First, assume that we know that if the open set $A$ is connected then $A$ is homologically trivial in dimension 0 (All closed 0-forms on $A$ are exact). I'd like to show that if open $A$ is homologically trivial in dimension 0 then $A$ is connected. So my approach is to do so via the contrapositive. Please comment on whether my approach is reasonable.

Question

anonymous · Answer

We wish to show that if open $A$ is disconnected there exists a closed 0-form that is not exact on $A$. If open $A$ is disconnected, we can decompose it into disjoint connected open sets $\{U_i\}$ where $\bigcup_i U_i = A$. By the previous assumption each $U_i$ is homologically trivial. Therefore any closed 0-form on a given $U_i$ will be exact (and therefore constant). So then we construct our closed but not exact 0-form as follows. Simply let it take on a distinct constant on each $U_i$ such that $f(x) = c_i$ when $x \in U_i$. Clearly it is not exact on $A$ since it is not constant on all of $A$. So we need only show that it is closed. Since $f$ is constant on each $U_i$ and therefore $df = 0$ on each $U_i$ and the union is $A$ we see that $df = 0$ on all of $A$ and therefore it is closed as we wanted.

Is this a reasonable approach to the problem?