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When I learned Sard's Theorem in differential topology by myself, I was thinking whether it would be possible to prove a converse version of the theorem. That is to say, can we somehow show that each (non-compact) manifold arises as the set of regular points of some smooth map? However, this claim seems too strong to be true, so I have made the claim become less restrictive by only requiring the differentials $df_{x}$ to be nonzero for each $x$, which yields the following claim:

Consider an arbitrary manifold $X$ with dimension $\dim X \geq 1$, then there always exists some map $f:X \rightarrow \mathbb{R}^2$, such that the differential $df_{x}$ is nonzero for any $x \in X$

I come up the claim above (simplified version of the converse) by testing on several examples. However, I haven't managed to prove its correctness (though I believe it should be true...) any hints/ideas on this? Also, I would appreciate it if someone can tell me more about developing a converse to the Sard's theorem.

A related question (which shows that some null set can't be expressed as a set of critical values of some map):

https://math.stackexchange.com/questions/210036/on-the-converse-of-sards-theorem

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    $\begingroup$ Any open manifold admits a $C^1$ map to $\mathbb R$ without critical points, see Hirsch, Theorem 4.8 in [On Imbedding Differentiable Manifolds in Euclidean Space, Ann. Math, 1961], doi.org/10.2307/1970318. Composing the map with an inclusion of $\mathbb R$ into $\mathbb R^2$ gives a desired map with everywhere nonzero differential. $\endgroup$
    – Igor Belegradek
    May 27, 2022 at 23:15
  • $\begingroup$ Hi Igor, thank you so much for your reference! However, I'm not familiar with the terminologies like smooth triangulation of manifolds...and I'm using the references in the paper to study it...Thus, is it possible to prove my claim above with only basic tools (like submersion) in differential topology? $\endgroup$
    – pureorapplied
    May 27, 2022 at 23:33
  • $\begingroup$ Reposted on MSE, here. $\endgroup$
    – Moishe Kohan
    Jun 1, 2022 at 23:01

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