Explanation of "inclusion reversing". So I am currently self-studying Galois theory with no one to ask questions to and
that is why I am here. Anyways,
I was reading about the fundamental theorem of Galois theory on this website here https://jhavaldar.github.io/notes/2017/10/19/galoistheory1.html and I am stuck on Proposition 4:
"The above association is inclusion reversing:
If $F_1 \subseteq F_2 \subseteq K$ then $Aut(K/F_2) \leq Aut(K/F_1)$
$\vdots$
"
The proposition continuous and states other things as well. The part that I do not understand is "inclusion reversing", what is that? And what does it mean for $Aut(K/F_2) \leq Aut(K/F_1)$? Is this stating that the cardinality of $Aut(K/F_2)$ is less than $Aut(K/F_1)$?
 A: Recall that a partially ordered set (poset) is a tuple $(X, \leq)$ of a set together with a reflexive, transitive and antisymmetric relation on $X\times X$.
We can now define the following:
DEFINITION If $(X, \leq)$ and $(Y, \preceq)$ are posets, then an anti-monotone map (or antitone map) is a map $f\colon X\to Y$ such that $x\leq x'$ implies $f(x')\preceq f(x)$.
In the case where the partial order on both sets stems from an inclusion, we call an antitone map inclusion-reversing.
In particular, the statement now becomes:
PROPOSITION
Let $K/F$ be a normal and separable[1] field extension.
The map $f\colon F_1 \mapsto \operatorname{Aut}(K|F_1)$ is a well-defined antitone map between the posets

*

*$(\{\text{Intermediate fields of}\ K/F\}, \subseteq)$

*$(\{\text{Subgroups of}\ \operatorname{Aut}(K)\ \text{fixing}\ F\}, \leq)$.

Here, $\leq$ is the relation “is a subgroup of”. However, since the subgroups we are concerned with inherit their group structure from the big group $\operatorname{Aut}(F)$ they are all contained in, they are a subgroup if and only if they are included, i.e. $\leq$ is the same as $\subseteq$.
[1] I'm Not sure how much that can be weakened here, the precise demands are however irrelevant for the point I'm making.

Note that we can formulate antitone maps a bit differently using the notion of a reverse order:
Observation If $(X, \leq)$ is a partial order, then the relation $$
\geq\colon x\geq x' \Leftrightarrow x'\leq x
$$ turns $(X, \geq)$ into a partial order as well. We will call that relation the reverse order or opposite order of $(X, \leq)$.
Proposition A map $f\colon (X, \leq) \to (Y, \preceq)$ is antitone if and only if $f\colon (X, \leq) \to (Y, \succeq)$ is monotone.
