conv ( { y 1 , . . . , y$_{n}$ } ) with ˆ y / ∈ ⋃ n i $_{=1}$Γ( y$_{i}$ ) , which (by DeMorgan’s law) is equiv- alent to ˆ y ∈ ⋂ n i $_{=1}$A − $^{1}$( y$_{i}$ ) ⇐⇒ { y 1 , . . . , y$_{n}$ } ⊆ A (ˆ y ) . Since A (ˆ y ) is convex, ˆ y ∈ A (ˆ y ) and the proof is complete. Remark 10 (i) The proof of Theorem 9 clearly establishes its equivalence with Theorem 6. (ii) Theorem 9 is a particular instance of the Browder-Ky Fan fixed point theorem (where the convexity of X \ A − $^{1}$( y ) in Definition 8 is dispensed with; see e.g., [2, 3, 4]). (iii) Note that if X is compact, the compactness condition in Theorem 9 is vacuously satisfied with K = X. To the best of our knowledge, in this generality and in the context of the Browder-Ky Fan fixed point theorem, this condition was first introduced in [2, 3] in 1982. It builds on the so-called Karamardian coercivity condition for complementarity problems (early seventies), taken up in 1977 by Allen in the context of fixed point theorems for set-valued maps (case where K = C ); see [2, 3, 9] for references and details. We end this section with a coincidence theorem between N and N − 1 maps with a direct proof based on Proposition 3. We shall make use of a well-known selection property enjoyed by F $^{∗}$− maps of [2], thus by N − maps (see [2, 3]). Lemma 11 Let A ∈ N ( X, Y ) with Y convex. For any compact subset K of X , there exist a continuous (single-valued) mapping s : K −→ Y and a convex compact finite polytope P ⊆ Y such that s ( x ) ∈ A ( x ) ∩ P for all x ∈ K. Proof. Since ∀ x ∈ X, A ( x ) =  , then X = ⋃ y ∈ Y A − $^{1}$( y ), a union of open sub- sets of X. By compactness, K ⊆ ⋃ n i $_{=1}$A − $^{1}$( y$_{i}$ ) for some finite subset { y 1 , . . . , y$_{n}$ } ⊂ Y. Let { λ$_{i}$ : K −→ [0 , 1] } n i =1 be a continuous partition of unity subordinated to the open cover { A − $^{1}$( y$_{i}$ ) } n i $_{=1}$, and define s : K −→ P = conv ( { y 1 , . . . , y$_{n}$ } ) ⊂ Y by putting s ( x ) := ∑ n i $_{=1}$λ$_{i}$ ( x ) y$_{i}$, x ∈ K. Clearly, for a given x ∈ K, λ$_{i}$ ( x ) = 10