Set $J = (0,1)$, $K_0 = (0, \tfrac{1}{2})$, and $K_1 = (\tfrac{1}{2}, 1)$. Then $K_0$ and $K_1$ are disjoint open subintervals of $J$. On $K_0$ the map is $F(x) = 2x$, which maps $(0,\tfrac{1}{2})$ onto $(0,1) = J$. On $K_1$ the map is $F(x) = 2x - 1$, which maps $(\tfrac{1}{2},1)$ onto $(0,1) = J$. Hence $F(K_0) = F(K_1) = J$, and this is a horseshoe for $F$ with $n = 1$.