**Proof.**
It is clear that (2) implies (1). To show that (1) implies (2) we argue by induction on $n$. The first case is $n = 2$ (since $n = 1$ is trivial). In this case we have $a_1f_1 + a_2f_2 = 1$ for some $a_1, a_2 \in \mathcal{O}(U)$. By assumption we can find a covering $\{ U_ j \to U\} $ such that for each $j$ either $a_1f_1|_{U_ j}$ is invertible or $a_2f_2|_{U_ j}$ is invertible. Hence either $f_1|_{U_ j}$ is invertible or $f_2|_{U_ j}$ is invertible as desired. For $n > 2$ we have $a_1f_1 + \ldots + a_ nf_ n = 1$ for some $a_1, \ldots , a_ n \in \mathcal{O}(U)$. By the case $n = 2$ we see that we have some covering $\{ U_ j \to U\} _{j \in J}$ such that for each $j$ either $f_ n|_{U_ j}$ is invertible or $a_1f_1 + \ldots + a_{n - 1}f_{n - 1}|_{U_ j}$ is invertible. Say the first case happens for $j \in J_ n$. Set $J' = J \setminus J_ n$. By induction hypothesis, for each $j \in J'$ we can find a covering $\{ U_{jk} \to U_ j\} _{k \in K_ j}$ such that for each $k \in K_ j$ there exists an $i \in \{ 1, \ldots , n - 1\} $ such that $f_ i|_{U_{jk}}$ is invertible. By the axioms of a site the family of morphisms $\{ U_ j \to U\} _{j \in J_ n} \cup \{ U_{jk} \to U\} _{j \in J', k \in K_ j}$ is a covering which has the desired property.

Assume (1). To see that the map in (3) is surjective, let $(f, c)$ be a section of $\mathcal{O} \times \mathcal{O}$ over $U$. By assumption there exists a covering $\{ U_ j \to U\} $ such that for each $j$ either $f$ or $1 - f$ restricts to an invertible section. In the first case we can take $a = c|_{U_ j} (f|_{U_ j})^{-1}$, and in the second case we can take $b = c|_{U_ j} (1 - f|_{U_ j})^{-1}$. Hence $(f, c)$ is in the image of the map on each of the members. Conversely, assume (3) holds. For any $U$ and $f \in \mathcal{O}(U)$ there exists a covering $\{ U_ j \to U\} $ of $U$ such that the section $(f, 1)|_{U_ j}$ is in the image of the map in (3) on sections over $U_ j$. This means precisely that either $f$ or $1 - f$ restricts to an invertible section over $U_ j$, and we see that (1) holds.
$\square$

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