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Suppose we have two path-connected spaces $G$ and $H$. Suppose also that $G$ is a topological group with an identity element $e$ and there is a covering $$ p: H \rightarrow G $$ The problem asks that we choose an element $f$ in the fiber of $e$ to be the identity element of $H$ and show $H$ to be a topological group. I was able to find a map $$f:H \times H \rightarrow G$$ via the composition of the multiplication on $G$ and $p \times p$. There is a theorem referred to within the book I am using which states that there is a map $H\times H \rightarrow H$ when $$f_{*}(\pi_{1}(H \times H, (f,f)) \subseteq p_{*}(\pi_{1}(H,f))$$

I am having trouble carefully showing this containment. I am wondering if this containment makes use of the continuous multiplication on $G$. This is a problem in Peter May's Concise Course on Algebraic Topology, so everything has some categorical framework. Any help on this would be appreciated.

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1 Answer 1

You have $f:H\times H\rightarrow G\times G\rightarrow G$, where the second map is the product in $G$. The first map will map $\pi_1(H\times H)$ into $p_*(\pi_1(H))\times p_*(\pi_1(H))$. Since $p_*(\pi_1(H))$ is a subgroup, the multiplication map will map $p_*(\pi_1(H))\times p_*(\pi_1(H))$ into $p_*(\pi_1(H))$.

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Wow, okay, I completely missed the closure property of subgroups. Thank you. –  user135520 Mar 15 at 18:56
    
A small detail to help someone in looking at this in the future: For point wise multiplication of paths is equivalent to composition of paths. In the above framework: For loops $\alpha$ and $\beta$ in $\pi_{1}(H)$. We have a point-wise multiplication of $\alpha*\beta(t)=\alpha(t)\beta(t)$ –  user135520 Mar 16 at 0:10

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