Using the notation of the five-lemma, give an example where the maps \(\alpha, \beta, \delta\), and \(\varepsilon\) are zero but \(\gamma\) is nonzero. This can be done with short exact sequences in which all the groups are either \(\mathbb{Z}\) or \(0\).
Solution
Use the five-lemma notation \[ \begin{array}{ccccccccc} A&\longrightarrow&B&\longrightarrow&C&\longrightarrow&D&\longrightarrow&E\\ \downarrow\alpha&&\downarrow\beta&&\downarrow\gamma&&\downarrow\delta&&\downarrow\varepsilon\\ A'&\longrightarrow&B'&\longrightarrow&C'&\longrightarrow&D'&\longrightarrow&E'. \end{array} \] Take the upper exact sequence to be \[ 0\longrightarrow0\longrightarrow\mathbb Z \xrightarrow{\operatorname{id}}\mathbb Z\longrightarrow0 \] and the lower exact sequence to be \[ 0\longrightarrow\mathbb Z \xrightarrow{\operatorname{id}}\mathbb Z \longrightarrow0\longrightarrow0. \] Thus \[ (A,B,C,D,E)=(0,0,\mathbb Z,\mathbb Z,0) \] and \[ (A',B',C',D',E')=(0,\mathbb Z,\mathbb Z,0,0). \] Define \[ \alpha=0, \qquad \beta=0, \qquad \gamma=\operatorname{id}_{\mathbb Z}, \qquad \delta=0, \qquad \varepsilon=0. \]
1. Both rows are exact.
In the upper row, exactness is immediate at the zero groups, while at the two copies of \(\mathbb Z\) it follows because the middle map is the identity. In the lower row the same argument applies to the identity map \(\mathbb Z\to\mathbb Z\).
2. The diagram commutes.
Every square involving one of \(\alpha,\beta,\delta,\varepsilon\) has both composites equal to zero. For the square containing \(\gamma\), the upper map into \(C\) is the zero map \(0\to\mathbb Z\), so both composites there are zero; for the square leaving \(C\), the lower map \(C'\to D'\) is zero, so both composites are again zero.
3. The four outer vertical maps are zero but the middle map is nonzero.
By construction \[ \alpha=\beta=\delta=\varepsilon=0, \] whereas \[ \gamma=\operatorname{id}_{\mathbb Z}\ne0. \]
This gives the requested example using only the groups \(\mathbb Z\) and \(0\).