MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Hi everybody,

I need to know if there is a notion of the product of two categories, or what could substitute for this in general. Do you have any references please?

The definition I have in mind is the following:

DEFINITION : Let $\mathcal{A}$ and $\mathcal{B}$ be two categories. The product category, if it exists, is a category $\mathcal{C}$ together with two functors $p:\mathcal{C}\to\mathcal{A}$, $q:\mathcal{C}\to\mathcal{B}$ such that for all other $(\mathcal{C}',p',q')$ as above there exists a unique functor $\pi:\mathcal{C}'\to\mathcal{C}$ such that $p\circ\pi=p'$ and $q\circ\pi=q'$.

I encounter the following (potential) counter-example to the existence of a product.

$\bullet$ Assume that both $\mathcal{A}$ and $\mathcal{B}$ are sets, and moreover that one does not have arrows at all, i.e. \begin{equation} \mathrm{Hom}(X,X')\;=\;\left\{\begin{array}{rcl} \{\mathrm{Id}_X\}&\textrm{ if }&X=X'\\ \emptyset&\textrm{ if }& X\neq X' \end{array}\right. \end{equation} What should be the product in this case? I think that the product should be the category whose objects are the elements of the product of sets $\mathcal{C}=\mathcal{A}\times\mathcal{B}$, and \begin{equation} \mathrm{Hom}_{\mathcal{C}}((X,Y),(X',Y'))\;=\;\mathrm{Hom}_{\mathcal{A}}(X,X')\times \mathrm{Hom}_{\mathcal{B}}(Y,Y') \end{equation} $\bullet$ But this does not work! In fact one has

LEMMA : With this definition $\mathrm{Hom}_{\mathcal{C}}((X,Y),(X',Y'))$ always have exactly a single element.

PROOF : If $X=X'$ or if $Y=Y'$ one has $\mathrm{Hom}_{\mathcal{C}}((X,Y),(X',Y'))=\mathrm{Hom}_{\mathcal{A}}(X,X')\times \emptyset=\mathrm{Hom}_{\mathcal{A}}(X,X')$ or $\mathrm{Hom}_{\mathcal{C}}((X,Y),(X',Y'))=\emptyset\times \mathrm{Hom}_{\mathcal{B}}(Y,Y')=\mathrm{Hom}_{\mathcal{B}}(X,X')$ respectively. Then compose the unique existing arrows $(X,Y)\to(X,Y')\to(X',Y')$. $\Box$

The Lemma implies that $p$ and $q$ do not exists.

share|cite|improve this question

closed as too localized by Martin Brandenburg, Dan Petersen, Neil Strickland, Qiaochu Yuan, Tom Leinster Feb 17 '12 at 16:53

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

GIve a set $X$ you have that $X\times \emptyset=\emptyset$ (may be you get a mistake considering $X\times \{*\}=X$ – Buschi Sergio Feb 17 '12 at 13:42
Please read a) the FAQ of mathoverflow, b) any introduction to category theory. – Martin Brandenburg Feb 17 '12 at 13:42
I've deleted the inappropriate tags (logic, set-theory, higher-category-theory). – Martin Brandenburg Feb 17 '12 at 13:43
Thank you Martin ! – PULITA ANDREA Feb 17 '12 at 13:50
Virtually every introduction to category theory will answer your question. In fact, for two categories $\mathcal{A}$, $\mathcal{B}$ there is always a product category $\mathcal{A} \times \mathcal{B}$ and its definition is as straightforward as it can possibly be. Wikipedia and nLab also know the answer: – Sebastian K. 0 secs ago – Niemi Feb 17 '12 at 14:04
up vote 1 down vote accepted

Your example should work and there is a glitch in your proof. The product with an empty set is always again empty. In fact, you can construct the product of two categories $\mathcal{C}$ and $\mathcal{D}$ by taking pairs of morphisms $(f,g)$ with $f$ in $\mathcal{C}$ and $g$ in $\mathcal{D}$ and using the obvious composition.

As a side remark, while this is the usual definition of the product of two categories, your universal property is "evil" in the sense that one should not require the existence of a unique functor and commutativity on the nose but really use the 2-categorical structure available in $\mathbf{Cat}$.

share|cite|improve this answer
Thanks ! Do you have any reference ? A Book about the 2-category theory ? – PULITA ANDREA Feb 17 '12 at 13:52
You should check out the usual sources (e.g. Mac Lane's book contains the definition of a 2-category). You find more references in the nLab (or even answers, e.g. in the examples section of <>;. – Kay Werndli Feb 17 '12 at 14:00
Thank you Kay ! – PULITA ANDREA Feb 17 '12 at 14:01

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