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text: 'In [-4c7], a **generalized element** of an object $X$ of a category is any morphism $x : A \\to X$ with [3lg codomain] $X$. In this situation, $A$ is called the **shape**, or **domain of definition**, of the element $x$. We'll unpack this.\n\n## Generalized elements generalize elements ##\n 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 $I = \\{*\\}$.\n 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).\n\n## Generalized elements in sets ##\n 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 $X$. \n[todo: is there more to say here? or less?]\n\n## Sometimes there is no `best shape' ##\nBased 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 $A$. \n\nHowever, 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])\nIntuitively, 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.\n\n## Morphisms are functions on generalized elements ##\n 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.',
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