# Fixed-point-like theorems on subspaces

Philippe Bich, Bernard Cornet

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Philippe Bich, Bernard Cornet. Fixed-point-like theorems on subspaces. 2004. halshs-03330770

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UNIVERSITÉ PARIS 1
- PANTHÉON - SORBONNE -

CERMSEM

UMR CNRS 8095

Cahiers de la MSE

Fixed-point-like theorems on subspaces

Philippe BICH

Bernard CORNET

2004.64

<figure><img src="image_3.png" /></figure>

CENTRE NATIONAL
DE LA RECHERCHE
SCIENTIFIQUE

Maison des Sciences Économiques, 106-112 boulevard de L'Hôpital, 75647 Paris Cedex 13

ISSN : 1624-0340
# Fixed-point-like theorems on subspaces

Philippe Bich and Bernard Cornet

**Abstract:** In this paper, we prove a fixed-point-like theorem for multi-valued mappings defined on the finite Cartesian product of Grassmannian manifolds and convex sets. Our result generalizes two different kinds of theorems: the fixed-point-like theorem by Hirsh, Magill, Mas-Colell (1990) or Husseiny, Lasry, Magill (1990) and the fixed-point theorem by Gale, Mas-Colell (1975) (which generalizes Kakutani's theorem (1941)).

**Key words:** fixed-point, Grassmannian manifold, degree theory.

**AMS classification:** 47H04,47H10,47H11.

## 1 Introduction

In this paper, we prove a fixed-point-like theorem for multi-valued mappings defined on the finite Cartesian product of Grassmannian manifolds and convex sets. Let $k$ be an integer and let $V$ be a Euclidean space such that $0 \le k \le \dim V$, then the $k$-Grassmannian manifold of $V$, denoted $G^k(V)$, is the set of all the $k$-dimensional subspaces of $V$. The set $G^k(V)$ is a smooth compact manifold but, in general, it does not satisfy properties such as convexity or acyclicity and its Euler characteristic may be null. This prevents from using classical fixed-point theorems as Brouwer's [2], Kakutani's [14], or Eilenberg-Montgomery's theorem [7].

Our main result generalizes two different kinds of theorems: the fixed-point-like theorem by Hirsh, Magill, Mas-Colell [12] or Husseiny, Lasry, Magill [13] and the fixed-point theorem by Gale, Mas-Colell [8] (which generalizes Kakutani's Theorem [14]). As in [12] and [13], we shall mainly use techniques from degree theory. As a consequence of our main result we first deduce the standard fixed-point theorems when the variable is in a convex domain (such as Brouwer's and Kakutani's theorem) and second Borsuk-Ulam's theorem.

The main result of this paper is directly motivated by the existence problem of equilibria in economic models with incomplete markets; in Bich-Cornet [1], it is used to extend the classical existence result by Duffie-Shaffer [6] to the non-transitive case.
The paper is organized as follows. The main result is stated in Section 2 together with some direct consequences of it, namely, the results by Hirsh, Magill, Mas-Colell [12], Gale, Mas-Colell [8] and Borsuk-Ulam's Theorem. The proof of the main result is given in Section 3 and the Appendix recalls the main properties of the Grassmannian manifold, used in this paper.

# 2 Statement of the results

## 2.1 Preliminaries

A correspondence $\Phi$ from a set $X$ to a set $Y$ is a map from $X$ to the set of all the subsets of $Y$ and the graph of $\Phi$, denoted $G(\Phi)$, is defined by $G(\Phi) = \{(x, y) \in X \times Y | y \in \Phi(x)\}$. A mapping $\varphi : X \to Y$ is said to be a selection of $\Phi$ if $\varphi(x) \in \Phi(x)$ for all $x \in X$. If $A$ is a subset of $X$, we let $\Phi(A) = \cup_{x \in A} \Phi(x)$ and the restriction of $\Phi$ to $A$, denoted $\Phi|_A$, is the correspondence from $A$ to $Y$ defined by $\Phi|_A(x) = \Phi(x)$ if $x \in A$. If $X$ and $Y$ are topological spaces, the correspondence $\Phi$ is said to be lower semi-continuous (l.s.c.) [resp. upper semi-continuous (u.s.c.)] if for every open set $U \subset Y$, the set $\{x \in X | \Phi(x) \cap U \neq \emptyset\}$ is open in $X$ [resp. the set $\{x \in X | \Phi(x) \subset U\}$ is open in $X$ and, for every $x \in X$, $\Phi(x)$ is compact].

If $x = (x_1, ..., x_n)$ and $y = (y_1, ..., y_n)$ belong to $R^n$, we denote by $x \cdot y = \sum_{i=1}^n x_i y_i$, the scalar product of $R^n$, $\|x\| = \sqrt{x \cdot x}$, the Euclidian norm. If $x \in R^n$ and $r \in R_+$, we let $B(x, r) = \{y \in R^n | \|x - y\| < r\}$ and $\bar{B}(x, r) = \{y \in R^n | \|x - y\| \le r\}$. If $E$ is a vector subspace of $R^n$, we denote by $E^{\perp} = \{u \in R^n | \forall x \in E, x \cdot u = 0\}$ the orthogonal space to $E$. If $u_1, ..., u_k$ belong to $E$, a vector space, we denote by $\text{span}\{u_1, ..., u_k\}$ the vector subspace of $E$ spanned by $u_1, ..., u_k$.

Let $V$ be a Euclidean space and let $k$ be an integer such that $0 \le k \le \dim V$, we denote by $G^k(V)$ the set consisting of all the linear subspaces of $V$ of dimension $k$, called the $(k)$-Grassmannian manifold of $V$. Then it is known that $G^k(V)$ is a smooth manifold of dimension $k(\dim V - k)$ and we refer to the Appendix for the properties we shall use hereafter, together with the precise definition of the manifold structure on $G^k(V)$.

## 2.2 The main result and some consequences

The aim of this paper is to prove the following result:

**Theorem 1** Let $I, J$ be two finite disjoint sets. For every $i \in I$, let $k_i$ be an integer and let $V_i$ be a Euclidean space such that $0 \le k_i \le \dim V_i$. For every $j \in J$, let $C_j$ be a nonempty, convex, compact subset of a Euclidean space $V_j$, and let $M = \prod_{i \in I} G^{k_i} \times \prod_{j \in J} C_j$.

