How to interpret a "generalized element"? I am somewhat confused by the concept of "generalized element" in category theory. It makes sense to me that a morphism from a singleton set to a set $X$ "picks out" an element of $X$ and therefore the data involved in a choice of one such morphism is equivalent to a choice of an element of $X$. But then what is the data of a choice of morphism from a 2 element set to $X$? It seems like there are two possible answers:

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*An ordered pair of elements of $X$. This is what Wikipedia says the answer is. And it makes some sense.


*An unordered pair, i.e. a subset of $X$ with size either 1 or 2. I think that this answer makes more sense because, given a morphism $f : \{a,b\} \to X$, how do you decide which is the "first" entry of $f$ — $f(a)$ or $f(b)$? You would have to pick whether $a$ or $b$ the "first" element of the set, but a mere set doesn't have that kind of structure.
Now, I can also see a good argument for the first option: the size of $X \times X$ is $|X|^2$, and the size of the set of all morphisms $\{a,b\} \to X$ is $|X|^2$, but the number of subsets of $X$ with size 1 or 2 isn't. So I don't really know which choice is most "natural".
And then if you consider the situation with the category of real vector spaces instead of the category of sets, it becomes even less clearer. A morphism $f : \mathbb R^1 \to X$ picks out a vector in $X$ — if you pick a basis of $\mathbb R^1$. If the basis of $\mathbb R^1$ is $\{e\}$, then the vector it picks out is $f(e)$. But it is not so clear what data is involved in a morphism $f : \mathbb R^1 \to X$ if you don't have a basis of $\mathbb R^1$.
So it seems like maybe the answer is that in order to be able to interpret a "generalized element" properly, you need to add some structure to the domain — in the case of sets, you need to pick an order, and in the case of vector spaces, you need to pick a basis. This is confusing me.
So my questions are:

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*Which interpretation of a generalized element that is a morphism from a 2-element set is more useful / correct / used in practice?


*Is there some concept that I could study that would clear up this confusion?
 A: A generalized element doesn't come with any preferred interpretation in terms of "actual" elements. You can give it one, but as you say, that is extra data. You are correct that in order to identify a function $\{ a, b \} \to X$ with an ordered pair in $X \times X$, strictly speaking you have to pick an ordering on $\{ a, b \}$. Another way to say this is that there are two functors $\text{Hom}(\{ a, b\}, X)$ and $X \times X$ of the variable $X$, and they are naturally isomorphic but not uniquely; there are two isomorphisms corresponding to the two possible orders.
A function $\{ a, b \} \to X$ is definitely not an unordered pair. If $X$ has two distinct elements $x_1, x_2$ then a function satisfying $f(a) = x_1, f(b) = x_2$ is not literally the same as a function satisfying $f(a) = x_2, f(b) = x_1$. Very strictly speaking it is neither an "ordered pair" nor an "unordered pair" but a third thing, which is naturally isomorphic to an ordered pair but not uniquely.
The case of $\mathbb{R}^1$ is similar. There is an object, namely the free vector space $\mathbb{R}[\bullet]$ on a point $\bullet$, that has a canonical basis by virtue of its universal property, namely $\{ \bullet \}$. That is, $\mathbb{R}[\bullet]$ is by definition the object representing the forgetful functor from vector spaces to sets, meaning $\text{Hom}(\mathbb{R}[\bullet], V) \cong V$ (where the $V$ on the RHS is the underlying set), and this identification is given by evaluating a linear map on $\bullet$. $\mathbb{R}^1$ is isomorphic to $\mathbb{R}[\bullet]$ and so the corresponding functors they represent are also isomorphic, but again not uniquely so: there is an $\mathbb{R}^{\times}$ worth of isomorphisms, corresponding to choices of a basis of $\mathbb{R}^1$.
I think most people would regard this example as somewhat pedantic compared to the first one, because it is standard practice to think of $\mathbb{R}^1$ as being equipped with a canonical basis, namely $\{ 1 \}$. In other words, it is standard practice to identify $\mathbb{R}^1$ with $\mathbb{R}[\bullet]$, and more generally to identify $\mathbb{R}^n$ with the free vector space on $n$ elements (but strictly speaking this identification also requires a choice of order of the $n$ elements). This use of "canonical" is slightly less canonical than the above one.
