Covering of a topological group is a topological group 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 $p^{-1}(p(x))$?

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 $x$.
