# Generalized element

In category theory, a **generalized element** of an object \(X\) 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 \(X\). is there more to say here? or less?

## 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 add link \(1\). However, in the category of groups, since the terminal object is also initial explain this somewhere, each object has a unique generalized element of shape \(1\). However, in this case, there is a single shape that suffices, namely the integers \(\mathbb{Z}\). A generalized element of shape \(\mathbb{Z}\) 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 \(\text{Set} \times \text{Set}\) link to a page about the product of two categories. 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 \((2^A)^X\times(2^{X+B})^Y \cong 2^{X\times A + Y \times (X + B)} \cong 2^{X \times A + Y \times B + X \times Y}\), and maps from \((X,Y)\) into the second are the same as elements of \((2^{Y+A})^X \times (2^B)^Y \cong 2^{X\times(Y+A) + Y \times B} \cong 2^{X \times A + Y \times B + X \times Y}\). 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. explain or avoid this terminology) 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 create Yoneda lemma page states that every function on generalized elements which commutes with reparameterization, i.e. \(f(xu) = f(x) u\), is actually given by a morphism in the category.

Parents:

- Mathematics
Mathematics is the study of numbers and other ideal objects that can be described by axioms.