I'm asked to show that in singular homology and cohomology $H_0(\mathbb{Q},\mathbb{Z}) \ncong H^0(\mathbb{Q},\mathbb{Z})$, $\mathbb{Q}$ being endowed with the subspace topology from $\mathbb{R}$.
Since $\mathbb{Q}$ is totally disconnected, the zeroth homology is just the free $\mathbb{Z}$-module on the points of $\mathbb{Q}$, while one can show either directly or via the universal coefficient theorem that $H^0$ is given by the functions from path components, i.e., points of $\mathbb{Q}$, to $\mathbb{Z}$. So…it looks to me like I could give an isomorphism mapping $\sum n_iq_i \mapsto f, f(q_i)=n_i, q_i \in \mathbb{Q}, n_i \in \mathbb{Z}$, extending to an isomorphism of modules/abelian groups by linearity. What am I missing here?
Best Answer
Essentially, $$H_0(\mathbb Q;\mathbb Z)=\oplus_{q\in\mathbb Q} \mathbb Z$$ while $$H^0(\mathbb Q;\mathbb Z)=\prod_{q\in\mathbb Q}\mathbb Z$$
In particular, $H_0$ is countable, while $H^0$ is uncountable.