# Saying $a \in b$ in category theory

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?

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You should make your question more precise. 1) Which properties of $\in$ do you demand? (Of course it shouldn't be an arbitrary relation) 2) What is a category of sets? Do you mean a concrete category? Or a (full) subcategory of the category of sets? 3) Motiviation, see Berci's comment below :). – Martin Brandenburg Feb 9 at 11:18

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Why would you want to talk about $a\in b$? Else your idea is correct. Unfortunately not so many categories will satisfy that $f$ is isomorphism/equivalence of categories. – Berci Feb 9 at 11:19
Martin: Is my clarification satisfactory? Berci: I assume there's some philosophical reason why I might not? E.g., that part of the point of thinking of the sets as a category is to forget many of their features? Would you like to say anything about this? As far as $f$ not being an isomorphism, do you mean that it won't be except in weird corner cases? I'm getting a sense that might be true from looking at the morphisms in the coslice category. – Nick Thomas Feb 9 at 11:33

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@NickThomas $a \in b$ is not expressible in category theory, for the simple reason that the category cannot distinguish between sets that have the same cardinality. – Zhen Lin Feb 9 at 11:36

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@NickThomas, Sec 2.3 (pp.34-38) of Lawvere & Rosebrugh's "Sets for Mathematics" shows how to express inclusion and membership categorically in terms of morphisms (as hinted by Martin in his answer). – alancalvitti Feb 11 at 22:57
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## 3 Answers

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.

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Well, it only shows that category equivalence is not in friendship with the 'real' membership relation. – Berci Feb 9 at 12:42
@Berci: Actually, this example is an isomorphism of categories! – Hurkyl Feb 9 at 14:19
Hurkyl: It is not an isomorphism. As I've said, there is no set in the image which contains $\emptyset$. @Berci: What do you mean by "only"? Can you imagine any reasonable "categorical definition" which is not invariant under equivalences? – Martin Brandenburg Feb 9 at 14:25
+1 Your example is very nice @MartinBrandenburg . Is this original to you or could you perhaps point to some reference which describes this and other proofs of "evil" mathematical definitions? – magma Feb 10 at 18:08

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 $\{ X \}$.

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
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@Martin: But if $X \in Z$, then setting $Y = Z \cup \mathcal{P}X$ will give a diagram. – Hurkyl Feb 9 at 14:51

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).

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The OP wants $x \in y$ for two objects of $C$. This is not possible with the definition, where an element is a morphism from the terminal object. – Martin Brandenburg Feb 9 at 12:07
Hmm.. I see. Then the isomorphism $f:(1\downarrow C)\to C$ is indeed needed.. Well, I thought about it as a more general question. – Berci Feb 9 at 12:11
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