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. 



What is the product of two categories? Or its substitute ?
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 $\mathbf{Cat}$.


