Algebraic topology ex. 6
Shaked Kara
April 27, 2022
Question 1. Product of CW spaces
Proof. a) Let X, Y be two CW-spaces, we will construct the CW-space using induction. For some k + m = n
denote X m’th skeleton by Xm and m’th cells by Dαm with the corresponding maps ϕα ∶ ∂Dαm → X and for
Y k’th by Yk , Dβk and ϕβ ∶ ∂Dβk → Y . Now, for n − 1 skeletons (X × Y )n−1 ⊆ ⊔l+t=n−1 Xl × Yt ⊆ X × Y . We
n
will construct n cell with the identification Dαβ
≃ Dαm × Dβk . The boundary is given by:
n
∂Dαβ
≃ ∂(Dαm × Dβk ) ≃ (∂Dαm × Dβk ) ⊔ (Dαm × ∂Dβk ) ⊆ (Xm−1 × Yk ) ⊔ (Xm × Yk−1 ) ⊆ (X × Y )n−1
and so they maps to neutrally (X × Y )n−1 . Thus we get the pushout:
n
∐αβ ∂Dαβ
n
∐αβ Dαβ
(X × Y )n−1
Sk((X × Y )n )
And also we have the maps:
(X × Y )n−1 → (X × Y )n
n
∐ Dαβ
αβ
= ∐ (Dαm × Dβk ) → (X × Y )n
αβ
Which agrees with the union of the boundaries and so we have a continues map f ∶ Sk((X ×Y )n ) → (X ×Y )n .
Now, for two finite CW-spaces for some point (x, y) ∈ X × Y there exists unique cells s.t x ∈ Dαm and y ∈ Dβk
n
and so, (x, y) ∈ Dαβ
and so f is a bijection. We know that a continuous bijection between compact and
Hausdorff spaces is a homeomorphism and so we get the same topology.
Proof. b) Let X be the CW-space consist 1 0-cell and unaccountably many 1-cells with the 0-cell as their
origin. Let Y be the same as X but with countably many 1-cells. Denote the interval [0, 1] by Is for
s ∈ {(s1 , s2 , ...)∣si ∈ N}. In that notation we can write X = ⋁s Is and Y = ⋁j Ij for j ∈ N with amalgamation
on the endpoints of I. A basic open set containing (x0 , y0 ) in the product topology has the form U × V
where U = ⋁s [0, as ) and V = ⋁j [0, bj ). Now, Let psj be the point (1/sj , 1/sj ) ∈ Is × Ij ⊂ X × Y and let P be
the union of all these points psj . Thus P consists of a single point in each 2-cell of X × Y , so P is closed in
the CW topology on X × Y . Now, let x0 is the common endpoint of the intervals Is and y0 is the common
endpoint of the intervals Ij .Choose a sequence t = (t1 , t2 , ⋯) with tj > j and tj > 1/bj for all j, and choose
an integer k > 1/at . Then tk > k > 1/at hence 1/tk < at . We also have 1/tk < bk . So (1/tk , 1/tk ) is a point of
P that lies in [0, at ) × [0, bk ) and hence in U × V . Thus, P has nonempty intersection with U × V and so it
is not closed in the product topology.
Proof. c) We know that
χ(X × Y ) = ∑(−1)n dim Hn (X × Y )
n
and
χ(X)χ(Y ) = ∑(−1)k dim Hk (X) ∑(−1)m dim Hm (Y )
m
k
1
Because X and Y are finite CW complexes we saw that each n-cell in X ×Y is the product of an i-cell in X and
an (ni)-cell in Y and so it sufficient to show that dimn (X × Y ) = ∑nk=0 dim Hk (X) ⋅ dim Hn−k (Y ). We know
that that Hn (X × Y ) = Hn (C●C∶W (X × Y )) = ker (∂n ) / Im (∂n+1 ) where ∂˙n ∶ Hn ((X × Y )n , (X × Y )n−1 ) →
Hn ((X × Y )n−1 , (X × Y )n−2 ). We saw that for CW the n’th homology is the free group on ∣An ∣ where An is
the set of n-cells. Because every cell added in (X × Y )n came from (X)k and (Y )n−k we get an isomorphism
on the generators h ∶ Hn ((X × Y )n , (X × Y )n−1 ) → ⊕nk=0 Hk ((X)k , (X)k−1 ) ⊕ Hn−k ((Y )n−k , (Y )n−k−1 ) and
so we get that dim Hn (X × Y ) = dim (⊕nk=0 Hk ((X)k ) ⊕ Hn−k ((Y )n−k )) and thus the equality needed.
