Exercise 7.70#

Suppose \(𝑗: \Omega β†’ \Omega\) is a morphism of sheaves on \(𝑋\), such that \(𝑝 ≀ 𝑗(𝑝)\) holds for all \(π‘ˆ βŠ† 𝑋\) and \(𝑝 ∈ \Omega(π‘ˆ)\). Show that for all \(π‘ž ∈ \Omega(π‘ˆ)\) we have \(𝑗(𝑗(π‘ž)) ≀ 𝑗(π‘ž)\) iff \(𝑗(𝑗(π‘ž)) = 𝑗(π‘ž)\).

Author’s solution#

It is clear that if \(𝑗(𝑗(π‘ž)) = 𝑗(π‘ž)\) then \(𝑗(𝑗(π‘ž)) ≀ 𝑗(π‘ž)\) by reflexivity. On the other hand, assume the hypothesis, that \(𝑝 ≀ 𝑗(𝑝)\) for all \(π‘ˆ βŠ† 𝑋\) and \(𝑝 ∈ \Omega(π‘ˆ)\). If \(𝑗(𝑗(π‘ž)) ≀ 𝑗(π‘ž)\), then letting \(𝑝 ≔ 𝑗(π‘ž)\) we have both \(𝑗(𝑝) ≀ 𝑝\) and \(𝑝 ≀ 𝑗(𝑝)\). This means \(p β‰… j(p)\), but \(\Omega\) is a poset (not just a preorder) so \(𝑝 = 𝑗(𝑝)\), i.e. \(𝑗(𝑗(π‘ž)) = 𝑗(π‘ž)\) 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).