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$~$. [todo: 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 [todo: add link] $~$1$~$. However, in the category of groups, since the terminal object is also initial [todo: 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}$~$ [todo: 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. [todo: 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 [todo: 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.