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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.

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    $\begingroup$ GIve a set $X$ you have that $X\times \emptyset=\emptyset$ (may be you get a mistake considering $X\times \{*\}=X$ $\endgroup$ Feb 17 '12 at 13:42
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    $\begingroup$ Please read a) the FAQ of mathoverflow, b) any introduction to category theory. $\endgroup$ Feb 17 '12 at 13:42
  • $\begingroup$ I've deleted the inappropriate tags (logic, set-theory, higher-category-theory). $\endgroup$ Feb 17 '12 at 13:43
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    $\begingroup$ 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: en.wikipedia.org/wiki/Product_category ncatlab.org/nlab/show/product+category – Sebastian K. 0 secs ago $\endgroup$
    – Niemi
    Feb 17 '12 at 14:04
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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 $\mathbf{Cat}$.

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  • $\begingroup$ Thanks ! Do you have any reference ? A Book about the 2-category theory ? $\endgroup$ Feb 17 '12 at 13:52
  • $\begingroup$ 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 <nlab.mathforge.org/nlab/show/2-limit>. $\endgroup$ Feb 17 '12 at 14:00

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