Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am confused about the definition of a category given in the Wikipedia article on Category theory:

It seems to me that the structure being described (the "arrows" between objects in some class) is just a binary relation that is both reflexive and transitive. If so, what is meant by the set of all morphisms (arrows) from one object to another? The definition says that every morphism (arrow) has a unique source object and target object. Is there not then at most one arrow from one object to another?

EDIT 1

I am looking for formal axioms of category theory expressed only in the notation of first-order logic and set theory -- no words, just variables, functions, logical connectors, quantifiers, predicates, '=' signs, '$\in$', etc.

EDIT 2

How about: class $ob$ is a category iff...

  1. $\forall a,b \in ob\exists hom \forall f(f\in hom \leftrightarrow \forall d\in a(f(d)\in b)) $

  2. $\forall a\in ob\exists i \forall b\in a(i(b)=b)$

Note that the required properties of composition are a direct result of functionality of each morphism.

EDIT 3

I must be trying the patience of the moderators here. Sorry guys! This has turned into a rather open-ended discussion. To be continued at the sci.logic and sci.math newsgroups:

https://groups.google.com/group/sci.logic/browse_frm/thread/bd8d78c920e1d28c#

EDIT 4

I'm not sure this is right either, but you might consider the following.

I define the Category and Arrow predicates as follows:

$\forall x (Category(x) \leftrightarrow \forall y\in x \exists f \forall z\in y (f(z)=z))$

This would be redundant if the elements of x were all sets because you can prove the existence of an identity function on every set.

$\forall x\forall a,b\in x (Arrow(x,a,b) \leftrightarrow a\in x \land b\in x \land \exists f \forall c\in a (f(c)\in b) \land \forall c\in b \rightarrow \exists d\in a (f(d)=c)$

It is then easy to prove that the Arrow relation is reflexive:

$\forall x (Category(x) \rightarrow \forall a\in x Arrow(x,a,a))$

And that the Arrow relation is transitive:

$\forall x (Category(x)\rightarrow \forall a,b,c\in x (Arrow(x,a,b) \land Arrow(x,b,c) \rightarrow Arrow(x,a,c)))$

Thanks all for your help.

share|improve this question
1  
Regarding your second edit: (1) note that "morphisms" need not be actual functions. (2) To quote Awodey: "One important slogan of category theory is It's the arrows [morphisms] that really matter!" –  Arthur Fischer Oct 10 '12 at 5:22
    
Maybe a drastically different concrete example would help: matrix algebra is a category. The morphisms are matrices. The product of morphisms is given by matrix multiplication. Objects are natural numbers. The source and target of a morphism are its dimensions. –  Hurkyl Oct 10 '12 at 5:35
    
Another standard example is the suspension of a monoid $M$: the category with a single object (it doesn't matter what the object is: we usually call it *), and $\hom(*, *) = M$. (composition of morphisms is given by the product in $M$) –  Hurkyl Oct 10 '12 at 5:55
    
For an axiomatisation as asked for in Edit 1, see proofwiki.org/wiki/Definition:Morphisms-Only_Metacategory –  Lord_Farin Oct 11 '12 at 14:07
add comment

5 Answers

You're misinterpreting the meaning of the word "unique" (which was poor word choice on the part of whoever wrote that, so I am removing it). It just means that an arrow doesn't have more than one source or target.

Here is a formal definition of a small category (this will allow me to ignore size issues which I think are irrelevant when first trying to understand category theory). I'm afraid I'm too attached to words to follow the "no words" edict, but I hope this will be formal enough. A category consists of the following data:

  • A set $\text{Ob}$ (objects),
  • For every $a, b \in \text{Ob}$, a set $\text{Hom}(a, b)$ (morphisms from $a$ to $b$),
  • For every $a \in \text{Ob}$, an element $\text{id}_a \in \text{Hom}(a, a)$ (identity),
  • For every $a, b, c \in \text{Ob}$, a function $\circ : \text{Hom}(a, b) \times \text{Hom}(b, c) \to \text{Hom}(a, c)$ (composition).

(I am writing function composition in the opposite of the usual order. You should think of a morphism $f \in \text{Hom}(a, b)$ as an arrow $f : a \to b$ pointing from $a$ on the left to $b$ on the right.)

This data is subject to the following axioms:

  • Identity: for every $f \in \text{Hom}(a, b)$, we have $\text{id}_a \circ f = f$, and for every $g \in \text{Hom}(b, a)$, we have $g \circ \text{id}_a = g$.
  • Associativity: for every $f \in \text{Hom}(a, b), g \in \text{Hom}(b, c), h \in \text{Hom}(c, d)$, we have $f \circ (g \circ h) = (f \circ g) \circ h$.

