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The question is let $f: X \rightarrow \mathbb{R}^2$, show that for almost every $c \in \mathbb{R}$, we have that $f^{-1}(\{c\}\times\mathbb{R})$ is a smooth submanifold of $X$.

I want to apply Sard's theorem. But the precise statement I plan to use is:

For any smooth map $f$ of a manifold $X$ with boundary into a boundaryless manifold $Y$, almost every point of $Y$ is regular value of both $f: X \rightarrow Y$ and $\partial f: \partial X \rightarrow Y$.

To my understanding, $\mathbb{R}^2$ is boundaryless under this situation according to Shaun Ault's respons at Is $\mathbb{R}^2$ boundaryless?

But the bigger problem is, I don't know if $X$ is with boundary. I hope the condition that $X$ is with boundary is to apply the later part of the theorem, $\partial f: \partial X \rightarrow Y$. So when we only need almost every point of $Y$ is regular value of $f: X \rightarrow Y$, we don't need to know if $X$ is with boundary.

This sounds make sense, but I really don't want to assume anything when I am starting to learn differential topology. So could anyone point me some directions?

Thank you very much

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1LiterTears
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    It sounds like that formulation of Sard's theorem isn't the best tool for this problem. Do you know any other formulations which might help? – Adam Saltz Jun 11 '13 at 23:49
  • Thanks Adam. This is the closest I have, others involve measure zero. – 1LiterTears Jun 11 '13 at 23:51
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    Just checking -- do you know the relationship between "measure zero" and "almost every"? – Adam Saltz Jun 11 '13 at 23:55
  • Not really - they are the same....? – 1LiterTears Jun 12 '13 at 00:06
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    You should definitely look this up before returning to Sard's theorem! Briefly: the sentence "almost every $x \in X$ has property $P$" is equivalent to "the set of $x \in X$ without property $P$ has measure zero." If you are working from a textbook or a set of notes, this should be addressed. – Adam Saltz Jun 12 '13 at 00:31
  • Hi Adam, thank you so much for your clarification. Yes, that is what I understood. Thanks so much for your guidance. – 1LiterTears Jun 12 '13 at 02:32

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