Since the space A is unravelable, there exists a covering E of A whose fibre $E_a$ is a principal homogeneous $\pi_1(A, a)$-set (IV, p. 342, Theorem 1). By assumption, this operation is obtained by means of the homomorphism $\pi_1(f, a)$ from an operation of $\pi_1(B, b)$ on $\pi_1(A, a)$. This implies the injectivity of the homomorphism $\pi_1(f, a)$.

(i) The homomorphism $\pi_1(f, a):\pi_1(A, a)\rightarrow \pi_1(B, b)$ is surjective.

(ii) For every pair $(E,E')$ of coverings of B, the mapping

which associates to a B-morphism $g: E\rightarrow E'$ the A-morphism

is bijective.

By means of the homomorphism $\pi_1(f, a)$, every $\pi_1(B, b)$-set is endowed with a structure of $\pi_1(A, a)$-set. Condition (i) is then equivalent to the following condition (i$'$):

(i$'$) For every pair $(F,F')$ of $\pi_1(B, b)$-sets, every $\pi_1(A, a)$-morphism of F into $F'$ is a $\pi_1(B, b)$-morphism.

In fact, if the homomorphism $\pi_1(f, a)$ is surjective, every $\pi_1(A, a)$-morphism of $\pi_1(B, b)$-sets is a $\pi_1(B, b)$-morphism. Conversely, let us take a $\pi_1(B, b)$-set F reduced to a point and let $F'=\pi_1(B, b)/f_*(\pi_1(A, a))$. The mapping of F into $F'$ whose image is $f_*(\pi_1(A, a))$ is a $\pi_1(A, a)$-morphism but is not a $\pi_1(B, b)$-morphism if $F'$ is not reduced to a point, that is to say if $\pi_1(f, a)$ is not surjective.

Since the space B is unrollable, every $\pi_1(B, b)$-set is isomorphic to the $\pi_1(B, b)$-set $E_b$, where E is a covering of B (IV, p. 344, remark 1). The equivalence of (i$'$) and (ii) therefore follows from prop. 2 of III, p. 310.

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

(ii) The image of the homomorphism from $\pi_1(A, a)$ into $\pi_1(B, a)$ induced by the canonical injection is reduced to the identity element.

The implication (ii)$\Rightarrow$(i) follows from corollary 3 (III, p. 309).

Conversely, suppose that every covering of B is trivializable over A. This is in particular the case for the simply arcwise connected covering $(\lambda_a(B), \varepsilon_B)$ (IV, p. 342, th. 1), so that the homomorphism $\pi_1(A, a)\rightarrow \pi_1(B, a)$ is trivial (III, p. 309, cor. 3 of prop. 1).

a) For every connected component U of X$,\overline{f(U)}$ is a

connected, open and closed subset of Y, and one has $\overset{-1}{f}(f(U)) = U$.

b) For every connected component V of Y, there exists a connected component U of X such that $V =f(U)$. The mapping of U into V induced by $f$ by passing to the subsets is relatively connected.

c) The connected components of X ( resp. of Y) are open and closed.

a) Let V be the set of $y\in Y$ which possess a neighbourhood W such that

$\overset{-1}{f}(W)$ is connected and meets U; this is an open set of Y. It contains $\overline{f(U)}$ since $f$ is relatively connected. Conversely, let $y\in V$; let

W be a neighbourhood of $y$ such that $\overset{-1}{f}(W)$ is connected and meets U. For

every neighbourhood $W'$ of $y$ such that $\overset{-1}{f}(W')$ is connected, $\overset{-1}{f}(W\cap W')$ is

not empty, hence $\overset{-1}{f}(W\cup W')$ is connected (TG, I, p. 81, prop. 2). By

hypothesis, $\overset{-1}{f}(W\cup W')$ meets U; hence one has $\overset{-1}{f}(W\cup W')\subset U$ and, a fortiori, $W'\cap f(U)=\not\emptyset$. This proves that $y\in \overline{f(U)}$; therefore, $V =\overline{f(U)}$. In particular, the set $\overline{f(U)}$ is open and closed in Y; prop. 1 (TG, I, p. 81) further implies that it is connected.

