# Are there Boolean homomorphisms from the degenerate (trivial) algebra to non-denegerate algebras?

Let's start with the definition of a constant morphism: If $\mathcal{C}$ has a terminal object, then $f:A\to B$ is constant iff it factors through this terminal object. Take $\mathcal C=BA$, the category of Boolean algebras and homomorphisms between them. The terminal object is $\mathbb D$, the one-element (or degenerate, or trivial) Boolean algebra.

Then a constant Boolean homomorphism $f:A\to B$ must factor through $\mathbb D$, i.e. there must exist two homomorphisms $g:A\to \mathbb D$, $e: \mathbb D\to B$ such that $f=e\circ g$.

But if $B\neq \mathbb D$, then $e$ does not exist (stated here at p. 330). It would follow that constant Boolean homomorphisms do not exist when their codomain is not degenerate, but this somehow does not feel right. For example, $e: \mathbb D\to B$ is actually the definition of a global element for any Boolean algebra $B$, so it should exist, as there are no empty Boolean algebras.

Not sure whether this reasoning is correct, help would be much appreciated.

I don't have access to the linked page of the book, but if $\mathbf{A} = \langle A, \wedge, \vee, ', 0_A , 1_A\rangle$ and $\mathbf{B} = \langle A, \wedge, \vee, ', 0_B , 1_B\rangle$ are Boolean algebras and $f : \mathbf{A} \to \mathbf{B}$ ia a homomorphism, then it must be $f(0_A) = 0_B$ and $f(1_A) = 1_B$.
If $\mathbf{A}$ is trivial, then $0_A = 1_A$, and so it must be $0_B = 1_B$, that is, $\mathbf{B}$ must also be trivial.
This means that there is no constant Boolean homomorphism, except from the trivial Boolean algebra to itself (at least if we consider homomorphisms in the sense that is usual in Universal Algebra, and these are the usual ones in Boolean algebras).

• Aren't all homomorphisms $f:A\to B$ constant also when $A$ is non-trivial and $B$ is trivial? $f(0_A)=0_B$ and $f(1_A)=1_B$ are trivially satisfied also when $A$ is non-trivial. – Dawid K Oct 2 '17 at 15:04
• @DawidK Yes, so it is a necessary and sufficient condition, that is, the morphism is constant iff $B$ is trivial. Notice also that this is valid for other types of algebras: all those that have two distinguished elements that are only equal if the algebra is trivial. Another example would be a ring with $1$. – amrsa Oct 2 '17 at 15:17
• All right, this clears things up. Thanks. – Dawid K Oct 2 '17 at 15:20

Global elements are not the same thing as elements of the underlying set; so the non-existence of global elements does not mean that the underlying set is empty. The same thing happen for unital rings : a global element is a ring homomorphism $0\to R$, so non-trivial ring never have global elements, but their underlying sets are non-empty.

Another case where the two concept are different is when the category is pointed, like the category of groups. There a global element is a morphism from the trivial group, so every group has exactly one global element, even though his underlying set can have as many element as you want.

If you want to recover the elements of the underlying set, you can use the free object on one element (assuming it exists) : indeed, if $U:\mathcal{C}\to \mathbf{Set}$ is the forgetful functor and $F:\mathbf{Set}\to \mathcal{C}$ is left adjoint to $U$, then for any object $X$ we have $$U(X)\cong \mathbf{Set}(1,U(X))\cong \mathcal{C}(F(1),X),$$ so elements of $U(X)$ correspond to arrows $F(1)\to X$ (here $1$ is a one-element set). For example, in the case of groups, an element of a group $G$ is the same thing as a morphism $\Bbb Z\to G$.

• Thanks for the answer. I would then conclude that if $f:A\to B$ is constant, then $B\cong\mathbb D$; would you agree? Also, could you please clarify what is $F(1)$ in your notation? – Dawid K Oct 2 '17 at 15:10
• I've clarified the notation a little bit, is that better? – Arnaud D. Oct 2 '17 at 18:47