54 2 Fundamental Results on Infinite-Dimensional Manifolds
In the theory of Q-manifolds, proper maps and proper homotopies are important. Due to Proposition 1.3.1, a proper map between metrizable spaces is identical to a perfect map. For proper maps, the following proposition is simple but important:
[2.9.6] Proposition 2.1.17 Let U be a locally finite open cover ofY such that every U ∈ U has the compact closure in Y (so Y is locally compact).4 If a map f :X→Y isU-close to a proper mapg, thenf is also proper.
Sketch of Proof.For each compact setA ⊂Y, f−1(A)⊂g−1(st(A,Ucl)), where st(A,Ucl) is compact because of local finiteness ofUcl.
It is said that proper mapsf, g:X →Y areproperly homotopic(resp.
properly U-homotopic) if there is a proper homotopy (resp. a proper U -homotopy)h:X×I→Y from f tog. Then, we denote
f ≃pg (
resp.f ≃pUg) .
Applying Proposition 2.1.17 above, to a homotopyh:X×I→Y and a constant homotopyh0prX:X×I→Y, we have the following:
Proposition 2.1.18 LetU be a locally finite open cover of Y such that every U ∈ U has the compact closure in Y. Then, aU-homotopy h:X×I→Y is proper if h0:X→Y is proper. ⊓⊔
2.2 The Toru´nczyk Factor Theorem 55 Πℓf
1L={
x∈LN(∥x(i)∥)i∈N∈ℓf1} with the norm∥x∥1=∑
i∈N∥x(i)∥. The space Πℓf
1Lis called theℓf1-product.
The following can be proved in the same way as the completeness ofℓ1. Proposition 2.2.1 IfL is a Banach space, then theℓ1-product Πℓ1Lis also a Banach space.
Sketch of Proof.For each Cauchy sequence (xn)n∈Nin Πℓ1L, we havex∈LN such thatx(i) = limn→∞xn(i) for eachi∈Nbecause (xn(i))n∈N is Cauchy inL. Then, show thatx∈Πℓ1Land limn→∞∥xn−x∥1= 0.
Notice that the ℓ1-product and the ℓf1-product of R = (R,| · |) are ℓ1
and ℓf1, respectively. It is easy to construct a linear isometry from the ℓ1 -product of ℓ1(Γ) onto ℓ1(N×Γ), which sends theℓf1-product of ℓf1(Γ) onto ℓf1(N×Γ). Thus, the pair (
Πℓ1ℓ1(Γ),Πℓf
1ℓf1(Γ))
is linearly isometric to the pair (ℓ1(Γ), ℓf1(Γ)) if Γ is infinite.
Remark 2.1 More generally, for a sequence (Li)i∈Nof normed linear spaces Li= (Li,∥ · ∥i),i∈N, we can similarly define theℓ1-product
Πℓ1Li={
x∈LN ∑i∈N∥x(i)∥i<∞} , with the norm ∥|x∥|1 =∑
i∈N∥x(i)∥i This definition is available even if Li, i ∈ N, are F-normed spaces (cf. Sect. 1.4). Using ℓp instead of ℓ1, the ℓp -product and theℓfp-product can also be defined.
Recall thatX is called a uniform retractof L if there is a retraction r:L→X that is uniformly continuous atX, that is, for eachε >0, there is δ >0 such that for eachx∈X andy∈L,∥x−y∥< δimplies∥x−r(y)∥< ε
(cf. Sect. 5.8 in [GAGT]). The following is the main theorem of this section: [§5.8]
Theorem 2.2.2 (Toru´nczyk) Let X be a retract of a normed linear space L= (L,∥ · ∥)with a retractionr:L→X that is uniformly continuous atX. Then, X×Πℓf
1L≈Πℓf
1L. If X is∥ · ∥-complete, then (X×Πℓ1L, X×Πℓf
1L)
≈(
Πℓ1L,Πℓf
1L) .
To prove Theorem 2.2.2, we may assume that 0 ∈ X. Since the case X = {0} is obvious, we may also assume that X ̸= {0}. We introduce a modification to theℓ1-product. A functionv:R+→R+ issub-additiveif
v(s+t)6v(s) +v(t) for each s, t∈R+.
