In this paper D. B. McAlister introduced a very interesting class of morphisms for inverse semigroups, which he called v-prehomomorphisms. For a such morphism $\theta : S \to T$ we have
- $\theta(ab)\le \theta(a)\theta(b)$
- $\theta(a^{\star})=\theta(a)^{\star}$
for all $a, b\in S.$
If $a^{\star}a\ge bb^{\star}$ (and similarly for $\le$), then $\theta$ preserve the product $ab.$ In addition, these maps preserve idempotents and the natural partial ordering of $S.$
This raises the question: Does it preserve restrictions?
In other words,
Given an idempotent $e$ and arbitrary $a,$ is it true that $\theta(ae)=\theta(a)\theta(e)$ and $\theta(ea)=\theta(e)\theta(a)$?
I believe this is true. Usually proofs of these type of facts involve some kind of cleaver trick, if it is true. I tried every thing came to my mind, but did not succeeded yet.
Does anyone know a proof, counter example or a reference to this simple conjecture?
Added: Assuming $\theta(ae)=\theta(a)\theta(e)$ for all $e, a$ we have $$\theta(ea)=\theta((a^{\star}e)^{\star})=(\theta(a^{\star})\theta(e))^{\star}=\theta(e)\theta(a).$$ Hence, two equalities are equivalent, and we only need to (dis)prove one of them.
This is not true. Let $S$ be the semilattice $\{a,b,c,d\}$ in which $c < a$, $c < b$ and $d < c$. Let $\theta: S \to S$ be the map defined by $\theta(a) = a$, $\theta(b) = b$ and $\theta(c) = \theta(d) = d$. Then $\theta$ is a prehomomorphism but $\theta(ab) = \theta(c) = d < \theta(a)\theta(b) = ab = c$.