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.