If v : R+ → R+ is a sub-additive homeomorphism, then v(0) = 0 and v is increasing (i.e., s < t⇒v(s)< v(t)). Moreover, the continuity at 0 implies
56 2 Fundamental Results on Infinite-Dimensional Manifolds
thattn→0⇒v(tn)→0). Then, for a norm normed linear spaceL= (L,∥·∥), we have anF-norm defined byv(∥x∥).
For a sub-additive homeomorphismv:R+→R+, we define Π(ℓ1,v)L={
x∈LN(2iv(∥x(i)∥))i∈N∈ℓ1
} and Π(ℓf
1,v)L={
x∈LN(2iv(∥x(i)∥))i∈N∈ℓf1}
=LNf ∩Π(ℓ1,v)L, and∥x∥v=∑
i∈N2iv(∥x(i)∥) for eachx∈Π(ℓ1,v)L(refer to Remark 2.1). For x, y∈Π(ℓ1,v)L, we have−x, x+y∈Π(ℓ1,v)Land
∥−x∥v=∥x∥v and ∥x+y∥v6∥x∥v+∥y∥v.
By induction, nx ∈ Π(ℓ1,v)L for every n ∈ N. Moreover, if 0 < t < 1, then tx∈Π(ℓ1,v)Land∥tx∥v<∥x∥v. For everya >0, choosen∈Nso thata < n.
Then, ax = (a/n)(nx) ∈ Π(ℓ1,v)L and (−a)x = −(ax) ∈ Π(ℓ1,v)L. Thus, Π(ℓ1,v)Lis a linear space. It should be remarked that∥ · ∥vis not a norm, but anF-norm (cf. Sect. 1.4), which induces the linear metricd(x, y) =∥x−y∥v. We regard Π(ℓ1,v)L as a metric linear space with this metric. Then, each projection prn : Π(ℓ1,v)L→L is continuous.
Lemma 2.2.3 For any sub-additive homeomorphismsv:R+→R+,i∈N, (Π(ℓ1,v)L,Π(ℓf
1,v)L)
≈(
Πℓ1L,Πℓf
1L) .6
Proof. We define φ:(
Π(ℓ1,v)L, Π(ℓf
1,v)L)
→(
Πℓ1L, Πℓf
1L) and ψ:(
Πℓ1L, Πℓf
1L)
→(
Π(ℓ1,v)L, Π(ℓf
1,v)L) as follows:
φ(x)(i) =
2iv(∥x(i)∥)
∥x(i)∥ x(i) if x(i)̸=0,
0 if x(i) =0;
ψ(x)(i) =
v−1(2−i∥x(i)∥)
∥x(i)∥ x(i) if x(i)̸=0,
0 if x(i) =0.
6 More generally, replacing 2iv, i ∈ N, with sub-additive homeomorphisms vi : R+ → R+, we haveF-normed spaces Li = (L, vi(∥ · ∥)) and their ℓ1- and ℓf1 -product (see Remark 2.1). Then, the following also holds:
(Πℓ1Li, Πℓf 1Li)
≈(
Πℓ1L, Πℓf 1L)
.
2.2 The Toru´nczyk Factor Theorem 57 Since ∥φ(x)(i)∥ = 2iv(∥x(i)∥) and 2iv(∥ψ(x)(i)∥) = ∥x(i)∥, it follows that φ and ψ are well-defined, φψ = id, and ψφ = id. It remains to verify the continuity ofφand ψ.
(Continuity of φ) For eachx∈Π(ℓ1,v)Landε >0, choosem∈Nso that
∑
i>m2iv(∥x(i)∥) < ε/4. From the continuity of priφ, we have δ > 0 such that
y∈Π(ℓ1,v)L, ∥x−y∥v< δ ⇒ ∑
i<m
∥φ(x)(i)−φ(y)(i)∥< ε/4.
Fory∈Π(ℓ1,v)L, if∥x−y∥v< ε/4 then
∑
i>m
2iv(∥y(i)∥)6∑
i>m
2iv(∥x(i)∥) +∑
i>m
2iv(∥x(i)−y(i)∥)
< ε/4 +∥x−y∥v< ε/2.
Hence,∥x−y∥v <min{δ, ε/4}implies
∥φ(x)−φ(y)∥16∑
i<m
∥φ(x)(i)−φ(y)(i)∥
+∑
i>m
∥φ(x)(i)∥+∑
i>m
∥φ(y)(i)∥
< ε/4 +∑
i>m
2iv(∥x(i)∥) +∑
i>m
2iv(∥y(i)∥)
< ε/4 +ε/4 +ε/2 =ε.
