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).
3.4 PL Maps and Simplicial Maps 159 3.4.2 Theorem LetKand Lbe cell complexes. A mapf :|K| → |L|is PL if and only if the graphG(f)off is a polyhedron.
Proof. To prove the “only if” part, letif :|K| →G(f) be the natural injec-tion and K′ be a subdivision of K such that f is affine on each cell in K′. Then, if(K′) ={if(C) |C ∈K′} is a cell complex (cf. 3.1.9(3)). Note that
|if(K′)|=G(f) as sets. For eachA⊂G(f),
Ais closed in G(f) ⇔ pr|K|(A) is closed in |K|=|K′|
⇔ ∀C∈K′, pr|K|(A)∩Cis closed inC
⇔ ∀C∈K′, A∩if(C) is closed inif(C)
⇔ Ais closed in|if(K′)|. Therefore,|if(K′)|=G(f) as spaces.
To prove the “if” part, letM be a cell complex with|M|=G(f). Then, pr|K|(M) = {pr|K|(D) | D ∈ M} is a cell complex (cf. 3.1.9(3)). Since pr|K||G(f) :G(f) =|M| → |pr|K|(M)|is a homeomorphism, it follows that
|pr|K|(M)|=|K|as spaces. By Theorem 3.2.11,Kand pr|K|(M) have a com-mon subdivisionK′. For eachC ∈K′,if(C) is a cell contained in some cell in M. Note that pr|K||if(C) is an affine homeomorphism, hence so is if|C.
Since pr|L|is affine on each cell in M, f = pr|L|◦if is affine on each cell in K′, that is,f is PL. ⊓⊔
3.4.3 Lemma For every PL mapf :|K| → |L|,Khas a subdivisionK′ such thatf is affine on each cellC∈K′ andf(K′)≺L(i.e., for each cellC∈K′, f(C)is contained in some cell inL).
Proof. By replacingK with a subdivision, we may assume thatf|C is affine for eachC∈K. According to Theorem 3.4.2, there is a cell complexM such that|M|=G(f), the graph off. By Proposition 3.2.12, the following is a cell complex:
M′=
C∩DC∈M, D∈K×cL such that C∩D=∅
≺M.
EachC∈M is covered by finitely many cells ofK×cLbecause it is compact in |K ×c L| by Lemma 3.4.1. Then, each cell of M is covered by finitely many cells of M′. Therefore,M′ is a subdivision ofM. We apply Theorem 3.2.11 to obtain a common subdivisionK′ ofKand pr|K|(M′). Observe that pr|L|(M′) ≺ L and f = pr|L|◦if, where if : |K| → G(f) is the natural injection. Then, we havef(K′)≺pr|L|(M′)≺L. ⊓⊔
Using Lemma 3.4.3, we can easily prove the following:
3.4.4 Proposition The composition of PL maps is also PL. ⊓⊔
Remark 3.In Proposition 3.3.4, if f and g are PL and h: |K| ×I→ |L| is the straight-line homotopy fromf to g, then eachht:|K| → |L|is PL. But, in general,h:|K×cI| → |L|is not PL. In fact, by Theorem 3.2.11,K has a subdivisionK′ such that bothf|Candg|Care affine for eachC∈K′. Then, ht|C is affine by definition. As an example of the straight-line homotopy h being non-PL, consider the affine mapsf, g:I→I2 defined byf(s) = (s,0) and g(s) = (0, s). In this case, the straight-line homotopy h is defined by h(s, t) = ((1−t)s, ts). Note that
h((1−t)(0,0) +t(1,1)) =h(t, t) = (t−t2, t2) for eacht∈I.
Any cell complexKwith|K|=I2has a cellCsuch thatA=C∩(0,0),(1,1) is a non-degenerate line segment. Then,h|Cis not affine.
Remark 4.It should be remarked that the image of a PL map is, in general, not a polyhedron. In fact, letf :R+ →I2 be the PL map defined as follows:
f(t) =
⎧⎪
⎨
⎪⎩
(t,0) ift∈I= [0,1],
(2−i+1, t−2i+ 1) ift∈[2i−1,2i], (2−i(2i+ 2−t),2i+ 1−t) ift∈[2i,2i+ 1].