The preceding arguments show moreover that $\overset{-1}{f}(f(U))\subset U$,

whence the equality $\overset{-1}{f}(f(U)) = U$, the other inclusion being obvious. In particular, U is open and closed in X.

b) Let V be the connected component of a point $y$ of Y, let W be a

neighbourhood of $y$ such that $\overset{-1}{f}(W)$ is connected and nonempty and let U be the

connected component of X containing $\overset{-1}{f}(W)$. Since $y\in \overline{f(U)}$, it follows from a) and from the definition of the connected component of $y$ that $V =\overline{f(U)}$. Hence V is open and closed in Y.

a) The image $f(X)$ is dense in Y;

b) The space Y is locally connected; $-_1$

c) For every open and connected set V of Y, the set $f(V)$ is connected.

Suppose that the mapping $f$ is relatively connected. The density of $f(X)$ in Y follows from the definition of a relatively

connected mapping. Let V be an open subset of Y; the mapping $f_V:\overset{-1}{f}(V)\rightarrow$ V is relatively connected. By Prop. 4, the connected components of V are open and closed in V. It follows that Y is locally connected (TG, I, p. 85, prop. 11). To prove assertion c), it is enough to prove that X is connected if Y is. By the lemma, there exists

a connected component U of X such that $Y =\overline{f(U)}$ and $\overset{-1}{f}(f(U)) = U$, whence U = X and X is connected.

Conversely, suppose that conditions a), b), c) are satisfied, and let us show that $f$ is relatively connected. Let $y$ be a point of Y; since Y is locally connected, $y$ has a fundamental system of connected open neighbourhoods. If W is such a neighbourhood, conditions

c) and a) imply that $\overset{-1}{f}(W)$ is connected and nonempty. Hence the mapping $f$ is relatively connected.

The restriction of $g$ to every connected component of X is constant. By Corollary 1, there exists a mapping $h: Y\rightarrow F$, constant on each connected component of Y, such that $g=h\circ f$. The mapping $h$ is locally constant, since the connected components of Y are open. The uniqueness of such a mapping follows from the fact that $f(X)$ is dense in Y.

Let $\mathscr{F}$ be the sheaf on X of X-morphisms from $X\times_YT$ into $X\times_YT'$ and let $\mathscr{G}$ be the sheaf on Y of Y-morphisms from T into $T'($I, p. 45, example 4). For every open set U of Y, let us put $\varphi_U= (f_U)^*:\mathscr{G}(U)\rightarrow$

$\mathscr{F}(\overset{-1}{f}(U))$. The mappings $\varphi_U$ define a sheaf morphism $\varphi :\mathscr{G}\rightarrow \varphi_*(\mathscr{F})$ and it is enough to prove that $\varphi$ is an isomorphism of sheaves.

Since Y is locally connected (IV, p. 354, cor. 2), the connected open sets over which T and $T'$ are trivializable form a basis of the topology of Y. By corollary 2 of I, p. 55, it is enough to prove that for such an open set U, the mapping $\varphi_U$ is bijective, which allows us to suppose that Y is connected and that the coverings T and $T'$ are the trivial coverings $Y\times F$ and $Y\times F'$ where F and $F'$ are sets endowed with the discrete topology. The mapping $(x,(y, t))\mapsto (x, t)$ identifies the X-space $X\times_Y(Y\times F)$ with $X\times F$ (resp. the X-space $X\times_Y(Y\times F')$ with $X\times F'$). Since the space X is connected (IV, p. 354, cor. 2), the sets $\mathscr{C}_Y(Y\times F; Y\times F')$ and $\mathscr{C}_X(X\times F; X\times F')$ are both identified with the set $\mathscr{F}(F; F')$ of mappings from F into $F'$, and the mapping $f^*$ is identified with the identity mapping of $\mathscr{F}(F; F')$. This concludes the proof.

Let in fact $h: X\times_YT\rightarrow X\times_YT'$ and $h': X\times_YT'\rightarrow X\times_YT$ be X-isomorphisms inverse to one another. By proposition 5, there exist Y-morphisms $g: T\rightarrow T'$ and $g': T'\rightarrow T$ such that $f^*(g) =h$ and $f^*(g') =h'$. We then have $f^*(g'\circ g) =f^*(g')\circ f^*(g) =$ Id$_{X\times_YT}$, therefore $g'\circ g=$ Id$_T$, because the mapping $f^*$ is injective. Analogously, $g\circ g'=$ Id$_{T'}$. The coverings T and $T'$ are therefore isomorphic.

