Let\(\pi: \mathbb{Z} \to \mathbb{Z}_2\) be the canonical epimorphism. Show that the induced map \(\overline{\pi}: \mathrm{Hom}(\mathbb Z_2, \mathbb Z) \to \mathrm{Hom}(\mathbb Z_2, \mathbb Z_2)\) is the zero map. Conclude that \(\overline{\pi}\) is not an epimorphism.
Thus the functor \(\operatorname{Hom}_\ZZ(\ZZ_2,-)\) applied to the short exact sequence \(0\to\ZZ\xrightarrow{2}\ZZ\xrightarrow{\pi}\ZZ_2\to0\) gives a sequence that is not exact at \(\operatorname{Hom}(\ZZ_2,\ZZ_2)\): the functor \(\operatorname{Hom}_\ZZ(M,-)\) is left exact but not right exact in general.
Solution
The induced map is post-composition, \(\overline{\pi}(\varphi) = \pi \circ \varphi\).
1. \(\operatorname{Hom}_\ZZ(\ZZ_2, \ZZ) = 0\).
Let \(\varphi \in \operatorname{Hom}_\ZZ(\ZZ_2, \ZZ)\). Since \(2 \cdot 1 = 0\) in \(\ZZ_2\), \[2 \varphi(1) = \varphi(2 \cdot 1) = \varphi(0) = 0 \quad \text{in } \ZZ.\] As \(\ZZ\) is torsion-free, \(\varphi(1) = 0\), and since \(1\) generates \(\ZZ_2\), \(\varphi = 0\).
2. \(\overline{\pi}\) is the zero map.
By step 1 the domain of \(\overline\pi\) is \(\{0\}\), and \(\overline{\pi}(0) = \pi \circ 0 = 0\).
3. \(\overline{\pi}\) is not an epimorphism.
The identity \(\operatorname{id}_{\ZZ_2}\) is a nonzero element of \(\operatorname{Hom}_\ZZ(\ZZ_2, \ZZ_2)\), since \(\operatorname{id}_{\ZZ_2}(1) = 1 \neq 0\). By step 2 the image of \(\overline{\pi}\) is \(\{0\}\), so \(\overline\pi\) is not surjective.