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Let $ U ,V \subset \mathbb R ^n$ two simply connected open sets such that $ \displaystyle{ U \cap V}$ is a connected set.

If $\omega$ is a closed 1-form wich is exact in $U$ and $V$ prove that: $\omega$ is exact in $ \displaystyle{ U \cup V}$

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I can't see why $f_1 - f_2$ is constant on $U \cap V$. – passenger Jun 17 '12 at 13:03
up vote 2 down vote accepted

We use the fact that a differential form on a simply connected open set is exact (result from Poincaré). Since $U$ and $V$ are simply connected open set, we can find functions $f_1$ and $f_2$ such that $df_1=\omega$ on $U$ and $df_2=\omega$ on $V$.

On $U\cap V$, we have $d(f_1-f_2)=df_1-df_2=\omega-\omega=0$, and since $U\cap V$ is connected, $f_1-f_2$ is constant on $U\cap V$. Therefore, we can define $F$ by $F(x)=f_1(x)-f_1(x_0)$ for $x\in U$, $F(x)=f_2(x)-f_2(x_0)$, where $x_0\in U\cap V$ (if $U$ and $V$ are disjoint, the result is clear).

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If $ x \in U \cap V$ what is F(x) ? – passenger Jun 17 '12 at 13:15
Silly question. Thank you very much for your reply! – passenger Jun 17 '12 at 13:20

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