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10 CHAPTER 1
There are many kinds of functions, but the most useful ones for
us are the morphisms from a source to a well-understood standard
target. We will call this a representation. It is implicit that the
target we choose is one that we know a lot about, so that from our
knowledge that there is a morphism, and better yet our knowledge
of some additional properties of the morphism, we can obtain new
knowledge about the source object.
DEFINITION: A representation is a morphism from a source
object to a standard target object.
Counting and Inequalities as Representations
Going back to the counting example, we think about ﬁnite sets—
for example, {sun , earth, moon, Jupiter} or {1, Kremlin, π} or any
set that contains a ﬁnite number of items. This collection of ﬁnite
sets contains the special sets {1}, {1, 2}, {1, 2, 3}, and so on. In the
context of counting, given any two ﬁnite sets A and B,amorphism
is a one-to-one correspondence from A to B. A representation in
this case is a morphism from the source (a given ﬁnite set, e.g., the
set of sheep in your ﬂock) to the target, which must be one of the
special sets {1}, {1, 2}, {1, 2, 3}, and so on. The special property that
we demand of the morphisms in the context of counting is that they
should be one-to-one correspondences. For example, if you have a
ﬂock of exactly three sheep for your source, a representation of that
ﬂock must have {1, 2, 3} as its target. Thus, the “essential nature”
of the source that is preserved by the morphism, in this context,is
the number of elements it contains.
There are a lot of possible morphisms—n! to be exact, where
n is the number of elements in the source and target.
5
When we
are counting the number of elements in a set, we do not actually
care about which morphism we grasp onto. But there is no choice
about the target: it is {1, 2, 3, ..., n} if and only if n is the number of
elements in our source.
5
The notation n!, pronounced n factorial, means the product of all of the numbers from
1throughn. For example, 5! is 1 · 2 · 3 · 4 · 5 = 120. For an exercise, you can ﬁnd the six
possible one-to-one correspondences from the set {red, blue, green} to the set {1, 2, 3}.

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