For $i \in I$ and $k = 1, ..., k_i$, let $F_i^k$ be a correspondence from $M$ to $V_i$ with convex values, for $j \in J$, let $F_j$ be a correspondence from $M$ to $C_j$
with convex values, and suppose that, for every $i \in I$ and $k = 1, \dots, k_i$ [resp. $j \in J$], the correspondence $F_i^k$ [resp $F_j$] is either l.s.c or u.s.c.

Then, there exists $\bar{x} = ((\bar{x}_i)_{i \in I}, (\bar{x}_j)_{j \in J}) \in M$ such that:

(i) [either $F_i^k(\bar{x}) \cap \bar{x}_i \neq \emptyset$ or $F_i^k(\bar{x}) = \emptyset$] for every $i \in I$ and $k = 1, \dots, k_i$;

(ii) [either $F_j(\bar{x}) \cap \{\bar{x}_j\} \neq \emptyset$ or $F_j(\bar{x}) = \emptyset$] for every $j \in J$.

The proof of Theorem 1 is given in Section 3. A first consequence of Theorem 1 is the following theorem by Hirsch et al. ([12]).

**Corollary 1** Let $V_1$ be a Euclidean space, let $k_1$ be an integer such that $0 \le k_1 \le \dim V_1$, and for every $k = 1, \dots, k_1$ let $f^k : G^{k_1}(V_1) \to V_1$ be a continuous mapping. Then, there exists $\bar{x} \in G^{k_1}(V_1)$ such that: for every $k = 1, \dots, k_1$, $f^k(\bar{x}) \in \bar{x}$.

**Proof.** Take $I = \{1\}$, $J = \emptyset$ and $F_1^k(x) = \{f^k(x)\}$ for every $x \in G^{k_1}(V_1)$ and for every $k = 1, \dots, k_1$. From Theorem 1, there exists $\bar{x} \in M = G^{k_1}(V_1)$ such that for every $k = 1, \dots, k_1$, $F_1^k(\bar{x}) \cap \bar{x} \neq \emptyset$, i.e., $f^k(\bar{x}) \in \bar{x}$. $\square$

A second consequence of Theorem 1 is the following generalization of Gale and Mas-Colell Theorem [8], which is also a generalization of Kakuani's Theorem. Hereafter, we use the formulation by Gourdel [9] allowing each correspondence to be either upper semi-continuous or lower semi-continuous.

**Corollary 2** Let $J$ be a finite set, for $j \in J$, let $C_j$ be a nonempty, convex, compact subset of a Euclidean space, let $F_j$ be a correspondence from $M := \prod_{j \in J} C_j$ to $C_j$ with convex values, such that the correspondence $F_j$ is either l.s.c or u.s.c.. Then, there exists $\bar{x} = (\bar{x}_j)_{j \in J} \in M$ such that: for every $j \in J$, [either $\bar{x}_j \in F_j(\bar{x})$ or $F_j(\bar{x}) = \emptyset$].

**Proof.** Take $I = \emptyset$ and apply Theorem 1. $\square$

**Remark 1** According to our definition, an u.s.c. correspondence has compact values and without this requirement, Theorem 1 may not be true, as we can see in the following counterexample. Let $M := G^1(R^2)$. Each element $D$ of $G^1(R^2)$ can be written as $D_t = \{\lambda(\text{cost}, \text{sint}) \mid \lambda \in \mathbb{R}\}$, for some $t \in [0, \pi[$. Let us define the correspondence $F$ from $M$ to $R^2$ by $F(D_0) = R \times \{1\}$ and $F(D_t) = D_t \cap (R \times \{1\}) + \{(1,0)\}$ if $t \in ]0, \pi[$. We let the reader check that for every open set $U \subset R^2$ the set $\{x \in M \mid F(x) \subset U\}$ is open in $M$ and that $F$ has nonempty, convex (and closed) values. Yet, it is straightforward that $F(x) \cap x = \emptyset$ for every $x \in G^1(R^2)$. $\square$

Another consequence of our main result is the following multi-valued version of Borsuk-Ulam's theorem. We denote by $S^n$ the unit sphere of $R^{n+1}$.
**Corollary 3** For $k = 1, ..., n$, let $F^k$ be a correspondence from $S^n$ to $R$ with nonempty and convex values, such that, for every $k = 1, ..., n$, $F^k$ is either l.s.c or u.s.c.. Then, there exists $\bar{x} \in S^n$ such that:

$$ \forall k \in \{1, ..., n\}, F^k(\bar{x}) \cap F^k(-\bar{x}) \neq \emptyset. $$

**Proof.** For every $k = 1, ..., n$, let $\hat{F}^k$ be the correspondence from $S^n$ to $R$ defined by

$$ \hat{F}^k(x) = \{u - v \mid u \in F^k(x), v \in F^k(-x)\}. $$

We let the reader check that for every $k = 1, ..., n$, the correspondence $\hat{F}^k$ has nonempty, convex values and that it is u.s.c. [resp. l.s.c.] if $F^k$ is u.s.c. [resp. l.s.c.]. So, to prove Corollary 3, it suffices to show the existence of $\bar{x} \in S^n$ such that $0 \in \hat{F}^k(\bar{x})$ for every $k = 1, ..., n$.

Let us define, for every $k = 1, ..., n$ the correspondence $H^k$, from $G^n(R^{n+1})$ to $R^{n+1}$, as follows: for every $E \in G^n(R^{n+1})$, we let $H^k(E) = \hat{F}^k(x)x$, where $x$ is an arbitrary element of $E^\perp \cap S^n$. The correspondence $H^k$ is well defined since $E^\perp \cap S^n = \{x, -x\}$ for some element $x \in S^n$ and since $\hat{F}^k(x)x = \hat{F}^k(-x)(-x)$.

Take $I = \{1\}$, $V_1 = R^{n+1}$, $k_1 = n$, $J = \emptyset$, and apply Theorem 1 to the correspondences $H^k$, which clearly satisfy the assumptions of Theorem 1. So there exists $\bar{E} \in G^n(R^{n+1})$ such that $\bar{E} \cap H^k(\bar{E}) \neq \emptyset$ for every $k = 1, ..., n$.

Now, if $\bar{x}$ is an arbitrary point of $\bar{E}^\perp \cap S^n$ then we have $\bar{E} \cap \hat{F}^k(\bar{x})\bar{x} \neq \emptyset$; from $\bar{x} \in \bar{E}^\perp$ and $\bar{x} \neq 0$, we finally obtain $0 \in \hat{F}^k(\bar{x})$ for every $k = 1, ..., n$, which ends the proof of Corollary 3. □

# 3 Proof of Theorem 1

The proof is given in three steps, corresponding to the following three subsections. The first step gives the proof under the additional assumption that $J = \emptyset$ and the correspondences $F_i^k$ are single-valued. The second step only assumes in addition that $J = \emptyset$. Finally, the third step gives the proof under the assumptions of Theorem 1.