(Continuity of ψ) For each x∈ Πℓ1L and ε > 0, choosem ∈ N so that
∑
i>m∥x(i)∥< ε/4. From the continuity of priψand 2iv, we haveδ >0 such that
y∈Πℓ1L, ∥x−y∥1< δ ⇒ ∑
i<m
2iv(∥ψ(x)(i)−ψ(y)(i)∥)< ε/4.
Fory∈Πℓ1L, if∥x−y∥1< ε/4 then
∑
i>m
∥y(i)∥6∑
i>m
∥x(i)∥+∑
i>m
∥x(i)−y(i)∥< ε/4 +∥x−y∥1< ε/2.
Hence,∥x−y∥1<min{δ, ε/4} implies
∥ψ(x)−ψ(y)∥v 6∑
i<m
2iv(∥ψ(x)(i)−ψ(y)(i)∥)
+∑
i>m
2iv(∥ψ(x)(i)∥) +∑
i>m
2iv(∥ψ(y)(i)∥)
< ε/4 +∑
i>m
∥x(i)∥+∑
i>m
∥y(i)∥
< ε/4 +ε/4 +ε/2 =ε.
The proof is completed. ⊓⊔
58 2 Fundamental Results on Infinite-Dimensional Manifolds
t=w(s) t=12s+12
1 s 0 γ
1 2
1 t
Fig. 2.4.Definition ofu
Recall thatr:L→X is a retraction that is uniformly continuous atX.
Lemma 2.2.4 There exists a sub-additive homeomorphismv:R+→R+ and γ >0such that the following condition (∗)is satisfied:
(∗)x∈X,y∈L,v(∥y∥)< γ ⇒ ∥y∥6v(∥y∥),v(∥x−r(x+y)∥)62v(∥y∥).
Proof. First, we definew:R+→R+ as follows:
w(s) = sup{
∥x−r(x+y)∥x∈X, ∥y∥6s} .
Then, w−1(0) = {0}. Indeed, w(0) = 0 is obvious. Recall that 0 ∈ X and X \ {0} ̸=∅. By the path-connectedness of X, for each s > 0, we can find y∈X\ {0}with∥y∥6s, whencew(s)>∥r(y)∥=∥y∥>0. Moreover, by the uniform continuity ofratX, it can be easily seen thatwis continuous at 0.
In particular, we haveγ ∈(0,1) such that w(s)<1/2 for 06s < γ. Let K be the convex hull of the following set inR2 (see Fig. 2.4):7
∪
s∈[0,γ)
({s} ×[0, w(s)])
∪(
{1} ×[0,1]) .
We defineu:R+→R+ as follows:
u(s) =
sup{
t|(s, t)∈K}
for 06s61, 1
2s+1
2 fors>1.
Then, u is a sub-additive homeomorphism. Indeed, by definition, we have u−1(0) ={0}. It is easy to see thatu|Iis strictly increasing and continuous,
7 Recall that the convex hull of a setAin a linear spaceEis denoted by⟨A⟩. Using this notation, we can write as follows:
K=⟨ ∪
s∈[0,γ)
({s} ×[0, w(s)])
∪(
{1} ×[0,1])⟩
.
2.2 The Toru´nczyk Factor Theorem 59 so it is a homeomorphism of I onto itself, hence u is a homeomorphism. It follows from the convexity of∪
s∈R+({s} ×[0, u(s)]) that u(s+t)
s+t s6u(s) and u(s+t)
s+t t6u(t),
henceu(s+t)6u(s) +u(t) for eachs, t∈R+. Moreover, observe that w(s)6u(s) for 06s < γ.
Sinces6u(s)61 for 06s61, it follows that
s6u(s)6u2(s)6· · ·61 for 06s61.
Sinces>u(s)>1 for s>1, it follows that
s>u(s)>u2(s)>· · ·>1 for s>1.
Now, we can define a homeomorphismv:R+ →R+ byv=∑
i∈N2−iui, whereu1=uandui+1=u◦ui. To see the condition (∗), letx∈X andy∈L withv(∥y∥)< γ(<1). Then ∥y∥6u(∥y∥)6v(∥y∥)< γ, so we have
∥x−r(x+y)∥6w(∥y∥)6u(∥y∥), which implies
v(∥x−r(x+y)∥)6v(u(∥y∥)) =∑
i∈N
2−iui+1(∥y∥)
= 2
∑∞ i=2
2−iui(∥y∥)62v(∥y∥).
