On the wikipedia page for conservative vector fields there is a proof sketch given for the following result:
Wikipedia: Let $U$ be open subset of $\mathbb{R}^2$. Then all closed differential $1$-forms are exact if and only if $U$ is simply connected.
I assume there is a combinatorial variant for a finite simplicial complex, but I am not really comfortable with fundamental groups or the Hurewicz isomorphism, and the part of their argument regarding torsion I don't follow, so I was hoping someone could help me fill in the blanks for the following claim:
Claim: Let $K$ be a finite simplicial complex. Then $K$ is simply connected if and only if $H^1_{\Delta}(K, \mathbb{R})=0$.
The two parts I'm having trouble formalizing are (i) why a trivial fundamental group yields $H_1(K,\mathbb{Z}) = 0$ (which I suspect is trivial), and how I can get the other direction, where I need to go from $H^1(K, \mathbb{R})=0$ to $H_1(K,\mathbb{Z})= 0$. Why can't torsion be an issue?
The Hurewicz theorem states that, for a connected complex $K$, $H_1(K,\Bbb Z)$ is the Abelianisation of the fundamental group $\pi_1(K)$. We can have $H_1(K,\Bbb Z)$ trivial if say $K$ has non-Abelian simple fundamental group.
In the original example for $U$ an open subset of $\Bbb R^2$, $H_i(U,\Bbb Z)$ never has torsion, so by then universal coefficient theorem, $H_i(U,\Bbb R)=H_i(U,\Bbb Z)\otimes \Bbb R$. Also, if $U$ is not simply connected, there is a point in its complement such that a cycle in $U$ winds about it. That cycle gives a non-zero element of $H_1(U,\Bbb Z)$.