Exercise B1 (a): (From Tony’s notes - Exercise 1A, page 3) Let C_{n} = ℤ/8 for n ≥ 0 and let C_{n} = 0 for n < 0. Let d_{n} : C_{n} → C_{n−1} be defined by \overline{x} ↦ \overline{4x} for each n. Show that \{C_{n}, d_{n}\} is a chain complex, and compute all the homology modules of this complex.

Solution: We need to show that d_{n−1} ∘ d_{n} = 0 for all n. If n > 1, then d_{n}(ℤ/8) = \{\overline{0}, \overline{4}\}. But then

d_{n−1} ∘ d_{n}(ℤ/8) = \{d_{n−1}(\overline{0}), d_{n−1}(\overline{4})\}
= \{\overline{0}\}

as desired. If n ≤ 1, d_{n−1} = 0, so we get the desired property that d_{n−1} ∘ d_{n} = 0.

We now compute the homology modules. For n > 1 we have

ker d_{n} = \{\overline{0}, \overline{2}, \overline{4}, \overline{6}\} ≅ ℤ/4
im d_{n+1} = \{\overline{0}, \overline{4}\} ≅ ℤ/2

Thus, for all n > 0, we have

H_{n}(ℤ/8) = \frac{ℤ/4}{ℤ/2} ≅ ℤ/2

For n = 0, ker d_{0} = ℤ/8 and im d_{1} = \{\overline{0}, \overline{4}\} ≅ ℤ/2. Hence, for n = 0, H_{0}(ℤ/8) = \frac{ℤ/8}{ℤ/2} ≅ ℤ/4. For n < 0 we have ker d_{n} = im d_{n+1} = 0, and thus H_{n}(ℤ/8) = 0. We summarize below:

H_{n}(ℤ/8) = \begin{cases}ℤ/2 & n > 0 \\ ℤ/4 & n = 0 \\ 0 & n < 0 \end{cases}