Does the cartesian product have a neutral element? Let $A$ be any set. Is there a set $E$ such that $A \times E = E \times A = A$?
I thought of the empty set, but Wikipedia says otherwise. This operation changes dimension, so an isomorphism might be needed for such element to exist.
 A: In some sense, the whole reason we have these things called addition and multiplication and the ring axioms is because of certain properties satisfied by the Cartesian product and disjoint union. Both are associative and commutative (up to natural isomorphism). One distributes over the other (up to natural isomorphism). Both have identity elements (up to natural isomorphism). Decategorify, restricting to finite sets, and you get the non-negative integers. Take the Grothendieck group, and you get the integers, and then at some point you are led to write down the ring axioms in general. But it's good to keep in mind where it all comes from. 
A: For "equality", no ... if $a \in A$ then it does not have the form $(a,e)$ of an element of $A \times E$, by the Axiom of Foundation.  
For "bijection", yes.  I leave that to you.
A: It's unfortunate (but not your fault at all) that one needs to place a lot of explanations and specifications in place to be able to give a definite answer to what seems like a reasonable question. Unfortunate because in fact there are a lot of assumptions behind the notation.
The first assumption is what is meant by equality? Do we mean set equality, or do we mean set equivalence (up to a bijection)? 
The second question is what is meant by $A\times B$; it is the set of ordered pairs, sure, but what is an ordered pair? There are many different definitions. One of the most standard ones is the Kuratowski definition of ordered pair:
$$(a,b) = \bigl\{ \{a\},\{a,b\}\bigr\}.$$
The precise definition of ordered pair will matter if we are asking about equality of sets rather than simple equivalence up to a bijection.
The third question is whether you meant to ask if there is a set $E$ such that for all sets $A$, $E\times A = A\times E = A$; or if you meant to ask if for every set $A$ there exists a set $E$ such that $E\times A = A\times E = A$. Formally, they are different questions (though an affirmative answer for the former implies an affirmative answer for the latter). 
And the final question is what Set Theory you are working in. 


*

*In its weakest version, if equality of sets means "up to a bijection", then the answer is yes: any singleton set $E$ will satisfy the stronger question: if $E$ is a singleton, then for every set $A$, there is a bijection between $A\times E$ and $E\times A$, and both are bijectable with $A$. The projection map onto $A$ is a bijection in this case, and the precise definition of ordered pair does not matter.

*What if by equality we mean true set equality, and we are asking for an identity element for the cartesian product? The answer is no, regardless of your precise definition of ordered pair, as long as your Set Theory has some of the basic axioms (Specification/Separation and Power Sets will do). If $E$ is empty, then for any nonempty $A$ we have $E\times A\neq A$. If $E$ is not empty, then let $A$ be a set that contains an element that is not an element of $E$ (such sets exist by the Axiom of Specification: if $X=\{e\in E\mid e\notin e\}$, then it is an easy exercise to show that $X$ is not an element of $E$; and then the element of the power set of $E$ that contains only $X$ is a set that contains an element that is not an element of $E$). Then $E\times A\neq A\times E$, because if $x\in A$ is an element not in $E$, and $e\in E$ is any element, then $(x,e)\in A\times E$, but $(x,e)\notin E\times A$, because $x\notin E$. 

*What if by equality we mean true set equality, but the set $E$ is allowed to depend on $A$? In Set Theory with the Axiom of Foundation and the Kuratowski definition of ordered pair, the only set $A$ for which there is a set $E$ with $A\times E = A$ is the empty set. Indeed, if $A$ is empty, then for any set $E$ we have $A\times E = \emptyset = A$. Now, assume the Axiom of Foundation/Regularity:

For every nonempty set $C$, there exists $x\in C$ such that $x\cap C = \emptyset$. 

Assume that $A$ is nonempty, and there exists a set $E$ such that $A\times E = A$. For every $a\in A$ there exists $a'\in A$ and $e\in E$ such that $(a',e)=a$. That means that $a = \bigl\{ \{a'\},\{a',e\}\bigr\}$. Hence, for every $a\in A$ there exists $a'\in A$ such that $\{a'\}\in a$. Let $C$ be the set
$$C = A\cup\bigl\{ x\in\mathcal{P}(A)\mid x\text{ is a singleton}\bigr\}.$$
By the Axiom of Foundation, there exists $c\in C$ such that $c\cap C=\emptyset$. If $c\in A$, then by the argument above there exists $a'\in A$ such that $\{a'\}\in c$; but $a'\in A$, so $\{a'\}\in C$, hence $\{a'\}\in c\cap C$, contradicting the choice of $c$. Therefore, we conclude that $c\in\mathcal{P}(A)$ and $c$ is a singleton, $c=\{a\}$ for some $a\in A$. But then $a\in c\cap C$, again contradicting the choice of $c$.
This contradiction arises from assuming that there exists a set $E$ such that $A\times E = A$ when $A$ is nonempty, so we conclude that if $A$ is nonempty, then no such $E$ exists.

*However, if you do not have the Axiom of Foundation, then it's possible that there exist some nonempty sets $A$ for which there exist sets $E$ with $A\times E = A$. For example, if you take ZF with Aczel's anti-foundation axiom, then there exists a set $A$ such that $A=\{A\}$. Then $A\times A = A$, since
$$(A,A) = \bigl\{ \{A\},\{A,A\}\bigr\} = \bigl\{\{A\},\{A\}\bigr\} = \bigl\{\{A\}\bigr\} = \{A\} = A.$$
So $A\times A = \{(A,A)\} = \{A\} = A$. 
But even in this situation there always exist sets that do not satisfy the desired property (that is, sets $A$ for which there does not exist any $E$ with $A\times E = E\times A = A$); simply take a well-founded non-empty set and proceed as above (a "well-founded set" just means a set that does satisfy the statement in the Axiom of Foundation). 