Some concrete classes of examples to keep in mind ("categories-as-mathematical-objects" rather than "categories-as-settings-to-study-mathematical-objects") are the following:

  • A monoid is a category with one object; that is, $\text{Ob}$ is a one-element set. The elements of the monoid are the morphisms from the unique object to itself.
  • A poset is a category in which $\text{Hom}(a, b)$ has either $1$ or $0$ elements (corresponding to whether $a \le b$ or not); moreover, if $\text{Hom}(a, b)$ and $\text{Hom}(b, a)$ both have one element, then $a = b$. The existence of identities expresses reflexivity, the composition law expresses transitivity, and associativity is automatic.
  • A groupoid is a category in which every morphism $f : a \to b$ has an inverse $g : b \to a$, which is a morphism satisfying $f \circ g = \text{id}_a, g \circ f = \text{id}_b$. Groupoids are a simultaneous generalization of groups, equivalence relations, and group actions. An important example is the fundamental groupoid $\Pi_1(X)$ of a topological space $X$, which is the groupoid whose objects are the points of $X$ and whose morphisms are the homotopy classes of paths between points in $X$; composition is given by concatenating paths.

One example of "categories-as-settings-to-study-mathematical-objects":

  • The "matrix category" $\text{Mat}$ is the category whose objects are the non-negative natural numbers $\mathbb{Z}_{\ge 0}$ and whose morphisms $\text{Hom}(n, m)$ are the $n \times m$ matrices, say over some ring (or $m \times n$; whichever convention makes composition correspond to matrix multiplication).
share|improve this answer
    
Alternately, you could talk about the set $\text{Mor}$ of all morphisms, but then composition becomes a partial function. If you don't want to deal with partial functions in whatever system you're working with, then composition can be thought of as a function $\text{Mor} \times \text{Mor} \times \text{Mor} \to 2$ describing the possible triples of composable arrows. –  Qiaochu Yuan Oct 9 '12 at 6:39
    
Would you check the first sentence of what you wrote about posets, starting at moreover? –  Brian M. Scott Oct 9 '12 at 6:47
    
@Brian: caught it. Thanks. –  Qiaochu Yuan Oct 9 '12 at 6:48
3  
@roman: in my version of set theory, it is not even possible to ask the question of whether two sets (which are not given as subsets of a common superset) are disjoint. (Alternately, in my version of set theory, sets are disjoint unless I say explicitly otherwise.) To me, intersection is an operation defined on subsets of a given set. It is not an operation defined on sets. –  Qiaochu Yuan Oct 9 '12 at 9:07
1  
@Dan: no. A morphism from $a$ to $b$ is an element of the set $\text{Hom}(a, b)$. Ignoring composition (but composition is the most important part!), you can think of the collection of objects and morphisms as the vertices and edges of a directed multigraph. Again, if you would prefer, I could write down an axiomatization that talks about the set of all morphisms, but stating some of the axioms would be more awkward, I think. –  Qiaochu Yuan Oct 9 '12 at 18:10
show 3 more comments

There can be many distinct arrows between two given objects. For example, consider the category of sets: The objects are all sets (or all sets in a given Grothendieck universe, to avoid foundational issues), and the arrows are functions between sets. If $X, Y$ are sets, there are usually many different functions $f: X \to Y$, and each function is an arrow from $X$ to $Y$. The situation is similar in many other commonly used categories.

Category theory is best learned by example; it's difficult to get intuition for it in the abstract. If you're having trouble understanding some categorical concept, see how it applies in a few familiar categories, like sets, groups, or modules.

share|improve this answer
1  
In the category of sets, the objects are sets and the morphisms are functions between sets. Many categories follow this pattern: the objects are sets with some additional structure (e.g., groups, rings, topological spaces) and the morphisms are structure-preserving maps (e.g., homomorphisms, linear maps, continuous functions). However, not all categories are like that, so just think of these as motivating examples. From the perspective of category theory, you can't look at the internal structure of an object (it might not have any); you can only analyze it by the structure of its arrows. –  Daniel Hast Oct 9 '12 at 4:15
1  
@DanChristensen: Here's an introductory book on category theory, made available online by the authors: Abstract and Concrete Categories: The Joy of Cats. The first chapter grounds it pretty solidly in set theory. However, category theory has the nickname "general abstract nonsense" for good reason; even if you ordinarily prefer a more abstract approach (as do I), speaking from personal experience, category theory is one area where you really won't get far without familiarity with lots of examples and motivation. –  Daniel Hast Oct 9 '12 at 5:09
2  
@DanChristensen: Your statement about requiring that a definition be capable of being read by a computer in order for something to be (mathematically) well-defined is much more akin to a philosophical assertion than the definition of a category provided by Wikipedia. –  Arthur Fischer Oct 9 '12 at 6:18
3  
@Dan: How much real mathematics have you ever seen that you could run through a formal proof-checking program? –  Brian M. Scott Oct 9 '12 at 6:42
2  
@DanChristensen: Sorry, but MacLane also uses words, and despite your apparent dislike of words, both books are still perfectly formal about it. If you want to have the definitions in purely symbolic form, you could always convert it to that notation yourself. I don't see the point of doing so, though. –  Daniel Hast Oct 9 '12 at 16:31
show 11 more comments

