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If we have a covering $p:H\rightarrow G$ where $G$ is a topological group, then $H$ is also a topological group. The multiplication function can be defined as follows. Consider the map $f:H\times H \rightarrow G$ which is a composition of the map $p\times p$ and the multiplication function on $G$. Choose $h\in p^{-1}(e)$ where $e$ is the identity element of $G$. If $$f_* (\pi_1(H\times H,(h,h))) \subset p_*(\pi_1(H,h)),$$ then $f$ can be lifted to a map $g:H\times H \rightarrow H$ such that $p\circ g = f$ and $g(h,h) = h$. Suppose we have shown the "if" part, then $g$ should function as our multiplication map on $H$. But given any $x\in H$, why do we know that $g(x,h) = x$ and that $g(x,h)$ does not equal any other element of $p^{-1}(p(x))$?

Down
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  • Perhaps you should lift $H \to G^G$ to $H^H$. – Martin Brandenburg Sep 24 '10 at 16:19
  • Martin, would you care to elaborate that? If I do what you say (I guess $G^G$ is the same as $G\times G$) then we would get a map $H\rightarrow H\times H$, but how would that help me? – Down Sep 24 '10 at 16:59
  • If $x \in H$, what do you mean by $p^{-1}(x)$? – Dylan Wilson Sep 24 '10 at 18:10
  • Dylan, see the update. – Down Sep 24 '10 at 19:01
  • GDown, $G^G$ is not $G\times G$ (why would we have two notiations for the same thing!): it is the space of continuous maps from $G$ to $G$. – Mariano Suárez-Álvarez Sep 24 '10 at 23:30
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    Though this question is old, anyone who reads it should note that the claim in the title, as written, is not necessarily true. It is certainly true when $G$ is path-connected and locally path-connected and presumably the OP has assumed this. The general statement for connected groups $G$ is (remarkably) an open question. – Jeremy Brazas Feb 01 '16 at 22:04

1 Answers1

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Consider the map $k: H\to H$ given by $k(x) = g(x,h)$. Then for any $x\in H$, we have $$p\circ k(x) = p\circ g(x,h) = m\circ (p\times p)(x,h) = m(p(x),e) = p(x),$$ which implies that $k$ is a lift of $p\colon H\to G$. Note also that $k(h) = g(h,h) = h$. Thus $k$ and the identity are both lifts of $p$ that agree at a point, so they are equal. This implies $g(x,h)=x$ for all $x$.

Michael Albanese
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Jack Lee
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    A follow up question: How do we know that $k(h) = h$ and that it just not equals any other element of $p^{-1}(e)$? – Down Sep 28 '10 at 10:54
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    Your original question said "Suppose we have shown the 'if' part," namely that $f$ can be lifted to a map $g:H\times H\to H$ such that $p\circ g=f$ and $g(h,h)=h$. So I was supposing that, which implies $k(h)=h$ by definition of $k$. – Jack Lee Sep 28 '10 at 14:13
  • You are correct, professor Lee, for some reason forgot that part... – Down Sep 28 '10 at 15:21
  • Don't we need to show that $k$ is a homeomorphism of $H$ before we can conclude that it is an automorphism of the covering? Or is it the case that (for $H$ connected) if $p\circ k = p$, then $k$ is automatically a homeomorphism? – Michael Albanese Jan 20 '22 at 18:30
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    @MichaelAlbanese: It's probably better just think of $k$ as a lift of $p:H\to G$. Two lifts that agree at a point are equal. I've edited my answer. – Jack Lee Jan 21 '22 at 16:13