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By A. I. Fetísov

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Second proof. Let H and S be as in the first proof. Observe that H is homeomorphic to the space of unordered pairs of points u, v ∈ X. 6). Therefore, the surface S is homeomorphic to RP2 and cannot be embedded into R3 . This shows that H is self-intersecting, which implies the result. 4. Climbing mountains together. Suppose two climbers stand on different sides at the foot of a two-dimensional (piecewise linear) mountain. As they move toward the top of the mountain, they can move up and down; they are also allowed to move forward or backtrack.

Two mountain climbers. 5 (Mountain climbing lemma). Let f1 , f2 : [0, 1] → [0, 1] be two continuous piecewise linear functions with f1 (0) = f2 (0) = 0 and f1 (1) = f2 (1) = 1. Then there exist two continuous piecewise linear functions g1 , g2 : [0, 1] → [0, 1], such that g1 (0) = g2 (0) = 0, g1 (1) = g2 (1) = 1, and f1 (g1 (t)) = f2 (g2 (t)) for every t ∈ [0, 1]. The mountain climbing lemma is a simple and at the same time a powerful tool. We will use it repeatedly in the next subsection to obtain various results on inscribed polygons.

Use the mountain climbing lemma to parameterize the curves C ′ = { h1 (τ ), t1 (τ ) , τ ∈ [0, 1]} and C ′′ = {(h2 (τ ), t2 (τ ) , τ ∈ [0, 1]}, so that t1 (τ ) = t2 (τ ). Now define the average curve C ∗ of C ′ and C ′′ as C∗ = h1 (τ )/2 + h2 (τ )/2, t1 (τ ) , τ ∈ [0, 1] . By construction, curve C ∗ is continuous, starts at t = 0 and ends at t = 1. Consider a curve H given by the function h(t) = ϕ(u1 (t)). 13. Since h(t) is a continuous function, h(0) = 0 and h(1) = 1, we conclude that C ∗ intersects H at some t = T .

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