Description

1. (i) We consider the sequence
(1)
p where p∗ is induced from the covering map X →− X. First we show exactness of each row. As mentioned in class, Π acts on Cn(Xb) by permuting generators, so the map Cn(Xb) −−t−→1 Cn(Xb) induced by the action of t − 1 must be injective. As X/b Π ∼= X, the covering map Xb →−p X is cellular, so the induced map Cn(Xb) → Cn(X) is surjective. Finally, if eni ∈ Cn(Xb) then

as eni and project to the same cell in X, and conversely if ) = 0 for some z ∈ Cn(X) then z must be of the form Pj(t − 1)enj as X/Π ∼= X, so ker(p∗) = im (t − 1).
To show (1) is exact it remains to show each square in the diagram below commutes (for each n). I don’t know how to do this.
t−1 p
Here is a statement of the Snake Lemma.
Lemma (snek) Let A be a commutative ring, and let
φ ψ

be a commutative diagram of A-modules with exact rows. Then there is an
A-linear map δ : kerγ → coker α such that the sequence
kerα kerβ ker coker α coker β coker γ
(3)
is exact. Moreover, if φ is injective then the first map kerα → kerβ is injective, and if ψ′ is surjective, the last map coker β → coker γ is surjective. The

exact sequence (3) is functorial in the diagram (2).
Expanding (1) to make things more visually clear
0 0 0
(4)
… … …
we apply the snake lemma to the 0th row to get an exact sequence
0 → Z0(Xb) −−t−→1 Z0(Xb) →Z0(X) →−∂
t−1

C1(Xb)/B0(Xb) −−→ C1(Xb)/B0(Xb) → C1(X)/B0(X) → 0
and in general apply the snake lemma to the nth row to get an exact sequence
0 → Zn(Xb) −−t−→1 Zn(Xb) →Zn(X) →−∂
Cn+1(Xb)/Bn(Xb) −−t−→1 Cn+1(Xb)/Bn(Xb) → Cn+1(X)/Bn(X) → 0
So, for each n we have a commutative diagram with exact rows
Cn+1(Xb)/Bn(Xb)Cn+1(Xb)/Bn(Xb)Cn+1(X)/Bn(X)0
0 b b Z (X)
(5)
and applying the snake lemma to (5) for each n yields the long exact sequence
(ii) Suppose that X has the homology of a circle. First, we have H0(Xb) ∼= k since Xb is connected, so H0(Xb) has no kΠ-free summands. If i ≥ 2 then Hi(X) = 0, so we get an exact sequence
t−1
Hi(Xb) −−→ Hi(Xb) → 0
which means we can write each x ∈ Hi(Xb) as (t − 1)y for some y ∈ Hi(Xb), so Hi(Xb) does not have any kΠ-free summand.
For i = 1, as X is a kHS1 the long exact sequence (6) ends like

so the situation is similar to the one described for i ≥ 2 above. From this Hi(Xb) is finitely generated as a k-vector space for all i.
(iii) As the maps induced on chains by t − 1 and p are injective and surjective, respectively, we break up the long exact sequence (6) into short exact sequences
…
t−1
0
0 Hi(Xb) Hi+1(Xb)
t−1 Hi(Xb)
Hi+1(Xb)
… Hi(X) Hi+1(X) 0
0
which split as the homology modules are finitely generated k-vector spaces by
(ii). So, for each i we have
rankHi(Xb) = rankHi(Xb) + rankHi(X)
thus χ(Xb) = χ(Xb) + χ(X), so χ(X) = 0.
2. (i) We write π1X with the presentation π1X = ⟨a,b | a2b−2⟩, with a and φ b pictured below. We define a surjective homomorphism π1X −→ Π by a 7→ t and b 7→ t. From this we see that
π1Xb ∼= kerφ = ⟨aib−i⟩i∈Z.

If we tile R2 with these squares to get the universal cover of X, we note that aib−i corresponds to moving the basepoint down by i squares. From covering space theory this means that Xb corresponds to a 1-by-Z vertical strip in the tiling, which we can think of as a Mobius strip with infinite width, thus having homotopy type S1. Therefore Hi(Xb) = 0 for i > 1. As a kΠ module
H0(Xb) ∼= k[t,t−1]/(t − 1)
and
H1(Xb) ∼= k[t,t−1]/(t + 1).
(ii) We know π1X = ⟨a⟩ where a is a generator for . So, we get the identity map , thus π1Xb is trivial so X is the universal cover of X, thus X has homotopy type S2. As X has the homology of S1 and
Π acts trivially on X we must have
H0(Xb) ∼= k[a,a−1]/(a − 1) ∼= H1(Xb)
as kΠ-modules.