Proposition 8

Let B be a connected nonempty topological space. Let G be a discrete group and let $(E, p)$ be a principal covering of B with group G. Suppose that E is connected. With the above notation, one has:

a) For every G-set F endowed with the discrete topology, the mapping $\alpha_F$ is an isomorphism of the G-set F onto the G-set $\mathscr{C}_B(E; E\times^GF)$.

b) Let X be a covering of B. The B-morphism $\beta_X$ is an isomorphism of $E\times^G\mathscr{C}_B(E; X)$ onto X if and only if the covering $(E\times_BX$, pr$_1)$ of E is trivializable.

a) Let $y$ and $y'$ be points of F such that $\alpha_F(y) =\alpha_F(y')$. The space E is not empty; let us choose a point $x$ in it. We have $\pi (x, y) =\pi (x, y')$ in $E\times^GF$. Hence there exists $g\in G$ such that $x\cdot g=x$ and $g^{-1}\cdot y=y'$. The first equality implies that $g$ is the identity element $e$ of G, hence $y=y'$. Thus the mapping $\alpha_F$ is injective. On the other hand, let $h\in$ $\mathscr{C}_B(E; E\times^GF)$ and let $x$ be a point of E. Let $x'\in E,y'\in F$ be such that $h(x) =\pi (x', y')$. In particular, $p(x) =p(x')$; there then exists an element $g$ of G such that $x'=x\cdot g$, and we also have $h(x) =\pi (x, y)$, where we have set $y=g\cdot y'$. The B-morphisms $h$ and $\alpha_F(y)$ coincide at the point $x$ of E; they are therefore equal since the space E is connected (I, p. 34, cor. 1 of prop. 11), and this proves that the mapping $\alpha_F$ is surjective.

b) By proposition 7, b) of I, p. 105 applied to $F =\mathscr{C}_B(E; X)$, the covering $p^*(E\times^G\mathscr{C}_B(E; X))$ of E is isomorphic to the trivial covering $E\times \mathscr{C}_B(E; X)$. If $\beta_X$ is an isomorphism, the covering $p^*(X)$ of E is therefore trivializable. Conversely, suppose that the covering $p^*(X)$ is trivializable and let us prove that the B-morphism $\beta_X$ is bijective; it will follow that $\beta_X$ is a B-isomorphism (I, p. 30, cor. 2 of prop. 6).

The mapping $\beta_X$ is deduced by passing to the quotient from the mapping $\gamma : E\times \mathscr{C}_B(E; X)\rightarrow X$ defined by $\gamma (x, h) =h(x)$. Let us prove that the mapping $\gamma$ is surjective. Let $x$ be a point of X. The projection of the B-space E is surjective, for it is a principal covering; there therefore exists a point $y$ of E such that $(y, x)\in E\times_BX$. There then exists a continuous section $s$ of the trivializable covering pr$_1: E\times_BX\rightarrow E$ such that $s(y) = (y, x)$. The mapping $h=$ pr$_2\circ s: E\rightarrow X$ is a B-morphism, and we have $h(y) =x$, whence the surjectivity of the mapping $\gamma$ and, consequently, that of the mapping $\beta_X$.

Let us prove finally that $\beta_X$ is injective. Let $(x, h)$ and $(x', h')$ be elements of $E\times \mathscr{C}_B(E; X)$ such that $h(x) =h'(x')$. Observe that $x$ and $x'$ have the same projection in B; there therefore exists an element $g$ of G such that $x'=x\cdot g$. We then have $h(x) =h'(x\cdot g) = (g\cdot h')(x)$. Since the space E is connected, we have $h=g\cdot h'($I, p. 34, cor. 1 of prop. 11). Thus, $(x, h)$ and $(x', h')$ have the same class in $E\times^G\mathscr{C}_B(E; X)$, which proves that the mapping $\beta_X$ is injective, and completes the proof.