M8130 Algebraic topology, tutorial 02, 2024 3.10.2024
Exercise 1. Show that CPn
is a CW-complex.
Solution. Clearly CP0
is a point. Next, we have
CPn
= Cn+1
\ {0}/{v ∼ λv, λ ∈ C \ {0}} ∼= S2n+1
\ {0}/{v ∼ λv, |λ| = 1} ∼=
∼= {(w, 1 − |w2|) ∈ Cn+1
, w ∈ D2n
}/{w ∼ λw for |w| = 1} ∼=
∼= (D2n
∪ S2n−1
)/{w ∼ λw for w ∈ S2n−1
} = D2n
∪f CPn−1
.
Now taking the canonical projection S2n−1
→ CPn−1 ∼= S2n−1
/ ∼ as the attaching map
f yields a CW-complex with one cell in every even dimension till 2n and none in the odd
ones.
Exercise 2. Let A = {1
n
∪ {0}} ⊆ R. Show that (I, A) does not have the HEP, i.e. the
inclusion A → I is not a cobration.
Solution. If A × J ∪ I × {0} was a retract of I × J, the retraction would have to preserve
connected subsets. But A×J∪I×{0} is not locally connected while I×J is, a contradiction.
Exercise 3. From the lecture we know that A := {1
n
, n ∈ N} ∪ {0} as a subspace of R is
not a CW-complex. Show that X := I × {0} ∪ A × I is not a CW-complex either.
Solution. Suppose that X is a CW-complex. Then it cannot contain cells of dimension
≥ 2, because it becomes disconnected after removing any point. In fact, the space obtained
after removing any point (a, 0) with a ∈ A has more than two connected components
(three for a > 0, to be exact), so these points cannot lie inside a 1-cell. Therefore these
points must form 0-cells, but we already know that A does not have discrete topology, a
contradiction.
Exercise 4. Prove that every compact set A in a CW-complex X can have a nonempty
intersection with only nitely many cells.
Solution. X is comprised of cells that are indexed by elemnts of some set J. Let B be a set
containing exactly one point from each intersection A ∩ eβ
, β ∈ J. We need to show that
B is closed and discrete, which will imply that B is compact (since B ⊆ A) and discrete,
hence nite. We know that a set C ⊆ Xn
is closed i both C ∩ Xn−1
and C ∩ en
α for each
α ∈ J are closed, because Dn
∪f Xn−1
is a pushout. Using induction, this implies that
C ⊆ X is closed i C ∩ eα is closed for each α ∈ J. Since B ∩ eα contains at most one
point for any α ∈ J and X is T1 (even Hausdor), this shows that B is closed. Using the
same argument, B with any one point removed is closed. Therefore B is also discrete and
we are done.
Exercise 5. Show that the Hawaiian earring given by
X = {(x, y) ∈ R2
, (x −
1
n
)2
+ y2
=
1
n2
for some n}
is not a CW-complex.
M8130 Algebraic topology, tutorial 02, 2024 3.10.2024
Solution. Suppose that X is a CW-complex. Using similar arguments as in the previous
exercise, we can see that (0, 0) must be a 0-cell and that X must have either innitely
many 0-cells, or innitely many 1-cells. But since X is compact, exercise 5 implies that X
can have only nitely many cells, a contradiction.
Exercise 6. Show that for a short exact sequence 0 → A
f
−→ B
g
−→ C → 0 of abelian groups
(or more generally modules over a commutative ring) the following are equivalent:
(1) There exists p : B → A such that pf = idA.
(2) There exists q : C → B such that gq = idC.
(3) There exist p : B → A and q : C → B such that fp + qg = idB.
(Another equivalent condition is B ∼= A ⊕ C, with (p, g) and f + q being the respective
inverse isomorphisms.)
Solution.
(1) =⇒ (2) and (3):
Since g is surjective, for any c ∈ C there is some b ∈ B such that g(b) = c. Moreover,
for any other b′
∈ B such that also g(b′
) = c, we have b − fp(b) = b′
− fp(b′
), since
b − b′
∈ ker g = imf, so that b − b′
= f(a) and
fp(b − b′
) = fpf(a) = f(a) = b − b′
.
This shows that we can correctly dene q(c) := b − fp(b) for any such b. Then we have
gq(c) = g(b) − gfp(b) = g(b) = c
(since gf = 0), which shows that gq = idC, and also qg(b) = b−fp(b), hence fp+qg = idB.
(3) =⇒ (1) and (2):
Applying f from the right to the equation fp + qg = idB yields fpf = f (since gf = 0),
which together with the fact that f is injective implies pf = idA. Similarly, applying
g from the left yields gqg = g, which together with the fact that g is surjective implies
gq = idC.
Exercise 7. Let 0 → A∗
f
−→ B∗
g
−→ C∗ → 0 be a short exact sequence of chain modules.
We have dened the connecting homomorphism ∂∗ : Hn(C) → Hn−1(A) by the formula
∂∗[c] = [a], where ∂c = 0, f(a) = ∂b and g(b) = c. Show that this denition does not
depend on a nor b.
Solution. We have ∂a = 0 i f(∂a) = 0 (using injectivity of f) i 0 = ∂f(a) = ∂∂b, and
the last condition is true.
Now let b, b′
∈ B be such that g(b) = g(b′
) = c with a, a′
∈ A such that f(a) = b, f(a′
) = b′
.
Then b − b′
∈ ker g = imf, so b − b′
= f(a) for some a ∈ A. Therefore f(∂a) = ∂b − ∂b′
=
f(a − a′
) and the injectivity of f implies ∂a = a − a′
, hence [0] = [∂a′
] = [a] − [a′
] and we
are done.