Proposition 1

Let B be a topological space and let $(E, p)$ be a covering of B. Let Y be a topological space and let $f: Y\rightarrow B$ be a continuous mapping. Let $y\in$ Y$,x\in$ E$,b\in$ B be points such that $f(y) =p(x) =b$. Suppose the space Y connected and locally arcwise connected. In order that there exist a continuous lifting $g: Y\rightarrow E$ of $f$ such that $g(y) =x$, it is necessary and sufficient that the image of the homomorphism $\pi_1(f, y):\pi_1(Y, y)\rightarrow \pi_1(B, b)$ be contained in the image of the homomorphism $\pi_1(p, x):\pi_1(E, x)\rightarrow \pi_1(B, b)$.

The condition is necessary without any assumption on the space Y. In fact, if such a lifting $g$ exists, we have $\pi_1(f, y) =\pi_1(p, x)\circ \pi_1(g, y)$.

Let us prove that it is sufficient. Let $s: \Lambda_b(B)\rightarrow \Lambda_x(E)$ denote the homeomorphism inverse to the homeomorphism $c\mapsto p\circ c($III, p. 302, cor. 2 of prop. 3) and let $\varphi : \Lambda_y(Y)\rightarrow \Lambda_x(E)$ be the mapping $d\mapsto$ $s(f\circ d)$. The mapping $\varphi$ is continuous (I, p. 132, lemma).

Let $d$ and $d'\in \Lambda_y(Y)$ be paths with origin $y$ having the same term; let us prove that the paths $\varphi (d)$ and $\varphi (d')$ have the same term. Put $c=f\circ d,c'=f\circ d'$. Since the path $d*\overline{d'}$ is a loop in Y at $y$, the path $c*\overline{c'}$ is a loop in B at $b$ and its class belongs to the image of the homomorphism $\pi_1(f, y)$, hence to the image of the homomorphism $\pi_1(p, x)$ by assumption. By cor. 2 of prop. 4 (III, p. 303), the path $s(c*\overline{c'})$ is a loop in E at $x$. The same is true of the path $s(c'*\overline{c})$ which, by uniqueness of the lifting of paths, is equal to $s(c*\overline{c'})$. Therefore $s(c'*\overline{c})(\frac{1}{2}) =s(c')(1) =s(c)(1)$, as was to be proved.

Let us denote by $e_E: \Lambda_x(E)\rightarrow E$ and $e_Y: \Lambda_y(Y)\rightarrow Y$ respectively the term mappings. Since the space Y is supposed connected and locally arcwise connected, the mapping $e_Y$ is surjective and open (III, p. 262, prop. 10). By the preceding paragraph, there exists a unique mapping $g: Y\rightarrow E$ such that $e_E\circ \varphi =g\circ e_Y$. It is continuous, for the mapping $e_Y$ is strict ( I, p. 18, example 2).

Let us verify finally that the mapping $g$ lifts the mapping $f$ and that $g(y) =x$. Every point $z$ of Y is the term of a path $c$ with origin $y$. The path $\varphi (c)$ is a lifting of $f\circ c$ with origin $x$ and with term the point $g(z)$. Therefore $p(g(z)) =f(z)$. For $z=y$ and $c=e_y$, we have $\varphi (e_y) =e_x$, whence $g(y) =x$.