MATH 5713 Section: 001 Topology II MWF 11:50 – 12:40 KIMP 308 Spring 2015 Prof. Matthew Clay Homework 6 – Solution 1. Let A ⊆ X. a. Show that H0 (X, A) = 0 if and only if A meets every path component of X. b. Show that H1 (X) → H1 (X, A) is surjective if and only if every path component of X contains at most one path component of A. Proof. Considering the long exact sequence of the pair (X, A) we have: · · · → H1 (X) → H1 (X, A) → H0 (A) → H0 (X) → H0 (X, A) → 0. Hence H0 (X, A) = 0 if and only if H0 (A) → H0 (X) is surjective. Also, H1 (X) → H1 (X, A) is surjective if and only if H0 (A) → H0 (X) is injective. The homology group H0 (X), respectively H0 (A), is a free abelian group with basis the path components of X, respectively A, and the map H0 (A) → H0 (X) is induced by inclusion of path components. Therefore H0 (A) → H0 (X) is surjective if and only if A meets every path component of X. This proves a. Further H0 (A) → H0 (X) is injective if and only if every path component of X meets at most one path component of A. This proves b. e n (X) ∼ e n+1 (SX) where SX = X × I/(X × {0}) ∪ (X × {1}) is the suspension of 2. Prove that H =H X. Proof. Consider the long exact sequence for the pair (CX, X) where CX = X × I/X × {1} and we identify X with the subspace X × {0} ⊂ CX: e n+1 (CX) → Hn+1 (CX, X) → H e n (X) → H e n (CX) → · · · ··· → H e n (CX) = 0 for all n and hence Hn+1 (CX, X) ∼ e n (X). Since As CX is contractible we have that H =H (CX, X) is a good pair (i.e., X is a deformation retract of an open set in CX, specifically the set e n+1 (SX) as CX/X = SX. X × [0, 1/2) ⊂ CX), we have that Hn+1 (CX, X) ∼ =H 3. Suppose X = _ Sα2 and A ⊂ X is a finite set of points. Compute Hn (X, A). α 1 e n (X) = ⊕α H e n (S 2 ) hence: Proof. We have that H α ( e n (X) = ⊕α Z if n = 2, H 0 else As X is path connected and A 6= ∅, we have H0 (X, A) = 0 by Problem 1. The long exact sequence for the pair (X, A) contains the exact sequence: e 1 (X) → H1 (X, A) → H e 0 (A) → H e 0 (X) = 0 0=H e 0 (A) = Zd−1 where A contains d points. For n > 1 the long exact sequence Therefore H1 (X, A) ∼ =H contains the exact sequences: e n (A) → H e n (X) → Hn (X, A) → H e n−1 (A) = 0 0=H e n (X) for n > 1. Thus we find: Therefore Hn (X, A) ∼ =H d−1 if n = 1, Z Hn (X, A) = ⊕α Z if n = 2, 0 else 4. Prove that if nonempty open sets U ⊆ Rm and V ⊆ Rn are homeomorphic, then m = n. Proof. For any x ∈ U , using X = Rm , A = Rm − {x} and Z = Rm − U , by excision we have Hk (U, U − {x}) = Hk (X − Z, A − Z) ∼ = Hk (X, A) = Hk (Rm , Rm − {x}). Considering the long e k−1 (Rm − {x}) as exact sequence for the pair (Rm , Rm − {x}) we see that Hk (Rm , Rm − {x}) ∼ =H m m m−1 e k (R ) = 0 for all k. Since R − {x} deformation retracts to S H we have Hk (U, U − {x}) ∼ = m m Hk (R , R −{x}) = Z if k = m and equals 0 otherwise. Similarly, for any y ∈ V , Hk (V, V −{y}) = Z if k = n and equals 0 otherwise. If h : U → V is a homeomorphism and x ∈ U , then h∗ : Hk (U, U − {x}) → Hk (V, V − {h(x)}) is an isomorphism for all k and thus m = n. 5. If X is an n dimensional CW–complex show that Hk (X) = 0 for k > n and that Hn (X) is free. Proof. If X is a 0 dimensional CW–complex then: ( ⊕x∈X Z if n = 0, Hn (X) = ⊕x∈X Hn (x) = 0 else Thus the statement holds in this case. Let n > 0 and assume that the statement holds for CW– complexes of dimension n − 1. We consider the long exact sequence of the pair (X, X (n−1 ). This contains the exact sequences: Hk (X (n−1) ) → Hk (X) → Hk (X, X (n−1) ) → Hk−1 (X (n−1) ) 2 e k (X/X (n−1) ). The quotient X/X (n−1) is a As (X, X (n−1) ) is a good pair we have Hk (X, X (n−1) ) = H e k (X/X (n−1) ) = ⊕α Z if k = n and equals 0 otherwise. wedge of n dimensional spheres ∨α Sαn and so H When k > n, we have Hk (X (n−1) ) = Hk−1 (X (n−1) ) = 0 by hypothesis and thus Hk (X) = Hk (X, X (n−1) ) = Hk (∨α Sαn ) = 0. When k = n we have that the map Hk (X) → Hk (X, X (n−1) ) is injective. As any subgroup of a free abelian group is free abelian, the group Hn (X) is free abelian. Thus by induction the statement holds for all n. 3

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