# If a covering space of a topological space X has a topological group structure, when we transfer this structure on X?

Let (G,e) be a topological group and p : G → X be a covering map. When p can transfer the group structure to make X a topological group? It is clear that if p is a group homomorphism it is done. But we want to find minimal conditions. Moreover, the converse statement happen without extra condition, i.e, if p : X → G is a covering map and G is a topological group, then X is also a topological group and p is a group homomorphism.

Your condition does not make sense : "if $$p$$ is a group morphism": $$X$$ has no group structure so $$p$$ can't be a group morphism !

A necessary condition is clearly : "if $$p(x)=p(x'), p(y)=p(y')$$, then $$p(xy)=p(x'y')$$".

Is it sufficient ? Assume we have this condition, then there is only one way to put a group structure on $$X$$ that makes $$p$$ a group morphism; but a priori this group structure need not be continuous. That's where the covering hypothesis comes into play.

Indeed let $$a,b\in X$$, and $$V$$ a neighbourhood of $$ab$$. Let also $$x,y\in G$$ with $$p(x)=a, p(y)=b$$ (thus $$p(xy) = ab$$).

Let $$W$$ be a smaller open neighbourhood of $$ab$$, and $$O$$ an open neighbourhood of $$xy$$ such that $$p$$ is a homeomorphism $$O\to W$$ (exists by the covering hypothesis). Multiplication is continuous on $$G$$, so we can find open neighbourhoods $$U, U'$$ of $$x,y$$ such that $$UU'\subset O$$. Now $$p$$ is an open map so $$p(U)$$ is an open neighbourhood of $$a$$, $$p(U')$$ of $$b$$ and so $$p(U)p(U') \subset W$$ proves that multiplication on $$X$$ is continuous at $$(a,b)$$ : hence it is continuous.

One can proceed similarly with the inverse map to show that it's also continuous.

Therefore, our condition was sufficient : it suffices for $$p$$ to induce a set-theoretic group structure on $$X$$.

Notice that I haven't actually used that $$p$$ was a covering map : all I needed was that $$p$$ be a surjective open map.

• Dear Max thanks for spending the time. You are right, if p has the condition it is done. My main question now is whether this condition can be obtained from evenly covered property or not? Can we find a counterexample? – Araz Binevli Oct 27 '18 at 16:59
• What do you mean "evenly covered"? – Max Oct 27 '18 at 18:20
• I mean the covering maps main property. Let p : X → Y be a surjective map. The space Y is evenly covered by p if for every y in Y, there is an open neighborhood U of y such that U is evenly covered by p (i.e, p^{-1}(U) is a union of disjoint open sets in X, each of which is mapped homeomorphically onto U by p). – Araz Binevli Oct 28 '18 at 18:47
• The condition doesn’t follow from that property; for example consider the torus double-covering the Klein bottle (which cannot admit a group structure). – Aleksandar Milivojevic Nov 5 '18 at 4:50