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# Exercise 81 - Chapter 1

edited June 2018

In Example 1.80 we found a left adjoint for the monotone map $$( 3 \times - ) : \mathbb{N} \rightarrow \mathbb{R}$$. Now find a right adjoint for the same map.

Example 1.80 Consider the map $$( 3 \times - ) : \mathbb{N} \rightarrow \mathbb{R}$$ which sends $$x \in \mathbb{N}$$ to $$3x$$, which we can consider as a real number $$3x \in \mathbb{N} \subseteq \mathbb{R}$$ .

Write $$\lceil z \rceil$$ for the smallest natural number greater than $$z \in \mathbb{R}$$, and write $$\lfloor z \rfloor$$ for the largest natural number smaller than $$z \in \mathbb{R}$$, e.g. $$\lceil 3.14 \rceil = 4$$ and $$\lfloor 3.14 \rfloor = 3$$. As the left adjoint $$\mathbb{R} \rightarrow \mathbb{N}$$ , let’s see if $$\lceil −/3 \rceil$$ works. It is easily checked that $$x \le 3y \text{ if and only if } \lceil x/3 \rceil \le y .$$ Success! Thus we have a Galois connection between $$\lceil −/3 \rceil$$ and $$3 \times -$$.

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1.

Since a right adjoint is a "best approximation from below", I'd guess $$\lfloor x/3\rfloor$$. Can someone check this?

Comment Source:Since a right adjoint is a "best approximation from below", I'd guess \$$\lfloor x/3\rfloor \$$. Can someone check this?
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2.
edited May 2018

I printed out a different version of the book (from the arxiv) and this is Exercise 1.75 there.

Notation: $$f\dashv g$$ means that $$f$$ is the left adjoint and $$g$$ is the right adjoint.

So here is what I got for this problem: $$\lceil -/3\rceil\dashv 3\times-$$ (this was the example given).

$$3\times-\dashv\lfloor -/3\rfloor$$ (Checked!)

Then I tried to find a right adjoint for $$\lfloor -/3\rfloor$$ and came up with the function $$3\times-+2$$.

Finally, $$3\times-+2$$ does not seem to have a right adjoint. Here's my argument. If $$g$$ were the right adjoint, then $$x\le g(0)$$ if and only if $$3x+2\le 0$$, and this cannot happen for any natural number $$x$$.

Comment Source:I printed out a different version of the book (from the arxiv) and this is Exercise 1.75 there. _Notation_: \$$f\dashv g\$$ means that \$$f\$$ is the left adjoint and \$$g\$$ is the right adjoint. So here is what I got for this problem: \$$\lceil -/3\rceil\dashv 3\times-\$$ (this was the example given). \$$3\times-\dashv\lfloor -/3\rfloor \$$ (Checked!) Then I tried to find a right adjoint for \$$\lfloor -/3\rfloor \$$ and came up with the function \$$3\times-+2 \$$. Finally, \$$3\times-+2 \$$ does not seem to have a right adjoint. Here's my argument. If \$$g\$$ were the right adjoint, then \$$x\le g(0)\$$ if and only if \$$3x+2\le 0\$$, and this cannot happen for any natural number \$$x\$$.
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3.

I think I saw elsewhere on this site that $$3\times-:\mathbb{N}\to\mathbb{R}$$ does not have a right adjoint. In any case, I wanted to make sure that that fact is mentioned here, and to say why.

By definition of right adjoint, if such a function $$R:\mathbb{R}\to\mathbb{N}$$ were to exist, it would have to send $$z\in\mathbb{R}$$ to the largest natural number smaller than $$\frac{z}{3}$$. But if $$z\leq-1$$ no such natural number exists, so there is no right adjoint.

If we restrict the codomain to $$3\times-:\mathbb{N}\to(-1,\infty)$$, then $$\lfloor\frac{-}{3}\rfloor$$ is the right adjoint.

Comment Source:I think I saw elsewhere on this site that \$$3\times-:\mathbb{N}\to\mathbb{R}\$$ does not have a right adjoint. In any case, I wanted to make sure that that fact is mentioned here, and to say why. By definition of right adjoint, if such a function \$$R:\mathbb{R}\to\mathbb{N}\$$ were to exist, it would have to send \$$z\in\mathbb{R}\$$ to the largest natural number smaller than \$$\frac{z}{3}\$$. But if \$$z\leq-1\$$ no such natural number exists, so there is no right adjoint. If we restrict the codomain to \$$3\times-:\mathbb{N}\to(-1,\infty)\$$, then \$$\lfloor\frac{-}{3}\rfloor\$$ is the right adjoint.