Question 2. Calculating homology groups
Proof. a) For X = S 2 × S 5 × S 13 we will use the first question and the CW-structure of S n - as a single 0-cell
and a single n-cell. Denote S 2 = {Dα0 , Dα2 }, S 5 = {Dβ0 , Dβ5 }, S 13 = {Dγ0 , Dγ13 } So, looking on: {Dα0 , Dα2 } ×
{Dβ0 , Dβ5 } × {Dγ0 , Dγ13 } We can deduce that X has the cells
m+k+l
{DX
}m,k,l
when m run on the cells in S 2 , k on S 5 and l on S 13 . So, we get the CW-structure containing cell with dim
for every sum, i.e cells with dim n ∈ {0, 5, 2, 7, 13, 18, 15, 20} ∶= Cdim . In addition, as we mentioned in Q.1.c)
for CW the n’th homology is the free group on ∣An ∣ where An is the set of n-cells. In our case we have only
one cell for n ∈ Cdim so we get:
Hn = Z for n ∈ {0, 5, 2, 7, 13, 18, 15, 20}
Hn = 0 otherwise.
Proof. b) Doing the same as in (a) for the 3-torus T 3 = S 1 × S 1 × S 1 we get the CW-structure with one 0-cell
v three 1-cells a, b, c three 2 cells α, β, γ and one 3 cell σ and we get the cellular chain complexes:
d2
d3
d1
⋯ → 0 → Z⟨σ⟩ Ð→ Z⟨α, β, γ⟩ Ð→ Z⟨a, b, c⟩ Ð→ Z⟨v⟩ → 0
The 3-torus T 3 can be constructed from the cube [0, 1] × [0, 1] × [0, 1] by identifying each pair of opposite
square faces. As in the TA for the 2-torus we can repeat the process and get that d2 is zero. For d3 we
get 3 maps from the surface of the cube to a S 2 obtained by composing the map from the boundary of
the 3-cell to the 2-skeleton with the quotient map collapsing the complement of the (α, β, γ) 2-cell to a
point. Denote them by ψσα , ψσβ , ψσγ . Now, the first is the projection map on the surface of the cube
that identifies opposite faces because both the boundary of the 3-cell and the 2-skeleton are the surface of
the cube. With the exception that in the 2-skeleton the opposite faces are identified. Now, we will show
that ψσα has degree 0. Let r be the reflection ∂D3 → ∂D3 that exchanges the two faces of the 2-cell. We
get ψσα ○ r = ψσα which implies on a generator z: (ψσα )⋆ (r⋆ (z)) = (ψσα )⋆ (z) since r is a reflection on a
sphere we get: − (ψσα )⋆ (r⋆ (z)) = (ψσα )⋆ (z) Hence the maps ψσα , ψσβ , ψσγ has degree zero. And so we get:
H0 (T 3 ) = H3 (T 3 ) = Z
H1 (T 3 ) = H2 (T 3 ) = Z3
Hn (T 3 ) = 0 for n > 3
Question 3. Tile Mg with squares.
Proof. We can look on n-cells as [0, 1]n and with that notation tiling is actually construct (specific) CWstructure for the space. Lets look on Mg . In the TA we saw the CW-structure of Mg as one vertex, 2g 1-cells
and one 2-cell. And so we will get that:
χ(Mg ) = ∑(−1)k dim Hk (X) = 1 − 2g + 1 = 2 − 2g
k
Assuming we succeeded in constructing a CW-structure as Miriam asked with g 0-cells. So, by the ”building”
instructions every 0-cell is the meeting point of 4 2-cells (squares) and 4 1-cells (the squares edges) so we get
that the number of 1-cells in this construction is
4 ⋅ g ⋅ 1/2
2
(every 0-cells give us 4 edges but each edge has 2 vertex and thus being counted 2 times). In the same way
the number of 2-cells in this construction is
4 ⋅ g ⋅ 1/4
(every 0-cells give us 4 square but each square has 4 vertex and thus being counted 4 times). Thus, we
get that the CW-structure in this construction containing g 0-cells, 2g 1-cells and g 2-cells. So: χ(Mg ) =
∑k (−1)k dim Hk (X) = g − 2g + g = 0. Now, χ of 2 CW-structures of the same space remain the same and
thus: 2 − 2g = 0 and we get that g must be 1.
