Cotorsion-free groups and related local properties of ...curtkent/arches/fischer.pdf · Part 1:...

Part 1: Cotorsion-free groupsPart 2: Local Properties of π1

Cotorsion-free groups and related local propertiesof fundamental groups

Hanspeter Fischer (Ball State University)joint work with

Part 1: Katsuya Eda (Waseda University, Japan)Part 2: Jeremy Brazas (West Chester University)

Arches Topology ConferenceMoab, UT

May 10, 2018

Part 1: K. Eda, H. Fischer / Part 2: J. Brazas, H. Fischer Local properties of fundamental groups

Generalized Covering SpacesGeneralizations of Slenderness

Definition

Let X be connected and locally path connected, x ∈ X .We endow π1(X , x) with the coarsest topology such that

π1(X , x) is a topological group

H ⩽ π1(X , x) is open if (and only if) there is a coveringprojection p ∶ (X , x)→ (X , x) with p#π1(X , x) = H.

π1(X , x) has basis g ⋅ π(U , x) ∣ g ∈ π1(X , x), U open cover of Xπ(U , x) ⩽ π1(X , x) is generated by all elements [αβα−] withα ∶ ([0,1],0)→ (X , x), β ∶ ([0,1],0,1)→ (U, α(1)), U ∈ U .

U ∈ U

Definition

π1(X , x) has basis g ⋅ π(U , x) ∣ g ∈ π1(X , x), U open cover of X

π(U , x) ⩽ π1(X , x) is generated by all elements [αβα−] withα ∶ ([0,1],0)→ (X , x), β ∶ ([0,1],0,1)→ (U, α(1)), U ∈ U .

U ∈ U

Definition

π1(X , x) has basis g ⋅ π(U , x) ∣ g ∈ π1(X , x), U open cover of Xπ(U , x) ⩽ π1(X , x) is generated by all elements [αβα−] withα ∶ ([0,1],0)→ (X , x), β ∶ ([0,1],0,1)→ (U, α(1)), U ∈ U .

U ∈ U

Properties

(1) π(U , x) is a normal subgroup of π1(X , x).

(2) If V refines U , then π(V, x) ⩽ π(U , x).

(3) Spanier group: πs(X , x) =⋂Uπ(U , x) = 1.

(a) πs(X , x) = 1 ⇔ π1(X , x) is Hausdorff

(b) πs(X , x) is open ⇔ X has a universal covering space

(c) 1 is open ⇔ π1(X , x) is discrete

⇔ X is semilocally simply connected

⇔ X has a simply connected covering space

Properties

Example: The Hawaiian Earring H =∞⋃k=1

πs(H,∗) = 1, but ∀ U ∶ π(U ,∗) /= 1

π1(H,∗) is Hausdorff

H has no universal covering space

H has no simply connected covering space

Example: The Griffiths twin cone G = Cone(H) ∨ Cone(H).

πs(G,∗) = π1(G,∗) /= 1

π1(G,∗) is not Hausdorff (indiscrete)

G has a universal covering space (id ∶ G→ G)

G has no simply connected covering space

Theorem [F-Zastrow 2007]

For every path-connected space X , there is a generalized coveringprojection p ∶ X → X with p#π1(X , x) = πs(X , x).

If πs(X , x) = 1, then p ∶ X → X is a generalized universal covering.

Examples with πs(X , x) = 1 include:

1-dimensional spaces [Eda-Kawamura 1998]

subsets of surfaces [F-Zastrow 2005]

certain “trees of manifolds” [F-Guilbault 2005]

X = lim←Ð (Mn1 ←Mn

1 #Mn2 ←Mn

1 #Mn2 #Mn

3 ← ⋯)

1 #Mn2 ←Mn

1 #Mn2 #Mn

3 ← ⋯)

1 #Mn2 ←Mn

1 #Mn2 #Mn

3 ← ⋯)

Defintion

We call p ∶ X → X a generalized covering projection if

(1) X is path connected (pc) and locally path connected (lpc).

(X , x)p

(Y pc,lpc , y)

∃! f99

∀f // (X , x)

⇐⇒ f#π1(Y , y) ⩽ p#π1(X , x)

p ∶ X → X is characterized by p#π1(X , x) = H ⩽ G = π1(X , x)(up to equivalence by the conjugacy class of H in G ).

