An example of a direct product that is not a Cartesian product Is there an (easy to understand) example of a product that is not a Cartesian product?
In modern mathematics one learns first that products of "spaces" are Cartesian products, for instance, products of topological spaces, vector spaces, modules, groups. The elements of these products have "coordinates". which are familiar from linear algebra.
Then one learns that Cartesian products are "direct products" which have a uniqueness-property (if you are unlucky, they call it the "universal mapping property".
Then one meets Category Theory and there one learns that direct products are products with the "Universal Mapping Property".
Of course one accepts this development. But is it ever made clear why there is this development? That is why:
I would like to have an example that shows that the Cartesian product is not enough to construct products in general and a more general definition of the product, that is the categorical product, is needed.
 A: Edited answer:
What you call "direct product" is usually just called a "product". The dual notion is that of a coproduct. Products and coproducts are the easiest examples of limits and colimits, in case you wanted to go more in depth. When you say "Cartesian product", you mean a product (in some concrete category) that has as underlying set the Cartesian product of the sets. A concrete category is a category in which the objects are sets with some extra structure and the morphisms are functions between sets that preserve this structure. These are the most common categories at the beginning, but they are by no means alone in the categorical world. Yours is a fairly common perplexity, because in all the easiest examples of concrete categories (sets, groups, $R−$modules, topological spaces,...) the product "is" a Cartesian product. So let us see an example where this is not the case. It will not really be a concrete category (there are some concrete categories with more interesting products, but they often require more background), but at least the objects will be sets. Here you can find more examples.
Relations are "generalizations" of functions: given sets $A$ and $B$, the graph of a function $f \colon A \to B$ is a subset of $A \times B$ with some specific properties: every element $a \in A$ appears in exactly one element $(a,b)$, namely the one with $b=f(a)$. Instead, a relation $r \colon A \to B$ is just any subset of $A \times B$. One might require other properties (you can have a look at equivalence relations or order relations, for instance), but we deal with all relations here. You might have noticed that a relation has no preferred direction: a subset of $A \times B$ is "the same" as a subset of $B \times A$. An element of $A$ might have multiple "images" under a relation (or one, or none at all), so instead of writing $r(a)=b$ (which would not make sense), we write $(a,b) \in r$. Composition works "exactly as for functions" (if you think about it): given a second relation $s \colon B \to C$ we say that $(a,c) \in s \circ r$ if and only if there is some $b \in B$ such that $(a,b) \in r$ and $(b,c) \in s$. Now we can consider the category with sets as objects and relations as morphisms. We call this (somewhat confusingly) the category of relations. This has the usual category of sets with functions as a subcategory (the objects are the same, but there are more relations than functions), but it behaves quite differently, as we are about to see.
Extra questions:


*

*what is the identity relation of a set $A$?

*when is a relation an isomorphism (i.e., when does it have a left and right inverse)?


I claim that the product of some sets $\{A_i\}_{i \in I}$ in the category of relations "is" the disjoint union $\sqcup A_i$ of all the $A_i$. The "product map" (or "projection") from $\sqcup A_i$ to a specific $A_j$ is $r_j := \{(a,a) \mid a \in A_j \}$. Note that the set $\sqcup A_i$ is in general much smaller than the Cartesian product: the disjoint union of two lines is just two lines, their Cartesian product is a whole plain. It is a very instructive exercise to convince yourself that this is indeed the product, i.e., that if a set $P$ has a given relation $p_j$ to each $A_j$, then there is a unique relation $q$ from $P$ to $\sqcup A_i$ such that $p_j = r_j \circ q$ for all $j \in I$.
Solution: $q = \cup p_i$.
Further exercises:


*

*show that $\sqcup A_i$ is also the coproduct in the category of relations (and, this time, this agrees with the coproduct of sets)

*explain (intuitively) why the Cartesian product is not the product in the category of relations, and/or why the disjoint union is not the product in the category of sets with functions

