By Christoph Schweigert

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**Extra info for Algebraic Topology [Lecture notes]**

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The restriction of the differential of Sn (X) gives a chain complex U . . → SnU (X) → Sn−1 (X) → . . We denote its homology by HnU (X). As we will see now, these chains suffice to detect everything in singular homology. 12. 1. e. B k (c) ∈ Sn (A). 2. If c ∈ Sn (X) is a cycle relative A ⊂ X, then B(c) is a cycle relative A as well that is homologous to c relative A. 3. Let U be an open covering of X. Then every cycle in Sn (X) is homologous to a cycle in SnU (X). Proof. 1. This follows at once from the definition of barycentric subdivision.

If Z is locally compact and all spaces are Hausdorff, there is a homeomorphism C(X × Z, W ) ∼ = C(X, C(Z, W )) 50 (∗) of topological spaces. Here, for f : X × Z → W we consider for any x ∈ X f # (x) : Z → W z → f (x, z) and send f to f # ∈ C(X, C(Z, W )). • Using these facts, we show the following Lemma: Let X, Y and Z be topological spaces satisfying the Hausdorff condition and suppose that π : X → Y gives Y the quotient topology and that Z is locally compact. Then π × id : X × Z → Y × Z gives Y × Z the quotient topology.

The map φˉk sends X+ and X− to [X+ ]. 4 deg(φˉk ) = deg(F ◦ (id ∨ A) ◦ T ) = deg(id) + deg(A) = 1 + (−1)i . and d is either zero or two. For the cellular complex, we find ∙2 0 0 ... → Z → Z → Z → 0 . Thus, depending on n we get k=0 Z n Hk (RP ) = Z/2Z k n, k odd 0 otherwise. for n even. For odd dimensions n we get k = 0, n Z n Hk (RP ) = Z/2Z 0 < k < n, k odd 0 otherwise. Note that RP 1 ∼ = S1 and RP 3 ∼ = SO(3). 13 Homology with coefficients Let G be an arbitrary abelian group.