uk∈σ(0), v1 · · · vk∈τ(0)
©
.
We will show this by induction. According to the proof, the simplicial subdivision of (K×cL)(n+1) is defined as the simplicial complex consisting of the simplexes(u1, v1), . . . ,(uk, vk),(uk+1, vk+1), whereu1 · · · uk∈ σ0(0),v1 · · · vk∈τ0(0) for someσ0×τ0 ∈(K×cL)(n) and (uk+1, vk+1) is the maximum vertex ofσ×τ ∈K×L(n+1) containing the barycenter of(u1, v1), . . . ,(uk, vk). Sinceσ∩rintσ0=∅andτ∩rintτ0=∅, we have σ0 σandτ0 τ, henceu1 · · · uk uk+1∈σ(0)andv1 · · · vk
vk+1∈τ(0).
Conversely, consider the simplex (u1, v1), . . . ,(uk, vk), where u1
· · · uk ∈ σ(0), v1 · · · vk ∈ τ(0) for some σ×τ ∈ (K×cL)(n+1). We may assume thatk >1 and (uk, vk)= (uk−1, vk−1). Letσ′ be the face ofσ with the vertices u1, . . . , uk−1 andτ′ the face of τ with the vertices v1, . . . , vk−1. Then,σ′×τ′∈(K×cL)(n), hence(u1, v1), . . . ,(uk−1, vk−1)
3.3 Product Complexes and Homotopy Extension 155 is a simplex of the simplicial subdivision of (K×cL)(n) by the inductive assumption. Since the barycenter of(u1, v1), . . . ,(uk−1, vk−1)is contained in the cellσ×τ,(u1, v1), . . . ,(uk, vk)is a simplex of the simplicial subdi-vision of (K×cL)(n+1).
(u1, v1) (u1, v2)
(u2, v1) (u3, v1) (u3, v2)
(u2, v2)
Fig. 3.1.A product simplicial complex
We now consider the following useful theorem:
3.3.3 Theorem (Homotopy Extension Theorem) Let L be a subcom-plex of a cell comsubcom-plexK and h: |L| ×I→X a homotopy into an arbitrary spaceX. Ifh0 extends to a mapf :|K| →X, thenhextends to a homotopy
¯h: |K| ×I→ X with ¯h0 =f. Moreover, if his a U-homotopy for an open coverU ofX, then¯hcan be taken as aU-homotopy.
Proof. Leth:|L| ×I→X be aU-homotopy such that h0 extends to a map f :|K| →X. For eachn∈ω, we define
Kn=L∪K(n) and Pn= (K×c{0})∪(Kn×cI).
Then,KnandPn are subcomplexes ofKandK×cI, respectively. Moreover,
|Pn| is a closed subspace of |K| ×I that contains |L| ×Ias a subspace (cf.
Theorem 3.3.1). Moreover,|K| ×I=
n∈ω|Pn| has the weak topology with respect to the tower|P0| ⊂ |P1| ⊂ |P2| ⊂ · · ·.
We can define the mapg0:|P0| →X as follows:
g0(x, t) =
h(x, t) for (x, t)∈ |L| ×I,
f(x) for (x, t)∈(|K| × {0})∪(|K(0)| \ |L|)×I.
It is obvious that g0||K0| ×Iis a U-homotopy. Assume that we have maps gi:|Pi| →X,i < n, such thatgi||Pi−1|=gi−1andgi||Ki|×Iis aU-homotopy.
LetC∈K\Lbe ann-cell. By takingvC∈rintC, eachx∈Ccan be written as
x= (1−s)y+svC, y∈∂C, s∈I.
We can choose 0< δC<1 so that{gn−1(C(y))|y∈∂C} ≺ U, where
s 1 0 2δC
t 1
s=tδC
δC
y
vC
C x
y vC
0
x= (1−s)y+svC
Fig. 3.2.C×I
C(y) ={y} ×I∪ {((1−s)y+svC, 0)|0s2δC}. Then, we can define a map
rC:C×I→C× {0} ∪∂C×I⊂ |Pn−1| as follows:
rC((1−s)y+svC, t)
=
⎧⎪
⎪⎪
⎨
⎪⎪
⎪⎩
(1−s)y+svC, 0
if 2δCs1, 1−2(s−tδC)
2−t
y+2(s−tδC) 2−t vC, 0
iftδCs2δC, y, t−δ−C1s
if 0stδC. For eachx∈C,rC({x} ×I) ={(x,0)}orrC({x} ×I)⊂C(y), wherey∈∂C withx∈ y, vC. It follows that
{gn−1(rC({x} ×I))|x∈C} ≺ U.
Observe that rC|C× {0} ∪∂C ×I = id. We can extend gn−1 to a map gn :|Pn| →X defined bygn|C×I=gn−1rC for eachn-cellC∈K\L. Then, gn||Kn| ×Iis aU-homotopy.