## 3.1 Proof when $J = \emptyset$ and $F_i^k$ is single-valued

We first prove Theorem 1 under the additional assumptions that $J = \emptyset$ and the $F_i^k$ are single-valued. This is exactly the statement below:

**Theorem 2** Let $I$ be a finite set and for $i \in I$, let $k_i$ be an integer and let $V_i$ be a Euclidean space such that $0 \le k_i \le \dim V_i$. Let $M := \prod_{i \in I} G^{k_i}(V_i)$ and for $i \in I$, let $f_i : M \to (V_i)^{k_i}$ be a continuous mapping. Then, there exists $\bar{x} = (\bar{x}_i)_{i \in I} \in M$ such that

$$ \forall i \in I, f_i(\bar{x}) \in (\bar{x}_i)^{k_i}. $$
The proof of Theorem 2 is given in two steps. In the first step we additionally assume that the mappings are smooth and the second step gives the proof in the general case.

### 3.1.1 Proof of Theorem 2 when the $f_i$ are smooth

Let $M := \prod_{i \in I} G^{k_i}(V_i)$ and define $f: M \to \prod_{i \in I} V_i^{k_i}$ by

$$f(x) = (\mathrm{proj}_{(x_i^{k_i})^{\perp}} f_i(x))_{i \in I} \quad \text{for } x = (x_i)_{i \in I} \in M$$

and the subsets $Z, Z_1$ and $Z_2$ of $M \times \prod_{i \in I} V_i^{k_i}$ by

$$Z = \{(x, y) \in M \times \prod_{i \in I} V_i^{k_i} \mid \forall i \in I, y_i \in (x_i^{k_i})^{\perp}\},$$

$$Z_1 = \{(x, y) \in M \times \prod_{i \in I} V_i^{k_i} \mid y = f(x)\},$$

$$Z_2 = \{(x, y) \in M \times \prod_{i \in I} V_i^{k_i} \mid y = 0\}.$$

Proving Theorem 2 amounts to show the existence of $\bar{x} \in M$ such that $f(\bar{x}) = 0$, or equivalently such that $Z_1 \cap Z_2 \neq \emptyset$. For this, we will use the following Intersection Theorem, which is a direct consequence of mod2 intersection theory (see, for example, [10] p.79 and [5] p.127).

**Intersection Theorem** Let $Z$ be a smooth boundaryless manifold of dimension $2m$, let $Z_1, Z_2$ be two compact boundaryless submanifolds of $Z$ of dimension $m$. If $\bar{Z}_1$ is a compact boundaryless $m$-submanifold of $Z$ homotopic to $Z_1$ and if the manifolds $\bar{Z}_1$ and $Z_2$ intersect transversally in a unique point $\bar{z}$ (which means that $T_{\bar{z}}\bar{Z}_1 + T_{\bar{z}}Z_2 = T_{\bar{z}}Z$), then $Z_1 \cap Z_2 \neq \emptyset$.

The proof of Theorem 2 consists to check that the above defined sets $Z, Z_1$ and $Z_2$ (together with the set $\bar{Z}_1$ defined below) satisfy the assumptions of the Intersection Theorem.

*The sets $Z, Z_1$ and $Z_2$ satisfy the assumptions of the Intersection Theorem.* Let us recall that for every $i \in I$, $G^{k_i}(V_i)$ is a smooth, boundaryless, compact, manifold of dimension $k_i(\dim V_i - k_i)$ (see Lemma 3 in the Appendix). Thus $M := \prod_{i \in I} G^{k_i}(V_i)$ is a boundaryless, smooth compact, manifold of dimension $m = \sum_{i \in I} k_i(\dim V_i - k_i)$. Clearly $Z$ is a fiber bundle whose base space is $M$ and whose fiber at $x = (x_i)_{i \in I} \in M$ is the vector space $\prod_{i \in I} (x_i^{k_i})^{\perp}$ which has the dimension of $M$. Hence, $Z$ is a smooth manifold of dimension $2m$.

The mapping $f: M \to \prod_{i \in I} V_i^{k_i}$ is a smooth mapping from Parts (c), (d) and (e) of Lemma 3 in the Appendix. Consequently, $Z_1$ is a smooth
compact boundaryless submanifold of Z of dimension m. Finally, Z₂ is
clearly a smooth boundaryless compact submanifold of Z of dimension m.

The manifold Z₁ is homotopic to the manifold ¯Z₁ that we now define. For every i ∈ I, let ¯xᵢ ∈ Gᵏᵢ(Vᵢ) and let {eᵢ¹, ..., eᵢᵏᵢ} be an orthonormal basis of ¯xᵢ. For every i ∈ I, let gᵢ : Gᵏᵢ(Vᵢ) → Vᵢᵏᵢ and g : M → Πᵢ∈I Vᵢᵏᵢ be the mappings defined by:

$$
\forall x_i \in G^{k_i}(V_i), g_i(x_i) = (\mathrm{proj}_{x_i \perp}(\bar{e}_i^1), \dots, \mathrm{proj}_{x_i \perp}(\bar{e}_i^{k_i})) \in (x_i^{\perp})^{k_i} = (x_i^{k_i})^{\perp}
$$

$$
\forall x = (x_i)_{i \in I} \in M, g(x) = (g_i(x_i))_{i \in I}.
$$

We let

$$
\bar{Z}_1 := \{(x, y) \in M \times \prod_{i \in I} V_i^{k_i} \mid y = g(x)\}.
$$

To show that the manifold Z₁ is homotopic to Z̄₁, we let H : [0, 1] ×
Z₁ → Z be the continuous mapping defined by H(t, (x, f(x))) = (x, (1 −
t)f(x) + tg(x)). Then H(0, .) is the canonical inclusion from Z₁ to Z, and
H(1, .)(Z₁) = Z̄₁.