This completes the proof. ⊓⊔
Using the sub-additive homeomorphismvwith the condition (∗), we define f :X×Π(ℓ1,v)L→Π(ℓ1,v)Las follows:
f(x, y) = (y(1) +x, y(2) +x−r(y(1) +x), . . . ,
y(i) +x−r(y(i−1) +x), . . .).
Then, the following is easily observed:
f(
X×Π(ℓf
1,v)L)
⊂Π(ℓf
1,v)L.
Lemma 2.2.5 The functionf is well-defined and continuous.
60 2 Fundamental Results on Infinite-Dimensional Manifolds
Proof. (Well-definedness off) Let (x, y)∈X×Π(ℓ1,v)L. We can takem∈N so thatv(∥y(i)∥)< γ fori>mbecause∑∞
i=12iv(∥y(i)∥)<∞. By condition (∗), we have
v(∥x−r(y(i−1) +x)∥)62v(∥y(i−1)∥) for eachi > m.
Then, it follows that 2v(∥y(1) +x∥) +
∑∞ i=2
2iv(∥y(i) +x−r(y(i−1) +x)∥)
62v(∥y(1) +x∥) +
∑m i=2
2iv(∥y(i) +x−r(y(i−1) +x)∥)
+
∑∞ i=m+1
2iv(∥y(i)∥) + 4
∑∞ i=m+1
2i−1v(∥y(i−1)∥)<∞, that is,f(x, y)∈Π(ℓ1,v)L.
(Continuity off) Let (x, y)∈X×Π(ℓ1,v)Landε >0. Sincey∈Π(ℓ1,v)L, we can choosem∈Nso that
v(∥y(i)∥)< γ/2 for each i>m and
∑∞ i=m
2iv(∥y(i)∥)< ε/30.
From the continuity of prif, we can find δ > 0 such that ∥x−x′∥ < δ and
∥y−y′∥v < δimply
2v(∥(y(1) +x)−(y′(1) +x′)∥) +
∑m i=2
2iv(∥(y(i) +x−r(y(i−1) +x))
−(y′(i) +x′−r(y′(i−1) +x′))∥)< ε/3.
Note that if∥y−y′∥v < γ/2 andi>m, then
v(∥y′(i)∥)6v(∥y(i)∥) +v(∥y(i)−y′(i)∥)< γ.
Now, assume that∥x−x′∥< δand ∥y−y′∥v<min{δ, γ/2, ε/15}. Then, by condition (∗), we can see that∥f(x, y)−f(x′, y′)∥v < εas follows:
2.2 The Toru´nczyk Factor Theorem 61
∥f(x, y)−f(x′, y′)∥v < ε/3 +
∑∞ i=m+1
2iv(∥y(i) +x−r(y(i−1) +x)∥)
+
∑∞ i=m+1
2iv(∥y′(i) +x′−r(y′(i−1) +x′)∥)
6ε/3 +
∑∞ i=m+1
2iv(∥y(i)∥) + 4
∑∞ i=m+1
2i−1v(∥y(i−1)∥)
+
∑∞ i=m+1
2iv(∥y′(i)∥) + 4
∑∞ i=m+1
2i−1v(∥y′(i−1)∥)
6ε/3 + 5
∑∞ i=m
2iv(∥y(i)∥) + 5
∑∞ i=m
2iv(∥y′(i)∥)
6ε/3 + 10
∑∞ i=m
2iv(∥y(i)∥) + 5
∑∞ i=m
2iv(∥y(i)−y′(i)∥)
< ε/3 +ε/3 +ε/3 =ε.
Therefore,f is continuous. ⊓⊔
By induction, we definegn: Π(ℓ1,v)L→L,n∈N, as follows:
g1(z) =z(1), g2(z) =z(2) +r(z(1)), . . . ,
gn(z) =z(n) +rgn−1(z), . . . , that is,g1= pr1andgn = prn+ r◦gn−1 forn >1.
