Here is an interesting question raised by Alice Rhyl.
Let $C$ be a category with a terminal object $1$ and finite coproducts. How many different morphisms $f : 1 \to 1 + 1$ can there be?
There are always two obvious morphisms $f : 1 \to 1 + 1$, coming from the definition of coproduct. But if $C$ is the category with one object and one morphism, $1 + 1 = 1$ so these two obvious morphisms are equal and there's really just one.
Can there be three different morphisms $f : 1 \to 1 + 1$? I don't know.
There can be four. Take $C = \mathrm{Set}^2$; then the terminal object in $C$ is $(1,1)$ (where $1$ is your favorite one-element set), and there are four different morphisms $f: (1,1) \to (1,1) + (1,1)$.
Indeed, any power of two is possible; just take $C = \mathrm{Set}^n$.
What other numbers are possible? (I find finite cardinals more interesting here.)
So far $C$ has just been any category with a terminal object and finite coproducts. In a paper I'm writing with Christian Williams, I'm more interested in the case where $C$ is a cartesian closed category with finite coproducts. How many morphisms $f : 1 \to 1 + 1$ can there be in this case?
(All the examples I've given above are categories of this sort.)
Best Answer
Let $V$ be any variety of idempotent operations, such as the varieties of idempotent groupoids (aka magmas), or idempotent semigroups, or semilattices, or lattices. Then $V$ is complete and cocomplete (if we include in $V$ an empty algebra); $1$ is just the $1$-element algebra, and morphisms $1\to A$ are in 1–1 correspondence with set-theoretical elements of $A$. The object $1$ is also the free algebra on $1$ generator, hence $1+1$ is the free algebra on $2$ generators. So, all in all, there are as many morphisms $1\to1+1$ as is the cardinality of the free $V$-algebra on two generators. For example:
If $V$ is the variety of semilattices, the number is $3$.
If $V$ is the variety of idempotent semigroups, the number is $6$.
If $V$ is the variety of idempotent groupoids, the number is $\aleph_0$. More generally, if $V$ is the variety of all algebras with $\kappa\ge\aleph_0$ idempotent binary operations, then the number is $\kappa$.