Aut(X p→ X ) ≈ NG(H)/HIf H = 1, then p ∶ X → X is a generalized universal covering.

p ∶ X → X is standardly constructed and fibers must be T1

(i.e. X must be homotopically Hausdorff relative to H)

Defintion

(X , x)p

(Y pc,lpc , y)

∃! f99

∀f // (X , x)

⇐⇒ f#π1(Y , y) ⩽ p#π1(X , x)

Defintion

(X , x)p

(Y pc,lpc , y)

∃! f99

∀f // (X , x)

⇐⇒ f#π1(Y , y) ⩽ p#π1(X , x)

Aut(X p→ X ) ≈ NG(H)/H

If H = 1, then p ∶ X → X is a generalized universal covering.

Defintion

(X , x)p

(Y pc,lpc , y)

∃! f99

∀f // (X , x)

⇐⇒ f#π1(Y , y) ⩽ p#π1(X , x)

Defintion

(X , x)p

(Y pc,lpc , y)

∃! f99

∀f // (X , x)

⇐⇒ f#π1(Y , y) ⩽ p#π1(X , x)

Definition

An abelian group A is called slender if for every homomorphismh ∶ ZN → A, ∃ n ∈ N ∀cn, cn+1, ⋅ ⋅ ⋅ ∈ Z: h(0, . . . ,0, cn, cn+1, . . . ) = 0.

Definition

A group G is called n-slender if for every homomorphismh ∶ π1(H,∗)→ G , ∃ n ∈ N ∀γ ⊆ Cn ∪ Cn+1 ∪⋯, we have h([γ]) = 1.

Examples: Free groups are n-slender [Higman, Griffiths 1952-56].Certain HNN extensions of n-slender groups are n-slender,including the Baumslag-Solitar groups [Nakamura 2015].

Theorems [Eda 1992, 2005]

(1) An abelian group A is n-slender ⇔ A is slender.

(2) A group G is n-slender ⇔ for every Peano continuum X andevery homomorphism h ∶ π1(X , x)→ G , ∃ U ∶ h(π(U , x)) = 1.

Definition

A group G is homomorphically Hausdorff relative to X iffor every homomorphism h ∶ π1(X , x)→ G , ⋂

Uh(π(U , x)) = 1

i.e. the image of h (viewed as a quotient map) is Hausdorff.

Examples: N-slender groups and residually n-slender groups arehomomorphically Hausdorff relative to every Peano continuum.

Examples of residually n-slender groups include π1 of

1-dimensional spaces

Planar sets

Pontryagin surface

∏2 = lim←Ð (P2 ← P2#P2 ← P2#P2#P2 ← ⋯)Pontryagin sphereX = lim←Ð (T 2 ← T 2#T 2 ← T 2#T 2#T 2 ← ⋯)

Definition

Uh(π(U , x)) = 1

Planar sets

Pontryagin surface

Definition

Uh(π(U , x)) = 1

Planar sets

Pontryagin surface

Definition

Uh(π(U , x)) = 1

Planar sets

Pontryagin surface

Theorem

Let A be an abelian group. Then A is cotorsion-free ⇔A is homomorphically Hausdorff relative to every Peano continuum.

Recall:An abelian group A is called cotorsion if(A ⩽ B abelian and B/A torsion-free )⇒ (B = A⊕ C for some C).

An abelian group A is called cotorsion-free if it does not containa non-zero cotorsion subgroup.

Facts:A is cotorsion-free ⇔ A is torsion-free, Q /⩽ A, Jp /⩽ A ∀ primes p.

A is slender ⇔ is cotorsion-free and ZN /⩽ A.

Theorem

Proposition

If G is residually n-slender, then G is homomorphically Hausdorffrelative to every Peano continuum X .

Proof: Let X be a Peano continuum, h ∶ π1(X , x)→ G a homom.G ⩽∏i∈I Gi with each Gi n-slender, pj ∶∏i∈I Gi → Gj projections.Since each Gi is homomorphically Hausdorff relative to X , we havepj(⋂U h(π(U , x))) ⩽ ⋂U pj h(π(U , x)) = 1 for all jHence, ⋂U h(π(U , x)) ⩽ ⋂j∈I ker(pj) = 1.

Corollary

If X is a Peano continuum with π1(X , x) residually n-slender,then X admits a generalized universal covering.

Proof: Consider id# ∶ π1(X , x)→ π1(X , x).Since π1(X , x) is homomorphically Hausdorff relative to X , wehave πs(X , x) = ⋂U id#(π(U , x)) = 1.

Proposition

Corollary

Proposition

Corollary

Proposition

Corollary

Switching Topologies π1(X , x)Test-maps and Closure Operators

Switch topologies: Start with the compact-open-topology on

Ω(X , x) = α ∶ ([0,1],0,1)→ (X , x).

