Let \(\pi: \mathbb{Z} \to \mathbb{Z}/2\mathbb{Z}\) be the canonical quotient homomorphism of \(\mathbb{Z}\)-modules. Show that the induced homomorphism on Hom groups: \[ \pi_*: \operatorname{Hom}_\mathbb{Z}(\mathbb{Z}/2\mathbb{Z}, \mathbb{Z}) \to \operatorname{Hom}_\mathbb{Z}(\mathbb{Z}/2\mathbb{Z}, \mathbb{Z}/2\mathbb{Z}), \quad f \mapsto \pi \circ f \] is not surjective, and conclude that the functor \(\operatorname{Hom}_\mathbb{Z}(\mathbb{Z}/2\mathbb{Z}, -)\) is not right exact.
Solution
1. \(\Hom_{\ZZ}(\ZZ/2\ZZ,\ZZ)=0\).
For \(f\colon\ZZ/2\ZZ\to\ZZ\), \(2f([1])=f([2])=f([0])=0\), and \(\ZZ\) has no nonzero element \(y\) with \(2y=0\), so \(f([1])=0\). Since \([1]\) generates \(\ZZ/2\ZZ\), \(f=0\).
2. \(\Hom_{\ZZ}(\ZZ/2\ZZ,\ZZ/2\ZZ)\neq0\).
It contains \(\operatorname{id}_{\ZZ/2\ZZ}\), which sends \([1]\) to \([1]\neq[0]\).
3. \(\pi_*\) is not surjective.
By step 1, the image of \(\pi_*\) is \(\{0\}\), and by step 2 the target is nonzero.
4. Q.E.D.
\(\pi\) is surjective, but by step 3 the induced map \(\pi_*\) is not, so \(\Hom_{\ZZ}(\ZZ/2\ZZ,-)\) does not preserve the exactness of \(\ZZ\xrightarrow{\pi}\ZZ/2\ZZ\to0\).