Options

Exercise 85 - Chapter 1

edited June 2018 in Exercises

Does \( F := \lfloor −/3\rfloor \) have a right adjoint \( R : \mathbb{N} \rightarrow \mathbb{N} \) ?

Figure

If not, why?

If so, does its right adjoint have a right adjoint?

Comments

  • 1.
    edited April 2018

    This seems to now be Exercise 80 in the latest draft (April 20th).

    \(F\) does indeed have a right adjoint \(R : \mathbb{N} \to \mathbb{N}\); it is defined by \(R(x) = 3(x + 1) - 1\). (One can alternately write \(3x + 2\), but this version changes less as you change the modulus.)

    \(R\) does not have a right adjoint of its own. Suppose it did, and call it \(R'\). Then \(0 \le R'(0)\), since 0 is the minimum of \(\mathbb{N}\). So we expect \(R(0) \le 0\); however, \(R(0) = 3(0 + 1) - 1 = 2\), which is strictly greater than \(0)\). Therefore, \(R\) cannot have a right adjoint.

    Comment Source:This seems to now be Exercise 80 in the latest draft (April 20th). \\(F\\) does indeed have a right adjoint \\(R : \mathbb{N} \to \mathbb{N}\\); it is defined by \\(R(x) = 3(x + 1) - 1\\). (One can alternately write \\(3x + 2\\), but this version changes less as you change the modulus.) \\(R\\) does _not_ have a right adjoint of its own. Suppose it did, and call it \\(R'\\). Then \\(0 \le R'(0)\\), since 0 is the minimum of \\(\mathbb{N}\\). So we expect \\(R(0) \le 0\\); however, \\(R(0) = 3(0 + 1) - 1 = 2\\), which is strictly greater than \\(0)\\). Therefore, \\(R\\) cannot have a right adjoint.
  • 2.
    edited April 2018

    I fixed the typo you caught; can renumber this and other exercises as needed when I have a bit of time.

    Comment Source:I fixed the typo you caught; can renumber this and other exercises as needed when I have a bit of time.
Sign In or Register to comment.