Category with no product?

Question

Is there a family of objects in some category which has no product? If so is there a simple reason for it?

Answer 1

Three examples for your enjoyment:

1. Let $C$ be a discrete category, so that the only morphisms are the identity morphisms. If $a, b \in C$ are distinct objects, then the product $a \times b$ does not exist. Indeed, we must have $C(d, a \times b) = C(d, a) \times C(d, b)$ for all objects $d$; but since $a\neq b$, at least one of the two sets $C(d, a)$ and $C(d, b)$ is empty, hence $C(d, a \times b)$ is empty for all $d$. This is impossible, since when $d=a\times b$ there should be the identity morphism.

2. Let $\mathbf Q$ be considered as a total order. If $S\subset \mathbf Q$, the product $\prod_{s \in S} s$ in this category is nothing but $\sup S$, when this $\sup$ is a rational number. When $\sup S$ is irrational or infinite, the product does not exist. Obviously, all finite nonempty products exist.

3. In any category, the empty product is a terminal object. Thus, when $C$ has no terminal object, the empty product does not exist in $C$.

Answer 2

There are lots of examples here. For example, let $\mathsf{Ab}_f$ be the category of finitely-generated abelian groups and consider the family of $\mathsf{Ab}_f$-objects $\{A_i:i\in\Bbb N\}$ where $A_i=\Bbb Z$. Then $\prod A_i$ does not exist in $\mathsf{Ab}_f$.

Note, however, that the Yoneda embedding $\mathscr{C}\to[\mathscr C^{\DeclareMathOperator{op}{op}\op},\mathsf{Set}]$ is a fully-faithful functor from any category $\mathscr C$ into the complete category $[\mathscr C^{\DeclareMathOperator{op}{op}\op},\mathsf{Set}]$.