The space E is locally connected, since B is. Let $E_0$ be the connected component of $x$ in E, so that the space $(E_0, x)$ is a pointed covering of $(B, b)$ (I, p. 80, Cor. 1 of Prop. 6). There then exists a unique morphism of coverings $f: E\rightarrow E_0$ such that $f(x) =x$. If $i$ denotes the canonical injection of $E_0$ into E, the mapping $i\circ f: E\rightarrow E$ is a morphism of coverings which maps $x$ to $x$, just as does the mapping Id$_E$; since $(E, x)$ is a universal covering of $(B, b)$, we therefore have $i\circ f=$ Id$_E$. It follows that $i$ is surjective, hence $E_0= E$. Consequently, E is connected.

Let $y$ be a point of $E_b$ and let us consider the pointed covering $(E, y)$ of $(B, b)$; by hypothesis there exists a unique morphism of coverings $f: E\rightarrow E$ such that $f(x) =y$. The mapping $s: E\rightarrow E\times_BE$ defined by $t\mapsto (t, f(t))$ is a continuous section of the mapping pr$_1: E\times_BE\rightarrow E$. By Corollary 4 of I, p. 81, this proves that the covering $E\times_BE$ of E given by the mapping pr$_1$ is trivializable. The covering E is therefore Galois (Theorem 2 of I, p. 102). It then follows from I, p. 112, corollary to Proposition 10, that every covering of B can be associated with E.

(i) The covering E is trivializable over A;

(ii) Every covering of B is trivializable over A.

Property (ii) obviously implies property (i). The converse follows from the fact that every covering of B can be associated with the covering E (I, p. 105, Proposition 7).

The mapping $p_A$ is finite, continuous, and has the following properties:

(i) For every $x\in E$ such that $x= 0\not$ , one has $p_A(x)>0$ ;

(ii) For every $s\in \mathbf{R}_+$ and every $x\in E$, one has $p_A(sx) =sp_A(x)$.

(iii) For all $x$ and $y\in E$, one has $p_A(x+y)\leqslant p_A(x) +p_A(y)$.

(iv) In order that a point $x$ of E belong to A, it is necessary and sufficient that $p_A(x)\leqslant 1$.

For $x\in E$, let us denote by $\|x\|$ the euclidean norm of $x$ (TG, VI, p. 7). Since A is compact, there exists a real number $M>0$ such that every point $x$ of A satisfies $\|x\|\leqslant M$. Since 0 is an interior point of A, there exists a real number $m >0$ such that every point $x$ of E such that $\|x\|\leqslant m$ belong to A. Consequently, one has the relation $\|x\|/M\leqslant p_A(x)\leqslant \|x\|/m$, for every $x\in E$. In particular, $p_A(0) = 0$ and $p_A(x)= 0\not$ if $x= 0\not$ .

Assertions (ii) and (iv) follow immediately from the definition of the mapping $p_A$.

Let $x$ and $y$ be points of E. Let $x'$ and $y'$ be points of A such that $x=p_A(x)x'$ and $y=p_A(y)y'$. If $x$ and $y$ are not both zero, $p_A(x) +p_A(y)>0$ and one has

Since A is convex, this proves that $x+y$ belongs to $(p_A(x) +$ $p_A(y))A$, whence $p_A(x+y)\leqslant p_A(x) +p_A(y)$. If $x=y= 0$, this inequality still holds, for $p_A(0) = 0$. This proves assertion (iii).

Applying this inequality to $x+y$ and $-y$, one deduces that, for every pair $(x, y)$ of points of E, one has

$|p_A(x+y)-p_A(x)|\leqslant$ max($p_A(y), p_A(-y)$)$\leqslant m^{-1}\|y\|$.

This proves that the mapping $p_A$ is continuous, whence the lemma.

The mapping $p_A$ is called the gauge of the convex subset A.

(i) For every $x\in E$ and every $t\in \mathbf{R}_+,u(tx) =tu(x)$;

(ii) For every $x\in E$, there exists $\lambda \in \mathbf{R}_+$ such that $u(x) =\lambda x$;

(iii) One has $u(A) =\mathbf{B}_n$.

The mapping $u$ is a homeomorphism and induces, by passing to subspaces, a homeomorphism of A onto $\mathbf{B}_n$, a homeomorphism of the interior of A onto the interior of $\mathbf{B}_n$ and a homeomorphism of the boundary of A in E onto the sphere $\mathbf{S}_{n-1}$.

Let $p_A$ be the gauge of the convex subset A. Let $x\in E$ and let $t\in \mathbf{R}_+$; in order that $x\in tA$, it is necessary and sufficient that $p_A(x)\leqslant t$. Since A is compact, there exists a real number M such that $\|x\|\leqslant M$ for every $x\in A$.

