Generalized element

In category theory, a generalized element of an object of a category is any morphism $x : A \to X$ with codomain $X$ . In this situation, $A$ is called the shape, or domain of definition, of the element $x$ . We'll unpack this.

Generalized elements generalize elements

We'll need a set with a single element: for concreteness, let us denote it $I$ , and say that its single element is $*$ . That is, let $I = {*}$ . For a given set $X$ , there is a natural correspondence between the following notions: an element of $X$ , and a function from the set $I$ to the set $X$ . On the one hand, if you have an element $x$ of $X$ , you can define a function from $I$ to $X$ by setting $f (i) = x$ for any $i \in I$ ; that is, by taking $f$ to be the constant function with value $x$ . On the other hand, if you have a function $f : I \to X$ , then since $*$ is an element of $I$ , $f (*)$ is an element of $X$ . So in the category of sets, generalized elements of a set $X$ that have shape $I$ , which are by definition maps $I \to X$ , are the same thing (at least up to isomorphism, which as usual is all we care about).

Generalized elements in sets

In the category of sets, if a set $A$ has $n$ elements, a generalized element of shape $A$ of a set $X$ is an $n$ -tuple of elements of $X$ .

Sometimes there is no `best shape'

Based on the case of sets, you might initially think that it suffices to consider generalized elements whose shape is the terminal object

1

. However, in the category of groups, since the terminal object is also initial

, each object has a unique generalized element of shape

1

. However, in this case, there is a single shape that suffices, namely the integers

Z

. A generalized element of shape

Z

of an abelian group

A

is just an ordinary element of

A

However, sometimes there is no single object whose generalized elements can distinguish everything up to isomorphism. For example, consider $Set \times Set$

. If we use generalized elements of shape

(X, Y)

, then they won't be able to distinguish between the objects

(2^{A}, 2^{X + B})

and

(2^{Y + A}, 2^{B})

, up to isomorphism, since maps from

(X, Y)

into the first are the same as elements of

(2^{A})^{X} \times (2^{X + B})^{Y} ≅ 2^{X \times A + Y \times (X + B)} ≅ 2^{X \times A + Y \times B + X \times Y}

, and maps from

(X, Y)

into the second are the same as elements of

(2^{Y + A})^{X} \times (2^{B})^{Y} ≅ 2^{X \times (Y + A) + Y \times B} ≅ 2^{X \times A + Y \times B + X \times Y}

. These objects will themselves be non-isomorphic as long as at least one of

X

and

Y

is not the empty set; if both are, then clearly the functor still fails to distinguish objects up to isomorphism. (More technically, it does not reflect isomorphisms.

) Intuitively, because objects of this category contain the data of two sets, the information cannot be captured by a single homset. This intuition is consistent with the fact that it can be captured with two: the generalized elements of shapes

(0, 1)

and

(1, 0)

together determine every object up to isomorphism.

Morphisms are functions on generalized elements

If $x$ is an $A$ -shaped element of $X$ , and $f$ is a morphism from $X$ to $Y$ , then $f (x) := f \circ x$ is an $A$ -shaped element of $Y$ . The Yoneda lemma

states that every function on generalized elements which commutes with reparameterization, i.e.

f (x u) = f (x) u

, is actually given by a morphism in the category.

LESSWRONG
LW