Since each projection prn: Π(ℓ1,v)L→L is continuous, the lemma below can be proved by induction:
Lemma 2.2.6 Each gn: Π(ℓ1,v)L→Lis continuous. ⊓⊔ For eachz ∈Π(ℓf
1,v)L, there is some m∈Nsuch thatz(i) = 0 fori>m, hencegm(z) =rgm−1(z)∈X andgn+1(z) =rgn(z) =gn(z) forn>m. Thus, (gn|Π(ℓf
1,v)L)
n∈Nconverges tog′∞: Π(ℓf
1,v)L→X. WhenX is∥ · ∥-complete, g∞′ can be extended tog∞ : Π(ℓ1,v)L →X. In fact, by the next lemma,g∞
can be defined as follows:
g∞(z) = lim
i→∞rgi(z) for each z∈Π(ℓ1,v)L.
Lemma 2.2.7 For eachz∈Π(ℓ1,v)L,(rgn(z))n∈N is∥ · ∥-Cauchy.
Proof. For eachε >0, choosem∈Nso that
∑∞ i=m+1
2iv(∥z(i)∥)<min{ε, γ}.
62 2 Fundamental Results on Infinite-Dimensional Manifolds Then,v(∥z(i)∥)< γ fori > m. For eachn′> n>m,
v(∥rgn(z)−rgn′(z)∥)6v(∥rgn(z)−rgn+1(z)∥) +· · · +v(∥rgn′−1(z)−rgn′(z)∥)
=v(∥rgn(z)−r(z(n+ 1) +rgn(z))∥) +· · · +v(∥rgn′−1(z)−r(z(n′) +rgn′−1(z))∥) 62v(∥z(n+ 1)∥) +· · ·+ 2v(∥z(n′)∥)
6
∑∞ i=m+1
2iv(∥z(i)∥)<min{ε, γ}, hence we have
∥rgn(z)−rgn′(z)∥6v(∥rgn(z)−rgn′(z)∥)< ε.
Therefore, (rgn(z))n∈Nis∥ · ∥-Cauchy. ⊓⊔
From the proof of Lemma 2.2.7, the following can also be obtained:
Lemma 2.2.8 WhenX is∥ · ∥-complete, if v(∥z(i)∥)< γ fori > m, then
∥rgm(z)−g∞(z)∥6v(∥rgm(z)−g∞(z)∥) 62
∑∞ i=m+1
v(∥z(i)∥) (
6
∑∞ i=m+1
2iv(∥z(i)∥) )
. Now, we can prove the continuity ofg′∞ andg∞.
Lemma 2.2.9 The functiong∞′ is continuous. WhenX is∥ · ∥-complete,g∞
is also continuous.
Proof. Because of their similarity, we only prove the continuity of g∞. Let z∈Π(ℓ1,v)Landε >0. Choosem∈Nso that
v(∥g∞(z)−rgm(z)∥),
∑∞ i=m
2iv(∥z(i)∥)<min{
ε/4, γ} .
Then, ∥g∞(z)−rgm(z)∥ < ε/4 and v(∥z(i)∥) < γ/2 for i > m. Since the continuity of rgm follows from Lemma 2.2.6, we can find δ > 0 such that
∥z−z′∥v< δimplies∥rgm(z)−rgm(z′)∥< ε/4. If∥z−z′∥v< γ/2 andi>m, then
v(∥z′(i)∥)6v(∥z(i)∥) +v(∥z(i)−z′(i)∥)< γ.
On the other hand, if∥z−z′∥v< ε/4 then
∑∞ i=m+1
2iv(∥z′(i)∥)6
∑∞ i=m+1
2iv(∥z(i)∥) +∥z−z′∥v< ε/4 +ε/4 =ε/2.
2.2 The Toru´nczyk Factor Theorem 63 Hence, by Lemma 2.2.8, if∥z−z′∥v<min{δ, γ/2, ε/4} then
∥g∞(z)−g∞(z′)∥6∥g∞(z)−rgm(z)∥+∥rgm(z)−rgm(z′)∥
+∥g∞(z′)−rgm(z′)∥
< ε/4 +ε/4 +
∑∞ i=m+1
2iv(∥z′(i)∥)
< ε/2 +ε/2 =ε.
Therefore,g∞ is continuous. ⊓⊔ We defineg′: Π(ℓf
1,v)L→X×Π(ℓf
1,v)Las follows:
g′(z) = (g′∞(z), g1(z)−g′∞(z), . . . , gi(z)−g∞′ (z), . . .).
WhenX is∥ · ∥-complete,g: Π(ℓ1,v)L→X×Π(ℓ1,v)Lis defined by g(z) = (g∞(z), g1(z)−g∞(z), . . . , gi(z)−g∞(z), . . .).
