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Proposition 4. If $(X,\tau)$ is an Alexandrov space, then $\End(X,\tau)\simeq\End(Y,\sigma)$ if and only if

 

• $(Y,\sigma)$ is homeomorphic to $(X,\tau)$, or

 
• $(Y,\sigma)$ is homeomorphic to the opposite of $(X,\tau)$ (i.e., the Alexandrov space corresponding to $\ge_\tau$), or

 
• the two spaces are the discrete and indiscrete topologies on sets of the same cardinality.

 

Consequently, $(X,\tau)$ has a unique endomorphism monoid iff $(X,\le_\tau)\simeq(X,\ge_\tau)$, and $\tau$ is neither discrete nor indiscrete unless $|X|\le1$.

Proposition 4. If $(X,\tau)$ is an Alexandrov space, then $\End(X,\tau)\simeq\End(Y,\sigma)$ if and only if

• $(Y,\sigma)$ is homeomorphic to $(X,\tau)$, or

• $(Y,\sigma)$ is homeomorphic to the opposite of $(X,\tau)$ (i.e., the Alexandrov space corresponding to $\ge_\tau$), or

• the two spaces are the discrete and indiscrete topologies on sets of the same cardinality.

Consequently, $(X,\tau)$ has a unique endomorphism monoid iff $(X,\le_\tau)\simeq(X,\ge_\tau)$, and $\tau$ is neither discrete nor indiscrete unless $|X|\le1$.

Proof: If, say, $a\sim_\tau b$ but $a\nsim_\sigma b$, then all mappings $X\to\{a,b\}$ are in $\End(X,\tau)$, hence in $\End(X,\sigma)$, hence $(X,\sigma)$ is discrete, hence all mappings $X\to X$ are in $\End(X,\tau)$, hence $(X,\tau)$ is indiscrete lest $a\nsim_\tau b$.

If $\le_\tau$ is not an equivalence, let us fix $a\lnsim_\tau b$. This also implies we can fix $V\in\tau$ whose complement is not in $\tau$. Then $\phi_{a,b,V}\in\End(X,\tau)$ and $\phi_{b,a,V}\notin\End(X,\tau)$, hence $a\lnsim_\sigma b$ or $b\lnsim_\sigma a$. W.lo.g., we assume the former (the other choice leads to the opposite order). Then for each $U\in\tau$, $\phi_{a,b,U}\in\End(X,\tau)$ implies $U\in\sigma$, i.e., $\tau\subseteq\sigma$. Since $\tau$ is the finest topology with specialization preorder $\le_\tau$, if $\tau\subsetneq\sigma$, then (in view of ${\sim_\tau}={\sim_\sigma}$) there are $x,y$ such that $x\lnsim_\tau y$ and $x\nleq_\sigma y\nleq_\sigma x$. But as above, this contradicts $\phi_{x,y,V}\notin\End(X,\sigma)$ for suitable $V\in\tau$. Thus, $\tau=\sigma$.

Proof: If, say, $a\sim_\tau b$ but $a\nsim_\sigma b$, then all mappings $X\to\{a,b\}$ are in $\End(X,\tau)$, hence in $\End(X,\sigma)$, hence $(X,\sigma)$ is discrete, hence all mappings $X\to X$ are in $\End(X,\tau)$, hence $(X,\tau)$ is indiscrete lest $a\nsim_\tau b$. QED

If $\le_\tau$ is not an equivalence, let us fix $a\lnsim_\tau b$. This also implies we can fix $V\in\tau$ whose complement is not in $\tau$. Then $\phi_{a,b,V}\in\End(X,\tau)$ and $\phi_{b,a,V}\notin\End(X,\tau)$, hence $a\lnsim_\sigma b$ or $b\lnsim_\sigma a$. W.lo.g., we assume the former (the other choice leads to the opposite order). Then for each $U\in\tau$, $\phi_{a,b,U}\in\End(X,\tau)$ implies $U\in\sigma$, i.e., $\tau\subseteq\sigma$. Since $\tau$ is the finest topology with specialization preorder $\le_\tau$, if $\tau\subsetneq\sigma$, then (in view of ${\sim_\tau}={\sim_\sigma}$) there are $x,y$ such that $x\lnsim_\tau y$ and $x\nleq_\sigma y\nleq_\sigma x$. But as above, this contradicts $\phi_{x,y,V}\notin\End(X,\sigma)$ for suitable $V\in\tau$. Thus, $\tau=\sigma$. QED

The characterization can be easily extended to all non-$R_0$ spaces. Recall that $(X,\tau)$ is $R_0$ if $\le_\tau$ is symmetric (i.e., ${\le_\tau}={\sim_\tau}$).