The manifolds $\bar{Z}_1$ and $Z_2$ intersect transversally in a unique point. First, notice that $\bar{Z}_1 \cap Z_2 = \{(x, 0) \in M \times \prod_{i \in I} V_i^{k_i} \mid g(x) = 0\}$ is the singleton $(\bar{x}, 0) = ((\bar{x}_i)_{i \in I}, 0)$. But $\bar{Z}_1$ and $Z_2$ intersects each other transversally in Z means that $T_{(\bar{x},0)}\bar{Z}_1 + T_{(\bar{x},0)}Z_2 = T_{(\bar{x},0)}Z$. Recalling that dim $T_{(\bar{x},0)}\bar{Z}_1$+dim $T_{(\bar{x},0)}Z_2$= dim $T_{(\bar{x},0)}Z = 2m$ we only have to show that $T_{(\bar{x},0)}\bar{Z}_1 \cap T_{(\bar{x},0)}Z_2 = \{0\}$. Finally, noticing that $T_{(\bar{x},0)}\bar{Z}_1 = \{(u, Dg(\bar{x})(u)) \mid u \in T_{\bar{x}}M\}$ and $T_{(\bar{x},0)}Z_2 = \{(u, 0) \mid u \in T_{\bar{x}}M\}$ we only have to prove that $Dg(\bar{x})$ is injective which is proved in the following lemma:

**Lemma 1** *Dg*(*x*) *is injective*.

**Proof.** Recalling that for every $x = (x_i)_{i \in I} \in M$, $g(x) = (g_i(x_i))_{i \in I}$, the mapping $Dg(\bar{x})$ is injective if and only if for every $i \in I$, $Dg_i(\bar{x}_i)$ is injective.

So, let $i \in I$, and let $(\varphi, U)$ be a local chart of $G^{k_i}(V_i)$ at $\bar{x}_i$ and let
$\psi : (\bar{x}_i^{\perp})^{k_i} \to G^{k_i}(V_i)$ be the inverse mapping of $\varphi : U \to (\bar{x}_i^{\perp})^{k_i}$.
From the Appendix, if $\{\bar{e}_1^1, \dots, \bar{e}_i^{k_i}\}$ is a given orthonormal basis of $\bar{x}_i$, $\psi$ can be defined
by:

$$
\psi(u^1, \dots, u^{k_i}) = \operatorname{span} \{\bar{e}_i^1 + u^1, \dots, \bar{e}_i^{k_i} + u^{k_i}\} \text{ for every } (u^1, \dots, u^{k_i}) \in (\bar{x}_i^{\perp})^{k_i}.
$$

The mapping $g_i \circ \psi$ being the local representation $g_i$ in the chart $(\varphi, U)$, proving that $Dg_i(\bar{x}_i)$ is injective amounts to proving that $D(g_i \circ \psi)(0)$ is injective. This is a consequence of the following claim.
**Claim 1** $\forall (h_1, ..., h_{k_i}) \in (\bar{x}_i^\perp)^{k_i}, D(g_i \circ \psi)(0)(h_1, ..., h_{k_i}) = -(h_1, ..., h_{k_i})$.

**Proof of Claim 1.** Let $p: V_i \times (\bar{x}_i^\perp)^{k_i} \to V_i$ be defined by

$$p(y, u) := \mathrm{proj}_{\psi(u)} y.$$

If we prove that for every $y \in V_i$, the derivative of the mapping $p_y: u \to p(y, u)$ is the linear mapping $Dp_y(0): (\bar{x}_i^\perp)^{k_i} \to V_i$ defined by

$$Dp_y(0)(h) = \sum_{k=1}^{k_i} (y \cdot \bar{e}_i^k)h_k, \forall h = (h_1, ..., h_{k_i}) \in (\bar{x}_i^\perp)^{k_i}$$

then Claim 1 will be proved. Indeed, taking $y = \bar{e}_i^k$ for every $k = 1, ..., k_i$, we would obtain $D_{\bar{e}_i^k}p(0)(h_1, ..., h_{k_i}) = h_k$. Thus, since $g_i \circ \psi(u) = (\bar{e}_i^1, ..., \bar{e}_i^{k_i}) - (p_{\bar{e}_i^1}(u), ..., p_{\bar{e}_i^{k_i}}(u))$, it would entail Claim 1.

Now, for every $u = (u_1, ..., u_{k_i}) \in (\bar{x}_i^\perp)^{k_i}$, there exists $\lambda(y, u) = (\lambda_k(y, u))_{k=1,...,k_i} \in R^{k_i}$ such that

$$p(y, u) = \mathrm{proj}_{\psi(u)} y = \sum_{k=1}^{k_i} \lambda_k(y, u)(\bar{e}_i^k + u^k),$$

with $(\lambda_k(y, u))$ satisfying

$$(-y + \sum_{k=1}^{k_i} \lambda_k(y, u)(\bar{e}_i^k + u^k)) \cdot (\bar{e}_i^j + u^j) = 0 \text{ for every } j = 1, ..., k_i.$$

This can be equivalently rewritten as follows

$$(I_{k_i} + G(u))\lambda(y, u) = (y \cdot (\bar{e}_i^1 + u^1), ..., y \cdot (\bar{e}_i^{k_i} + u^{k_i})),$$

where $I_{k_i}$ is the $k_i \times k_i$ identity matrix, and $G(u)$ is the $k_i \times k_i$ matrix $G(u) = (u^j \cdot u^k)_{j,k=1,...,k_i}$. Besides, for $u$ in a neighborhood $\mathcal{N}$ of 0 small enough, the matrix $(I_{k_i} + G(u))$ is invertible. Consequently, the mapping $\lambda(.,.)$ is smooth on $V \times \mathcal{N}$, which implies that the mapping $p(.,.)$ is smooth on $V \times \mathcal{N}$. Differentiating with respect to $u$, the above equality at $u = 0$, we obtain, for every $h = (h_1, ..., h_{k_i}) \in (\bar{x}_i^\perp)^{k_i}$:

$$DG(0)(h)\lambda(y, 0) + D_u\lambda(y, 0)(h) = 0.$$

But it is clear that $DG(0) = 0$. Consequently, $D_u\lambda(y, 0) = 0$.

