Remark 2

Let B be a topological space such that every point has a simply connected neighbourhood; then every covering of a covering of B is a covering of B. Consider in fact a covering $(E, p)$ of B together with a covering $(F, q)$ of E. We shall prove that $(F, p\circ q)$ is a covering of B. Since the question is local on B, we may suppose that the space B is simply connected and therefore that E is a trivializable covering of B. Let V be a connected component of E. It is open and closed and $p|V: V\rightarrow B$ is a homeomorphism (I, p. 69, proposition 1); it follows that the space V is simply connected. Every connected component W of $\overset{-1}{q}(V)$ is open and closed in $\overset{-1}{q}(V)$, hence in F, and the mapping $q$ induces a homeomorphism of W onto V. The mapping $p\circ q$ therefore makes F a covering of B, trivializable (loc. cit.).