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?


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.


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.


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:



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.

  • 2
    $\begingroup$ 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!" $\endgroup$
    – user642796
    Oct 10 '12 at 5:22
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    $\begingroup$ 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. $\endgroup$
    – user14972
    Oct 10 '12 at 5:35
  • $\begingroup$ 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 $M$) $\endgroup$
    – user14972
    Oct 10 '12 at 5:55
  • $\begingroup$ For an axiomatisation as asked for in Edit 1, see proofwiki.org/wiki/Definition:Morphisms-Only_Metacategory $\endgroup$
    – Lord_Farin
    Oct 11 '12 at 14:07
  • $\begingroup$ For an axiomatization without words and using the usual definition with objects (as opposed to morphisms-only) versions, see us.metamath.org/mpegif/df-cat.html $\endgroup$
    – Mark S.
    Jun 3 '18 at 17:26

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).
  • $\begingroup$ 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. $\endgroup$ Oct 9 '12 at 6:39
  • $\begingroup$ Would you check the first sentence of what you wrote about posets, starting at moreover? $\endgroup$ Oct 9 '12 at 6:47
  • $\begingroup$ @Brian: caught it. Thanks. $\endgroup$ Oct 9 '12 at 6:48
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    $\begingroup$ @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. $\endgroup$ Oct 9 '12 at 9:07
  • 1
    $\begingroup$ @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. $\endgroup$ Oct 9 '12 at 18:10

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.

  • 3
    $\begingroup$ No, it's very much pure mathematics. I'm not sure what you mean by "philosophy", but examples and motivation are certainly part of mathematics; otherwise, you'd just be manipulating strings of symbols without any ideas behind it. Looking at your edit, what you want is perhaps more along the lines of formal proof-checking programs or something, because mathematicians don't work purely in low-level symbols like that — leave that sort of thing to computers. $\endgroup$ Oct 9 '12 at 5:40
  • 3
    $\begingroup$ @DanChristensen: If everything is well-defined, then in principle, you could run it through a sufficiently advanced proof-checking program. However, that would usually be impractical, so it's generally sufficient that everything could be made as formal as necessary, without actually writing it down in that sort of ultra-formal notation. You're unlikely to find most mathematical ideas presented with that level of formality, because it's assumed that a knowledgeable reader will be able to fill in the gaps and formalize as needed. $\endgroup$ Oct 9 '12 at 6:09
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    $\begingroup$ @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. $\endgroup$
    – user642796
    Oct 9 '12 at 6:18
  • 4
    $\begingroup$ @Dan: How much real mathematics have you ever seen that you could run through a formal proof-checking program? $\endgroup$ Oct 9 '12 at 6:42
  • 4
    $\begingroup$ @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. $\endgroup$ Oct 9 '12 at 16:31

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 $\square x$.

  • $\begingroup$ This axiomatization doesn't seem to include identities. $\endgroup$ Oct 9 '12 at 18:18
  • 2
    $\begingroup$ @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. $\endgroup$
    – user14972
    Oct 9 '12 at 21:26

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


This is an "arrows only" axiomatization.


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.

  • 7
    $\begingroup$ @Dan: you understand incorrectly. $\endgroup$ Oct 11 '12 at 5:55
  • 4
    $\begingroup$ @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. $\endgroup$ Oct 11 '12 at 6:10
  • 3
    $\begingroup$ @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. $\endgroup$ Oct 11 '12 at 6:20
  • 2
    $\begingroup$ The homotopy category is known to be non-concrete and has multiple morphisms between objects. $\endgroup$
    – Zhen Lin
    Oct 11 '12 at 12:41
  • 6
    $\begingroup$ @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. $\endgroup$ Oct 11 '12 at 13:03

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