Finally, differentiating the equality $p(y, u) = \sum_{k=1}^{k_i} \lambda_k(y, u)(\bar{e}_i^k + u^k)$ at $(y, 0)$, one obtain, for every $h = (h_k)_{k=1}^{k_i} \in (\bar{x}_i^\perp)^k$

$$D_u p(y, 0)(h) = \sum_{k=1}^{k_i} \lambda_k(y, 0)h_k = \sum_{k=1}^{k_i} (y \cdot \bar{e}_i^k)h_k,$$

which ends the proof of Claim 1.□
### 3.1.2 Proof of Theorem 2 in the general case

Since $M$ is a compact manifold and $V_i^{k_i}$ is a Euclidean space, for every $i \in I$, each continuous mapping $f_i: M \to V_i^{k_i}$ can be approximated by a sequence of smooth mappings $f_i^n: M \to V_i^{k_i}$ converging to $f_i$, in the sense that $\lim_{n \to \infty} \|f_i^n - f_i\|_{\infty} = 0$ (see, for example, Hirsch [11]). Applying Step 1 to the smooth mappings $f_i^n$, we deduce the existence of $(x_i^n)_{i \in I} \in M$ such that

$$ \forall i \in I, f_i^n(x_i^n) \in (x_i^n)^{k_i} $$

or equivalently

$$ \mathrm{proj}_{(x_i^{n\perp})^{k_i}} f_i(x_i^n) = 0. $$

From the compactness of $M$, without any loss of generality, one can suppose that the sequence $(x_i^n)_{i \in I}$ converges to some element $(\bar{x}_i)_{i \in I} \in M$. We have:

$$ \| \mathrm{proj}_{(\bar{x}_i^{\perp})^{k_i}} f_i(\bar{x}_i) - \mathrm{proj}_{(\bar{x}_i^{n\perp})^{k_i}} f_i^n(x_i^n) \| \le \| \mathrm{proj}_{(\bar{x}_i^{\perp})^{k_i}} f_i(\bar{x}_i) - \mathrm{proj}_{(\bar{x}_i^{n\perp})^{k_i}} f_i(x_i^n) \| + \\ \| f_i^n - f_i \|_{\infty}. $$

Consequently, from the convergence of $f_i^n$ to $f_i$ and the continuity of the mapping $(u, v) \to \mathrm{proj}_{(u^{\perp})^{k_i}} v$ (see Lemma 3 in the Appendix), we obtain

$$ \mathrm{proj}_{(\bar{x}_i^{\perp})^{k_i}} f_i(\bar{x}_i) = 0, $$

or equivalently

$$ \forall i \in I, f_i(\bar{x}_i) \in ((\bar{x}_i^{\perp})^{k_i})^{\perp} = (\bar{x}_i)^{k_i}, $$

which ends the proof of Theorem 2.□

## 3.2 Proof of Theorem 1 when $J = \emptyset$

We now prove Theorem 1 when $J = \emptyset$. The proof rests on the following claim:

**Claim 2** For every $i \in I$ and every $k \in \{1, ..., k_i\}$, there exists an u.s.c. correspondence $\hat{F}_i^k$ from $M$ to $V_i$, with nonempty convex values, such that:

$$ \forall x \in M, [F_i^k(x) \neq \emptyset] \Rightarrow [\forall y \in \hat{F}_i^k(x), \exists \lambda \in R, \lambda y \in F_i^k(x)]. $$

**Proof of Claim 2** Let $i \in I$ and $k \in \{1, ..., k_i\}$. We distinguish two cases.

Assume first that $F_i^k$ is lower semi-continuous. Let $U_i^k = \{x \in M | F_i^k(x) \neq \emptyset\}$. Then $U_i^k$ is an open subset of $M$ and $F_i^k|_{U_i^k}$ is a lower semi-continuous correspondence with non-empty convex values. From Michael [15], there exists a continuous selection $f_i^k$ of $F_i^k|_{U_i^k}$, i.e., $f_i^k: U_i^k \to V_i$ is a continuous mapping such that $f_i^k(x) \in F_i^k(x)$ for every $x \in U_i^k$. Let $B_i$ be the closed unit ball of $V_i$, and let us define the correspondence $\hat{F}_i^k$
from $M$ to $B_i$ by $\hat{F}_i^k(x) = \{f_i^k(x)/\|f_i^k(x)\|\}$ if $[x \in U_i^k$ and $f_i^k(x) \neq 0]$ and $\hat{F}_i^k(x) = B_i$ otherwise. We let the reader check that the correspondence $\hat{F}_i^k$ satisfies the conclusion of Claim 2.

We now consider the case where $F_i^k$ is upper semi-continuous. Let $U_i^k = \{x \in M \mid F_i^k(x) \neq \emptyset\}$. Then $U_i^k$ is a closed subset of $M$. From Cellina [4], one can extend $F_i^k|_{U_i}$ as follows: there exists a correspondence $\hat{F}_i^k$ from $M$ to $V_i$ which is upper semi-continuous, with non-empty, convex and compact values, such that for every $x \in U_i^k$, $F_i^k(x) = \hat{F}_i^k(x)$. $\square$

We now come back to the proof of Theorem 1 when $J = \emptyset$. For every $i \in I$ and $k = 1, ..., k_i$, let $\hat{F}_i^k$ be the u.s.c. correspondence from $M$ to $V_i$ with nonempty convex (compact) values defined in Claim 2. From Cellina ([3]), for every integer $n$ there exists a continuous mapping $f_i^{k,n} : M \to V_i$ such that

$$G(f_i^{k,n}) \subset G(\hat{F}_i^k) + B(0, \frac{1}{n}).$$

Now, for $i \in I$, let $f_i^n : M \to (V_i)^{k_i}$ be defined by

$$\forall x \in M, f_i^n(x) = (f_i^{1,n}(x), ..., f_i^{k_i,n}(x)).$$

Applying Theorem 2 to the mappings $f_i^n$, we deduce the existence of $(\bar{x}_i^n)_{i \in I} \in M$ such that for every $i \in I$, $f_i^n(\bar{x}^n) \in (\bar{x}_i^n)^{k_i}$, hence

$$y_i^{k,n} := f_i^{k,n}(\bar{x}^n) \in \bar{x}_i^n.$$

Since the correspondence $\hat{F}_i^k$ is bounded ($M$ is compact and $\hat{F}_i^k$ is u.s.c.), the sequence $(y_i^{k,n})$ is bounded. Thus, without any loss of generality, one can suppose that the sequence $(y_i^{k,n})$ converges to some $y_i^k \in V_i$ when $n$ tends to $+\infty$.

Besides, from the compactness of $M$, without any loss of generality, one can suppose that $(\bar{x}_i^n)_{i \in I}$ converges to $\bar{x} = (\bar{x}_i)_{i \in I} \in M$ when $n$ tends to $+\infty$.