Then,g is an extension ofg′.
Lemma 2.2.10 The function g′ is continuous. When X is∥ · ∥-complete, g is well-defined and continuous.
Proof. Since the continuity of g′ can be proved in the same way as for g, we only prove that g is well-defined and continuous. Evidently, it suffices to show the well-definedness and continuity of pg : Π(ℓ1,v)L→ Π(ℓ1,v)L, where p:X×Π(ℓ1,v)L→Π(ℓ1,v)L is the projection.
(Well-definedness of pg) For each z ∈ Π(ℓ1,v)L, choose m ∈ N so that
∑∞
i=m2iv(∥z(i)∥)< γ. Sincev(∥z(i)∥)< γ fori>m, it follows from Lemma 2.2.8 that
∑∞ i=m
2iv(∥gi(z)−g∞(z)∥)6
∑∞ i=m
2iv(∥gi(z)−rgi−1(z)∥)
+
∑∞ i=m
2iv(∥g∞(z)−rgi−1(z)∥)
6
∑∞ i=m
2iv(∥z(i)∥) + 2
∑∞ i=m
2i (∑∞
j=i
v(∥z(j)∥) )
< γ+ 2
∑∞ j=m
(∑j
i=m
2i )
v(∥z(j)∥)
6γ+ 4
∑∞ j=m
2jv(∥z(j)∥)< γ+ 4γ= 5γ <∞.
64 2 Fundamental Results on Infinite-Dimensional Manifolds Therefore,pgis well-defined.
(Continuity ofpg) For eachz∈Π(ℓ1,v)Landε >0, choosem∈Nso that
∑∞ i=m
2iv(∥z(i)∥)<min{
ε/20, γ} .
Then, v(∥z(i)∥)< γ/2 fori>m. Replacingγ by min{ε/20, γ} in the proof of the well-definedness ofpg above, we have
∑∞ i=m
2iv(∥gi(z)−g∞(z)∥)< ε/4.
If∥z−z′∥v< ε/20, then
∑∞ i=m
2iv(∥z′(i)∥)6
∑∞ i=m
2iv(∥z(i)∥) +
∑∞ i=m
2iv(∥z(i)−z′(i)∥)
< ε/20 +∥z−z′∥v< ε/10.
If∥z−z′∥v< γ/2 andi>m, then
v(∥z′(i)∥)6v(∥z(i)∥) +v(∥z(i)−z′(i)∥)< γ/2 +∥z−z′∥v< γ.
Thus, if∥z−z′∥v <min{ε/20, γ/2}, then
∑∞ i=m
2iv(∥z′(i)∥)< ε/10 and v(∥z′(i)∥)< γ for i>m.
By the same way as in the above, we have
∑∞ i=m
2iv(∥gi(z′)−g∞(z′)∥)< ε/2.
From the continuity of v, g∞ and gi−g∞, i = 1, . . . , m−1, there is some δ >0 such that∥z−z′∥v < δ implies
m−1∑
i=1
2iv(∥(gi(z)−g∞(z))−(gi(z′)−g∞(z′))∥)< ε/4.
Hence, if∥z−z′∥v<min{δ, ε/20, γ/2}then
∥pg(z)−pg(z′)∥v6
m−1∑
i=1
2iv(
∥(gi(z)−g∞(z))−(gi(z′)−g∞(z′))∥)
+
∑∞ i=m
2iv(∥gi(z)−g∞(z)∥) +
∑∞ i=m
2iv(∥gi(z′)−g∞(z′)∥)
< ε/4 +ε/4 +ε/2 =ε.
Therefore,pgis continuous. ⊓⊔
2.2 The Toru´nczyk Factor Theorem 65 Proof of Theorem 2.2.2. For eachx∈Π(ℓf
1,v)L,
f g′(x) =f(g∞′ (x), g1(x)−g∞′ (x), g2(x)−g′∞(x), . . .)
= (g1(x), g2(x)−rg1(x), g3(x)−rg2(x), . . .)
= (x(1), x(2), x(3), . . .) =x, that is,f g′ = id. For each (x, y)∈X×Π(ℓf
1,v)L,
g1(f(x, y)) =y(1) +x, g2(f(x, y)) =y(2) +x, . . . ,
so g′∞(f(x, y)) = gn(f(x, y)) = y(n) +x for sufficiently large n ∈ N. Since y(n) = 0 for sufficiently large n ∈ N, we have g∞′ (f(x, y)) = x. Then, it follows that
g′f(x, y) = (g∞′ (f(x, y)), g1(f(x, y))−g∞′ (f(x, y)),
g2(f(x, y))−g∞′ (f(x, y)), . . .)
