Why $f^{-1} (y)$ is a closed set

Question

Let $M$ be a compact smooth manifold, $f$ is a smooth map between $M$ and $N$. If $y \in N$ is a regular value, then $f^{-1} (y)$ is a closed set.

I don't know why $f^{-1} (y)$ is a closed set.

compactness is not even needed, just separation axioms :D – Sep 20 '13 at 06:46 — , Sep 20 '13 at 06:46

score 4 · Answer 1 · answered Sep 20 '13 at 06:38

4

Smooth maps are continuous an manifolds are Hausdorff. So $\{y\} \subseteq N$ is closed, hence its continuous preimage $f^{-1}(\{y\}) \subseteq M$ is.

answered Sep 20 '13 at 06:38

martini

84,101

Why $f^{-1} (y)$ is a closed set

1 Answers1