Proposition 5. If $(X,\tau)$ is a non-Alexandrov non-$R_0$ space, then $(X,\tau)$ has a unique endomorphism monoid.

Proof: Assume that $\End(X,\tau)=\End(X,\sigma)$. Let us fix $a\lnsim_\tau b$. There exists $V\in\tau$ whose complement is not in $\tau$ (e.g., any open set separating $b$ from $a$); then $\phi_{a,b,V}\in\End(X,\tau)$ and $\phi_{b,a,V}\notin\End(X,\tau)$, hence (1) $a\lnsim_\sigma b$ or (2) $b\lnsim_\sigma a$. (In particular, $(X,\sigma)$ is not $R_0$.) If (1) holds, then for every $U\in\tau$, $\phi_{a,b,U}\in\End(X,\tau)$ implies $U=\phi_{a,b,U}^{-1}[b]\in\sigma$, i.e., $\tau\subseteq\sigma$. If (2) holds, then the same argument gives $\{X\smallsetminus U:U\in\tau\}\subseteq\sigma$.

Since $(X,\sigma)$ is not $R_0$ either, a symmetric argument implies that (1') $\sigma\subseteq\tau$, or (2') $\{X\smallsetminus U:U\in\sigma\}\subseteq\tau$. It is impossible that (1) and (2') hold together: this would imply that $\tau$ is closed under complement, whence it is $R_0$. Likewise, (2) and (1') are incompatible. Thus, the only two possibilities are that either (1) and (2) hold, in which case $\tau=\sigma$, or (1') and (2') hold, in which case $\tau$ and $\sigma$ are mutually opposite Alexandrov spaces. QED

Notice that $(X,\tau)$ is $R_0$ iff the Kolmogorov quotient $X/{\sim_\tau}$ is $T_1$. It is easy to see that:

Lemma 6. If $(X,\tau)$ and $(X,\sigma)$ are spaces such that ${\sim_\tau}={\sim_\sigma}$, then $\End(X,\tau)=\End(X,\sigma)$ iff $\End((X,\tau)/{\sim_\tau})=\End((X,\sigma)/{\sim_\sigma})$.

In view of Lemma 3, this gives a reduction of the remaining classification to $T_1$ spaces. Observe that an $R_0$ space $(X,\tau)$ is Alexandrov iff $(X,\tau)/{\sim_\tau}$ is discrete.

Corollary 7. If $(X,\tau)$ is an $R_0$ non-Alexandrov space, then $(X,\tau)$ has a unique mononorphism monoid iff the $T_1$ space $(X,\tau)/{\sim_\tau}$ has a unique monomorphism monoid.

Assume first that $\le_\tau$ is an equivalence (i.e., ${\le_\tau}={\sim_\tau}$). By Lemma 3, we may assume that ${\sim_\tau}={\sim_\sigma}$ and $\tau$ is not indiscrete. Since $\tau$ is the finest topology with indistinguishability relation $\sim_\tau$, this implies $\sigma\subseteq\tau$; on the other hand, if we fix $a\nsim_\tau b$, and w.l.o.g. $a\lnsim_\sigma b$, then $\phi_{a,b,U}$ is in $\End(X,\sigma)$ for all $U\in\tau$, hence $U\in\sigma$, i.e., $\sigma=\tau$.

Assume first that $\le_\tau$ is an equivalence (i.e., ${\le_\tau}={\sim_\tau}$). By Lemma 3, we may assume that ${\sim_\tau}={\sim_\sigma}$ and $\tau$ is not indiscrete. (If $\tau$ is indiscrete, then either $\sigma$ is discrete and we are done, or ${\sim}_\sigma={\sim}_\tau$, hence $\sigma$ is indiscrete, i.e., $\sigma=\tau$, and we are also done.) Since $\tau$ is the finest topology with indistinguishability relation $\sim_\tau$, this implies $\sigma\subseteq\tau$; on the other hand, if we fix $a\nsim_\tau b$, and w.l.o.g. $a\lnsim_\sigma b$, then $\phi_{a,b,U}$ is in $\End(X,\sigma)$ for all $U\in\tau$, hence $U\in\sigma$, i.e., $\sigma=\tau$.