Exercise 82 - Chapter 1

Fredrick Eisele

Consider the preorder \( P = Q = \underline{3} \).

1) Let \( f , g \) be the monotone maps shown below:

Is it the case that \(f\) is left adjoint to \(g\)?

Check that for each \( 1 \le p, q \le 3 \), one has \( f(p) \le q \text{ iff } p \le g(q) \).

2) Let \( f , g \) be the monotone maps shown below:

Is it the case that \(f\) is left adjoint to \(g\)?

Ignacio Viglizzo

My answer is yes for both pairs of adjoints.

Edit: And it is wrong, as Jonathan points below. My method of checking wasn't good enough.

Jonathan Castello

For the second part, we have \(2 \le g(1)\) but not \(f(2) \le 1\), since \(2 = g(1)\) and \(f(2) = 2\). So it can’t be adjoint.

Fredrick Eisele

Consider Remark 1.78 where a claim is made about crossing arrows, if they are 'bending'. In the book it is not clear that the arrows are crossing, in the diagram above it is pretty clear that they cross.

What does arrow crossing signify?

Jonathan Castello

I don't find the "arrow crossing" principle very convincing -- it isn't difficult to adjust the figure so that no arrows cross, even though there's no adjoint. And when the preorders are more complicated than simple total orders, I expect there to be cases where arrows must cross even though there is an adjunction.

It works here because the arrow crossing principle is presumably valid on figures drawn by the authors, else the authors would not provide said principle (or said figures).

EDIT: Perhaps it might work more generally if one adds the proviso that in-arrows and out-arrows are sorted in where they attach to the points. But I still suspect there will be edge cases.

Fredrick Eisele

I think the authors were trying to motivate Proposition 1.86 by their remark. As an alternative to the 'crossing' arrows idea is that of approaching from above [or as drawn from the right] by the right adjoint and from below [or as drawn from the left] by the left.