Saying $a \in b$ in category theory

Question

Suppose I have a category $C$ of sets, and $a,b \in C$. How can I express, in the language of category theory, that $a \in b$? (To clarify: the objects of $C$ are actually sets, and I want to express that $a$ actually is a member of $b$, in the underlying set theory.) I am aware of the construction of an "element of a set" as a morphism $f : 1 \rightarrow b$. But I don't know how to say of an object $a$ that it is an element of $b$.

One idea would be to form the coslice category $1 \downarrow C$ (the category of morphisms $g : 1 \rightarrow x$ in $C$) and assume we have an isomorphism $f : (1 \downarrow C) \rightarrow C$ which does the right thing. As a newbie, I'm not positive that works. But even if it does, it has the disadvantage that we need to introduce the functor $f$. Is there a better way?

Answer 1

Any "categorical definition" should be invariant under equivalences of categories. If $p$ is any set with only one element $\star$, then $X \mapsto X \cdot p$ is an auto-equivalence of $\mathsf{Set}$. Here, we have $X \cdot p := \coprod_{x \in X} p := \bigcup_{x \in X} p \times \{x\} = \{(\star,x) : x \in X\}$. If $X$ is empty, then $X \cdot p$ is also empty. But the elements of $X \cdot p$ are never empty. Thus, if $\emptyset \in X$, then $\emptyset \cdot p = \emptyset \notin X \cdot p$.

Therefore, there is no categorical definition of $\in$. This is not really a coincidence or even a defect of category theory. Instead, category theory systematically replaces membership by morphisms. See also ETCS.

Answer 2

If your category has a terminal object and power objects, then for each object $X$ there is a distinguished element $\{ X \} : 1 \to \mathcal{P} X$.

If you had a monomorphism $\mathcal{P} X \to Y$, then one might opt to use it to interpret $\mathcal{P}X$ as a subobject of $Y$. In this context, it would make sense to call $X$ an element of $Y$. But it only makes sense in this context. We could extend it to subobjects $Z \subseteq Y$ and say that $X \in Z$ if and only if our chosen morphism $\mathcal{P} X \to Y$ factors through a representative of the subobject $Z$.

This might be more meaningful if you equipped your category with extra structure: a distinguished class of monomorphisms that you call "$\subseteq$". Presumably, you'd want the objects together with all of the distinguished morphisms to form a poset, and probably other features. With this structure in place, we might now define $X \in Z$ to mean that there exists a commutative diagram

$$ \begin{matrix} 1 & \to & Z \\ \downarrow & & {\small|}\!\cap \\ \mathcal{P}X &\subseteq& Y \end{matrix}$$

where the left arrow is $\{ X \}$.

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That said, it would be a very unusual situation for any of this to be of use. Other notions of element find much more utility, such as:

• The notion of a (global) element $1 \to X$
• The notion of a generalized element: any morphism at all with codomain $X$
  • Restrictions of which objects can be used as the domain are sometimes useful
• The relation $\in$ related to power objects

Answer 3

However, if one insists on expressing $\in$, it is possible (and sometimes it is indeed done), in exactly the same manner as you described: an element of an object $x$ could be defined as a morphism $1\to x$, just as in $\Bbb{Set}$.

But in general, what these common source object, "$1$" would be? The terminal object? For example in the category of groups, these $1\to G$ homomorphisms are all trivial. On the other hand, the elements of $G$ are represented by $\Bbb Z\to G$ homomorphisms (identifying the element as the image of $1\in\Bbb Z$). In general universal algebra, the free algebra on $1$ generator will play the role of the common source "$1$".

This way formulas like $u\in x$ are expressible, however, as Martin commented below, while $x$ is an object, $u$ is an other kind of entity (an arrow to $x$). For the wished interpretation where $u$ is also an object, a canonical (or at least, fixed) functor $f:C\to (1\downarrow C)$ is indeed needed, with the predescribed $1$ object, (but it usually makes no sense).