I'm trying to wrap my brain around adjoint functors. One of the examples I've seen is the categories $\bf IntLE \bf = (\mathbb{Z}, ≤)$ and $\bf RealLE \bf = (\mathbb{R}, ≤)$, where the ceiling functor $ceil : \bf RealLE \rightarrow IntLE$ is left adjoint to the inclusion functor $incl : \bf IntLE \rightarrow RealLE$. I want to check that the following are true, as they seem to be:

1. $floor : \bf RealLE \rightarrow IntLE$ would be right adjoint to $incl$
2. Between the dual categories of $\bf IntGE \bf = (\mathbb{Z}, ≥)$ and $\bf RealGE \bf = (\mathbb{R}, ≥)$, $ceil$ would be right adjoint to $incl$
3. Between $\bf RealGE$ and $\bf IntGE$, $floor$ would be left adjoint to $incl$

Is my understanding correct on these points?

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In a word, yes. – Robin Chapman Oct 14 '10 at 6:40
If you want to understand the idea of adjointness, I think it will be much much more useful to focus on "real" categories and natural examples of functors between them (groups, vector spaces, algebras; forgetfut functors and their adjoints, the free functors) – Mariano Suárez-Alvarez Oct 14 '10 at 14:42
Yeah, I started with the free / forgetful monoid adjunction and then saw this example, which kind of blew my mind that they were related in such an abstract way :) This seems to be a common experience in category theory. – Tom Crockett Oct 14 '10 at 20:26
I repeat my suggestion from comments below to take a look at George Bergman's Universal Algebra book, especially section 7.4 (for adjunctions). A postscript version is available at math.berkeley.edu/~gbergman/245/Ch.7.ps. – Arturo Magidin Oct 14 '10 at 21:47

Given categories $\mathcal{C}$ and $\mathcal{D}$, and functors $\mathbf{F}\colon\mathcal{C}\to\mathcal{D}$ and $\mathbf{U}\colon\mathcal{D}\to\mathcal{C}$, $\mathbf{F}$ is the left adjoint of $\mathbf{U}$ if and only if for every objects $C\in\mathcal{C}$ and $D\in\mathcal{D}$ there is a natural bijection between $\mathcal{C}(C,\mathbf{U}(D))$ and $\mathcal{D}(\mathbf{F}(C),D)$.

Let me use $\leq$ and $\geq$ for the relation among reals, and $\preceq, \succeq$ for the relation among integers.

For $ceil$ to be the left adjoint of the inclusion functor, you would need that for all real numbers $r$ and all integers $z$, $\lceil r\rceil \preceq z$ if and only if $r\leq z$. This holds, so you do have an ajunction (this works because in these categories, the morphism set IntLE$(a,b)$ is empty if $a\not\preceq b$, and contains a unique arrow if $a\preceq b$; and similarly with RealLE; so you get a natural bijection between the sets of arrows if and only if they are either both empty or both are singletons at the same time).

For $floor$ to be a right adjoint to $incl$, you would need that for all real numbers $r$ and all integers $z$, $z\preceq \lfloor r\rfloor$ if and only if $z\leq r$, which again is true; so $floor$ is a right adjoint to the inclusion functor.

For $ceil$ to be the right adjoint to the inclusion functor in the dual categories, you would need $z\succeq \lceil r \rceil$ if and only if $z\geq r$; and for $floor$ to be the left adjoint, you would need $\lfloor r\rfloor \succeq z$ if and only if $r\geq z$. Both hold, so your assertions 1 through 3 are correct.

P.S. Let me second Mariano's suggestion in the comments to keep in mind the case of the underlying set functor and the free group functor for thinking about right and left adjoints. I find myself going back to those two every time I need to remind myself of how things work with adjoints, what adjoints respect or don't respect, and especially when thinking about some of the other equivalent definitions, in particular the one in terms of the unit and co-unit of the adjunction (which are natural transformations between the identity functors and the functors $\mathbf{FU}$ and $\mathbf{UF}$).

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Thanks, one of the hardest things for me to understand about adjoints has been the equivalence of the definition by natural transformations from and to the unit and co-unit, and the definition by universal morphisms. – Tom Crockett Oct 14 '10 at 20:21
In the "underlying set functor" and "free group functor", the unit maps a set to the set of all "group words" in the elements of the set. You can think of the counit as mapping all group words on X to the reduced group words on X. Consider looking at the discussion in <href="math.berkeley.edu/~gbergman/245/Ch.7.ps">Section 7.4 of George Bergman's book</a>. – Arturo Magidin Oct 14 '10 at 21:42

Arturo has already posted a nice answer. I'd merely like to emphasize that such universal definitions often enable slick proofs, e.g. see below. For a much more striking example see the theorem in my post here, which presents a slick one-line proof of the LCM * GCD law via their universal definitions.

LEMMA $\rm\: \ \lfloor x/(mn)\rfloor\ =\ \lfloor{\lfloor x/m\rfloor}/n\rfloor\ \$ for $\rm\ \ n > 0$

Proof $\rm\quad\quad\quad\quad\quad\quad\quad k\ \le \lfloor{\lfloor x/m\rfloor}/n\rfloor$

$\rm\quad\quad\quad\quad\quad\iff\quad\ \ k\ \le\ \:{\lfloor x/m\rfloor}/n$

$\rm\quad\quad\quad\quad\quad\iff\ \ nk\ \le\ \ \lfloor x/m\rfloor$

$\rm\quad\quad\quad\quad\quad\iff\ \ nk\ \le\:\ \ \ x/m$

$\rm\quad\quad\quad\quad\quad\iff\ \ \ \ k\ \le\:\ \ \ x/(mn)$

$\rm\quad\quad\quad\quad\quad\iff\ \ \ \ k\ \le\ \ \lfloor x/(mn)\rfloor$

Compare the above trivial proof to more traditional proofs, e.g. the special case $\rm\ m = 1\$ here.

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Thanks, cool stuff! – Tom Crockett Oct 14 '10 at 20:27
For another example of a non-universal proof see this answer. – Bill Dubuque May 21 '12 at 19:31