Moreover, from Part (d) of Lemma 3 in the Appendix and from $y_i^{k,n} \in \bar{x}_i^{k,n}$, at the limit we have

$$\forall i \in I, \forall k = 1, ..., k_i, y_i^k \in \bar{x}_i.$$

Since the graph of $\hat{F}_i^k$ is closed (it is upper semi-continuous with compact values) and from $G(f_i^{k,n}) \subset G(\hat{F}_i^k) + B(0, \frac{1}{n})$, one obtain

$$y_i^k \in \hat{F}_i^k(\bar{x}).$$

To end the proof we assume that $F_i^k(\bar{x}) \neq \emptyset$. Since $y_i^k \in \hat{F}_i^k(\bar{x})$, by Claim 2, there exists $\lambda \in R$ such that $\lambda y_i^k \in F_i^k(\bar{x})$. Hence $\lambda y_i^k \in F_i^k(\bar{x}) \cap \bar{x}_i \neq \emptyset$ (since $y_i^k \in \bar{x}_i$). This ends the proof of Theorem 1. $\square$
### 3.3 Proof of Theorem 1 in the general case

We first prove the following lemma:

**Lemma 2** Let $C$ be a nonempty, convex, compact subset of a Euclidean space $V$. Then, there exists a continuous mapping $\rho: G^1(V \times R) \to C$ such that:

$$ \forall x \in G^1(V \times R), x \cap [C \times \{1\}] \neq \emptyset \Rightarrow x \cap [C \times \{1\}] = \{(\rho(x), 1)\}. $$

**Proof.** Since $C$ is compact, it is included in a closed ball $\bar{B}(0, k)$ of $V$. We let $r: V \to \bar{B}(0, k+1)$ be defined by $r(u) = \alpha(\|u\|)u$ where $\alpha: R_+ \to R_+$ is defined by

$$ \alpha(t) = 1 \text{ if } t \in [0, k], \alpha(t) = k + 1 - t \text{ if } t \in [k, k+1], \alpha(t) = 0 \text{ if } t \ge k+1. $$

Let $\pi_1: V \times R \to V$ and $\rho: G^1(V \times R) \to C$ be defined by $\pi_1(x, t) = x$ and

$$ \rho(x) = \begin{cases} \mathrm{proj}_C \circ r \circ \pi_1(x \cap [V \times \{1\}]) & \text{if } x \cap [V \times \{1\}] \neq \emptyset, \\ \mathrm{proj}_C(0) & \text{if } x \cap [V \times \{1\}] = \emptyset. \end{cases} $$

where $\mathrm{proj}_C: V \to C$ denotes the projection from $V$ to $C$. Then, one easily sees that $\rho$ satisfies the conclusion of Lemma 2. □

**Proof of Theorem 1.** Using Lemma 2, we first modify the correspondences $F_j$ for every $j \in J$ and replace each nonempty compact convex set $C_j \subset V_j$ by the Grassmannian manifold $G^1(V_j \times R)$. For every $j \in J$ let $\rho_j: G^1(V_j \times R) \to C_j$ be the mapping associated to $C_j \subset V_j$ by Lemma 2. Let

$$ \rho: \tilde{M} := \prod_{i \in I} G^{k_i}(V_i) \times \prod_{j \in J} G^1(V_j \times R) \to M := \prod_{i \in I} G^{k_i}(V_i) \times \prod_{j \in J} C_j $$

defined by

$$ \rho(x) = ((x_i)_{i \in I}, \rho_j(x_j)_{j \in J}), \text{ for } x = ((x_i)_{i \in I}, (x_j)_{j \in J}) $$

For $i \in I$ and $k = 1, \dots, k_i$, let $\tilde{F}_i^k$ be the correspondence from $\tilde{M}$ to $V_i$, defined by

$$ \tilde{F}_i^k(x) = F_i^k(\rho(x)). $$

For $j \in J$, let $\tilde{F}_j$ be the correspondence, from $\tilde{M}$ to $V_j \times R$, defined by

$$ \tilde{F}_j(x) = F_j(\rho(x)) \times \{1\}. $$

Now, applying the result proved in Section 3.2 (that is, Theorem 1 when $J = \emptyset$) to the correspondences $\tilde{F}_i^k$ and $\tilde{F}_j$, there exists $x = ((x_i)_{i \in I}, (x_j)_{j \in J}) \in \tilde{M}$ such that:

(i) [either $\tilde{F}_i^k(x) \cap x_i \neq \emptyset$ or $\tilde{F}_i^k(x) = \emptyset$] for every $i \in I$ and $i = 1, \dots, k_i$,
(ii) [either $\tilde{F}_j(x) \cap x_j \neq \emptyset$ or $\tilde{F}_j(x) = \emptyset$] for every $j \in J$.

Let $\bar{x} = \rho(x) \in M$, we end the proof by showing that it satisfies the conclusion of Theorem 1. From above, it is clearly the case for $i \in I$ and $k = 1, \dots, k_i$, i.e., we have:

(i) [either $F_i^k(x) \cap x_i \neq \emptyset$ or $F_i^k(x) = \emptyset$] for every $i \in I$ and $i = 1, \dots, k_i$.

Now, let $j \in J$. We first notice that $\tilde{F}_j(x) = \emptyset$ if and only if $F_j(x) = \emptyset$. Assume now that $\tilde{F}_j(x) \cap x_j \neq \emptyset$ and recall that $x_j \cap \tilde{F}_j(x) = x_j \cap (F_j(\bar{x}) \times \{1\})$ and $F_j(\bar{x}) \subset C_j$. Consequently, $x_j \cap (C_j \times \{1\}) \neq \emptyset$ and from Lemma 2 we get

$$ \emptyset \neq x_j \cap (F_j(\bar{x}) \times \{1\}) \subset x_j \cap (C_j \times \{1\}) = \{(\rho_j(x_j), 1)\}. $$

Hence, the equality holds and $\bar{x}_j = \rho_j(x_j) \in F_j(\bar{x})$. This ends the proof of Theorem 1. $\square$

# 4 Appendix: the Grassmannian manifold $G^k(V)$

Let $V$ be a Euclidean space and let $k$ be an integer such that $0 \le k \le \dim V$. In this section, we recall the properties of $G^k(V)$ which are used in this paper.