Let $u$ be a mapping satisfying the conditions of the proposition. We have $u(0) = 0$. Let $x\in E-\{0\}$ and let $\lambda \in \mathbf{R}_+$ be such that $u(x) =\lambda x$. For every $t\in \mathbf{R}_+$ such that $tx\in A$, we have $u(tx) =t\lambda x$. Since $u$ is injective, $\lambda = 0\not$ . Since $u(A)$ is contained in $\mathbf{B}_n$, we also have $\lambda \leqslant p_A(x)/\|x\|$. Put $z=x/\|x\|$; this is a point of $\mathbf{S}_{n-1}$. In order that $z$ should have a preimage in A under $u$, it is necessary and sufficient that the point $(\lambda \|x\|)^{-1}x$ should belong to A, that is, that $\lambda \|x\|\geqslant p_A(x)$, i.e. $\lambda \geqslant p_A(x)/\|x\|$. This implies the uniqueness of such a mapping $u$.

Let us now denote by $u$ the mapping of E into itself which maps 0 to 0 and $x$ to $(p_A(x)/\|x\|)x$, for every $x\in E-\{0\}$.

By Lemma $1,u$ is continuous at every point of E $-\{0\}$. We have $\|u(x)\|=p_A(x)$ for every $x\in E$ and $p_A(x)\rightarrow 0$ when $x\rightarrow 0$ (loc. cit.); consequently, $u$ is continuous at 0. Therefore $u$ is continuous.

The only preimage of 0 under $u$ is 0. Let $y\in E-\{0\}$. In order that an element $x$ of E should satisfy $u(x) =y$, it is necessary and sufficient that $x= (\|y\|/p_A(y))y$. It follows that $u$ is a continuous bijection of E onto itself. Its inverse is the mapping $v:y\mapsto (\|y\|/p_A(y))y$. Since $p_A$ is continuous and vanishes only at 0, the mapping $v$ is continuous at every point of E $-\{0\}$; the inequality $p_A(y)\geqslant \|y\|/M$ implies that $\|v(y)\|\leqslant M\|y\|$ for every $y\in E$. It follows that $v$ is continuous. Therefore the mapping $u$ is a homeomorphism of E onto itself. Since the relations $p_A(x)\leqslant 1$ and $x\in A$ are equivalent, we also have $u(X) =\mathbf{B}_n$. The proposition follows.

The mappings $g$ sought correspond bijectively (I, p. 9, prop. 3) to the continuous sections $s$ of the covering pr$_1: X\times_BE\rightarrow X$ such that $s(x) = (x, y)$. Such a section exists because the space X is simply connected; it is unique because the space X is connected (I, p. 34, cor. 1 of prop. 11).

Let X and Y be simply connected spaces, and suppose that Y is locally connected. The space $X\times Y$ is connected, since X and Y are so (I, p. 124). Let $(Z, f)$ be a nonempty covering of $X\times Y$; let us prove that it is trivializable. By corollary 2 of I, p. 70, it is enough to prove that for every point $z_0$ of Z, there exists a continuous section $s$ of $f$ such that $s(f(z_0)) =z_0$. Let therefore $z_0$ be a point of Z. Put $(x_0, y_0) =f(z_0)$. The subspace $X\times \{y_0\}$ of $X\times Y$ is homeomorphic to X. Hence the covering deduced from Z over $X\times \{y_0\}$ is trivializable and has a continuous section $\sigma$ such that $\sigma (x_0, y_0) =z_0$. Analogously, for every point $x$ of X, there exists a continuous section $\tau_x$ of $f$ over $\{x\}\times Y$ such that $\tau_x(x, y_0) =\sigma (x, y_0)$. For $(x, y)\in X\times Y$, put $s(x, y) =\tau_x(x, y)$. The mapping $s$ is a section of $f$, and one has $s(x_0, y_0) =z_0$. Since the space Y is connected and locally connected, it follows from theorem 1 of I, p. 35 that the mapping $s$ is continuous.

