Algebraic topology by Morgan J.W., Lamberson P.J.

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

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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 . So we have b′n = bn + fn (an ) for some an ∈ An . Following this through our construction, we see that this changes an−1 to an−1 + ∂an , and thus does not change our map on homology. Now, suppose we choose a different cycle representative for [c], say c + ∂c′n+1 for some ′ cn+1 ∈ Cn+1 .

We will construct H(∆n ) ∈ Sn+1 (∆n ) by induction on n, and then define H( aσ σ) = aσ σ∗ (H(∆n )). For the initial case n = 0, we have sd(∆0 ) = ∆0 and H(∆0 ) = 0. Suppose that n > 0 and that for all k < n we have H(∆k ) defined with ∂H(∆k ) = sd(∆k ) − ∆k − H(∂∆k ). Consider sd(∆n ) − ∆n − H(∂∆n ) ∈ Sn (∆n ). We want to show that this is a cycle. We have, ∂ sd(∆n ) − ∆n − H(∂∆n ) = sd(∂∆n ) − ∂∆n − ∂H(∂∆n ). The inductive hypothesis gives ∂H(∂∆n ) = sd(∂∆n ) − ∂∆n − H(∂∂∆n ) = sd(∂∆n ) − ∂∆n Thus, ∂ sd(∆n ) − ∆n − H(∂∆n ) = sd(∂∆n ) − ∂∆n − sd(∂∆n ) + ∂∆n = 0.

We make use of the following lemma. 5. The diameter of every simplex in sd(∆n ) ≤ n n+1 diam(∆n ). Proof. The details of the proof are left to the reader. The general idea is to first reduce to showing that the distance from any vertex vi of ∆n = [v0 , . . , vn ] to the barycenter b n diam(∆n ). Then, let bi be the barycenter of the face fi = is less than or equal to n+1 1 n [v0 , . . , vˆ1 , . . , vn ] of ∆n . Then, b = n+1 vi + n+1 bi . Notice that this implies b lies on the line segment between vi and bi , and the distance from vi to b is n/n + 1 times the length of [vi , bi ].

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