Exercise 7.70

Suppose \(𝑗: \Omega → \Omega\) is a morphism of sheaves on \(𝑋\), such that \(𝑝 ≤ 𝑗(𝑝)\) holds for all \(𝑈 ⊆ 𝑋\) and \(𝑝 ∈ \Omega(𝑈)\). Show that for all \(𝑞 ∈ \Omega(𝑈)\) we have \(𝑗(𝑗(𝑞)) ≤ 𝑗(𝑞)\) iff \(𝑗(𝑗(𝑞)) = 𝑗(𝑞)\).

Author’s solution

It is clear that if \(𝑗(𝑗(𝑞)) = 𝑗(𝑞)\) then \(𝑗(𝑗(𝑞)) ≤ 𝑗(𝑞)\) by reflexivity. On the other hand, assume the hypothesis, that \(𝑝 ≤ 𝑗(𝑝)\) for all \(𝑈 ⊆ 𝑋\) and \(𝑝 ∈ \Omega(𝑈)\). If \(𝑗(𝑗(𝑞)) ≤ 𝑗(𝑞)\), then letting \(𝑝 ≔ 𝑗(𝑞)\) we have both \(𝑗(𝑝) ≤ 𝑝\) and \(𝑝 ≤ 𝑗(𝑝)\). This means \(p ≅ j(p)\), but \(\Omega\) is a poset (not just a preorder) so \(𝑝 = 𝑗(𝑝)\), i.e. \(𝑗(𝑗(𝑞)) = 𝑗(𝑞)\) as desired.

Alternative answer

Assume \(j(j(q)) ≤ j(q)\); we must show \(j(j(q)) = j(q)\). Call \(r = j(q)\) so that we know \(j(r) ≤ r\). We were also given \(p ≤ j(p)\) for any \(p\), so substituting \(r\) for \(p\) we know \(r ≤ j(r)\). Therefore \(r = j(r)\) or \(j(q) = j(j(q))\).

We must also show that with the assumption \(j(j(q)) = j(q)\) we also have that \(j(j(q)) ≤ j(q)\). This is rather trivial; if the two terms are equal then they are also less than or equal to each other (in either direction).