By corollary 1, every covering of Y is in fact trivializable.

By IV, p. 353, prop. 4, one may suppose that the spaces X and Y are connected. The corollary then follows from proposition 5 and proposition 2 of IV, p. 352.

Let $\mathscr{U}$ be the set of open sets of Y whose inverse image in X

is simply connected. For every $V\in \mathscr{U}$, the covering $\overset{-1}{f}(V)\times_XZ$

of $\overset{-1}{f}(V)$ is trivializable; there thus exists a discrete space $F_V$ and an

isomorphism of coverings $g_V:\overset{-1}{f}(V)\times_XZ\rightarrow \overset{-1}{f}(V)\times F_V$. For every

pair $(V,V')$ of open sets belonging to $\mathscr{U}$, the mapping $f_{V\cap V'}:\overset{-1}{f}(V\cap$ $V')\rightarrow V\cap V'$ is relatively connected. By proposition 5 of IV, p. 356, there exists a unique isomorphism of coverings of $V\cap V'$, $h_{V',V}: (V\cap V')\times F_V\rightarrow (V\cap V')\times F_{V'}$, such that one has $f^*(h_{V',V})(x, t) =$ $g_{V'}(g_V^{-1}(x, t))$ for every $x\in V\cap V'$ and every $t\in F_V$. If V, $V',V''$ are elements of $\mathscr{U}$, one has $h_{V'',V}(x, t) =h_{V'',V'}(h_{V',V}(x, t))$ for every $x\in V\cap V'\cap V''$ and every $t\in F_V$. There then exists a unique Y-space T and, for every $V\in \mathscr{U}$, an isomorphism $h_V: T_V\rightarrow V\times F_V$, such that one has $h_{V',V}(x, t) =h_{V'}\circ h^{-1}_V(x, t)$ for every pair $(V,V')$ of open sets belonging to $\mathscr{U}$, every $x\in V\cap V'$ and every $t\in F_V($cf. TG, I, p. 16). The space T is in particular a covering of Y. There moreover exists a

unique mapping of $X\times_YT$ onto Z whose restriction to $\overset{-1}{f}(V)\times_YT$ is given by $g^{-1}_V\circ f^*(h_V)$ and this is an isomorphism of X-spaces, whence the proposition.

image $\overset{-1}{f}(V)$ is simply connected. Then, for every point $x$ of X, the homomorphism $\pi_1(f, x):\pi_1(X, x)\rightarrow \pi_1(Y, f(x))$ is bijective.

One may suppose the spaces X and Y connected (IV, p. 353, prop. 4). By corollary 3 (IV, p. 357), the homomorphism $\pi_1(f, x)$ is surjective. Proposition 6 and proposition 1 of IV, p. 351 imply that it is injective.

The arcwise connected component of $x$ in X is open, closed and contractible; this makes it possible to suppose that the space X is connected. Since the space X is contractible, the X-space $E =\lambda_x(X)$, endowed with the terminal mapping, is a nonempty arcwise simply connected covering (IV, p. 342, th. 1). The group G = Aut$_X(E)$ is isomorphic to $\pi_1(X, x)$; it therefore suffices to prove that the group G has a finite presentation.

Since X is compact, every point $x$ of X possesses a compact neighbourhood $K_x$ over which the covering E is trivializable (TG, I, p. 65, corollary). Since X is locally connected, every $x\in X$ possesses an open connected neighbourhood $W_x$, contained in $K_x$. Let F be a finite subset of X such that the $W_x$, for $x\in F$, cover X. Let $n$ be the cardinal of F.

We shall prove by induction that, for every integer $k$ such that 1 $\leqslant k\leqslant n$, there exist a subset A of cardinal $k$ contained in F and, for $x\in A$, a section $s_x$ of $p$ over $K_x$, such that the union of the $s_x(W_x)$, for $x\in A$, is a connected subset of E. The assertion is true for $k= 1$. Suppose it true for an integer $k$ such that $1\leqslant k < n$ and let us prove that it is true for $k+ 1$. Let A be a subset of F of cardinal $k$ and, for every $x\in A$, let $s_x$ be a section of E over $K_x$, such that $\bigcup_{x\in A}s_x(W_x)$ is connected. The open sets $\bigcup_{x\in A}W_x$ and $\bigcup_{x\in F-A}W_x$ are nonempty and cover X; their intersection is therefore nonempty, since X is connected. There thus exist $x\in A,y\in F-A$ and $z\in W_x\cap W_y$. Put $A'= A\cup \{y\}$ and choose a section $s_y$ of E over $K_y$ such that $s_y(z) =s_x(z)$. The connected open sets $\bigcup_{p\in A}s_p(W_p)$ and $s_y(W_y)$ are nonempty and have a common point; their union is therefore connected. This proves the assertion for $k+ 1$. By induction, it is therefore true for every integer $k\in \{1, . . . , n\}$.

