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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 $p^{-1}(p(x))$?

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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-Alvarez Sep 24 '10 at 23:30
up vote 7 down vote accepted

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 an automorphism of the covering $p$ (which you can also think of as a lift of $p$). Note also that $k(h) = g(h,h) = h$. Thus $k$ and the identity are both automorphisms of $p$ that agree at a point, so they are equal. This implies $g(x,h)=x$ for all $x$.

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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
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

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