Let πq1(X , x) be π1(X , x) with the quotient topology induced by

[⋅] ∶ Ω(X , x)→ π1(X , x).

Then πq1(X , x) is a quasi-topological group.

Theorem [Brazas-Fabel 2015]

Let H ⩽ π1(X , x).

1. If X is locally path connected, then H ⩽ πq1(X , x) is closed ⇔X is homotopically path-Hausdorff relative to H.

2. If X is homotopically path-Hausdorff relative to H, then Xadmits a generalized covering p ∶ X → X with p#π1(X , x) = H.

Switch topologies: Start with the compact-open-topology on

Ω(X , x) = α ∶ ([0,1],0,1)→ (X , x).

Let πq1(X , x) be π1(X , x) with the quotient topology induced by

[⋅] ∶ Ω(X , x)→ π1(X , x).

Then πq1(X , x) is a quasi-topological group.

Theorem [Brazas-Fabel 2015]

Let H ⩽ π1(X , x).

1. If X is locally path connected, then H ⩽ πq1(X , x) is closed ⇔X is homotopically path-Hausdorff relative to H.

2. If X is homotopically path-Hausdorff relative to H, then Xadmits a generalized covering p ∶ X → X with p#π1(X , x) = H.

Definition

X is homotopically path-Hausdorff relative to H ⩽ π1(X , x) if

∀ α,β ∶ ([0,1],0)→ (X , x) with α(1) = β(1) and [α ⋅ β−] /∈ H∃ 0 = t0 < t1 < ⋯ < tn = 1 and open subsets U1,U2, . . . ,Un withα([ti−1, ti ]) ⊆ Ui for i = 1,2, . . . ,n, such that

∀ γ ∶ [0,1]→ X with γ(ti) = α(ti) for i = 0,1, . . . ,n andγ([ti−1, ti ]) ⊆ Ui for i = 1,2, . . . ,n, we have [γ ⋅ β−] /∈ H.

Definition

Fix a “test space” (T, t) and T ⩽ π1(T, t), g ∈ π1(T, t).We say that H ⩽ π1(X , x) is (T ,g)-closed if for every “test map”f ∶ (T, t)→ (X , x) with f#(T ) ⩽ H, we have f#(g) ∈ H.

Theorem

Suppose X is metrizable and H ⩽ π1(X , x).

H is (S ,d∞)-closed ⇔ X is homotopically path-Hausdorff rel. H

⇓H is (D,d∞)-closed ⇔ X admits a generalized covering p ∶ X → X

with p#π1(X , x) = H⇓

H is (C , c∞)-closed ⇔ X is homotopically Hausdorff rel. H

Definition

Fix a “test space” (T, t) and T ⩽ π1(T, t), g ∈ π1(T, t).We say that H ⩽ π1(X , x) is (T ,g)-closed if for every “test map”f ∶ (T, t)→ (X , x) with f#(T ) ⩽ H, we have f#(g) ∈ H.

Theorem

Suppose X is metrizable and H ⩽ π1(X , x).

H is (S ,d∞)-closed ⇔ X is homotopically path-Hausdorff rel. H

⇓H is (D,d∞)-closed ⇔ X admits a generalized covering p ∶ X → X

with p#π1(X , x) = H⇓

H is (C , c∞)-closed ⇔ X is homotopically Hausdorff rel. H

The dyadic arc space D

0 j−12n−1

j2n−1

λn,j = `n,1`n,2⋯`n,j

λn = λn,2n−1

λ∞(t) = (t,0)

S = ⟨[λnλ−n+1] ∣ n ∈ N⟩ ⩽ π1(D,0)

D = ⟨[λn,jλ−n+1,2j] ∣ n ∈ N,1 ⩽ j ⩽ 2n−1⟩ ⩽ π1(D,0)

d∞ = [λ1λ−∞] ∈ π1(D,0)

0 j−12n−1

j2n−1

λn,j = `n,1`n,2⋯`n,j

λn = λn,2n−1

λ∞(t) = (t,0)

S = ⟨[λnλ−n+1] ∣ n ∈ N⟩ ⩽ π1(D,0)

D = ⟨[λn,jλ−n+1,2j] ∣ n ∈ N,1 ⩽ j ⩽ 2n−1⟩ ⩽ π1(D,0)

d∞ = [λ1λ−∞] ∈ π1(D,0)

0 j−12n−1

j2n−1

λn,j = `n,1`n,2⋯`n,j

λn = λn,2n−1

λ∞(t) = (t,0)

S = ⟨[λnλ−n+1] ∣ n ∈ N⟩ ⩽ π1(D,0)

