Consider the complex $\lbrace 1, 2, 3, 4, 12, 23, 34, 41, 13, 123 \rbrace$. Visually, it's a square with a diagonal edge, with one hole and one face.
In computing the homologies, I ended up getting that $H_{0} = R^{2}$, where $R$ is the ring of coefficients over the bases of the chain groups. I thought thatthe dimension of $H_{0}$ should always give us the number of connected components, which in this case is one. Why am I getting two?
We have $$H_0=\frac{\ker\partial_0}{\operatorname{im}\partial_1}.$$ Here, $\ker\partial_0$ is the free $R$-module generated by $1,2,3,4$ and $\operatorname{im}\partial_1$ is the submodule of $\ker\partial_0$ generated by $2-1,3-1$ and $4-1$. Therefore the elements of the quotient have the form $$\alpha 1+\beta2+\gamma3+\delta4+\operatorname{im}\partial_1=(\alpha+\beta+\gamma+\delta)1+\operatorname{im}\partial_1,\qquad\alpha,\beta,\gamma,\delta\in R.$$ Using this, we can see that $R\cong H_0$, where the isomorphism is given by $\lambda\mapsto\lambda 1+\operatorname{im}\partial_1$. (So, instead of $R^2$ you should be getting $R$.)