Paraphrased from Categories, Allegories by Freyd and Scedrov:

The theory of CATEGORIES is given by two unary operations and a binary partial operation. The elements are called "morphisms" or "maps". The operations are pronounced as

  • $\square x$ : the source of $x$
  • $x \square$ : the target of $x$
  • $xy$ : the composition of $x$ and $y$

the axioms are

  • $xy$ is defined iff $x \square = \square y$
  • $(\square x)\square = \square x$
  • $\square(x\square) = x \square$
  • $(\square x)x = x$
  • $x(x\square) = x$
  • $\square(xy) = \square(x(\square y))$
  • $(xy)\square = ((x\square)y)\square$
  • $x(yz) = (xy)z$

(in the above, in any equation, one side is defined iff the other side is defined)

Note that this set of axioms is modeled after syntax where functions act on the right, rather than on the left as usual. Note that if the target of $x$ is the source of $y$, then $xy$ is the product guaranteed to be defined, rather than $yx$.

In this definition, the notion of "object" is defined to be any morphism of the form $\square x$.

share|improve this answer
    
This axiomatization doesn't seem to include identities. –  Qiaochu Yuan Oct 9 '12 at 18:18
2  
@QiaochuYuan: If $x (y \square)$ is defined, $x \square = \square(y\square) = y\square$. Thus, $x (y \square) = x (x \square) = x$, making every element of the form $y \square$ a right identity. Similarly, every element of the form $\square x$ is a left identity. Finally, if $(y \square) z$ is defined, $(y \square) z = (\square (y \square)) z = z$, meaning $y \square$ is a two-sided identity morphism. –  Hurkyl Oct 9 '12 at 21:26
add comment

You can find find a fully formal theory of categories in nlab:

http://ncatlab.org/nlab/show/fully+formal+ETCS

This is an "arrows only" axiomatization.

share|improve this answer
add comment

Concrete Categories

Let $ob$ be a class such that

$\forall a (a\in ob \leftrightarrow P(a))$

for some property $P$.

Then $ob$ is a concrete category iff...

(1) $\forall a,b \in ob(\exists hom (\forall f(f\in hom \leftrightarrow (\forall d\in a(f(d)\in b) \land \exists d\in ob \forall e\in d\exists g\in a (f(g)=e)))))$

where the $f$'s are morphisms (functions) that map set $a$ to set $b$ preserving property P, and $hom$ is the set of all such morphisms.

(2) $\forall a\in ob(\exists i \forall b\in a(i(b)=b))$

where the $i$'s correspond to identity morphisms.

share|improve this answer
5  
@Dan: you understand incorrectly. –  Qiaochu Yuan Oct 11 '12 at 5:55
3  
@Dan: whatever you mean by the term "abstract category," it doesn't seem to be what anyone else means by the term. You are free to study sets with reflexive and transitive relations on them to your heart's content. They are called preorders: en.wikipedia.org/wiki/Preorder. Most categories are not preorders. –  Qiaochu Yuan Oct 11 '12 at 6:10
3  
@Dan: yes. Take any group $G$ with more than two elements. There is a category $BG$ with one object and morphisms the elements of $G$ (from that one object to itself). Composition is given by the group operation in $G$. All of this is in the answer I wrote. –  Qiaochu Yuan Oct 11 '12 at 6:20
2  
The homotopy category is known to be non-concrete and has multiple morphisms between objects. –  Zhen Lin Oct 11 '12 at 12:41
5  
@DanChristensen it isn't a matter of opinion. Here is a reprint of the paper in which Zhen's result was established by Freyd. There is no faithful functor for the homotopy category to Set, which is even a stronger result. tac.mta.ca/tac/reprints/articles/6/tr6.pdf You're still making all your morphisms surjective, which I still don't understand, and your axiom still permits the function $f(0)=f(1)=1$ from the cyclic group of order 2 to itself in the category of groups. –  Kevin Carlson Oct 11 '12 at 13:03
show 13 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.