By Andrey Lazarev

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Induces isomorphisms q∗ : Hn (X, A) → Hn (X/A, A/A) ∼ =H Proof. Consider the following commutative diagram: / Hn (X, V ) o Hn (X, A) q∗ Hn (X \ A, V \ A) q∗ / Hn (X/A, V /A) o Hn (X/A, A/A) q∗ Hn (X/A \ A/A, V /A \ A/A) The upper left horizontal map is an isomorphism since in the long exact sequence of the triple (X, V, A) the groups Hn (V, A) are all zero because a deformation retraction of V onto A gives a chain equivalence of complexes C∗ (V )/C∗ (A) and C∗ (A)/C∗ (A) = 0. The same argument shows that the lower left horizontal map is an isomorphism as well.

Tn−1 ). 16. If k < j the face maps satisfy n+1 n k j = n+1 n j−1 k : ∆n−1 −→ ∆n+1 . Proof. Just evaluate these affine maps on every vertex ei for 0 ≤ i ≤ n − 1. 2. Singular complex. 17. Let X be a topological space. A singular n-simplex in X is a continuous map σ : ∆n −→ X where ∆n is the standard n-simplex. 18. A singular 0-simplex in X is just a point x ∈ X. A singular 1-simplex is a path I = [0, 1] −→ X. 19. For a topological space X and an integer n ≥ 0 we define the group Cn (X) of singular n-chains in X as the free abelian group generated by all singular n-simplices in X.

V2 g g · f v0 f · v1 Further we have f + f −1 is homologous to f f −1 which is homologous to zero and it follows that f −1 is homologous to −f . Now show that h is surjective (if X is path-connected). Let ni σi be a 1-cycle representing a given homology class. After relabeling we can assume that in fact all ni s are ±1 and since inverse paths correspond to negative of the corresponding chains we can assume that all ni s are 1. If some of the σi is not a loop then since ∂( σi ) = 0 there must be another σj in the sum such that combined path σi σj is defined and we can, therefore, decrease the number of summands until all of them will be loops.