. The composition f : S

USTC, School of Mathematical Sciences Winter semester 2018/12/29 Algebraic topology by Prof. Mao Sheng Hints to exercise sheet 10 MA04311 Tutor: Lihao Huang, Han Wu Posted online by Dr. Muxi Li Hint1. If f (x) ̸= x, ∀x, then H(x, t) = ((1 − t)f(x) + tx)/|(1 − t)f(x) − tx| gives a ho- motopy connecting f and A, where A(x) := −x. Similarly, f ≃ Id if f(x) ̸= −x, ∀x. Neither f ≃ A nor f ≃ Id holds for deg f = 0 here. Hence there exist points x, y with f(x) = x and f (y) = −y.

Let G := F/ |F | : D

→ S

. The composition f : S

↩ → D

−→ S

is nullhomotopy so deg f = 0. There exist points x, y ∈ S

with f (x) = x and f (y) = −y, i.e. F (x) = |F (x)|x and F (y) = −|F (y)|y, according to the above discussion.

Hint2. Each even map f can be decomposed to f : S

− → RP

→ S

, where q is the quotient map. H

(RP

) = 0, hence deg f = 0, when n is even.

We knew that the degree of the composition

S

− → RP

→ RP

/ RP

∼ = S

is 1 + (−1)

when we calculated the cellular homology groups of projective spaces. Therefore H

( RP

) ≈ Z and q

: H

(S

) → H

( RP

) is multiplication by 2 when n is odd, since the homomorphism H

( RP

) → H

( RP

/ RP

) induced by quotient map is an isomorphism when n is odd, which can be prove by using the long exact sequences of homology groups of good pairs. Then it is distinct that the degree of f : S

− → RP

→ S

is even.

To give a even map with degree 2k, we construct g : S

→ S

with degree k, then the composition S

− → RP

→ RP

/RP

∼ = S

− → S

has the degree 2k.

Hint3. We first emphasize that the cellular chain map f

is defined as f

: H

(X

, X

) → H

(Y

, Y

), although we leave out the verifying that it is a chain map indeed. We have the following commutative graph

H

(X

) ^//

f_∗



H

(X

) ^//

f_∗



H

(X)

f_∗



H

(X

) ^// H

(X

) ^// H

(X),

where all the vertical arrows are induced by inclusion, the arrows in left side are surjective, and arrows in right side are isomorphic in both rows. The isomorphism I : H

(X) → H

(X) is given by the following process: for a given element α in H

(X), choose a preimage α

in H

(X

) along the first row in the above diagram, then I(α) := [j

(α

)], where j

: H

(X

) → H

(X

, X

) = C

(X) is the homomorphism induced by the quotient homomorphism of chain complex, and [ −] means the homology class in the cellular homology. Notice that we have known that j

(α

) is closed and uniquely determined by α up to homology, i.e. I is well defined when we proved H

≈ H

. We can choose f

(α

) as one of the preimages of f

(α) for the above commutative diagram. Hence I(f

(α)) = [j

◦ f

(α

)] = [f

◦ j

(α

)] for the following obvious commutative diagram

H

(X

)

Xn

//

f_∗



H

(X

, X

)

f_∗



H

(Y

)

nY

// H

(Y

, Y

),

while f

(I(α)) = [f

(j

(α

))] = I(f

(α)). So we have proved f

◦ I = I ◦ f

.

1

Moreover, we want to illustrate the cellular chain map f

when we treat C

(X) = H

(X

, X

) as the free abelian group generated by n-cells of X. Precisely, we want to prove that the coefficient a

in the expansion f

(e

) = ∑

j

a

c

is the degree of the composition D

/S

−→ X

/X

− → Y

/(Y

− c

)

ψ¯_j⁻¹

−−→ D

/S

,

where {e

}

and {c

}

are the n-cells of X and Y , φ

and ψ

are the characteristic maps of e

and c

respectively, and ¯ f means the map induced by f in the quotient spaces, etc. Recall that H

(X

, X

) is a free abelian group generated by {φ

(o

) }

, where o

is a generator of H

(D

, S

). Hence e

just means {φ

(o

)}

when we treat H

(X

, X

) as the free abelian group generated by n-cells of X. Now what we want results from the following diagram:

H

(D

, S

)

^//



H

(X

, X

)

^//



H

(Y

, Y

− c

)



H

(D

, S

)

ψj

oo



H e

(D

/S

)

^// H e

(X

/X

)

^// H e

(Y

/(Y

− c

)) H e

(D

/S

),

ψ¯j

oo

where all the vertical arrows are the isomorphism for the good pairs.

Hint4. Let i

(g)(k) = kg, ∀g ∈ C

(X) and d

(α) = ∂

◦ α, ∀α ∈ Hom(Z, C

(X)), then it can be verified that i

: C

(X) → Hom(Z, C

(X)) is an isomorphism and the diagram

C

(X)

^//

in



C

(X)

i_n−1



Hom( Z, C

(X))

^// Hom( Z, C

(X))

is commutative, ∀n. Hence the chain complexes (C

(X), ∂

) is isomorphic to (Hom( Z, C

(X)), d

).

So h

(X; Z) ≈ H

(X).

We show that Hom( Z

, H) = Hom( Q, H) = 0 when H is a free alelian group, hence h

(X; Z

) = h

(X; Q) = 0. For an arbitrary f ∈ Hom(Z

, H), mf (k) = f (mk) = f (0) = 0, so f (k) = 0 since H is free, ∀k ∈ Z

. For an arbitrary f ∈ Hom(Q, H), there is one non-zero coeffi- cient at least when we expand f (1) by a basis if H, i.e. f (1) = ae

+ ∑

i̸=i0

a

e

, a ̸= 0 and {e

} is a basis of H. Let f (1/(2a)) = be

+ ∑

i̸=i0

b

e

, then f (1) = 2af (1/(2a)) = 2abe

+ ∑

i̸=i0

2ab

e

. But the coefficients are uniquely determined by f (1), so we obtain a absurdity that a = 2ab ̸= 0.

Hence f (1) = 0, then f = 0.

Hint5. It is easy to calculate the cellular homology groups: H

(X) ≈ Z

, e H

(X) = 0, ∀i ̸= n.

And H

(X) ≈ Z

, e H

(X) = 0, ∀i ̸= n follow from the universal coefficient theorem.

(a) It is obvious that the quotient map q : X → X/S

= S

induces the trivial map on H e

(−; Z). We have the following exact sequence

H

(X/S

) → H

(X) → H

(S

) = 0

for good pair (X, S

), hence q

: H

(X/S

) → H

(X) ≈ Z

is surjective, and nontrivial.

If the splitting in the universal coefficient theorem for cohomology were natural, the diagram H

(X/S

)

^//

q^∗



Hom(H

(X/S

), Z) ⊕ Ext(H

(X/S

), Z)



H

(X)

^// Hom(H

(X), Z) ⊕ Ext(H

(X), Z)

2

would be commutative. The vertical arrow in the right side is the direct of Hom(H

(Y ), Z) → Hom(H

(X), Z) and Ext(H

(Y ), Z) → Ext(H

(X), Z), hence trivial, (for Hom(H

(X), Z) = 0 and Ext(H

(X/S

), Z) = 0. ) But q

is nontrivial so the diagram is not commutative.

(b) Similarly.

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