Corollary 1

Let B be an unwindable topological space and let $b$ be a point of B. Let E be a connected covering of B and let $x$ be a point of the fibre $E_b$. The following properties are equivalent:

(i) The group $\pi_1(E, x)$ is trivial.

(ii) The space E is simply connected.

(iii) The pointed space $(E, x)$ is a universal covering of $(B, b)$.

The implication (i)$\Rightarrow$(ii) has already been proved under the sole assumption that the space B is connected and locally arcwise connected (III, p. 309, cor. 2 of prop. 1), and the implication (ii)$\Rightarrow$(iii) without any assumption (I, p. 126, corollary).

Let us prove (iii)$\Rightarrow$(i). By Theorem 1 of IV, p. 342, there exists a covering $E'$ of B and a point $x'$ of the fibre $E'_b$ such that the group $\pi_1(E', x')$ is reduced to the identity element. Under assumption (iii), there exists a pointed B-morphism $f: (E, x)\rightarrow (E', x')$. Let $p: E\rightarrow$ B and $p': E'\rightarrow B$ denote the projections of the coverings E and $E'$; we have $p=p'\circ f$, hence $\pi_1(p, x) =\pi_1(p', x')\circ \pi_1(f, x)$. Since the group $\pi_1(E', x')$ is trivial, the injective homomorphism $\pi_1(p, x)$ has the identity element for image. This proves that the group $\pi_1(E, x)$ is reduced to the identity element.