First, we recall that $G^k(V)$ is a smooth boundaryless manifold of dimension $k(\dim V - k)$ (see, for example, Hirsh [11] and Lemma 3 below). The local charts can be defined as follows. Let $\bar{E} \in G^k(V)$ and let $\{\bar{e}_1, \dots, \bar{e}_k\}$ be some given orthonormal basis of $\bar{E}$, we define the mapping $\psi_{\bar{E}} : (\bar{E}^{\perp})^k \to G^k(V)$ by

$$ \psi_{\bar{E}}(u) = \operatorname{span} \{\bar{e}_1 + u_1, \dots, \bar{e}_k + u_k\}, \text{ for } u = (u_1, \dots, u_k) \in (\bar{E}^{\perp})^k. $$

Then it is easy to check that the mapping $\psi_{\bar{E}}$ is injective (see Claim 3 below), so $\psi_{\bar{E}}$ is a bijection from $(\bar{E}^{\perp})^k$ onto $U_{\bar{E}} = \psi_{\bar{E}}((\bar{E}^{\perp})^k)$. We can now consider the inverse mapping $\varphi_{\bar{E}} : U_{\bar{E}} \to (\bar{E}^{\perp})^k$ defined by $\varphi_{\bar{E}}(E) = \psi_{\bar{E}}^{-1}(E)$, which is clearly a bijection.

**Lemma 3** (a) $G^k(V)$ is a smooth boundaryless (i.e., $C^\infty$) manifold without boundary of dimension $k(\dim V - k)$ and $(U_{\bar{E}}, \varphi_{\bar{E}})_{\bar{E} \in G^k(V)}$ defines a $C^\infty$ atlas of $G^k(V)$.

(b) The set $G^k(V)$ is compact.

(c) The mapping $E \to E^{\perp}$ from $G^k(V)$ to $G^l(V)$ ($l = \dim V - k$) is a smooth diffeomorphism.

(d) The mapping $p : V \times G^k(V) \to V$ defined by $p(x, E) = \operatorname{proj}_E(x)$ is smooth. Hence, the set $\{(x, E) \in V \times G^k(V) | x \in E\}$ is a closed subset of $V \times G^k(V)$.

(e) The mapping $x \to x^p$ from $G^k(V)$ to $(G^k(V))^p$ is smooth.
We prepare the proof of the lemma with a claim.

**Claim 3** Let $\bar{E} \in G^k(V)$ and let $\{\bar{e}_1, ..., \bar{e}_k\}$ be an orthonormal basis of $\bar{E}$.

(a) The mapping $\psi_{\bar{E}}$ is injective.

(b) For every $u \in (\bar{E}^{\perp})^k$, $\psi(0) \cap \psi(u)^{\perp} = \psi(0)^{\perp} \cap \psi(u) = \{0\}$.

**Proof of Claim 3.** Part (a). Let $u = (u_1, ..., u_k)$ and $v = (v_1, ..., v_k)$ in $(\bar{E}^{\perp})^k$ such that

$$ \psi_{\bar{E}}(u) = \operatorname{span} \{\bar{e}_1 + u_1, ..., \bar{e}_k + u_k\} = \psi_{\bar{E}}(v) = \operatorname{span} \{\bar{e}_1 + v_1, ..., \bar{e}_k + v_k\}. $$

Then, there exist some real numbers $\lambda_{ij}$ ($i = 1, ..., k, j = 1, ..., k$) such that

$$ \bar{e}_i + u_i = \sum_{j=1}^{k} \lambda_{ij}(\bar{e}_j + v_j) \text{ for every } i = 1, ..., k. $$

Taking, for each inequality, the scalar product with $\bar{e}_l$, where $l = 1, ..., k$, we obtain $\lambda_{il} = 0$ if $i \neq l$ and $\lambda_{ll} = 1$. Hence $u_l = v_l$ for every $l = 1, ..., k$, and finally $u = v$.

Part (b). Let $x \in \bar{E} \cap \psi(u)^{\perp}$. Then there exists some real numbers $\lambda_i$ ($i = 1, ..., k$) such that $x = \sum_{i=1}^k \lambda_i \bar{e}_i$. Taking the scalar product with $\bar{e}_j + u_j$ ($j = 1, ..., k$), we obtain: $\lambda_j = 0$ for every $j = 1, ..., k$ which proves $\psi(0) \cap \psi(u)^{\perp} = \{0\}$. Similarly, let $x \in \bar{E}^{\perp} \cap \psi(u)$. Then there exists some real numbers $\lambda_i$ ($i = 1, ..., k$) such that $x = \sum_{i=1}^k \lambda_i (\bar{e}_i + u_i)$. Taking the scalar product with $\bar{e}_j$ ($j = 1, ..., k$), we obtain: $\lambda_j = 0$ for every $j = 1, ..., k$ which proves $\psi(0) \cap \psi(u)^{\perp} = \{0\}$ and ends the proof of the claim. $\square$

**Proof of Lemma 3.** Part (a). We prove that $(U_E, \varphi_E)_{E \in G^k(V)}$ is a smooth (i.e., $C^{\infty}$) atlas of $G^k(V)$ and it is then clear that $\dim M = \dim(E^{\perp})^k = k(\dim V - k)$. Let $(U_E, \varphi_E)$ and $(U_F, \varphi_F)$ be two local charts at $E$ and $F$, respectively, such that $U_E \cap U_F \neq \emptyset$. We shall prove that $\varphi_F \circ \varphi_E^{-1}$ is smooth (i.e., $C^{\infty}$). We let $\{e_1, ..., e_k\}$, and $\{f_1, ..., f_k\}$, be two orthonormal basis, respectively of $E$ and $F$. Let $(v_1, ..., v_k) = \varphi_F \circ \varphi_E^{-1}(u_1, ..., u_k)$ and $u = (u_1, ..., u_k)$, then there exist real numbers $\lambda_{ij}(u)$ ($i, j = 1, ..., k$) such that

$$ f_i + v_i = \sum_{j=1}^{k} \lambda_{ij}(u)(e_i + u_i) \quad (i = 1, ..., k). $$

The proof will be complete by showing that the real-valued functions $\lambda_{ij}(u)$ are differentiable with respect to $u$. Taking the scalar product with $f_l$ ($l = 1, ..., k$) we obtain

$$ f_i \cdot f_l = \sum_{j=1}^{k} \lambda_{ij}(u)(e_j + u_j) \cdot f_l \quad (l = 1, ..., k). $$
Thus for every $i = 1, ..., k$, the vector $\lambda_i(u) = (\lambda_{ij}(u))_{j=1}^k$ is the solution of a linear system whose matrix $G(u) = ((e_j + u_j) \cdot f_l)_{j,l=1,...,k}$ is now shown to be invertible (which clearly implies the differentiability of $\lambda_i(u)$). Indeed, if $G(u)\lambda = 0$ for some $\lambda \in \mathbb{R}^k$ then $\sum_{j=1}^k \lambda_j(e_j + u_j) \cdot f_l = 0$ (for $l = 1, ..., k$), thus $\sum_{j=1}^k \lambda_j(e_j+u_j) \in F^\perp$. Besides, since $\sum_{j=1}^k \lambda_j(e_j+u_j) \in \psi_E(u_1, ..., u_j) = \psi_F(v_1, ..., v_j)$, we finally obtain $\sum_{j=1}^k \lambda_j(e_j + u_j) \in F^\perp \cap \psi_F(v_1, ..., v_k) = \{0\}$ [from Claim 3]. Now, since $(e_j + u_j)_{j=1,...,k}$ is a basis we obtain $\lambda_j = 0$ for every $j = 1, ..., k$. $\square$

