Is $H^0(S^0;G)\simeq G\oplus G$ or $G$?

Question

In the article on topospaces for the (co)homology of spheres, it says $H^0(S^n,G)\simeq H^n(S^n,G)\simeq G$. Is this true when $n=0$?

I think not, for if we view $S^0$ as the union of two $0$-simplices $u$ and $v$, we get the simplicial chain $$ \cdots\to 0\to 0\to\langle u,v\rangle\to 0 $$ which dualizes to $$ \cdots\leftarrow 0\leftarrow 0\leftarrow\mathrm{hom}(\langle u,v\rangle, G)\leftarrow 0 $$ where all maps everywhere are $0$. Then wouldn't $H^0(S^0)=\mathrm{hom}(\langle u,v\rangle,G)\simeq G\oplus G$?

Answer 1

$H^0(X;G)$ is a direct product of as many copies of $G$ as path-connected components $X$ has.

Your computation is correct.