Let B be such a topological space. Let $(E, p)$ be a covering of B, let $x$ be a point of E and write $b=p(x)$. We have to prove that there exists a continuous section $s$ of $p$ such that $s(b) =x$ (cor. 2, I, p. 70). Let $\mathscr{S}$ be the set of pairs $(U, s_U)$ where U is a connected open subset of B containing $b$ and $s_U$ a continuous section of $p_U:\overset{-1}{p}(U)\rightarrow U$ such that $s_U(b) =x$. The set $\mathscr{S}$ is not empty ( I, p. 34, prop. 10). Let $(U, s_U)$ and $(V, s_V)$ be elements of $\mathscr{S}$. Then $s_U|U\cap V$ and $s_V|U\cap V$ are continuous sections of $p_{U\cap V}$ taking the value $x$ at $b$. By hypothesis, $U\cap V$ is connected; hence $s_U|U\cap V =s_V|U\cap V$ (I, p. 34, cor. 1 of prop. 11). Let A be the union of the open sets U when $(U, s_U)$ ranges over $\mathscr{S}$ and let $s: A\rightarrow E$ be the unique mapping such that $s|U =s_U$ for every pair $(U, s_U)\in \mathscr{S}$. The set A is open and connected (TG, I, p. 81, prop. 2), it contains $b$, and $s$ is a continuous section of $p_A$ such that $s(b) =x$. It now suffices to prove that the set A is closed, which will imply that it is equal to B.

Let $a$ be a point of B adherent to A and let V be a connected open neighbourhood of $a$ such that the covering E is trivializable over V. There exists a point $c$ in $A\cap V$ and a continuous section $s_V$ of $p_V$ such that $s_V(c) =s(c)$. Let $A'$ be the open set $A\cup V$. Since $A\cap V$ is connected, there exists a continuous section $s'$ of $p_{A'}$ extending $s$ and $s_V($I, p. 35, cor. 3 of prop. 11); the pair $(A', s')$ belongs to $\mathscr{S}$ and therefore $A'$ is contained in A. Hence $a$ belongs to A and A is closed.

In fact, the connected subspaces of $\mathbf{R}$ are the intervals (TG, IV, p. 8, th. 4) and the intersection of two intervals is an interval.

Let $(E, p)$ be a covering of X and let $y$ be a point of E. It is enough to prove that there exists a continuous section $s$ of $p$ such that $s(p(y)) =y$ (I, p. 70, cor. 2 of prop. 1). Suppose for example that $p(y)$ belongs to $U_1$. There then exists a continuous section $s_1: U_1\rightarrow E$ of $p_{U_1}$ such that $s_1(p(y)) =y$. Let $x$ be a point of $U_1\cap U_2$; there exists a continuous section $s_2$ of $p_{U_2}$ such that $s_2(x) =s_1(x)$. By corollary 3 of proposition 11 of I, p. 34, there exists a continuous section $s$ of $p$ extending both $s_1$ and $s_2$.

Put $U = U_1\cap U_2$ and suppose, arguing by contradiction, that the space U is not connected. We shall construct a connected covering of X of infinite degree; such a covering is not trivializable.

By assumption, the set U is the union of two disjoint, non-empty sets A and B, open (resp. closed) in X. For $i\in \{1,2\}$, let $Y_i$ be the $U_i$-space $(U_i\times \mathbf{Z}$, pr$_1)$ and let $Z_i$ be the subspace $U\times \mathbf{Z}$ of $Y_i$. The mapping $h: Z_1\rightarrow Z_2$ defined by

$(x, n)$ if $x\in A$ and $n\in \mathbf{Z}$

$(x, n+ 1)$ if $x\in B$ and $n\in \mathbf{Z}$

is a homeomorphism. Let Y be the space obtained by gluing $Y_1$ and $Y_2$ along $Z_1$ and $Z_2$ by means of the homeomorphism $h$ (TG, I, p. 17).

For $i\in \{1,2\}$, the canonical mapping $g_i$ of $Y_i$ into Y is a homeomorphism of $Y_i$ onto an open (resp. closed) subset of Y (loc. cit., prop. 9). For every integer $n\in \mathbf{Z}$, the sets $g_i(U_i\times \{n\})$, $i\in \{1,2\}$, are connected; since A and B are non-empty, $g_1(U_1\times \{n\})$ meets $g_2(U_2\times \{n\})$ and $g_2(U_2\times \{n+1\})$. It follows that the space Y is connected (TG, I, p. 81, corollary).

For $x\in U$ and $n\in \mathbf{Z}$, one has (pr$_1\circ h$)$(x, n) =x=$ pr$_1(x, n)$. There exists therefore a unique continuous mapping $p: Y\rightarrow X$ such that $p\circ g_i=$ pr$_1$ for $i\in \{1,2\}$. Let us prove that the X-space $(Y, p)$ is a covering. By construction, the fibres of the mapping $p$ are homeomorphic to the discrete space $\mathbf{Z}$.

For $i\in \{1,2\}$, the mapping $g_i$ defines by passing to subspaces an isomorphism of $U_i$-spaces of $U_i\times \mathbf{Z}$ onto $\overset{-1}{p}(U_i)$.

