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.
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 $p_*(\pi_1(H))$.