By Morgan J.W., Lamberson P.J.

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Then by exactness at Bn−1 , since bn−1 ∈ Ker(gn−1 ) = Im(fn−1 ) there is an element an−1 ∈ An−1 so that fn−1 (an−1 ) = ∂bn . We need to show that an−1 is a cycle. By commutativity of the diagram fn−2 (∂an−1 ) = ∂fn (an−1 ) = ∂∂bn−1 = 0. Then by exactness at An−1 , fn−1 is an injection, so ∂an−1 = 0. Thus, an−1 is a cycle. We made a choice when we picked a bn in the pre-image of c. If we choose another element b′n so that gn (b′n ) = c, we have gn (b′n − bn ) = gn (b′n ) − gn (bn ) = c − c = 0. That is, bn and b′n differ by an element in the Kergn =Imfn .

Furthermore, if S n − ψ(S n−1 ) = X ∐ Y with X and Y open, then ψ(S n−1 ) = X ∩ Y . ˜ 0 (S n − ψ(S n−1 )) ∼ Proof. By the theorem, H = Z and hence, H0 (S n − ψ(S n−1 )) ∼ = Z ⊕ Z. n n−1 Therefore S − ψ(S ) has two path components, but since S n is locally path connected, so is the open subset S n − ψ(S n−1 ) and thus path components are components. ¯ ⊂ S n − Y since Now, suppose S n − ψ(S n−1 ) = X ∐ Y with X and Y open, then X ¯ ⊂ [X ∪ ψ(S n−1 )] X ⊂ S n − Y and Y is open. Similarly, Y¯ ⊂ S n − X.

First we will define a map sd, which given an n-simplex ∆n , assigns a chain in Sn (∆n ). In order to define this map it will be convinient to think of the n-simplex in three equivalent ways. We can write the n-simplex as a sequence of inclusions of faces ∆n = τ 0 ⊂ τ 1 ⊂ · · · ⊂ τ n , where τ i is an i-face of ∆n . There are exactly (n + 1)! ways of doing this corresponding to the (n + 1)! elements of the permutation group Σn+1 of the n + 1 vertices of ∆n . To see this, we give a correspondance between ordered lists of the n + 1 vertices, and the expression of ∆n as a sequence of inclusions of faces.