D = ⟨[λn,jλ−n+1,2j] ∣ n ∈ N,1 ⩽ j ⩽ 2n−1⟩ ⩽ π1(D,0)

d∞ = [λ1λ−∞] ∈ π1(D,0)

0 j−12n−1

j2n−1

λn,j = `n,1`n,2⋯`n,j

λn = λn,2n−1

λ∞(t) = (t,0)

S = ⟨[λnλ−n+1] ∣ n ∈ N⟩ ⩽ π1(D,0)

D = ⟨[λn,jλ−n+1,2j] ∣ n ∈ N,1 ⩽ j ⩽ 2n−1⟩ ⩽ π1(D,0)

d∞ = [λ1λ−∞] ∈ π1(D,0)

0 j−12n−1

j2n−1

λn,j = `n,1`n,2⋯`n,j

λn = λn,2n−1

λ∞(t) = (t,0)

S = ⟨[λnλ−n+1] ∣ n ∈ N⟩ ⩽ π1(D,0)

D = ⟨[λn,jλ−n+1,2j] ∣ n ∈ N,1 ⩽ j ⩽ 2n−1⟩ ⩽ π1(D,0)

d∞ = [λ1λ−∞] ∈ π1(D,0)

0 j−12n−1

j2n−1

λn,j = `n,1`n,2⋯`n,j

λn = λn,2n−1

λ∞(t) = (t,0)

S = ⟨[λnλ−n+1] ∣ n ∈ N⟩ ⩽ π1(D,0)

D = ⟨[λn,jλ−n+1,2j] ∣ n ∈ N,1 ⩽ j ⩽ 2n−1⟩ ⩽ π1(D,0)

d∞ = [λ1λ−∞] ∈ π1(D,0)

Theorem

Suppose X is metrizable and π1(X , x) is residually n-slender.Then X is homotopically path-Hausdorff relative to H = 1.In particular, X admits a generalized universal covering.

Proof Know: π1(X , x) is homomorphically Hausdorff relative to D.

Show: H = 1 is (S ,d∞)-closed.

Let f ∶ (D,0)→ (X , x) with f#(S) = 1.

Let U be an open cover of D.

Choose n ∈ N with [λnλ−∞] =∏2n−1

j=1 [αjβjα−j ] ∈ π(U ,0),

where αj = λ∞∣[0, j−1

2n−1] and βj = `m,jλ∞∣[ j

2n−1, j−1

2n−1]. Then

f#(d∞) = f#([λ1λ−2 ])´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

⋯ f#([λn−1λ−n])´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

f#([λnλ−∞]) ∈ f#(π(U ,0)).

Consequently, f#(d∞) ∈ ⋂U f#(π(U ,0)) = 1

Theorem

Let f ∶ (D,0)→ (X , x) with f#(S) = 1.

j=1 [αjβjα−j ] ∈ π(U ,0),

2n−1, j−1

2n−1]. Then

f#(d∞) = f#([λ1λ−2 ])´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

⋯ f#([λn−1λ−n])´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

f#([λnλ−∞]) ∈ f#(π(U ,0)).

Theorem

Let f ∶ (D,0)→ (X , x) with f#(S) = 1.

j=1 [αjβjα−j ] ∈ π(U ,0),

2n−1, j−1

2n−1]. Then

f#(d∞) = f#([λ1λ−2 ])´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

⋯ f#([λn−1λ−n])´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

f#([λnλ−∞]) ∈ f#(π(U ,0)).

Theorem

Let f ∶ (D,0)→ (X , x) with f#(S) = 1.

j=1 [αjβjα−j ] ∈ π(U ,0),

2n−1, j−1

2n−1]. Then

f#(d∞) = f#([λ1λ−2 ])´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

⋯ f#([λn−1λ−n])´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

f#([λnλ−∞]) ∈ f#(π(U ,0)).

Theorem

Let f ∶ (D,0)→ (X , x) with f#(S) = 1.

j=1 [αjβjα−j ] ∈ π(U ,0),

2n−1, j−1

2n−1].

f#(d∞) = f#([λ1λ−2 ])´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

⋯ f#([λn−1λ−n])´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

f#([λnλ−∞]) ∈ f#(π(U ,0)).

Theorem

Let f ∶ (D,0)→ (X , x) with f#(S) = 1.

j=1 [αjβjα−j ] ∈ π(U ,0),

2n−1, j−1

2n−1]. Then

f#(d∞) = f#([λ1λ−2 ])´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

⋯ f#([λn−1λ−n])´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

f#([λnλ−∞]) ∈ f#(π(U ,0)).