= (x, y(1), y(2), . . .) = (x, y).
Hence,g′f|Π(ℓf
1,v)L= id. Thus, we haveX×Π(ℓf
1,v)L≈Π(ℓf
1,v)L.
IfX is∥ · ∥-complete, thenf g= id andgf= id because Π(ℓf
1,v)Lis dense in Π(ℓ1,v)L. In this case, we have
(X×Π(ℓ1,v)L, X×Π(ℓf
1,v)L)
≈(
Π(ℓ1,v)L,Π(ℓf
1,v)L) . Then, Theorem 2.2.2 follows from Lemma 2.2.3. ⊓⊔
As a corollary of Theorem 2.2.2, we have the following result on completely metrizable ARs, which is called theToru´nczyk Factor Theorem:
Theorem 2.2.11 (Toru´nczyk) For every non-degenerate completely metriz-able AR X, X×ℓ1(Γ)≈ℓ1(Γ), wheredensX = cardΓ. In general, for an arbitrary non-degenerate AR X, there exists a normed linear space L such that densX = densL andX×L≈L.
Proof. In the completely metrizable case, by virtue of Theorem 1.8.25, we have an admissible complete metricdofX such that (X, d) is a uniform AR.
By the Arens–Eells Embedding Theorem 1.8.2,X = (X, d) can be embedded isometrically into some Banach space L with densX = densL as a closed set, where densX is infinite because X is non-degenerate. Then, there is a retraction r:L→X that is uniformly continuous at X. By Theorem 2.2.2, X×Πℓ1L≈Πℓ1L. Since Πℓ1Lis a Banach space, we can apply Corollary 1.4.9 to obtain a Banach space F such that(
Πℓ1L)
×F ≈ℓ1(Γ), where cardΓ = dens Πℓ1L= densL= densX.
Hence, we have the result.
66 2 Fundamental Results on Infinite-Dimensional Manifolds
Without assuming the completeness ofX = (X, d),X can be embedded isometrically into some normed linear space L with densX = densL as a closed set. In this case, we have X×Πℓf
1L ≈Πℓf
1L. Thus, the general case also holds. ⊓⊔
As a corollary, we have the following:
Corollary 2.2.12 For an arbitrary infinite set Γ,ℓ1(Γ)N≈ℓ1(Γ).
Proof. Sinceℓ1(Γ)Nis a completely metrizable AR, we have ℓ1(Γ)N≈ℓ1(Γ)N×ℓ1(Γ)≈ℓ1(Γ).
Recall that the metrizable cone C(X) over a metrizable space X is the space {0} ∪ (X ×(0,1]) with the topology generated by open sets in the product space X×(0,1] and sets {0} ∪(X×(0, ε)), ε∈ (0,1). If X is compact, thenC(X) is homeomorphic to the quotient space (X×I)/(X×{0}), which is the usual cone. Themetrizable open coneoverX is the subspace {0} ∪(X×(0,1)) of C(X) and denoted byCo(X). It should be noted that if X is completely metrizable, then so are C(X) andCo(X). Moreover, the coneC(X) and the open coneCo(X) over an ANRX are ARs by Corollary 1.8.14.
By using the metrizable open cone, we can easily prove the following the-orem, which is also called the Toru´nczyk Factor Theorem:
Theorem 2.2.13 (Toru´nczyk) For every completely metrizable ANR X withw(X)6τ,X×ℓ1(Γ)is anℓ1(Γ)-manifold. In fact,X×ℓ1(Γ)is home-omorphic to an open set inℓ1(Γ).
Proof. As mentioned above, the open coneCo(X) is a completely metrizable AR and clearly w(Co(X))6τ. Then, Co(X)×ℓ1(Γ)≈ℓ1(Γ) by the above Theorem 2.2.11. Sinceℓ1(Γ)×(0,1)≈ℓ1(Γ),X×ℓ1(Γ) is homeomorphic to the open set (0,1)×X ×ℓ1(Γ) in Co(X)×ℓ1(Γ). Thus,X ×ℓ1(Γ) can be embedded inℓ1(Γ) as an open set, hence it is anℓ1(Γ)-manifold. ⊓⊔