# Is the product of two objects in a concrete category always a sub-object of the product of these objects viewed as sets?

Is the product of two objects in a concrete category always a sub-object of the product of these objects in the category of sets, i.e. is their product in the concrete category a subset of their cartesian product?

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The word "is" here rubs me the wrong way. I'd rather you rephrase the question as follows: for $F : C \to \text{Set}$ any functor, there's a natural morphism in $\text{Set}$ from $F(X \times Y)$ to $F(X) \times F(Y)$ given by the universal property of $F(X) \times F(Y)$. The question is whether this morphism is a monomorphism if $F$ is faithful. – Qiaochu Yuan Feb 27 '12 at 20:57

Let $P : \text{Set}^{op} \to \text{Set}$ be the powerset functor that sends a set $X$ to the set of its subsets and sends a function $f : X \to Y$ to the inverse image function $f^{-1} : PY \to PX$. Since $P$ is faithful we see that $\text{Set}^{op}$ is concrete but also that given any concrete $C$, $C^{op}$ is concrete: if $F : C \to \text{Set}$ is faithful, so is $P \circ F : C^{op} \to \text{Set}$. These $C^{op}$ tend to produce counterexamples (that answer your question negatively) when $C$ is some sort of algebraic category --in which case $C$ itself would not give counterexamples.

To give one specific example, take $C$ to be the category of Abelian groups. So we're making $C^{op}$ concrete by means of the functor $A \mapsto PA=\{$ subsets of $A \}$. The product in $C^{op}$ is the coproduct in $C$ which is the direct sum of Abelian groups. The canonical map $P(A \oplus B) \to PA \times PB$ is just $S \mapsto (S \cap A, S \cap B)$ which is not injective unless $A$ or $B$ is trivial.

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Nice and easy example. – Martin Brandenburg May 17 '12 at 18:12

I haven't found a counterexample yet, but I'm sure that there will one where $C$ has only one object (isn't this funny?):

A category with one object is just a monoid $M$. It has binary products if and only if there are $p_1,p_2 \in M$ such that $M \to M^2$, $m \mapsto (p_1 m,p_2 m)$ is a bijection, in other words: There is an isomorphism of right-$M$-sets $M \cong M^2$; an example is $\mathrm{End}(S)$ for an infinite set $S$.

A functor $M \to \mathrm{Set}$ is just a set $X$ on which $M$ acts on the left, also called an $M$-set. This functor is faithful iff the action is faithful, i.e. $\forall m,n \in M \forall x \in X : mx=nx \Rightarrow m=n$. The map $F(*) \to F(*) \times F(*)$ is just $X \to X \times X$, $x \mapsto (p_1 x,p_2 x)$. Is this injective, i.e. do we have $p_{0,1} x = p_{0,1} y \Rightarrow x = y$? Probably not!

I'm sure that someone can find a counterexample.

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