I would like prove that, under the conditions described below, no non-trivial variety exists.



> 
> Let $\mathcal{V}$ be a variety of algebras e.g. rings, semigroups, semilattices.
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Further suppose that:



> 
> 1. The empty algebra exists i.e. $\mathcal{V}$ has no constants.
> 2. The dual condition also holds. That is, let $\mathtt{1} \in \mathcal{V}$ be the one element terminal algebra. Then $id\_\mathtt{1}$ is the only homomorphism whose domain is $\mathtt{1}$.
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Any help much appreciated.



A variety of algebras satisfying some dual conditions
It's false. Take the theory generated by unary operations $\zeta$ and $\iota$ and a binary operation $\cdot$, subject to the equations
$$
\zeta(x) = \zeta(y),
\qquad
\iota(x) = \iota(y),
\qquad
\zeta(x) \cdot y = \zeta(x),
\qquad
\iota(x) \cdot y = y.
$$
An example of an algebra is any ring, with $\zeta(x) = 0$, $\iota(x) = 1$, and the usual $\cdot$. So the theory is not trivial.


Obviously $\emptyset$ is an algebra. Now suppose we have a homomorphism from $1$ to an algebra $A$, picking out an element $a \in A$. The equation $\iota(x) = \zeta(x)$ holds in $1$, so $\iota(a) = \zeta(a)$ too. Hence for all $b \in A$,
$$
b = \iota(a) \cdot b = \zeta(a) \cdot b = \zeta(a),
$$
giving $A = \{\zeta(a)\} \cong 1$. 


The trick is that although the theory has no *actual* constants, it has two "pseudo-constants", namely, $\zeta$ and $\iota$.


