Left & right adjoints in the context of posets. Definition 1: A function $\theta: X\to Y$ between posets is monotone if whenever $x\le y$, we have $\theta (x)\le\theta (y)$.
Definition 2: For any pair $f:A\to B$ and $g:B\to A$ of monotone maps, we say $g$ is a left adjoint of $f$ and $f$ is a right adjoint of $g$ if $$g(b)\le a\Leftrightarrow b\le f(a).$$

Theorem: Left adjoints preserve all joins; right adjoints preserve all meets.

I'm not at all sure how to prove this theorem (and it's actually a lemma in a paper). It looks like (it should be easy and) we can cut it in half since the argument for left adjoints looks set to be dual to that of right adjoints. I've taken an arbitrary join then hit it with $g$ (for such a pair $f, g$ as in Definition 2). Now what?
Please help :)
 A: It boils down to category theory, and the fact that if $L:\mathcal{C}\leftrightarrows\mathcal{D}: R$ is a pair of adjoint functors, then one of them (I can never remember which, but it's easy to figure out) preserves all limits, while the other preserves all colimits.
A poset $(P,\prec)$ can be viewed as a category $\mathcal{P}$ with object set $P$ and for all $x,y\in P$,
$$\mathcal{P}(x,y)=\begin{cases}\lbrace *\rbrace & \text{if }x\leq y\\\emptyset &\text{otherwise}\end{cases}$$
(Composition need not be defined as there is only one way to do it). This way, monotone maps between two posets are exactly the functors between the associated categories, and Galois correspondences (pairs of monotone maps that satisfy the condition from your second difinition) are exactly pairs of adjoint functors.
Here you'll take $\mathcal{C}=A$ and $\mathcal{D}=B$. To finish, you only need to understand that the meet of a subset $A\subset P$ is the categorical limit of the (inclusion) functor $\iota:\mathcal{A}\hookrightarrow \mathcal{P}$, while the join is its the colimit.
A: I know this question had been answered, but I'd like to point out that there is a fairly easy proof that does not involve category theory. We only need the following characterisation of joints:
Lemma: Let $(X, \leq)$ be a poset, $Y \subseteq X$ a subset. Then an element $z \in X$ is the joint of $Y$ if and only if for any $x \in X$ we have
$$
(\forall y \in Y : y \leq x) \Leftrightarrow z \leq x
$$
This lemma is easily proved by using the definition of joint as smallest upper bound.
Back to the question at hand (using your definition of $A$, $B$, $f$ and $g$): Let $Z \subseteq A$ be a set such that $z = \sup Z$ exists in $A$. To show that $f$ preserves joints, we must have $\sup f(Z) = f(z)$, or equivalently, we must have for all $b \in B$
$$
(\forall y \in Y: f(y) \leq b) \Leftrightarrow f(z) \leq b
$$
by the lemma above. And this we have, as the following calculation shows:
$$
\forall y \in Y : f(y) \leq b \\
\Leftrightarrow \forall y \in Y : y \leq g(b) \\
\Leftrightarrow z \leq g(b) \\
\Leftrightarrow f(z) \leq b
$$
where we again used the lemma for the second equivalence.
