H S class operator and its equality

$A \in S(K)$ iff $A$ is a subalgebra of some member of $K$

$A \in H(K)$ iff $A$ is a homomorphic image of some member of $K$

It is trivial to see the containment $SH \leq HS$. Taking a simple example of rings also proves $HS \neq SH$. Currently I am stuck with proving $SH(K) = HS(K)$ where K is a monounary algebra $(Z,successor).$

This question might sound belonging to stack exchange, but there isn't a large community of universal algebraists over there.

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Is $Z$ the integers, or an arbitrary set with a "successor" function? – Arturo Magidin Apr 14 '14 at 21:06
You assumed correctly – Anurag Sharma Apr 14 '14 at 22:34
So then... does the answer I give below settle it? – Arturo Magidin Apr 15 '14 at 3:30
yeah it certainly does. I didn't accept it earlier because I wasn't sure somewhere. But now it seems clear. – Anurag Sharma Apr 15 '14 at 3:43

I'm assuming that $(Z,successor)$ represents the set of integers with the successor function $s\colon Z\to Z$, $s(a)=a+1$.
I think the following works: we want to show that any homomorphic image of a subalgebra of $(Z,s)$ is a subalgebra of a homomorphic image of $(Z,s)$. A subalgebra of $(Z,s)$ is either the whole of $Z$ (in which case we don't have a problem showing any homomorphic image lies in $SH(Z)$) or else it is of the form $\{k\in Z\mid k\geq n_0\}$ for some integer $n_0$. These are all isomorphic to $(\mathbb{N},s)$, so we just need to consider homomorphic images of $(\mathbb{N},s)$.
Let $\Phi$ be a congruence on $(\mathbb{N},s)$, and let $\Psi = \Phi\cup\{(a,a)\mid a\in\mathbb{Z}, a\lt 0\}$ (my natural numbers include $0$). Then it is easy to verify that $\Psi$ is a congruence on $\mathbb{Z}$, and that if $x\in\mathbb{Z}$ and $a\in\mathbb{Z}$, $a\lt 0$, then $[x]=[a]$ if and only if $x=a$. Thus, the subalgebra of $Z/\Psi$ consisting of the equivalence classes represented by nonnegative integers form a subalgebra that is isomorphic to $\mathbb{N}/\Phi$, showing that every homomorphic image of a subalgebra may be realized as a subalgebra of a homomorphic image in this situation.
@AnuragSharma: For one thing, $(\mathbb{N},s)$ does not have $\mathbb{Z}$ as an image, but $(\mathbb{Z},s)$ certainly does... – Arturo Magidin Apr 15 '14 at 14:58
I have no idea what you mean by "structure wise". Clearly, there are homomorphic images of $(\mathbb{Z},s)$ that cannot be realized as (isomorphic to) homomorphic images of $(\mathbb{N},s)$, so that means that $\mathbf{H}(\mathbb{Z},s)$ cannot be equal to $\mathbf{H}(\mathbb{N},s)$, whatever "structurewise" is supposed to mean. – Arturo Magidin Apr 15 '14 at 17:27