Theorem

Let f ∶ (D,0)→ (X , x) with f#(S) = 1.

j=1 [αjβjα−j ] ∈ π(U ,0),

2n−1, j−1

2n−1]. Then

f#(d∞) = f#([λ1λ−2 ])´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

⋯ f#([λn−1λ−n])´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

f#([λnλ−∞]) ∈ f#(π(U ,0)).

Definition

We say X is 1-UV0 if for every x ∈ X and every open set U withx ∈ U there is an open set V with x ∈ V ⊆ U such that for everyf ∶ D2 → X with f (S1) ⊆ V there is a g ∶ D2 → U with f ∣S1 = g ∣S1 .

Examples

The following spaces are 1-UV0:

1-dimensional spaces (loops contraction within image)

Planar sets (loops contract without increasing diameter)

Sine wave spaces [Conner-Meilstrup-Repovs-Zastrow-Zeljko].

Definition

We say X is 1-UV0 if for every x ∈ X and every open set U withx ∈ U there is an open set V with x ∈ V ⊆ U such that for everyf ∶ D2 → X with f (S1) ⊆ V there is a g ∶ D2 → U with f ∣S1 = g ∣S1 .

Examples

The following spaces are 1-UV0:

1-dimensional spaces (loops contraction within image)

Planar sets (loops contract without increasing diameter)

Sine wave spaces [Conner-Meilstrup-Repovs-Zastrow-Zeljko].

Space A: Space B:

[Conner-Meilstrup-Repovs-Zastrow-Zeljko 2008]

Both A and B are 1-UV0.

If X is metrizable and 1-UV0,then X is homotopically Hausdorff rel. H = 1.

[F.-Repovs-Virk-Zastrow 2011]

A is homotopically path-Hausdorff rel. H = 1.B is not homotopically path-Hausdorff rel. H = 1.

Space A: Space B:

Both A and B are 1-UV0.If X is metrizable and 1-UV0,then X is homotopically Hausdorff rel. H = 1.

Space A: Space B:

Both A and B are 1-UV0.If X is metrizable and 1-UV0,then X is homotopically Hausdorff rel. H = 1.

Theorem

Suppose that X is a metrizable 1-UV0 space.Then X admits a generalized universal covering.

Show: H = 1 is (D,d∞)-closed.

Let f ∶ (D,0)→ (X , x) with f#(D) = 1.

Let α1, α2, α3, . . . be the lexicographical enumeration ofβn,j = `n,j`−n+1,2j`−n+1,2j−1. Then every loop f αi contracts in X .

Claim: ∀ε > 0 ∃N ∈ N ∀i ≥ N ∶ f αi has an ε-contraction in X .

Otherwise ∃ε > 0 ∃(αik ) with Image(αik )→ p ∈ [0,1]and f αik has no ε-contraction in X .

Define h ∶ (H,∗)→ (D,p) by h `k = sikαik s−ik

.Considering f h ∶ H→ X , we see that X is not 1-UV0 at f (p).

Theorem

Let f ∶ (D,0)→ (X , x) with f#(D) = 1.

Theorem

Let f ∶ (D,0)→ (X , x) with f#(D) = 1.

Theorem

Let f ∶ (D,0)→ (X , x) with f#(D) = 1.

Theorem

Let f ∶ (D,0)→ (X , x) with f#(D) = 1.

Theorem

Let f ∶ (D,0)→ (X , x) with f#(D) = 1.

Considering f h ∶ H→ X , we see that X is not 1-UV0 at f (p).

Theorem

Let f ∶ (D,0)→ (X , x) with f#(D) = 1.

Note: If (Hj)j∈J are (T ,g)-closed, then ⋂j∈J Hj is (T ,g)-closed.

Closure Operator

For H ⩽ π1(X , x), T ⩽ π1(T, t) and g ∈ π1(T, t), we may define

CLT ,g(H) = ⋂K ⩽ π1(X , x) ∣ K is (T ,g)-closed and H ⩽ K.Then

CLT ,g(H) = H if and only if H is (T ,g)-closed.

H ⩽ CLT ,g(H)If H ⩽ K , then CLT ,g(H) ⩽ CLT ,g(K)CLT ,g(CLT ,g(H)) = CLT ,g(H)If f ∶ (X , x)→ (Y , y) is continuous, thenf#(CLT ,g(H)) ⩽ CLT ,g(f#(H)).

Suppose t→ T is a cofibration.If N P π1(X , x), then CLT ,g(N) P π1(X , x).

If (T, t) is any space, then 0→ (T, t)∨ ([0,1],1) is a cofibration

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