**Part (b).** Let $(E^\nu)$ be a sequence in $G^k(V)$ and, for every $\nu$, let $\{e_1^\nu, ..., e_k^\nu\}$ an orthonormal basis of $E^\nu$. Without any loss of generality, we can assume that, for every $i = 1, ..., k$ the sequence $(e_i^\nu)$ converges to some element $e_i$ in $V$. Clearly, $\{e_1, ..., e_k\}$ is an orthonormal family in $V$ and we let $E = \text{span}\{e_1, ..., e_k\}$. We shall now prove that the sequence $(E^\nu)$ converges to $E$. Indeed, for $\nu$ large enough, there exists $u^\nu = (u_1^\nu, ..., u_k^\nu) \in (E^\perp)^k$ such that $E^\nu = \psi_E(u_1^\nu, ..., u_k^\nu)$. It can be written $e_i + u_i^\nu = \sum_{j=1}^k \lambda_{ij}^\nu e_j^\nu (i = 1, ..., k)$. Multiplying these equalities by $e_l(l = 1, ..., k)$ we obtain: $\sum_{j=1}^k \lambda_{ij}^\nu e_j^\nu \cdot e_l = 0$ if $i \neq l$ and $\sum_{j=1}^k \lambda_{ij}^\nu e_j^\nu \cdot e_i = 1$. This can be written (for every $i = 1, ..., k$) $(e_j^\nu \cdot e_l)_{j,l=1,...,k} (\lambda_{ij}^\nu)_{j=1}^k = (e_i \cdot e_l)_{l=1,...,k}$. If $\nu$ is large enough then $(e_j^\nu \cdot e_l)_{j,l=1,...,k}$ is invertible and converges to $Id$, which prove that the sequence $(\lambda_{ij}^\nu)_{j=1}^k$ converges to $(e_i \cdot e_l)_{l=1,...,k}$ for every $i = 1, ..., k$, i.e $(\lambda_{ii}^\nu)$ converges to 1 and $(\lambda_{ij}^\nu)$ converges to 0 for $i \neq j$. We finally obtain that $(u_i^\nu)$ converges to 0, which proves that $E^\nu$ converges to $E$. $\square$

**Part (c).** Let $\bar{E} \in G^k(V)$, and let $(\bar{e}_1, ..., \bar{e}_k)$, and $(\bar{f}_1, ..., \bar{f}_l)$ be orthonormal basis, respectively, of $\bar{E}$ and $\bar{E}^\perp$. Let $(u_1, ..., u_k) \in (\bar{E}^\perp)^k$ and let $E = \psi(u)$. Then it is easy to see that $E^\perp = \text{span}\{\bar{f}_1 + v_1, ..., \bar{f}_l + v_l\}$ where $v_i = -\sum_{j=1}^k (f_i \cdot u_j)\bar{e}_j$. So each $v_i$ is a smooth mapping with respect to the $u_i$ and conversely, from $(E^\perp)^\perp = E$, each $u_i$ is a smooth mapping with respect to the $v_i$. This ends the proof of Part (c). $\square$

**Part (d).** The differentiability of the mapping $p$ is left to the reader. Then notice that $\{(x, E) \in V \times G^k(V) | x \in E\} = \{(x, E) \in V \times G^k(V) | x = \text{proj}_E(x)\}$, which is clearly closed since the mapping $p : (x, E) \to \text{proj}_E(x)$ is continuous. $\square$

## References

[1] Bich, P., Cornet, B., Existence of financial equilibria: space of transfers of fixed dimension, Document de travail, Université de Paris 1 (1998).

[2] Brouwer, L., Über Abbildung von Mannigfaltigkeiten, Math. Ann. 71 (1912) 97-115.
[3] Cellina, A., Approximation of set-valued functions and fixed-point theorems, Annali di Mathematica Pura ed Applicata 4, (1969) 17-24.

[4] Cellina, A., A theorem on the approximation of compact multi-valued mappings, Rendiconti di scienze fisiche, mathematiche & naturale, Vol. XLVII, fasc. 6, (1969) 429-433.

[5] Doubrovine, B., Novikov, S., Fomenko, A., Géométrie contemporaine : méthodes et applications II, 1982, Edition MIR.

[6] Duffie, D, Shaffer, W, Equilibrium in incomplete markets I: basic model of generic existence, Journal of Mathematical Economics 14, (1985) 285-300.

[7] Eilenberg, S., Montgomery, D., Fixed Point Theorems for Multivalued Transformations, American Journal of mathematics 68, (1946) 214-222.

[8] Gale, D., Mas-Colell, A., An equilibrium existence theorem for a general model without orderd preferences, Journal of Mathematical Economics 2, (1975) 9-15.

[9] Gourdel, P., Existence of Intransitive Equilibria in Nonconvex Economies, Set-Valued Analysis 3, (1995) 307-337.

[10] Guillemin, V., Pollack, A., Differential topology, 1974, Prentice Hall.

[11] Hirsch, M.W., Differential Topology, 1976, Springer-Verlag.

[12] Hirsch, M., Magill, M., Mas-Colell, A., A geometric approach to a class of equilibrium existence theorems, Journal of Mathematical Economics 19, (1990) 95-106.

[13] Husseini, S. Y., Lasry, J.M., Magill, M., Existence of equilibrium with incomplete markets, Journal of Mathematical Economics 19, (1990) 39-67.

[14] Kakutani, S., A generalization of Brouwer's Fixed Point Theorem, Duke Math. Journal 8, (1941) 457-459.

[15] Michael, E., Continuous selection *I*, Annals of Mathematics 63, (1951) 361-382.

Philippe Bich, CEREMADE, Université Paris Dauphine, place du Maréchal de Lattre de Tassigny, 75016 Paris. Email: bich@ceremade.dauphine.fr
Bernard Cornet, CERMSEM, Université Paris 1, 106-112 boulevard de l'Hôpital, 75013 Paris. Email: cornet@univ-paris1.fr