By Christoph Schweigert

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

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To establish that X × Y has the weak topology, we can show that X × Y carries the quotient topology with respect to this map. We know that each Dnσ is locally compact, thus so is the disjoint union of closed discs. The map id Dnσ × Ψ gives ( Dnσ ) × Y the quotient topology and by assumption Y is locally compact and therefore, by the result of the previous point, Φ × idY induces the quotient topology on X × Y . 15. If D is a subset of a CW complex X and D intersects each cell in at most one point, then D is discrete.

Let f (n) : Sn → Sn be the map (x0 , x1 , . . , xn ) → (−x0 , x1 , . . , xn ). Then f (n) has degree −1. Proof. We prove the claim by induction. μ0 was the difference class [+1] − [−1], and f (0) ([+1] − [−1]) = [−1] − [+1] = −μ0 . We defined μn in such a way that Dμn = μn−1 . Therefore, as D is natural, Hn (f (n) )μn = Hn (f (n) )D−1 μn−1 = D−1 Hn−1 (f (n−1) )μn−1 = D−1 (−μn−1 ) = −μn . 6. The antipodal map A: has degree (−1)n+1 . S n → Sn x → −x 43 Proof. (n) Let fi : Sn → Sn be the map (x0 , .

Proof. • Consider first the case when X = Dn and A = ∂Dn = Sn−1 . For r : Dn × [0, 1] → Dn × {0} ∪ Sn−1 × [0, 1] we can choose the standard retraction of a cylinder onto its bottom and sides, cf. Figure VII-6 in Bredon, p. 451. • We inductively construct retractions ρr : X × {0} ∪ (A × I ∪ X r × I) → X × {0} ∪ A × [0, 1] , where ρr+1 extends ρr . Suppose that ρr−1 is given. Then extending to an r-cell of X amounts to extending on Dr × [0, 1] along Dr × {0} ∪ Sr−1 × [0, 1]. As we have seen, this can be done.