The Kuratowski Monoid I have been reading the paper "The Kuratowski closure complement theorem" by "B. J. Gardener and M. Jackson". In that the author discusses the 6 different monoid structures as follows: Extremally disconnected (ED), Open Unresolvable (OU), ED and OU, Partition space, Discrete space and the Kuratowski space.
He says the above six different Kuratowski monoids of n.e. topological spaces admit a natural partial order. It is Defined as follows:
$\mathbb{M}_1 \leq \mathbb{M}_2$ if $\exists$ a monoid homomorphism from $\mathbb{M}_2$ onto $\mathbb{M}_1$, Now he gives that there is a monoid homomorphism between $ED$ and $OU$ space. I define it as:
$f:OU \rightarrow ED\text{ and }OU $ as follows which is an homomorphism.
$f(c)=c;f(i)=i;f(ic)=f(cic)=f(ici)=f(ci)=f(ic)=ic;f(id)=id$
I have to rule out $\nexists$ any homomorphism between OU and partition space??? Can someone help me with this how to go about it as to how to rule out $\nexists$ any map between these spaces??
 A: There exists a homomorphism between any two monoids (send everything to the identity element).  What the authors are claiming instead is that no homomorphism maps the OU monoid $onto$ the partition monoid.
$Proof.$ Using the authors' notation, let $a$ denote complement, $b$ closure and $i$ interior (we use these same symbols for both monoids, as the context will always be clear).
The OU monoid has ten distinct elements: $\tt id$, $b$, $i$, $bi$ ($=bib$), $ib$ ($=ibi$), and $a$ applied to each of these five.  The partition monoid has six distinct elements: $\tt id$, $b$ ($=ib$), $i$ ($=bi$), and $a$ applied to each of these three (operators get applied in right-to-left order in the paper; for example $ai$ represents the complement of the interior).
$Lemma.$ In the partition monoid, $\left(\{x,y\}\cap\{a,\tt id\}=\emptyset\right)\implies\left(\{y,xy\}\subset\{s,as\}\right)$ for some
$s\in\{b,i\}$.
(See below for an exhaustive proof of the lemma.)
Suppose $g$ is a monoid homomorphism from the OU monoid into the partition monoid.
We have $g(a)g(a)=g(aa)=g({\tt id})=\tt id$.  Thus $g(a)\in\{a,\tt id\}$.
If $g(a)=\tt id$, it follows that $g$ maps the OU monoid into $\{g(b),g(i),g(bi),g(ib),\tt id\}$, a set of at most five elements, hence $g$ is not surjective.
We can therefore assume $g(a)=a$.
Case 1: $g(i)\in\{a,\tt id\}$.
If $g(i)=\tt id$, then $g(b)=g(aia)=g(a)g(i)g(a)=a({\tt id})a=\tt id$.
If $g(i)=a$, then $g(b)=g(aia)=g(a)g(i)g(a)=aaa=a$.
It follows that $g$ maps the OU monoid
into $\{a,\tt id\}$ and is therefore not surjective.
Case 2: $g(b)\in\{a,\tt id\}$.
Interchanging $i$ and $b$ in the proof of Case 1
shows that $g$ again maps the OU monoid into $\{a,\tt id\}$.
Case 3: $\{g(i),g(b)\}\cap\{a,\tt id\}=\emptyset$.
Since the OU monoid satisfies $i(bi)=ib$, we have $g(i)g(bi)=g(i(bi))=g(ib)$. We also have $g(b)g(i)=g(bi)$. By the lemma, it follows that
$\{g(i),g(b),g(ib),g(bi)\}\subset\{s,as\}$ for some $s\in\{b,i\}$. Noting that $g(at)=g(a)g(t)=ag(t)$ for all $t$, this implies that $g$ maps
the OU monoid into $\{s,as,a,\tt id\}$, hence $g$ is not surjective.
Conclude there is no surjective homomorphism from the OU monoid onto the partition monoid. $\square$
Conversely it is trivial that no surjective mapping exists from the partition monoid onto the OU monoid.  The OU and partition monoids are therefore not related under the partial order.
$Proof\ of\ Lemma.$ Here is a multiplication table for the subset $\{b,i,ab,ai\}$ of the partition monoid:
$\matrix{
ib&=&b&&&&&&&&\cr
bb&=&b&&&&&&&&\cr
ii&=&i&&&&&&&&\cr
bi&=&i&&&&&&&&\cr
\cr
i(ab)&=&aba(ab)&=&ab&&&&&&\cr
b(ab)&=&a(aba)b&=&a(ib)&=&ab&&&&\cr
i(ai)&=&(aba)a(aba)&=&ababa&=&a(bi)&=&ai&&\cr
b(ai)&=&ba(aba)&=&b(ba)&=&ba&=&a(aba)&=&ai\cr
\cr
(ai)b&=&ab&&&&&&&&\cr
(ab)b&=&ab&&&&&&&&\cr
(ai)i&=&ai&&&&&&&&\cr
(ab)i&=&ai&&&&&&&&\cr
\cr
(ai)(ab)&=&b&&&&&&&&\cr
(ab)(ab)&=&b&&&&&&&&\cr
(ai)(ai)&=&i&&&&&&&&\cr
(ab)(ai)&=&i&&&&&&&&\cr}$
