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I'm studying some problems related to differential topology and I came across the following exercise: if $f:M\rightarrow N$ is a surjective smooth (i.e., $C^\infty)$ function, $\dim(M)>\dim(N)$ and $R_f$ is the set of regular points of $f$, then the interior of $f(R_f)$ is dense in $N$.

I was trying to use Sard's Theorem, to guarantee the density of $V_f$, the set of regular values of $f$, and $V_f\subset f(R_f)$. However, I'm not able to prove that $V_f$ is an open subset of $f(R_f)$ nor to find a different approach.

So, I would like to know if this strategy is correct or there is a better way to prove the statement.

Renan Mezabarba
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  • If you have a regular value you want to find a neighborhood which consists of regular values. You should look for help in your analysis books. – Daniel Valenzuela Apr 06 '15 at 21:28

1 Answers1

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The set of regular points $R_f$ is open in $M$ and $f|_{R_f}:R_f\to N$ is an open mapping (essentially the Rank Theorem). Then $f(R_f)$ is open and coincides with its interior. Now the famous Sard-Brown theorem says that the set or regular values $RV_f=N\setminus f(M\setminus R_f)$ is dense in $N$. Since $f$ is onto, $RV_f\subset f(R_f)$, hence $f(R_f)$ is dense in $N$.

Jesus RS
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