By definition of the space Y, there exists a unique mapping $k$ of $(X-A)\times \mathbf{Z}$ into Y such that $k(x, n) =g_1(x, n)$ for $x\in (U_1-A)\times \mathbf{Z}$ and $k(x, n) =g_2(x, n-1)$ for $x\in (U_2-A)\times \mathbf{Z}$; it is an isomorphism of (X-A)-spaces of $(X-A)\times \mathbf{Z}$ onto $\overset{-1}{p}(X-A)$. Analogously, there exists a unique mapping $k'$ of $(X-B)\times \mathbf{Z}$ into Y which coincides with $g_1$ on $(U_1-B)\times \mathbf{Z}$ and with $g_2$ on $(U_2-B)\times \mathbf{Z}$ and it is an isomorphism of (X-B)-spaces of $(X-B)\times \mathbf{Z}$ onto $\overset{-1}{p}(X-B)$.

This proves that the X-space $(Y, p)$ is trivializable over the subsets $U_1,U_2$, X-A and X-B. One has $U_1\cup U_2= X$; if $U_1$ and $U_2$ are open in X, this proves that $(Y, p)$ is a covering of X. The same holds when $U_1$ and $U_2$ are closed in X, for then X-A and X-B are open subsets of X whose union is X. The proposition is thus proved.

Let $X_1$ and $X_2$ be the closures of A and of X-A respectively. The sets $X_1$ and $X_2$ are closed and connected (TG, I, p. 81, prop. 1); one has $X_1\cup X_2= X$ and their intersection $X_1\cap X_2$ is equal to the boundary of A. It is then enough to apply proposition 7.

One may suppose that B is not empty. Let $x$ be a point of $E_t$. By proposition 3 of I, p. 125 applied to the covering E and to the continuous mapping $T\rightarrow B\times T,u\mapsto (p_t(x), u)$, there exists a unique continuous mapping $f_x: T\rightarrow E$ such that $f_x(t) =x$ and $p(f_x(u)) =$ $(p_t(x), u)$ for every $u\in T$. Let $h: E_t\times T\rightarrow E$ be the mapping defined by $h(x, u) =f_x(u)$. We have $h(x, t) =x$ and $p\circ h=p_t\times$ Id$_T$. The mapping $h$ is a lifting to E of the mapping $p_t\times$ Id$_T$. The restriction of $h$ to $E_t\times \{t\}$ is continuous, as is also the restriction of $h$ to $\{x\} \times T$ for every point $x$ of $E_t$. Since the space T is locally connected, the mapping $h$ is continuous ( I, p. 37, cor. 1 of th. 1).

Let $b$ be a point of B. By construction, the mapping $h$ induces a bijection of the fibre $\overset{-1}{p_{t}}(b)\times \{t\}$ of $E_t\times T$ onto the fibre $\overset{-1}{p}(b, t)$ of E at $(b, t)$. Since the space T is connected and locally connected, the mapping $h$ is bijective (I, p. 84, cor. of prop. 7). It is therefore a $B\times T$-isomorphism (I, p. 30, cor. 2 of prop. 6).

Let $h'$ be a $(B\times T)$-isomorphism of the covering $(E_t\times T, p_t\times$ Id$_T)$ onto the covering E which maps $(x, t)$ to $x$ for every point $x\in E_t$. For every $x\in E_t$, the mappings $u\mapsto h(x, u)$ and $u\mapsto h'(x, u)$ are equal (I, p. 34, cor. 1 of prop. 11). Hence $h=h'$.

By proposition 8, one may suppose that there exist coverings F and $F'$ of B such that E and $E'$ are respectively the $(B\times T)$-spaces $F\times T$ and $F'\times T$. One then has $E_t= F\times \{t\},E'_t= F'\times \{t\}$ and the mapping $k$ is written $(x, t)\mapsto (k'(x), t)$, where $k'$ is a B-morphism of F into $F'$. The mapping $k'\times$Id$_T$ is a $B\times T$-morphism extending $k$; it is an isomorphism if $k$ is one.

Let $\widetilde{k}: E\rightarrow E'$ be a $(B\times T)$-morphism extending $k$. Let $x$ be a point of F. Denote by $q$ the projection of F and put $b=q(x)$. The mappings $u\mapsto \widetilde{k}(x, u)$ and $u\mapsto (k'(x), u)$ are liftings in $E'$ of the mapping $u\mapsto (b, u)$ of T into $B\times T$. They coincide at $t$. Since the space T is connected, they are equal. Thus, $\widetilde{k}$ is the $(B\times T)$-morphism $k'\times$ Id$_T: F\times T\rightarrow F'\times T$.

In view of prop. 3 of I, p. 9, it is a matter of proving that every continuous section of $f^*(E)$ over $B'\times \{t\}$ extends uniquely to a continuous section of $f^*(E)$. Now this follows from cor. 2 applied to the coverings $(B'\times T$, Id$_{B'\times T})$ and $f^*E$ of $B'\times T$.