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Math 215a Homework #4 solutions 1. Hatcher 2.1.14. Let’s do the last part first. I claim that if A is an abelian group, then there exists a short exact sequence 0 −→ Z −→ A −→ Zn −→ 0 iff A ' Z ⊕ Zm where m is a positive integer dividing n. To see this, suppose first that such an exact sequence exists. Let x ∈ A be the image of a generator of Z. Let y ∈ A map to a generator of Zn . Then ny maps to 0 ∈ Zn , so ny = kx for some integer k. It follows that the map Z2 → A sending (1, 0) 7→ x and (0, 1) 7→ y induces an isomorphism Z2 . A' Z{(−k, n)} We can find a change of basis of Z2 sending (−k, n) to (0, m) where m is the greatest common divisor of k and n. So the right hand side of the above equation is isomorphic to Z ⊕ Zm , and by definition m is a divisor of n. Conversely, if n = md, then we can define a short exact sequence f g 0 −→ Z −→ Z ⊕ Zm −→ Zn −→ 0 by f (1) := (d, −1) and g(a, b) := a + bd. By a similar but slightly more complicated argument, if A is an abelian group, then there exists a short exact sequence 0 −→ Zpm −→ A −→ Zpn −→ 0 iff A ' Zpa ⊕ Zpb where a and b are nonnegative integers such that a ≥ max(m, n) and a + b = m + n. 2. Hatcher 2.1.17(b). It follows easily from the long exact sequence that Hi (X, A) = Hi (X, B) = 0 for i 6= 1, 2. From the long exact sequence of the pair (X, A), we obtain an exact sequence 0 → H2 (X) → H2 (X, A) → H1 (A) → H1 (X) → H1 (X, A) → 0. Since A separates X, it follows that H1 (A) maps to zero1 in H1 (X). By the exact sequence, H2 (X, A) ' Z2 and H1 (X, A) ' Z 2g . 1

One way to see this using what we know so far is to consider the Mayer-Vietoris sequence relating the homology of X to the homology of the left and right halves and the homology of A.

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Likewise, there is an exact sequence 0 → H2 (X) → H2 (X, B) → H1 (B) → H1 (X) → H1 (X, B) → 0. In this case, one can choose identifications H1 (B) ' Z and H1 (X) ' Z2g so that the map H1 (B) → H1 (X) sends2 1 7→ (1, 0, . . . , 0). It then follows from the exact sequence that H1 (X, B) ' Z2g−1 and H2 (X, B) ' Z. 3. Hatcher 2.1.18. To compute H1 (R, Q), we use the long exact sequence i

∗ H0 (R) 0 = H1 (R) −→ H1 (R, Q) −→ H0 (Q) −→

where i : Q → R is the inclusion map. It follows from this exact sequence that H1 (R, Q) = Ker(i∗ ). Now H0 (Q) = ⊕Q Z. Moreover if σq denotes the 0-simplex mapping to q ∈ Q, then i∗ σq = 1 ∈ Z = H0 (R). So the kernel of i∗ consists of finite integer linear combinations of σq ’s with total coefficient zero. A basis for this is given by {σ0 − σq | q ∈ Q \ {0}}. 4. Hatcher 2.1.27. (a) This follows by applying the five-lemma to the diagram Hn (A) −−−→ Hn (X) −−−→ Hn (X, A) −−−→ Hn−1 (A) −−−→ Hn−1 (X)      ' '  ' ' y y y y y Hn (B) −−−→ Hn (Y ) −−−→ Hn (Y, B) −−−→ Hn−1 (B) −−−→ Hn−1 (Y ). (b) If g : (Dn , Dn \ {0}) → (Dn , S n−1 ), let g1 : Dn \ {0} → S n−1 denote the restriction. Since g is continuous at 0, it follows that e n−1 , since the g maps all of Dn to S n−1 . Hence g1 induces 0 on H n n−1 e map g1 factors through D . Since Hn−1 (S ) 6= 0, it follows that g1 does not induce an isomorphism on reduced homology, so g1 cannot be a homotopy equivalence. 2

One way to show this is to identify X with the quotient of a 4g-gon so that B corresponds to one of the edges. We will see a more general way to prove this after introducing the intersection pairing.

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5. Hatcher 2.2.2. Let f : S 2n → S 2n . If f (x) 6= x for all x, then f is homotopic to the antipodal map, so deg(f ) = (−1)2n+1 . If f (x) 6= −x for all x, then f is homotopic to the identity, so deg(f ) = 1. Since (−1)2n+1 6= 1, these possibilities are mutually exclusive, so there must exist x ∈ S 2n with f (x) = x or f (x) = −x. Next let f : RP2n → RP2n . Since S 2n is the universal covering space of RP2n , it follows from the Lifting Criterion that f lifts to a map RP2n → S 2n . We can then pull this back to a map fe : S 2n → S 2n , so that the diagram fe

S 2n −−−→ S 2n     y y f

RP2n −−−→ RP2n commutes, i.e. f ([x]) = [fe(x)] ∈ RP2n for all x ∈ S 2n . By the previous paragraph, there exists x ∈ S 2n such that fe(x) = ±x, which means that f ([x]) = [x].   0 −1 Taking the direct sum of n copies of the 2 × 2 matrix gives 1 0 a linear map R2n → R2n with no real eigenvalue, and hence a map RP2n−1 → RP2n−1 with no fixed point. 6. Extra problem. (a) The proof that ∂T = T ∂ is a straightforward matter of comparing signs. (b) We will construct a natural chain homotopy K : C∗ (X) → C∗+1 (X) such that ∂K + K∂ = 1 − T. (1) Let In ∈ Cn (∆n ) denote the identity map of the standard nsimplex ∆n . It is enough to define KIn ∈ Cn+1 (∆n ) such that ∂(KIn ) = In − T In − K∂In .

(2)

If we have solved equation (2), then for σ : ∆n → X we can define Kσ := σ# KIn , 3

and equation (1) will hold. We solve equation (2) by induction on n. Since Hn+1 (∆n ) = 0, a solution KIn to equation (2) exists iff the right hand side is a cycle. To check this, we compute ∂(In − T In − K∂In ) = ∂In − ∂T In − ∂K∂In = ∂In − ∂T In − (1 − T − K∂)∂In = −∂T In + T ∂In = 0. (This argument, and similar arguments which we did in class, can be formalized as the “acyclic models theorem”.)

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