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Question

Embedding a Smooth Compact Manifold into Rn Ivan Yaschenko

Question. Let M be a smooth compact manifold, which admits a continuous map f : M --> Rⁿ with the set of none-one-to-one points at most countable. Can one embed M into this Rⁿ?

A point x in M is said to be a none-one-to-one point with respect to map f iff f(x) = f(y) for some y in M different from x. I can prove the following.

Theorem 1.[Y] If the compact space (not necessarily a smooth manifold) M is cleavable over Rⁿ ( i.e. for every subset A of M there is a continuous mapping f : M --> Rⁿ so that f(A) and f(X\A) are disjoint [ASh, Tk]) then M admits a continuous map f into that Rⁿ with the set of none-one-to-one points at most countable.

Hence, one can easily find an embedding of M into Rⁿ⁺¹. The affirmative answer to the question implies that if M is a compact smooth manifold then moreover there is an embedding of M into Rⁿ.

In the case of a compact complex the answer is negative. Let X be the n-skeleton of 2n+2 - simplex. Then X is cleavable over R²ⁿ [Tk]. So from Theorem 1 it follows that X admits a map into R²ⁿ with the set of none-one-to-one points at most countable. But there is no embedding of X into R²ⁿ [Fl].

[Ash] Arhangel'skii A.V., Shakhmatov D.B., On pointwise approximation of arbitrary functions by countable families of continuous functions (in Russian), Trudy seminara imeni I.G.Petrovskogo 13 (1988), 206-227; English transl., Journal of Soviet Math. 50 (1990), 1497-1512.

[Fl] Flores A.I., Ubern-dimensionaler Komplexe, die nicht in den R²ⁿ topologisch einbettbar sind, Erg. Math. Kollog. 5 (1933), 17-22.

[Tk] Tkachuk V.V., A note on splittable spaces, Comm. Math. Univ. Carolinae 33 (1992), 551-555.

[Y] Yaschenko I.V., Ph.D. Thesis (In Russian), Moscow State University 1994.