Return to Answer

I think that we have

$$H_2(\mathbf{Z} \times \mathbf{Z}, \, \mathbf{Z}) \simeq \mathbf{Z}, \quad H^2(\mathbf{Z} \times \mathbf{Z}, \, \mathbf{C}^{\times})\simeq \mathbf{C}^{\times}.$$

In general, by the Universal Coefficient Theorem there is an isomorphism $$H^2(G, \, \mathbf{C}^{\times})=H_2(G, \, \mathbf{Z})^*$$ Now, it can happen that an infinite group is not isomorphic to its dual. For instance, in your case we have $$\mathbf{C}^{\times}=H^2(\mathbf{Z} \times \mathbf{Z}, \, \mathbf{C}^{\times})=H_2(\mathbf{Z} \times \mathbf{Z}, \, \mathbf{Z})^*=\mathrm{Hom}_{\mathbf{Z}}(\mathbf{Z}, \, \mathbf{C}^{\times}).$$

The isomorphism $\mathrm{Hom}_{\mathbf{Z}}(\mathbf{Z}, \, \mathbf{C}^{\times}) \simeq \mathbf{C}^{\times}$ is explicitly given by associating to every element $h \in \mathrm{Hom}_{\mathbf{Z}}(\mathbf{Z}, \, \mathbf{C}^{\times})$ its value in $1$, i.e., $h \mapsto h(1)$.

I think that we have

$$H_2(\mathbf{Z} \times \mathbf{Z}, \, \mathbf{Z}) \simeq \mathbf{Z}, \quad H^2(\mathbf{Z} \times \mathbf{Z}, \, \mathbf{C}^{\times})\simeq \mathbf{C}^{\times}.$$

In general, by the Universal Coefficient Theorem there is an isomorphism $$H^2(G, \, \mathbf{C}^{\times})=H_2(G, \, \mathbf{Z})^*$$ Now, it can happen that an infinite group is not isomorphic to its dual. For instance, in your case we have $$\mathbf{C}^{\times}=H^2(\mathbf{Z} \times \mathbf{Z}, \, \mathbf{C}^{\times})=H_2(\mathbf{Z} \times \mathbf{Z}, \, \mathbf{Z})^*=\mathrm{Hom}_{\mathbf{Z}}(\mathbf{Z}, \, \mathbf{C}^{\times}).$$

The isomorphism $\mathrm{Hom}_{\mathbf{Z}}(\mathbf{Z}, \, \mathbf{C}^{\times}) \simeq \mathbf{C}^{\times}$ is explicitly given by associating to every element $h \in \mathrm{Hom}_{\mathbf{Z}}(\mathbf{Z}, \, \mathbf{C}^{\times})$ its value in $1$, namely $h \mapsto h(1)$.

Both calculations are correct.

In general, there is an isomorphism $$H^2(G, \, \mathbf{C}^{\times})=H_2(G, \, \mathbf{Z})^*$$ Now, it can happen that an infinite group is not isomorphic to its dual. For instance, in your case we have $$\mathbf{C}^{\times}=H^2(\mathbf{Z} \times \mathbf{Z}, \, \mathbf{C}^{\times})=H_2(\mathbf{Z} \times \mathbf{Z}, \, \mathbf{Z})^*=\mathrm{Hom}_{\mathbf{Z}}(\mathbf{Z}, \, \mathbf{C}^{\times}).$$

I think that we have

$$H_2(\mathbf{Z} \times \mathbf{Z}, \, \mathbf{Z}) \simeq \mathbf{Z}, \quad H^2(\mathbf{Z} \times \mathbf{Z}, \, \mathbf{C}^{\times})\simeq \mathbf{C}^{\times}.$$

In general, by the Universal Coefficient Theorem there is an isomorphism $$H^2(G, \, \mathbf{C}^{\times})=H_2(G, \, \mathbf{Z})^*$$ Now, it can happen that an infinite group is not isomorphic to its dual. For instance, in your case we have $$\mathbf{C}^{\times}=H^2(\mathbf{Z} \times \mathbf{Z}, \, \mathbf{C}^{\times})=H_2(\mathbf{Z} \times \mathbf{Z}, \, \mathbf{Z})^*=\mathrm{Hom}_{\mathbf{Z}}(\mathbf{Z}, \, \mathbf{C}^{\times}).$$

The isomorphism $\mathrm{Hom}_{\mathbf{Z}}(\mathbf{Z}, \, \mathbf{C}^{\times}) \simeq \mathbf{C}^{\times}$ is explicitly given by associating to every element $h \in \mathrm{Hom}_{\mathbf{Z}}(\mathbf{Z}, \, \mathbf{C}^{\times})$ its value in $1$, i.e., $h \mapsto h(1)$.

Both calculations are correct.

In general, there is an isomorphism $$H^2(G, \, \mathbb{C}^{\times})=H_2(G, \, \mathbf{Z})^*$$ Now, it can happen that an infinite group is not isomorphic to its dual. For instance, in your case we have $$\mathbf{C}^{\times}=H^2(\mathbf{Z} \times \mathbf{Z}, \, \mathbf{C}^{\times})=H_2(\mathbf{Z} \times \mathbf{Z}, \, \mathbf{Z})^*=\mathrm{Hom}_{\mathbf{Z}}(\mathbf{Z}, \, \mathbf{C}^{\times}).$$

Both calculations are correct.

In general, there is an isomorphism $$H^2(G, \, \mathbf{C}^{\times})=H_2(G, \, \mathbf{Z})^*$$ Now, it can happen that an infinite group is not isomorphic to its dual. For instance, in your case we have $$\mathbf{C}^{\times}=H^2(\mathbf{Z} \times \mathbf{Z}, \, \mathbf{C}^{\times})=H_2(\mathbf{Z} \times \mathbf{Z}, \, \mathbf{Z})^*=\mathrm{Hom}_{\mathbf{Z}}(\mathbf{Z}, \, \mathbf{C}^{\times}).$$