Question 4. Gluing is well defined up to homotopy
Proof. Let φ, ψ ∶ S n−1 → X be two homotopic maps and denote the homotopy between them by F . we
want to prove that X ⊔φ Dn and X ⊔ψ Dn are homotopy equivalent. Its enough to show that they are both
deformation retracts of X ⊔F (Dn × I). In TA3 we saw that (Dn , S n−1 ) is a good pair. In particular we sat
that there is a retraction r ∶ Dn × I → Dn × (0) ∪ S n−1 × I. Now, lets construct homotopic map between the
above retraction and the identity map:
d ∶ Dn × I × I → Dn × I
d(x, t, s) = (s)r(x, t) + (1 − s)(x, t)
This is obviously an homotopic map. For the pushout square we know that a retraction on Dn × I induces
a retraction on the pushout (Dn × I) ∪F X. The space (Dn × {0} ∪ S n−1 × I) ∪F X is homeomorphic to
(Dn × {0}) ∪φ X by the map sending Dn × {0} and X to itself and S n−1 × I to F (S n−1 × I) and so we get
that X ⊔F (Dn × I) deformation retracts onto (Dn × {0} ∪ S n−1 × I) ∪F X = (Dn × {0}) ∪φ X. We can show
it for ψ in the exact same way and so we are done.
Question 5. CPn
Proof. a) In a similar way as in the TA we can obtain CPn as a quotient space√of the disk D2n under
the identifications v ∼ λv for v ∈ ∂D2n : First, the graph of the function w ↦ 1 − ∣w∣2 is a disk D+2n
bounded by the sphere S 2n−1 ⊂ S 2n+1 consisting of vectors (w, 0) ∈ Cn × C with ∣w∣ = 1. Also, Note that
the √
vectors in S 2n+1 ⊂ Cn+1 with last coordinate real and non-negative are precisely the vectors of the form
(w, 1 − ∣w∣2 ) ∈ Cn × C with ∣w∣ ≤ 1. Such vectors form exactly the graph mentioned above. Now, using
induction, for v ∈ S 2n+1 s.t v last coordinate is nonzero we can map it to a unique vector v ′ ∈ D+2n using the
identifications v ∼ λv. For v ∈ S 2n+1 s.t v last coordinate is zero and we can use the identifications v ∼ λv for
v ∈ S 2n−1 . So, we get description of CPn as the quotient of D+2n under the identifications v ∼ λv for v ∈ S 2n−1 .
So, the map CPn−1 ≅ S n−1 / ∼↪ Dn / ∼↪ S n / ∼≅ CPn gives us the pushout square:
∂Dn
Dn
CPn−1
CPn
And so the construction of CPn can be described as attaching 2n-cells to the boundary of the map ∂Dn →
CPn−1 from the pushout square and so the we get that the CW-structure on CPn containing 1 k-cell for
k ∈ [n] s.t k is even.
Proof. b) As in the TA, for A an abelian group, we will look on the chain complex:
dk+1
dk−1
dk
d1
k−1
) ≃ A Ð→ ⋯ Ð→ Ccell
⋯ Ð→ Ccell
(CPn ) ≃ A Ð→ Ccell
(CPn ) ≃ A → 0
k
k−1 (CP
0
Like in the TA look on the push-outs squares:
S k−1
CPk−1
∗
Dk
CPk
CPk /CPk−1
3
And so we can compute the degree:
q
π
dk = deg (S k−1 Ð→ CPk−2 → Ck−1 /CPk−2 ≅ S k−1 )
And we know that if k is even we have the 0-map (because there’s no k-1 cells as k-1 is odd) and otherwise
the degree of dk is 1 and we get: ... → 0 → A → 0 → ... → A → 0 I.e:
̃ k (CPn ; A) = A for k odd, k ≤ n
H
̃
Hk (CPn ; A) = 0 otherwise.
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