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Let $M$ be a (multiplicative) monoid with a topology $\tau$. I'd like some simple equivalent conditions for $(M,\tau)$ to be a topological monoid.

For example, a group $G$ with a topology is a topological group iff the "translations" $L_a:x\in G\mapsto ax\in G$ and $R_a:x\in G\mapsto xa\in G$ are continuous, the inversion $x\in G\mapsto x^{-1}\in G$ is continuous and the operation $(x,y)\in G\times G\mapsto xy\in G$ in continuous in the identities $(e,e)\in G\times G$.

In the case of monoids, if we assume that the translations are open mappings then the same arguments as in the case of groups can be used. But I think that this condition is way too strong. I don't want to use the inverse element also.

Thanks.

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1 Answer 1

up vote 2 down vote accepted

There is little hope in the general case. In particular, $1$ can be the unique inversible element of a monoid (think of $\mathbb{N}$ under addition) and hence monoids can be radically different from groups. The compact case offers a few more results, but not much. Let me extract a few basic results given in the introduction of this book:

W. Ruppert, Compact Semitopological Semigroups: An Intrinsic Theory, Lecture Notes in Mathematics 1079 (1984)

Let $M$ be a compact semitopological monoid.
(1) The multiplication is jointly continuous at all points $(1,x)$ and $(x,1)$, for $x \in M$.
(2) Suppose that $M$ has a dense group of units $H$. Then for every $x \in M$ the restriction of the multiplication to the set $xM \times M$ is jointly continuous at all points $(x,s)$, for $s \in M$.

Some consequences:
(1) Every maximal subgroup of a compact semitopological semigroup is a topological group;
(2) every compact semitopological semilattice is a topological semilattice;
(3) every compact semitopological semigroup with dense subgroup contains only one minimal idempotent.

The chapter Joint continuity offers some further results that are too specialized to be reproduced here.

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