# Is the free R-module on a sheaf of sets still a sheaf?

Let $L$ be a sheaf of sets on some site $S$. Let $F$ be the presheaf obtained by composing $L$ with the free R-module functor, i.e. for any object $U$, we define $F(U)$ to be the free $R$-module on the set $L(U)$. Is $F$ a sheaf?

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The second sheaf condition fails in general. –  Martin Brandenburg Nov 14 '11 at 16:53

Suppose $S$ is a topological space which has two non-empty open subsets $U$ and $V$ with an empty intersection.
Let $L$ be the sheaf of sets that assigns to each open set the singleton set $\lbrace *\rbrace$.
Let $R$ be the integers. Then the presheaf you describe assigns the integers to every open set, but this is not a sheaf because, for example, the section 2 (over $U$) and the section 3 (over $V$) agreee by default on $U\cap V$ but can't be patched together.