# Zeroth homology and cohomology of the rationals

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?

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What is the cardinality of $H_0$? The cardinality of $H^0$? Also, for your $f$, for how many $q$ is $f(q)\neq 0$? So is your map onto? – Thomas Andrews Sep 5 '12 at 16:57
OH! Yes, OK, it's that $H_0$ is a direct sum of countably many $\mathbb{Z}$s and $H^0$ a direct product, right? If only I'd written indices on that sum. – Kevin Carlson Sep 5 '12 at 17:05
Bingo. Exactly. – Thomas Andrews Sep 5 '12 at 17:07
@KevinCarlson For simplicial homology, don't we have that it is the free $\Bbb{Z}$ - module on all maps from $\Delta ^0$ to $\Bbb{Q}$? Or have you identified each map with a point in $\Bbb{Q}$? I guess this gives us an injective map from the number of maps of $\Delta ^0$'s into $\Bbb{Q}$, so $H_0(\Bbb{Q},\Bbb{Z})$ is countable. – user38268 Sep 5 '12 at 22:22
@BenjaLim, yes, I'm just identifying the maps and their images. – Kevin Carlson Sep 6 '12 at 1:01

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.