Chain Complexes and Exact Sequences

Chain complexes and exact sequences

Definition (Chain complex)

A chain complex is a set \(\mathcal{S} = (G_{i}, \phi_{i})_{i \in \mathbb{Z}}\) where:

1. For every \(i \in \mathbb{Z}\), \(G_{i}\) is a group,
2. For every \(i \in \mathbb{Z}\), \(\phi_{i}: G_{i} \to G_{i-1}\) is a group homomorphism.
3. \(\phi_i \circ \phi_{i+1}: G_{i+1} \to G_{i-1}\) is such that \((\phi_{i} \circ \phi_{i+1})(x) = 1_{G_{i-1}}\) for every \(x \in G_{i+1}\).

\[\cdots \to G_{3} \overset{\phi_{3}}{\to} G_{2} \overset{\phi_{2}}{\to} G_{1} \overset{\phi_{1}}{\to} G_{0} \overset{\phi_{-1}}{\to} G_{-1} \overset{\phi_{-2}}{\to} \cdots\]

Note that \((\phi_{i} \circ \phi_{i+1})(x) = \phi_{i} \big(\phi_{i+1}(x)\big)= 1_{G_{i-1}}\), hence \(\phi_{i+1}(x) \in \operatorname{Ker}(\phi_{i})\) for every \(i \in \mathbb{N}\). Let \(Z_{n}(\mathcal{S}) = \operatorname{Ker}(\phi_{n}) \subseteq G_{n}\). \(B_{n}(\mathcal{S})=\operatorname{Im}(\phi_{n+1}) \subseteq G_{n}\). Note that

\[B_{n}(\mathcal{S}) \leq Z_{n}(\mathcal{S}) \leq G_{n}\]

Definition (Exact sequence)

Take \(\mathcal{S} = (G_{i}, \phi_{i})_{i \in \mathbb{Z}}\) to be a chain complex. If for every \(i\) we have that \(\operatorname{Im}(\phi_{i+1}) = \operatorname{Ker}(\phi_{i})\). We say that \(\mathcal{S}\) is an exact sequence.

Definition (Short sequence)

A sequence is short if it is finite.

Example (Short exact sequence)

Consider

\[0 \rightarrow G_{2} \overset{ f }{\rightarrow} G_{1} \overset{ g }{\rightarrow} G_{0} \rightarrow 0,\]

where \(0\) denotes the trivial group. The unnamed homomorphisms are the trivial ones.

Theorem (Exactness of the sequence and injectivity/surjectivity)

Let \(\phi_{0}:G_{0}\to G_{1}\) be a homomorphism:

1. If \(0 \overset{ f }{\rightarrow} G_{0} \overset{ \phi_{0} }{\rightarrow} G_{1}\) is exact, then \(\phi_{0}\) is injective.
2. If \(G_{0} \overset{ \phi_{0} }{\rightarrow} G_{1} \overset{ f }{\rightarrow} 0\) is exact, then \(\phi_{0}\) is surjective.
3. If \(0 \overset{ f }{\rightarrow} G_{0} \overset{ \phi_{0} }{\rightarrow} G_{1} \overset{ g }{\rightarrow} 0\), then \(\phi_{0}\) is an isomorphism.

Proof: For (1), note that \(\operatorname{Ker}(\phi_{0}) = \operatorname{Im}(f) = \{ 1_{G_{0}} \}\), so \(\phi_{0}\) is injective. For (2), note that \(\operatorname{Im}(\phi_{0}) = \operatorname{Ker} f = G_{1}\). We combine both results for (3).

Example (Exact sequence \(\mathbb{Z} \to \mathbb{Z}_2\))

Consider

\[0 \rightarrow 2\mathbb{Z} \overset{ i }{\rightarrow} \mathbb{Z} \overset{ \pi }{\rightarrow} \mathbb{Z}_{2} \rightarrow 0,\]

where \(i(x) = x\) and \(\pi(x) = [x]\) for every \(x \in \mathbb{Z}\). It suffices to show that \(\operatorname{Im}(i) = \operatorname{Ker}(\pi)\).