1. Let \\(x\in\uparrow p\\), \\(y\in P\\), and \\(x\leq y\\). By definition of \\(\uparrow p\\), \\(p\leq x\\). By transitivity, this implies that \\(p\leq y\\). So, \\(y\in\uparrow p\\).
2. \\(P^\mathrm{op}\\) has the same elements as \\(P\\), so \\(\uparrow\\) maps \\(P^\mathrm{op}\\) to \\(\mathcal{U}(P)\\). Let \\(p,q\in P\\) with \\(q\leq_P p\\). By definition of \\(^\mathrm{op}\\), \\(p\leq_{P^\mathrm{op}}q\\). For every \\(x\in\uparrow q\\), \\(q\leq x\\). By transitivity, \\(p\leq x\\). So, \\(\uparrow q\subseteq\uparrow p\\). Thus, \\(\uparrow q\leq\uparrow p\\). Conclude that \\(\uparrow:P^\mathrm{op}\to\mathcal{U}(P)\\) is a monotone map.
3. \\(\uparrow(a)=\\{a,b,c\\}\\), \\(\uparrow(b)=\\{b\\}\\), \\(\uparrow(c)=\\{c\\}\\), \\(\varnothing\leq\uparrow(b)\leq\uparrow(a)\\), and \\(\varnothing\leq\uparrow(c)\leq\uparrow(a)\\).