Applying the preceding to $k=n$, there then exists, for every $x\in F$, a section $s_x$ of E over $K_x$, in such a way that $U =\bigcup_{x\in F}s_x(W_x)$ is a connected open subset of E. We have $p(U) = X$. Since the group G operates transitively in each fibre of the covering E, we have GU = E.

The closure of U is contained in $\bigcup_{x\in F}s_x(K_x)$ and therefore is compact. The operation of G in E is proper (I, p. 96, cor. 1), hence the set of pairs $(g, x)\in G\times E$ such that $x\in \overline{U}$ and $gx\in \overline{U}$ is a compact subset of $G\times E$ (TG, I, p. 77, prop. 6). It follows that the set of $g\in G$ such that $\overline{U}\cap g\overline{U}=\not\emptyset$ is compact. Since it is discrete, it is finite. A fortiori, the set of $g\in G$ such that $U\cap gU=\not\emptyset$ is finite. The theorem thus follows from prop. 10 of I, p. 136.

Let $\mathscr{O}$ be the set of open subsets U of X such that $A\cap U$ be meagre. The set of subsets of $\mathscr{O}$ consisting of pairwise disjoint open subsets, ordered by inclusion, is inductive. Let $\mathfrak{U}$ be a maximal element; there exists one by E, III, p. 20, th. 2.

Let O be the union of the open subsets of X belonging to $\mathfrak{U}$. For every open subset $U\in \mathfrak{U}$, let $(B_{U,n})$ be a sequence of rare subsets of U whose union is $A\cap U$. For every integer $n$, the union $B_n$ of the subsets $B_{U,n}$, with U ranging over $\mathfrak{U}$, is rare relative to O (TG, IX, p. 52, prop. 1). By TG, IX, p. 53, prop. 2, $B_n$ is a rare subset of X, since O is open in X. Consequently, the set $A\cap O$, equal to the union of the $B_n$, is a meagre subset of X.

Let F be the complement of O. It is a closed subset of X; let us prove that it has empty interior. In the contrary case, let $x$ be an interior point of F. By assumption, there exists an open neighbourhood V of $x$, which may be supposed contained in F, such that $A\cap V$ be meagre. Then V is disjoint from the open subsets of X belonging to $\mathfrak{U}$ and $\mathfrak{U}\cup \{V\}$ is a set of pairwise disjoint open subsets belonging to $\mathscr{O}$, which contradicts the maximality of $\mathfrak{U}$. The relation

then implies that A is meagre, which was to be proved.

Let X be a topological space.

Recall (TG, IX, p. 69) that a subset A of X is said to be approachable if there exists an open subset U of X such that $U\cap \complement A$ and $A\cap \complement U$ be meagre in X. The set of approachable subsets of X is a tribe which contains the Borel tribe (TG, IX, p. 69, lemma 8 and its proof). A meagre subset is approachable.

For every subset A of X, let D(A) be the set of points $x\in X$ such that for every neighbourhood V of $x,A\cap V$ is not meagre. One also sets $D^*(A) = A\cup D(A)$.

a) The set D(A) is closed in X; its complement is the largest open subset U of X such that $A\cap U$ be a meagre set.

b) In order that A be meagre, it is necessary and sufficient that D(A) be empty.

c) The set $D^*(A)$ is approachable.

d) For every approachable set B of X containing A, $D^*(A)\cap$ $\complement B$ is meagre.

