M8130 Algebraic topology, tutorial 08, 2017 13. 4. 2017
Exercise 1 (Five lemma). Let
A //
∼=
B //
∼=
C //
f

D //
∼=
E
∼=
A // B // C // D // E
be a commutative diagram of modules with exact rows. Show that the middle homomorphism
f is an isomorphism.
Solution. We will show the surjectivity of f, injectivity is dual. Let c ∈ C be arbitrary
and suppose it maps to d ∈ D. This d corresponds to some d ∈ D, and since d maps to
0 in E by exactness, d also has to map to 0 ∈ E by commutativity. By exactness, there
exists c ∈ C which maps to d. Since c and f(c) both map to d by commutativity, there
is (by exactness) some b ∈ B which maps to f(c) − c. This b corresponds to some b ∈ B,
which maps to some c ∈ C. Then by commutativity, f(c − c ) = f(c) − (f(c) − c) = c, as
desired. (Note that instead of the four vertical maps being isomorphisms, we only needed
the surjectivity of B → B, D → D and the injectivity of E → E).
Exercise 2. Show that for a nite CW-complex X and H∗
(Y ) being nitely generated free
group in all dimensions, the cross product
H∗
(X) ⊗ H∗
(Y )
µ
−→ H∗
(X × Y )
is an isomorphism. (In fact, the same is true for X being an innite CW-complex.)
Solution. First let X = pt be a point. Then H∗
(pt) = Z with 1 ∈ H0
(pt) and pt × Y is
homeomorphic to Y , hence
H∗
(pt) ⊗ H∗
(Y ) = Z ⊗ H∗
(Y ) ∼= H∗
(Y ) ∼= H∗
(pt × Y ).
Now let X = p1 p2 · · · pk be a nite disjoint union of points (i.e, a discrete set). Then
H∗
(X) = k
i=1 Z, hence
H∗
(X) ⊗ H∗
(Y ) =
k
i=1
Z ⊗ H∗
(Y ) ∼=
k
i=1
(Z ⊗ H∗
(Y )) ∼=
k
i=1
H∗
(Y )
∼= H∗
(Y Y · · · Y
n times
) ∼= H∗
(X × Y )
(We should also show that the isomorphism is indeed given by µ, but if e1, . . . , ek ∈ H0
(X)
are such that
ei(pj) =
1 if i = j
0 if i = j
,
it's not hard to see that
µ(ei ⊗ a) = (0, 0, . . . , a
i-th place
, 0, . . . , 0)
using projections and the denition of the cup product.)
M8130 Algebraic topology, tutorial 08, 2017 13. 4. 2017
Now we can proceed inductively:
i) We now know that the theorem is true for X of dimension 0.
ii) Since Dn
is homotopy equivalent to a point, the theorem is also true for X = k
α=1 Dn
α.
iii) Suppose that the theorem holds for nite CW-complexes of dimension n − 1. Then it
is also true for pairs ( Dn
α, Sn−1
α ).
iv) Since H∗
(X, A) ∼= H∗
(X/A) for a subcomplex A ⊆ X, the theorem also holds for
Sn
α
∼= Dn
α/ Sn−1
α .
v) Now let X = X(n)
be an n-dimensional CW-complex and consider the diagram
H∗
(X, A) ⊗ H∗
(Y ) //
µ

H∗
(X) ⊗ H∗
(Y )
µ

uu
H∗
(A) ⊗ H∗
(Y )
µ

δ∗⊗id
ii
H∗
(X × Y, A × Y ) // H∗
(X × Y )
uu
H∗
(A × Y )
δ∗
ii
,
where the lower triangle represents a long exact cohomology sequence by denition,
and the same is true for the upper triangle (since free modules are at). We know that
the theorem holds for X(n−1)
, and also for X(n)
/X(n−1) ∼= Sn−1
α . Therefore we can
use the Five lemma after unfolding the diagram in the appropriate dimensions, from
which it follows that the theorem also holds for X(n)
. This completes the induction.
Exercise 3. Compute the cohomology rings of CP2
× S6
and CP2
∨ S6
.
Solution. We have H∗
(CP2
) = Z[w]/ w3
for w ∈ H2
and H∗
(S6
) = Z[a]/ a2
for a ∈ H6
,
hence
H∗
(CP2
× S6
) = Z[w]/ w3
⊗ Z[a]/ a2 ∼= Z[w, a]/ w3
, a2
.
Next, it is true in general that H
∗
(X ∧ Y ) ∼= H
∗
(X) ⊕ H
∗
(Y ) is an isomoprhism of graded
rings (this can be proven proven straight from the denitions, but it takes some time).
Since CP2
∨ S6
is connected, we have H∗
(CP2
∨ S6
) ∼= H
∗
(CP2
) ⊕ H
∗
(S6
) ⊕ Z. Now
w ∪ a ∈ H8
= 0 (more generally, we could use that fact that (w, 0) ∪ (0, a) = (0, 0)).
Therefore
H∗
(CP2
∨ S6
) ∼= Z[w, a]/ w3
, a2
, wa .
M8130 Algebraic topology, tutorial 08, 2017 13. 4. 2017
Exercise 4. Show that the CP2
∨ S6
is not homotopy equivalent to CP3
.
Solution. It suces to show that the cohomology rings of these spaces are not isomorphic
(note that the additive group structure is not enough to distinguish them). We already
know that
H∗
(CP2
∨ S6
) ∼= Z[w, a]/ w3
, a2
, wa
and we have H∗
(CP3
) = Z[b]/ b4
for b ∈ H2
. Any isomorphism would have to map w and
b to ± each other (these are the respective generators in dimension 2), but w3
= 0 while
b3
= 0, so this is not possible.