# Exercise 7.70#

Suppose $$π: \Omega β \Omega$$ is a morphism of sheaves on $$π$$, such that $$π β€ π(π)$$ holds for all $$π β π$$ and $$π β \Omega(π)$$. Show that for all $$π β \Omega(π)$$ we have $$π(π(π)) β€ π(π)$$ iff $$π(π(π)) = π(π)$$.

## Authorβs solution#

It is clear that if $$π(π(π)) = π(π)$$ then $$π(π(π)) β€ π(π)$$ by reflexivity. On the other hand, assume the hypothesis, that $$π β€ π(π)$$ for all $$π β π$$ and $$π β \Omega(π)$$. If $$π(π(π)) β€ π(π)$$, then letting $$π β π(π)$$ we have both $$π(π) β€ π$$ and $$π β€ π(π)$$. This means $$p β j(p)$$, but $$\Omega$$ is a poset (not just a preorder) so $$π = π(π)$$, i.e. $$π(π(π)) = π(π)$$ as desired.

Assume $$j(j(q)) β€ j(q)$$; we must show $$j(j(q)) = j(q)$$. Call $$r = j(q)$$ so that we know $$j(r) β€ r$$. We were also given $$p β€ j(p)$$ for any $$p$$, so substituting $$r$$ for $$p$$ we know $$r β€ j(r)$$. 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 $$j(j(q)) β€ j(q)$$. This is rather trivial; if the two terms are equal then they are also less than or equal to each other (in either direction).