Then, f(R+) is not a polyhedron. Indeed, if f(R+) =|L| for a cell complex L, thenf(2n−1),n∈ω, should be vertices ofL, which are contained in the compact setf(I).
f(0) f(1)
f(2)
f(3) f(4)
Fig. 3.3. The image of the PL mapf
LetK andLbe simplicial complexes. A functionf :|K| → |L|is called a simplicial mapfromKtoL(or with respect toKandL) iff|σis affine and f(σ)∈Lfor eachσ∈K, where dimf(σ)dimσ. Evidently,f(K(0))⊂L(0) andf(K) ={f(σ)|σ∈K}is a subcomplex ofL. Whenσ=v1, . . . , vn ∈K, we havef(σ) =f(v1), . . . , f(vn) ∈Land
f ni=1tivi
=
n
i=1
tif(vi) for eachti0 with
n
i=1
ti = 1,
3.4 PL Maps and Simplicial Maps 161 where it is possible thatf(vi) = f(vj) for some i=j. Every simplicial map f : |K| → |L| is PL, so it is continuous (Corollary 2.5.4). For a simplicial map from K to L, we may write f :K→L. In fact, although it is actually a function from|K|to|L|,f induces a function fromKtoLbecause f(σ)∈ L for each σ ∈ K. Note that the composition of simplicial maps and the restriction of a simplicial map to a subcomplex are also simplicial.
3.4.5 Proposition Let K and L be simplicial complexes. For a function f0:K(0) →L(0), the following are equivalent:
(a) f0 extends to a simplicial mapf :K→L;
(b) f0(σ(0)) ∈Lfor eachσ∈K;
(c)
v∈σ(0)OL(f0(v))=∅ for eachσ∈K.
In this case, the simplicial extensionf off0 is unique.
Proof. The implication (a)⇒(b) follows from the definition. By Proposition 3.2.3, we have (b)⇔ (c). It remains to show the implication (b)⇒ (a). For eachσ∈K, the functionf0|σ(0) uniquely extends to an affine mapfσ:σ→ f0(σ(0)) ⊂ |L|. Becausefσ|σ∩τ=fτ|σ∩τ for eachσ, τ ∈K, we can define f : |K| → |L| by f|σ =fσ. Then, f is simplicial with respect toK and L.
The uniqueness off follows from the uniqueness of fσ,σ∈K. ⊓⊔ With regard to subdivisions of simplicial maps, we have the following:
3.4.6 Proposition LetK and L be simplicial complexes andf :K →L a simplicial map. For each simplicial subdivisionL′⊳L, there exists a simplicial subdivisionK′⊳K such thatf :K′ →L′ is simplicial.
Proof. We define
K0={σ∩f−1(τ)|σ∈K, τ ∈L′, τ ⊂f(σ)}.
Then,K0is a cell complex subdividingK. Indeed, letσ∈Kandτ∈L′with τ ⊂f(σ) and x∈σ∩f−1(τ). By 3.1.9(3), (f|σ)−1(τ) =σ∩f−1(τ) is a cell with (σ∩f−1(τ))x=σx∩f−1(τf(x)). Since rintf(σx)∩rintτf(x)=∅, we have τf(x) ⊂ f(σx) because L′ ⊳L. Thus, K0 satisfies (C1). To show (C2′′), let σ, σ′ ∈Kandτ, τ′ ∈L′ withτ ⊂f(σ),τ′ ⊂f(σ′), and
x∈rint(σ∩f−1(τ))∩rint(σ′∩f−1(τ′))
⊂σ∩σ′∩f−1(τ∩τ′) . Since rintσx∩rintσ′x=∅and rintτf(x)∩rintτf(x)′ =∅, we haveσx=σ′xand τf(x)=τf(x)′ . Then, from 3.1.9(3), it follows that
σ∩f−1(τ) = (σ∩f−1(τ))x=σx∩f−1(τf(x))
=σ′x∩f−1(τf(x)′ ) = (σ′∩f−1(τ′))x=σ′∩f−1(τ′).
