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In his 1969 paper “Locally flat imbeddings of topological manifolds” Lees proved that a closed oriented second countable topological manifold admits a locally flat embedding into some R^n.

Does the same result hold if the manifold is not required to be closed or oriented? If so, is there a published reference?

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    $\begingroup$ Remark 5.21.(2) in sciencedirect.com/science/article/pii/S0166864100000134 states "Results of Dancis show that a proper continuous map of a q-manifold to an n-manifold can be approximated, taking boundaries into account, by locally flat embeddings if $q$ is much less than $n$". The paper linked above is "Topologically, quasiconformally or Lipschitz locally flat approximation of embeddings" by Jouni Luukkainen. Thus start from a proper function on your manifold (eg, distance to a point), multiply the codomain by some $R^N$ and approximate. $\endgroup$
    – Igor Belegradek
    Commented Jan 29, 2017 at 23:56
  • $\begingroup$ @IgorBelegradek: Indeed, the cited paper by Dancis does not make any assumptions about the manifold. What does “distance to a point” mean in the context of arbitrary topological manifolds? $\endgroup$
    – Dmitri Pavlov
    Commented Jan 30, 2017 at 16:11
  • $\begingroup$ Disclaimer: I did not read Dancis paper carefully which is why I am not posting this as an answer. Regarding "distance to a point": any topological manifold is a finite dimensional locally compact metric ANR, and hence it embeds into some $\mathbb R^n$ as a closed subset (google the above phrase to locate a reference). The distance to a point in $\mathbb R^n$ defines a proper continuous function on any closed subset. $\endgroup$
    – Igor Belegradek
    Commented Jan 30, 2017 at 16:25
  • $\begingroup$ @IgorBelegradek: Many thanks for the clarification. Isn't such a closed embedding into R^n itself a proper map? (Proper means preimages of compact subsets are compact, in R^n compact subsets are precisely bounded closed subsets, and intersecting a bounded closed subset witha closed subset again gives a closed subset.) This seems to make the other two steps (distance function and multiplication by some R^N) redundant. $\endgroup$
    – Dmitri Pavlov
    Commented Jan 31, 2017 at 14:50
  • $\begingroup$ Yes, this works too. $\endgroup$
    – Igor Belegradek
    Commented Jan 31, 2017 at 16:48

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