By induction, we can obtain maps gn : |Pn| → X, n ∈ ω, such that gn||Pn−1|=gn−1 andgn||Kn| ×Iis aU-homotopy. The desired U-homotopy
¯h:|K| ×I→X can be defined by ¯h||Pn|=gn. ⊓⊔
For a cell complexL, two mapsf, g:X → |L|are said to becontiguous (with respect to L) iff, g areL-close, that is, for eachx∈X, there is some C∈Lsuch thatf(x), g(x)∈C.
3.3.4 Proposition LetK andLbe cell complexes. If two mapsf, g:|K| →
|L| are contiguous (with respect to L) then f ≃L g by the homotopy h :
|K| ×I→ |L| defined as follows:
h(x, t) = (1−t)f(x) +tg(x) for each(x, t)∈ |K| ×I.
3.3 Product Complexes and Homotopy Extension 157 Proof. Because f and g are contiguous, h is well-defined. We need to prove the continuity of h. Since |K| ×I= |K×cI|(Theorem 3.3.1), it suffices to show that h|C×Iis continuous for each C ∈ K. According to Proposition 3.2.6,f(C)∪g(C)⊂ |L0|for some finite subcomplex L0 ofL. For each pair (D1, D2)∈ L2, let L(D1, D2) be the subcomplex of L consisting of faces of the minimal cell of L containing D1 ∪D2, where L(D1, D2) = ∅ if L has no cells containing D1∪D2. Then, L1 =
(D1,D2)∈L20L(D1, D2) is a finite subcomplex of Lwith L0 ⊂L1 and h(C×I)⊂ |L1|. Note that the flat hull fl|L1| is finite-dimensional. Due to Proposition 2.5.1, fl|L1| has the unique topology for which the following operation is continuous:
fl|L1| ×fl|L1| ×I∋(x, y, t)→(1−t)x+ty∈fl|L1|.
The topology of |L1| is equal to the relative topology with respect to this topology. Indeed, due to Proposition 2.5.8, eachD∈L1has the unique topol-ogy for which the operation (x, y, t) → (1−t)x+ty is continuous. Hence, the inclusion |L1| ⊂ fl|L1| is continuous. By the compactness of |L1|, this inclusion is a closed embedding. Then, it follows thath|C×I:C×I→ |L1| is continuous. ⊓⊔
In Proposition 3.3.4,his called thestraight-line homotopy.
Remark 2.Using the same arguments, we can prove that Proposition 3.3.4 is valid even if |K| is replaced by a locally compact space X.8 It will be shown in§3.9 that every two contiguous maps defined on an arbitrary space are homotopic, where the homotopy is not always given by the straight-line homotopy. In fact, there are some cases where the straight-line homotopy is not continuous. Such an example can be obtained by reforming the example given in Proposition 3.3.2.
LetΓ =NN∪N. We defineK={0,eγ,0,eγ |γ∈Γ} and L=K∪
eγ,eγ′, 0,eγ,eγ′γ=γ′∈Γ ,
where eγ ∈ RΓ is the unit vector in RΓ (i.e., eγ(γ) = 1 andeγ(γ′) = 0 for γ′=γ). Then,K⊂Lare cell complexes inRΓ with dimK= 1 and dimL= 2. Let f, g : |K|2 → |L| be maps defined by f(x, y) = x and g(x, y) = y for each (x, y) ∈ |K|2, where |K|2 = |K ×cK| as spaces (see the proof of Proposition 3.3.2). Evidently, these maps f,g are contiguous. We can define h:|K|2×I→ |L|as follows:
h(x, y, t) = (1−t)f(x, y) +tg(x, y) = (1−t)x+ty.
We will prove that h is not continuous at (0,0,12) ∈ |K|2×I. For each a∈NNand i∈N, let
8More generally, it can be replaced by ak-spaceX. Indeed, to show the continuity of the straight-line homotopyh, it suffices to prove the continuity ofh|C×Ifor every compact setCinX.
va,i =12a(i)−1ea+12a(i)−1ei∈ 0,ea,ei ⊂ |L|.
As is easily observed,U =|L| \ {va,i|a∈NN, i∈N}is an open neighborhood of 0 =h(0,0,12) in|L|. Then,h(W2× {12})⊂U for any neighborhoodW of 0 in |K|. Indeed, for eachγ∈Γ, chooseδγ >0 so that [0, δγ]eγ ⊂W. We have a0 ∈NN such thata0(i)>max{i, δi−1} for everyi∈N. Takei0 ∈Nso that i0 > δa−01. Since a0(i0)−1 < i−01 < δa0, we have x0 = a0(i0)−1ea0 ∈ W. On the other hand, sincea0(i0)−1< δi0, we havey0=a0(i0)−1ei0 ∈W. Then, it follows that
h(x0, y0,12) = 12a0(i0)−1ea0+12a0(i0)−1ei0=vai0,i0 ∈U.
Therefore,his not continuous at (0,0,12).