[Sophie](https://forum.azimuthproject.org/profile/2225/Sophie%20Libkind), you gave the following example in favor of the definition from Lecture 26:

> Define \\(f: \mathbb Z \to \mathbb R\\) by \\(f(x) = x+ 1\\). It has a right adjoint \\(g: \mathbb R \to \mathbb Z\\) defined by \\(g(y) = \lfloor y-1 \rfloor\\).

But it seems to me that for this particular example both definitions are consistent with the super-theorem:

- According to the definitions from lecture 26, \\(f\\) is oplax and \\(g\\) is lax (as you have just showed).

- According to the definitions from lecture 27, \\(f\\) is not oplax, \\(f(0) = 1 \nleq 0\\), and \\(g\\) is not lax, \\(0 \nleq -1 = g(0)\\).

> Define \\(f: \mathbb Z \to \mathbb R\\) by \\(f(x) = x+ 1\\). It has a right adjoint \\(g: \mathbb R \to \mathbb Z\\) defined by \\(g(y) = \lfloor y-1 \rfloor\\).

But it seems to me that for this particular example both definitions are consistent with the super-theorem:

- According to the definitions from lecture 26, \\(f\\) is oplax and \\(g\\) is lax (as you have just showed).

- According to the definitions from lecture 27, \\(f\\) is not oplax, \\(f(0) = 1 \nleq 0\\), and \\(g\\) is not lax, \\(0 \nleq -1 = g(0)\\).