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# Exercise 85 - Chapter 1

edited June 2018

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

If not, why?

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

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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.
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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.
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