Let U be the union of the open subsets V of X such that $A\cap V$ be meagre. In order that a point $x$ belong to U, it is necessary and sufficient that it possess a neighbourhood V such that $A\cap V$ be meagre. Thus one has $U =\complement D(A)$ and therefore D(A) is a closed subset of X. By construction, the subspace U has a covering by open subsets whose intersection with A is meagre. By lemma 1, applied to the topological space U and to the subset $A\cap U$, $A\cap U$ is meagre in U, hence also in X. This proves assertion a), since, by construction, every open subset V of X such that $A\cap V$ be meagre is contained in U.

Assertion b) follows immediately from this.

c) One has $D^*(A) = A\cup D(A) = (A\cap U)\cup D(A)$. The set D(A) is closed, hence approachable (IV, p. 361); the meagre subset $A\cap U$ is so as well. Hence $D^*(A)$ is approachable (loc. cit.).

d) Let B be an approachable subset of X containing A. Its complement $\complement B$ is then an approachable subset of X (loc. cit.) and there therefore exists an open subset V of X such that $V\cap B$ and $\complement V\cap \complement B$ are meagre. Since $A\subset B$, $V\cap A$ is still meagre, whence $V\subset U$. Since $\complement U\cap \complement B$ is contained in $\complement V\cap \complement B$, it is also a meagre subset of X. Finally, the inclusions

show that $D^*(A)\cap \complement B$ is a meagre subset of X.

Let S be a Souslinian subspace of X. By definition, there exists a complete metric space P of countable type and a continuous surjective mapping $g$ of P onto S. By TG, IX, p. 64, lemma 3, there exists a sieve $C = (C_n, p_n, \varphi_n)_{n\in\mathbf{N}}$ of the metric space P.

For every integer $n$ and every $c\in C_n$, let us denote by $F_n(c)$ the image of the set $\varphi_n(c)$ under $g$; let us also put $F^*_n(c) = D^*(F_n(c))$ and

For every $c\in C_n,F^*_n(c)$ is approachable (IV, p. 361, lemma 2, c)). Since $C_{n+1}$ is countable, the union of the approachable subsets $F^*_{n+1}(c')$, where $c'$ runs through $\overset{-1}{p_{n}}(c)$, is again an approachable subset of X. It contains the union of the $F_{n+1}(c')$, which is equal to $F_n(c)$. Hence (loc. cit., d)), $G_n(c)$ is a meagre subset of X. The union G of the subsets $G_n(c)$, for $n\in \mathbf{N}$ and $c\in C_n$, is therefore a meagre subset of X.

Let $c_0\in C_0$ and let $x$ be an element of $F^*_0(c_0)\cap \complement G$; let us prove that $x\in F_0(c_0)$. Since $x\notin G_0(c_0)$ and $x\in F^*_0(c_0)$, there exists $c_1\in \overset{-1}{p_{0}}(c_0)$ such that $x\in F^*_1(c_1)$, by relation (1). By recurrence, there exists an element $c= (c_n)_n$ of $\prod_nC_n$ such that, for every integer $n\in \mathbf{N}$, one has $x\in F^*_n(c_n)$ and $p_n(c_{n+1}) =c_n$.

The subsets $\varphi_n(c_n)$ form a basis of a Cauchy filter in P; this filter converges to a point $p$ of P, since P is complete. The image of this filter under $g$ is a filter F on X, having as basis the set of the $F_n(c_n)$ for $n\in \mathbf{N}$, which converges to $g(p)$. Since $F^*_n(c_n)$ is contained in $\overline{F_n(c_n)}$ and $x$ belongs to each of the $F_n(c_n)$, $x$ is an adherent point of F, hence $x=g(p)$ because the space X is separated (TG, I, p. 52, prop. 1). The point $p$ belongs to $\overline{\varphi_1(c_1)}$, hence to $\varphi_0(c_0)$. Consequently, $x\in F_0(c_0)$.

Consequently, the set $F^*_0(c_0)$ - $F_0(c_0)$ is contained in G. It is therefore a meagre subset of X, hence also an approachable subset. Since $F_0(c_0) = F^*_0(c_0)$ - $(F^*_0(c_0)$ - $F_0(c_0))$ and $F^*_0(c_0)$ is approachable, $F_0(c_0)$ is approachable, because the approachable subsets of X form a tribe. Since $C_0$ is countable, the set S which is the union of the $F_0(c_0)$, for $c_0\in C_0$, is again approachable.

