Weibel 1.1 Complexes of $R$-Modules
Definition 1.1.1 (Chain complex)
[definition] A chain complex $C_\bullet$ of $R$-modules is a family $\{C_n\}_{n\in \ZZ}$ of $R$-modules, together with $R$-module maps $d = d_n:C_n \to C_{n-1}$ such that $d^2 = 0$.
The maps $d_n$ are called the differentials of $C_\bullet$.
The kernel of $d_n$ is the module of $n$-cycles of $C_\bullet$, denoted $Z_n = Z_n(C_\bullet)$.
The image of $d_{n+1}$ is the module of $n$-boundaries of $C_\bullet$, denoted $B_n(C_\bullet)$. Notice that since we have $d^2 = 0$,
\[0 \subseteq B_n \subseteq Z_n \subseteq C_n.\]Finally, the $n$th homology module of $C$ is the quotient $H_n(C_\bullet) = Z_n/B_n$ of $C_n$. [/definition]
Typically, in practice, when we write out a chain complex, we will write it as
\[C_\bullet: \quad \cdots \to C_{n+1} \xrightarrow{d_{n+1}} C_n \xrightarrow{d_n} C_{n-1} \to \cdots\]Exercise 1.1.1
[exercise] Set
\[C_n = \begin{cases} \ZZ_8 & \text{ if } n \ge 0, \\ 0 & \text{ if } n < 0, \end{cases}\]and
\[d_n(x \mod 8) = \begin{cases} 4x \pmod 8 & \text{ if } n > 0, \\ 0 \pmod 8 & \text{ if } n \le 0. \end{cases}\]Show that $C_\bullet$ is a chain complex of $\ZZ_8$ modules and compute its homology. [/exercise]
[proof=Solution] The chain complex we have here is
\[\cdots \xrightarrow{\times 4} \ZZ_8 \xrightarrow{\times 4} \ZZ_8 \xrightarrow{0} 0 \xrightarrow{0} 0 \to \cdots.\]Obviously $d^2 = 0$ (regardless of what $n$ is) since $4^2$ is a multiple of $8$. There are two homology modules to consider. At $n = 0$, we have are focusing on the degrees
\[\cdots \to \ZZ_8 \xrightarrow{\times 4} \ZZ_8 \to 0 \to \cdots.\]So then $Z_0 = \ker(0) = \ZZ_8$ and $B_0 = \{0, 4\}$. So then we have
\[H_0 = \frac{Z_0}{B_0} = \frac{\ZZ/8\ZZ}{4\ZZ/8\ZZ} \cong \ZZ/4\ZZ\]by the third isomorphism theorem.
At $n > 0$, we are focusing on the degrees
\[\cdots \to \ZZ_8 \xrightarrow{\times 4} \ZZ_8 \xrightarrow{\times 4} \ZZ_8 \to \cdots.\]Thus,
\[H_{n} = \frac{\ker (\times 4)}{\im (\times 4)} = \frac{2\ZZ/8\ZZ}{4\ZZ/8\ZZ} \cong \ZZ_2.\][/proof]
Definition (Category of chain complexes)
The category $\mathcal{Ch}(\text{mod } R)$ of chain complexes of (right) $R$-modules has as objects the chain complexes and morphisms $u:C_\bullet \to D_\bullet$, called chain maps, which are families of $R$-modules homomorphisms $u_n:C_n \to D_n$ that commute with the differential.
Exercise 1.1.2 (Homology functor)
[exercise] Show that a morphism $u:C_\bullet \to D_\bullet$ of chain complexes sends boundaries to boundaries and cycles to cycles, hence maps $H_n(C_\bullet) \to H_n(D_\bullet)$. Prove that each $H_n$ is a functor from $\mathcal{Ch}(\text{mod }R)$ to $\text{mod } R$. [/exercise]
[proof] ($Z_n(C) \to Z_n(D)$) Let $x \in Z_n(C)$. By definition, $d_n(x) = 0$ and so
\[d_n\circ u_n (x) = u_n \circ d_n(x) = 0.\]Thus, $u_n(x) \in Z_n(D)$.
($B_n(C) \to B_n(D)$) Similar argument.
($H_n$ is a functor) By the two above parts, $u$ induces a map on homology which is $H_n(u)$. Clearly this map respects compositions and preserves the identity. [/proof]
Exercise 1.1.3 (Split exact sequences of vector spaces)
[exercise] Choose vector spaces $\{B_n, H_n\}_{n\in \ZZ}$ over a field, and set $C_n = B_n \oplus H_n \oplus B_{n-1}$. Show that the projection-inclusions $C_n \to B_{n-1} \subseteq C_{n-1}$ make $\{C_n\}$ into a chain complex and that every chain complex of vector spaces is isomorphic to a complex of this form. [/exercise]
As very suggested by the choice of notation, we will define
\[d_n:C_n \to C_{n-1}\]by
\[(b_n, h_n, b_{n-1}) \mapsto (0, 0, b_{n-1}).\]Then $B_n$