Say I have an open cover of the interval $(0,1)$. I'll call this open cover $\{V_j\}_{j \in J}$. On each of these open sets there is a continuous, nonzero, complex-valued function $f_j: V_j \to \mathbb{C} \setminus \{0\}$. Is it possible to find a partition of unity $\{p_j\}_{j \in J}$ subordinate to the open cover $\{V_j\}_{j \in J}$ such that the continuous complex-valued function $\sum_j p_j\, f_j $ is never zero?
If it's possible but not easy to see, then I wouldn't mind if someone just provides reference(s)?
It's not always possible. Here's a counterexample. Take $V_1 = (0,\frac23)$, $V_2 = (\frac13,1)$, $f_1 \equiv 1$, and $f_2\equiv -1$. Then, whatever $p_1$ and $p_2$ are, we have $\sum_j p_j f_j = p_1 - p_2$, which is real-valued. Since $p_1-p_2 \equiv p_1\equiv 1$ on $(0,\frac13)$ and $p_1 - p_2 \equiv -p_2 = -1$ on $(\frac23,1)$, by the intermediate value theorem there must be a point $x\in (\frac13,\frac23)$ where $p_1(x)-p_2(x)=0$.