For every pair $(j_1, j_2)$ of distinct elements of I, the set of families $(x_i)\in X^I$ such that $(x_{j_1}, x_{j_2})\in O$ is an open dense subset of $X^I$. The intersection Ω of these open sets, when $(j_1, j_2)$ runs through the finite set of pairs of distinct elements of I, is therefore an open dense subset of $X^I$. Consequently, $\Omega \cap \prod_{i\in I}U_i$ is a nonempty open subset of $X^I$, hence contains an open subset of $X^I$ of the form $\prod_{i\in I}V_i$, where for every $i\in I$, $V_i$ is an open, nonempty, subset of $U_i$. This proves the lemma.

Let the space P be endowed with a metric $d$ compatible with its topology for which it is complete. Let $(A_n)_{n\in\mathbf{N}}$ be an increasing sequence of closed subsets of $P\times P$, with empty interiors, such that the graph $\Gamma_R$ of the relation R is contained in the union of the $A_n$.

Let $\mathscr{O}$ be the set of nonempty open subsets of P. We shall construct by induction a sequence $(f_n)_{n\in\mathbf{N}}$, where for every $n$, $f_n$ is a mapping of $\{0,1\}^n$ into $\mathscr{O}$, satisfying the following properties :

(i) For every $n\geqslant 1$, every $x\in \{0,1\}^{n-1}$ and every $t\in \{0,1\}$, the closure of the open set $f_n(x, t)$ is contained in $f_{n-1}(x)$;

(ii) For every $x\in \{0,1\}^n$, one has diam($f_n(x)$)$\leqslant 2^{-n}$;

(iii) For every $n\geqslant 1$ and every pair $(x, x')$ of distinct elements of $\{0,1\}^n,f_n(x)\times f_n(x')$ does not meet $A_{n-1}$.

Choose a nonempty open set U of X of diameter $\leqslant 1$ and define $f_0$ as the constant mapping with image $\{U\}$. Suppose that the mappings $f_0, . . . , f_n$ have been constructed.

Let $p:\{0,1\}^{n+1}\rightarrow \{0,1\}^n$ be the mapping defined by $p(x_0, . . . , x_n) =$ $(x_0, . . . , x_{n-1})$. By lemma 3 above, applied to the family of open sets $(f_n(p(x)))$ for $x\in \{0,1\}^{n+1}$ and to the dense open set $\complement A_n$ of $P\times P$, there exists a family $(g(x))_{x\in \{0,1\}^{n+1}}$ of nonempty open sets of P such that $g(x)\subset f_n(p(x))$ for every $x$ and such that $(g(x)\times g(x'))\cap A_n$ be empty for every pair $(x, x')$ of distinct elements of $\{0,1\}^{n+1}$. One then defines the mapping $f_{n+1}$ by choosing, for each element $x\in \{0,1\}^{n+1}$, a nonempty open subset of $g(x)$ whose diameter is $\leqslant 2^{-n-1}$ and whose closure is contained in $g(x)$.

For every element $x= (x_n)_{n\in\mathbf{N}}$ of $\{0,1\}^{\mathbf{N}}$, the sequence of sets $(f_n(x_0, . . . , x_{n-1}))_{n\in\mathbf{N}}$ is a decreasing sequence of open subsets of X each of which contains the closure of the next and whose diameter tends to 0; the intersection of this sequence of sets is therefore reduced to a point (TG, II, p. 15), which we denote by $f(x)$. If two points $x,x'$ of $\{0,1\}^{\mathbf{N}}$ satisfy $x_i=x'_i$ for $i\leqslant n$, one has $d(f(x), f(x'))\leqslant 2^{-n}$. Consequently, the mapping $f:\{0,1\}^{\mathbf{N}}\rightarrow P$ is continuous. Let $x$ and $x'$ be distinct elements of $\{0,1\}^{\mathbf{N}}$. For $n\in \mathbf{N}$ such that $(x_0, . . . , x_n)=\not$ $(x'_0, . . . , x'_n)$, the open set $f_{n+1}(x_0, . . . , x_n)\times f_{n+1}(x'_0, . . . , x'_n)$ is disjoint from $A_n$, by definition of $f_{n+1}$, hence the pair $(f(x), f(x'))$ does not belong to $A_n$. It follows that $f(x)$ and $f(x')$ are not equivalent for the relation R. Consequently, $f$ is injective and its image meets each equivalence class for R in at most one point. The proposition is thus proved.