All in all, those who were chained would consider nothing besides the shadows of the artifacts.
Plato, The Allegory of the Cave
Pushforwards and Pullbacks
A function gives us a way of taking elements in
and pushing them over to
. We call
the image of
under
. In fact, we can also think of
as giving a way to push a set of elements of
over to
.
Definition. Given and
, we define
. We call
the image of
under
or the pushforward of
by
.
Notice that by definition, is a subset of the target
.
E.g. If , and
, then
.
Proof. () Let
. Then
. So
.
If , then
is between
and
. So
.
If , then
is between
and
. So
.
() Let
. Then
. Let
. So
. Then
. Since
,
.
Notation. Most sources write for the image of
under
. This is not an entirely unreasonable choice, but it means that the symbol
has two interpretations -- if
is an element of
, then
is an element of
; if
is a subset of
, then
is a subset of
. Our notation
is always going to be a set. But you should know, for your future mathematics courses, that
is a standard notation.
Theorem/Realization. If , then
.
That is, takes subsets of
and yields subsets of
.
Definition. Given and
, we define
. We call
the preimage of
under
or the pullback of
by
.
Note that is a subset of the domain
.
E.g. If is given by
, then
and
.
Theorem/Realization. If , then
.
Notation. Just like with the image, there is a standard-but-terrible notation for the preimage. Most sources write for the preimage of
under
. This is just nutty, because the object in question doesn't have anything to do with inverses. Nevertheless, you should expect at least once in your mathematical life to have to deal with the string of symbols
, where
is some set. We live in a fallen world.
Theorem. Let ,
,
,
,
. Then
Proof. This is an exercise, but it's just an exercise in proving set equalities.
Characteristic Functions
Recall that for to be a set means that ``
" is an open sentence. We can view this in terms of functions.
Definition. Given a set , the characteristic function of
is the function
Technically in order to define this function, we need to specify a universe for -- but notice that it really doesn't much matter, because as long as
,
.
Exercise. If and
are subsets of some common universe
, then for any
, we have:
,
.
Theorem. .
This says that determines its characteristic function, and
determines
as well.