Therefore,K0satisfies (C2′′).
Letσ∈Kandτ∈L′ withτ⊂f(σ). Then,f(σ∩f−1(τ)) =f(σ)∩τ =τ.
For eachx∈σ∩f−1(τ), sinceτf(x)⊂f(σx) as seen in the verification of (C1) above, it follows that
f((σ∩f−1(τ))x) =f(σx∩f−1(τf(x))) =f(σx)∩τf(x)=τf(x). In particular, if v∈(σ∩f−1(τ))(0) then f(v)∈τ(0). Consequently, we have f(K0(0))⊂L′(0).
By Theorem 3.2.10, we have a simplicial subdivisionK′ ⊳K0 such that K′(0)=K0(0). Then,f(K′(0))⊂L′(0). For each simplexσ′ ∈K′, we haveσ∈ Kandτ∈L′ such thatτ ⊂f(σ),σ′⊂σ∩f−1(τ), andσ′(0) ⊂(σ∩f−1(τ))(0). Since f|σ′ is affine and f(σ′(0)) ⊂ τ(0), it follows that f(σ′) τ, hence f(σ′)∈L′. Thus,f :K′→L′ is simplicial. ⊓⊔
For homeomorphisms that are PL, we have the following:
3.4.7 Theorem Let K and L be cell complexes. If a homeomorphism f :
|K| → |L| is PL, then the inverse f−1 : |L| → |K| is also PL andK has a subdivision K∗ such that f|C : C → f(C) is an affine homeomorphism for eachC∈K∗andL∗={f(C)|C∈K∗} is a cell complex subdividingL.
Proof. Because the graph G(f) of f can be regarded as the graph of f−1 by changing the first and the second factors, the first assertion follows from Theorem 3.4.2.
Letif :|K| →G(f) andif−1 :|L| →G(f) be the natural injections, where if−1(y) = (f−1(y), y) for eachy∈ |L|. LetK′be a subdivision ofKsuch that f is affine on each cellC∈K′, andL′ be a subdivision ofLsuch thatf−1is affine on each cellD∈L′. As observed in the proof of Theorem 3.4.2,if(K′) and if−1(L′) are cell complexes with|if(K′)|=|if−1(L′)|=G(f) as spaces.
By virtue of Theorem 3.2.11,if(K′) andif−1(L′) have a common subdivision M. Then, K∗ = pr|K|(M) and L∗ = pr|L|(M) are subdivisions of K′ and L′, respectively. In addition,f(K∗) = pr|L|if(K∗) = pr|L|(M) =L∗, that is, L∗ ={f(C)|C ∈K∗}. For eachC∈K∗,f|C= pr|L|◦if|C :C →f(C) is an affine homeomorphism. ⊓⊔
A piecewise linear (PL) homeomorphism is literally defined as a homeomorphism being PL. Due to Theorem 3.4.7, the inverse of a PL home-omorphism is also a PL homehome-omorphism. For cell complexes K and L, the polyhedra|K|isPL homeomorphicto|L|if there exists a PL homeomor-phismf :|K| → |L|.
Remark 5.Every PL bijection between compact polyhedra is a PL homeo-morphism. However, a bijective PL mapf :|K| → |L|is, in general, not a PL homeomorphism. For example, definef :R+→∂I2as follows:
3.4 PL Maps and Simplicial Maps 163
f(t) =
⎧⎪
⎪⎪
⎨
⎪⎪
⎪⎩
(t,0) ift∈I= [0,1], (1, t−1) ift∈[1,2], (3−t,1) ift∈[2,3],
(0,2−n+2(n+ 2−t)) ift∈[n, n+ 1],n3.
Then,f is a PL bijection that is not a PL homeomorphism.
For simplicial complexesK andL, if a bijectionf :|K| → |L|is simplicial with respect to K andL, then the inversef−1 is also a simplicial map from L to K. A simplicial bijection is called a simplicial isomorphism. This is a homeomorphism, so it is also called a simplicial homeomorphism.
