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When defining a functor (between categories), I am usually told that it assigns to each object of the source category an object of the target category. I do not find this very satisfactory since we are dealing with proper classes here. Judging by the definition, it must be possible to have the concept of a "map" between proper classes. I would like to know what exactly that is and how it is defined.

I have attempted to read some books on set theory in search for an answer, but they all treat classes very briefly and never mention the possibility of having anything like a map between two of them. I would be just as happy if you could point me to a book where this is explained.

flag	asked Apr 28 2011 at 7:35 Jesko Hüttenhain 1,120●1●8

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This is becoming a very much non-research question. I think perhaps we can close the question. On the other hands, foundational questions on MO seem to be generally below research level and have mostly educational value, which would be a reason to keep this one. – Andrej Bauer Apr 28 2011 at 8:35

A class is given by a formula (which it defines). If $C,D$ are classes, then a map $C \to D$ is simply a class $f$ which defines a "map" $C \to D$ in the obvious sense: For all $x \in C$, i.e. all elements satisfying the formula defining $C$, there is exactly one $y \in D$, i.e. an element satisfying the formula defining $D$, such that $y = f(x)$, i.e. $(x,y)$ satisfies the formula defining $f$. As an exercise, prove that $V \to V, x \mapsto x + 1 := x \cup \{x\}$ is a map, where $V$ is the universe. – Martin Brandenburg Apr 28 2011 at 8:46

My problem was that I simply did not know you could easily define ordered pairs of classes. – Jesko Hüttenhain Apr 28 2011 at 8:51

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-1. I agree with Andrej Bauer. – Sergei Tropanets Apr 30 2011 at 14:16

Ricky Demer · Accepted Answer · 2011-04-28 08:00:57Z UTC

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http://en.wikipedia.org/wiki/Ordered_pair#Morse_definition

Definition:
A relation $R$ is functional if and only if for all ordered pairs $\langle x,z\rangle$ and $\langle y,w\rangle$ in $R$, if $x=y$ then $z=w$.

Definition: If $R$ is a relation, $\operatorname{Range}(R) = \{y : (\exists x)(\langle x,y\rangle \in R)\}$ .

Definition: A map is an ordered pair $\langle R,C\rangle$ such that $R$ is a functional relation and $\operatorname{Range}(R) \subseteq C$.

answered Apr 28 2011 at 8:00

Ricky Demer
5,257●5●24

	I suppose it wouldn't hurt to mention that all of what you've written applies to classes equally well as to sets. – Andrej Bauer Apr 28 2011 at 8:12
	I suppose you're right. – Ricky Demer Apr 28 2011 at 8:15
	So, I am guessing, this is only possible if I use Morse-Kelley and not with ZFC alone? My problem was that I thought $C\times D$ was not really defined if $C$ and $D$ were proper classes. – Jesko Hüttenhain Apr 28 2011 at 8:18
	It's possible with ZFC as well. Of couse $C \times D$ is defined for classes, just use the same definition as for sets. – Andrej Bauer Apr 28 2011 at 8:34
	Well, now, that helped clarify a lot. Thank you both a lot, that trivial little miscomprehension caused me years of agony. I would really like to accept both answers, but apparently you can't do that. – Jesko Hüttenhain Apr 28 2011 at 8:37

Andrej Bauer · Answer 2 · 2011-04-28 08:12:04Z UTC

Your question sounds like your preferred foundation is set theory, so let me speak in terms of set theory. A map $f : A \to B$ between sets is a functional relation, i.e., a subset $f \subseteq A \times B$ satisfying:

Totality: $\forall x \in A . \exists y \in B . (x,y) \in f$
Single-valuedness: $\forall x \in A . \forall y, z \in B . ((x,y) \in f \land (x,z) \in f \implies y = z)$.

We usually write $f(x) = y$ instead of $(x,y) \in f$.

The same definition applies to classes. A map $F : C \to B$ between classes $C$ and $D$ is a subclass of $C \times D$ which is total and single-valued.

Exercise (allowed since this is not a research question): the domain and codomain of a function $F : C \to D$ cannot be recovered from the functional relation $F$. (If $C$ and $F$ are empty, how do we recover $D$?) Therefore, the object part of a functor must be a triple $(C,D,F)$ rather than just $F$. But how can we form ordered triples of classes?

What are “maps” between proper classes?

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