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Let $Y\subset X$ a submanifold. Show that there is a manifold $Z$ with a regular value $z_0\in Z$ and a map $C^r, r\ge 1$, $f:X\longrightarrow Z$ such that $Y=f^{-1}(z_0)$.

Matematico
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  • As stated this is wrong. Take $\mathbb{R}^n$ in $S^n$. Maybe you want $Y$ to be closed? – ThorbenK Oct 17 '16 at 22:37
  • The teacher wrote that way. But I think he wants to show that every submanifold is inverse image of a regular value of an map $C^r$, with $r>0$. – Matematico Oct 17 '16 at 22:49
  • Even assuming $Y$ is closed in $X$, this is false. It can only be true if $Y$ has trivial normal bundle in $X$. – Ted Shifrin Oct 17 '16 at 23:03
  • Could enunciate and prove it? – Matematico Oct 17 '16 at 23:08
  • @LucianoGauss: Do you know about normal bundles? If $f\colon X\to Z$ is transverse to $W\subset Z$, then the normal bundle of $f^{-1}(W)$ in $X$ is the pullback of the normal bundle of $W$ in $Z$. This follows immediately from transversality. (Example: You should be able to show easily that you cannot have $f\colon\text{Möbius strip}\to (-1,1)$ with $0$ a regular value and $f^{-1}(0)$ the center circle of the Möbius strip.) – Ted Shifrin Oct 18 '16 at 01:36
  • I don't know. The ploblem written better: "Let $Y\subset X$ a submanifold. Show that there is a manifold $Z$ with a regular value $z_0\in Z$ and a map $C^r, r\ge 1$, $f:X\longrightarrow Z$ such that $Y=f^{-1}(z_0)$." How is solution of this problem? – Matematico Oct 18 '16 at 17:04
  • One point is the definition of a submanifold. There are different definitions of that around. Since manifolds are Hausdorff, for a continuous map $f \colon X \to Z$, the fibre $f^{-1}(z_0)$ is closed for every $z_0 \in Z$. So we have the necessary condition that $Y$ must be closed for the existence of a $C^r$ map $f \colon X \to Z$ and a $z_0$ with $Y = f^{-1}(z_0)$. Probably the definition of a submanifold used in your course includes closedness, then that's okay. But it's still better to include the definition of submanifold you were given to settle any doubts about that. – Daniel Fischer Oct 18 '16 at 19:01
  • The next point is more serious. As @TedShifrin mentioned, even if $Y$ is a closed submanifold, such a function only exists if a certain geometric condition is satisfied. Maybe the problem asks for the local existence of such functions, for every $y\in Y$ there exists an open $U\subset X$ with $y\in U$ and a $C^r$ function $f \colon U \to Z$ such that $U\cap Y = f^{-1}(z_0)$ where $z_0$ is a regular value of $f$? That would be a rather trivial exercise with most definitions of submanifold, but at least it would be true without additional conditions. – Daniel Fischer Oct 18 '16 at 19:07

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