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HOMEWORK 7 Problem 1: Let X be an arbitrary nonempty set. Compute the singular homology of X equipped with the trivial topology {∅, X} and of X equipped with the discrete topology. Solution: Let Xtriv and Xdisc denote the topological space that consists of X as a set together with the trivial and the discrete topology, respectively. For Xtriv observe that Xtriv is homotopy equivalent to a singleton space, so H0 (Xtriv ) = Z whereas Hi (Xtriv ) = 0 for i > 0. Indeed, fix any element x0 ∈ X and define ρ : X → X, ρ(x) = x0 for every x ∈ X. Then ρ is homotopic to idX (which proves that ρ is a deformation retraction onto its image) through the homotopy F : [0, 1] × X → X defined by F (0, x) = x0 and F (t, x) = x for t > 0; this is continuous as a map of topological spaces because every map Y → Xtriv is continuous. For Xdisc note that a discrete topological space is homeomorphic to the disjoint union of its points viewed as singleton spaces. Then Hatcher Proposition 2.6 tells us that (L M x∈X Z (i = 0) Hi (Xdisc ) = Hi ({x}) = {0} (i > 0). x∈X Observe that H0 (Xdisc ) is nothing else but the free abelian group generated by X, i.e. the group of all finite formal integer linear combinations of elements of X. Said differently, H0 (Xdisc ) = C0 (Xτ ) where Xτ denotes the set X together with an arbitrary topology τ . Remark: The discussion of Xtriv here is meant to illustrate a comment from lectures—if we drop the assumption of continuity from the definition of singular simplices, then the resulting “homology theory” is unable to distinguish any nonempty space from a singleton, in essence because the proof of homotopy invariance of H• does not use continuity in any meaningful way. Problem 2: Let U and V be two path-connected open subsets of Rn such that U ∪ V = Rn . Show that U ∩ V is path-connected. Solution: This is a straight-up Mayer-Vietoris question. We have a long exact sequence · · · → H1 (Rn ) → H0 (U ∩ V ) → H0 (U ) ⊕ H0 (V ) → H0 (Rn ) → 0. Now H1 (Rn ) = 0 since Rn is contractible (or even only since Rn is simply-connected) so H0 (U ∩ V ) injects into H0 (U ) ⊕ H0 (V ) with image equal to the kernel of the surjection φ : H0 (U ) ⊕ H0 (V ) → H0 (Rn ). Now H0 (U ) ∼ = H0 (V ) ∼ = H0 (Rn ) ∼ = Z since all three spaces are path-connected; conversely H0 (U ∩ V ) ∼ = Z would imply that U ∩ V is path-connected by Hatcher Proposition 2.6. It seems intuitively obvious that a surjection Z ⊕ Z → Z must have a “1-dimensional” kernel. In fact, this would be obvious from the standard rank-nullity theorem if we were using homology with coefficients in the abelian group G = (k, +), where k is a field—for example k = Q (so this is then one way of answering the question, assuming that Mayer-Vietoris works with G-coefficients). Alternatively “rank-nullity for abelian groups” (Google!) shows that H0 (U ∩ V ) has rank 1, and since H0 (U ∩ V ) is torsion-free by Hatcher Proposition 2.6, H0 (U ∩ V ) ∼ = Z. Arguably the least pretentious way out is to note that φ(a, b) = a − b if we use augmentation to identify H0 of each of U, V, Rn with Z, and so then obviously ker(φ) = {(a, a) : a ∈ Z} ∼ = Z. Note to self: There has to be some way of writing this solution without MV/homology. Problem 3: Let X = S 1 × S 1 be the torus and Y = S 1 × B 2 the solid torus. Compute the induced homomorphisms on H1 of the following two continuous maps: (a) f : X → X, f (z, w) = (z a wb , z c wd ) with a, b, c, d ∈ Z. (b) The inclusion map i : X → Y of X as the boundary of Y . 2 HOMEWORK 7 Solution: It suffices to compute the induced maps on π1ab , or equivalently on π1 since all fundamental groups in this question are abelian. (a) We know that π1 (S 1 × S 1 ) ∼ = Z2 via the isomorphism whose inverse sends (m, n) ∈ Z2 to the homotopy class of the loop t 7→ (e2πimt , e2πint ). Thus (exactly as in Homework 3) it follows straight from the definitions that f∗ (m, n) = (am + bn, cm + dn), or in other words f∗ = ( ac db ). (b) Similarly, what we have proved about fundamental groups is enough to deduce that the map Z → π1 (S 1 × B 2 ), m 7→ [t 7→ (e2πimt , 1)], is a group isomorphism (using that B 2 is contractible). It is then once again immediate from the definitions that i∗ (m, n) = m.