Obviously, a simplicial isomorphism is a PL homeomorphism. The inverse of a simplicial isomorphism and the composition of simplicial isomorphisms are also simplicial isomorphisms. It is said that K is simplicially isomorphic toL(denoted byK≡L) if there exists a simplicial isomorphismf :K→L.
Obviously,≡is an equivalence relation among simplicial complexes.
There exists a weaker equivalence relation among simplicial complexes. It is said that K iscombinatorially equivalent toL (denoted byK ∼=L) if they have simplicial subdivisions that are simplicially isomorphic. Then,∼= is an equivalence relation among simplicial complexes.
It is obvious that∼= is reflective and symmetric. To see that∼= is tran-sitive, let K1 ∼= K2 and K2 ∼= K3. Then, K1′ ≡ K2′ for some K1′ K1
and K2′ K2, and K2′′ ≡ K3′′ for some K2′′ K2 and K3′′ K3. By virtue of Corollary 3.2.13,K2′ andK2′′have a common simplicial subdivi-sionK2′′′, which induces K1′′′ K1′ and K3′′′ K3′′ such that K1′′′ ≡ K2′′′
andK3′′′≡K2′′′. Hence,K1′′′≡K3′′′, which meansK1∼=K3. Therefore,∼= is an equivalence relation among simplicial complexes.
This fact also follows from Theorem 3.4.8 below, Theorem 3.4.7, and Proposition 3.4.4.
3.4.8 Theorem Two simplicial complexes K and L are combinatorially equivalent to each other if and only if |K| and |L| are PL homeomorphic to each other, that is, there exists a PL homeomorphismf :|K| → |L|. Proof. If K ∼= L, then K and L have simplicial subdivisions K′ and L′, respectively, such that K′ ≡L′, hence there is a simplicial isomorphism f : K′→L′. Then,f :|K| → |L|is a PL homeomorphism.
Conversely, letf :|K| → |L|be a PL homeomorphism. By Theorem 3.4.7, there is a cell complexK′subdividingKsuch thatf|C:C→f(C) is an affine homeomorphism for eachC ∈K′ andL′={f(C)|C∈K′}is a subdivision of L. By Theorem 3.2.10, we have a simplicial subdivision K′′ of K′ with the same vertices. Then, L′′ ={f(σ) | σ ∈ K′′} is a simplicial subdivision of L. Observe thatf :K′′→L′′ is a simplicial isomorphism. Thus, we have K′′≡L′′, that is,K∼=L. ⊓⊔
For simplicial complexesK andL, the following implications are trivial:
K≡L⇒K∼=L⇒ |K| ≈ |L|.
Although it goes without saying that the converse of the first implication does not hold, the converse of the second does not either. It should be noted that
|K|=|L| impliesK∼=Lby Theorems 3.2.11 and 3.2.10. The converse of the second implication is calledHauptvermutung(the fundamental conjecture).
It took a long time to find finite simplicial complexesKandLsuch thatK∼=L but |K| ≈ |L|. It is known that this conjecture does not hold even if |K| and|L|aren-manifolds (i.e., there exists an n-manifold that has topological triangulationsK∼=L).9
Remark 6.By Theorem 3.4.8, it might be expected that every PL map f :
|K| → |L|is simplicial with respect to some simplicial subdivisions ofKandL.
However, this is not the case. For example, letKbe the natural triangulation of R+, that is, K =ω∪ {[n−1, n]| n ∈N}, and let L=I ={0,1,I}. We define a PL map f:|K| → |L| as follows:
f(2n) = 2−n−1 and f(2n+ 1) = 1−2−n−1 for eachn∈ω,
andf is affine on each [n, n+ 1]. Since every subdivision ofK containsω as vertices but every subdivision ofLhas only finitely many vertices, then f is not simplicial with respect to any simplicial subdivisions ofKandL.
In §3.6, it will be proved that every proper PL map f : |K| → |L| is simplicial with respect to some simplicial subdivisions ofKandL. According to Proposition 3.2.6, a (PL) map f : |K| → |L| is proper if and only if, for eachσ∈L, there is a finite subcomplex Kσ⊂K such thatf−1(σ)⊂ |Kσ|.