I've got a question about a lemma in Milnors "TOPOLOGY FROM THE DIFFERENTIABLE VIEWPOINT".

LEMMA 4: If $y\in N$ is a regular value, both for $f$ and for the restriction $f|_{\partial X}$, then $f^{-1}(y)\subset X$ is a smooth $m-n$-manifold with boundary. Furthermore the boundary $\partial (f^{-1}(y))$ is precisely equal to the intersection of $f^{-1}(y)$ with $\partial X$

Maybe this lemma can be called Submersion Theorem for manifold with boundary. So, I want to find a example and take $X=D^2$, $N=\mathbb{R}$ and $f^{-1}(y)$ should be $D^1$. $D^n$ is the unit disk. But I am not able to find the explicit function $f$ and the $y$.

Can someone help me?

  • $\begingroup$ What about projection?? $\endgroup$ – Anubhav Mukherjee Oct 27 '16 at 16:53
  • $\begingroup$ So, you mean $f:D^2\rightarrow\mathbb{R}\;(x_1,x_2)\mapsto x_1$? But what is $y\in\mathbb{R}$ to get $D^1=f^{-1}(y)$? $\endgroup$ – Thorben Oct 27 '16 at 16:56
  • $\begingroup$ Think... ${}{}{}{}{$ $\endgroup$ – Anubhav Mukherjee Oct 27 '16 at 17:00
  • $\begingroup$ Ah, okay. $D^1=\{x_1\in\mathbb{R}\;|\;1-{x_1}^2\geq0\}$, so $y$ has to be 0, right? Because $f^{-1}(0)$ are the $(x_1,x_2)$ with $x_1=0$ and $-1\leq x_2\leq 1$ $\endgroup$ – Thorben Oct 27 '16 at 17:06
  • $\begingroup$ What is $D^1$?? Read the theorem and think properly... $\endgroup$ – Anubhav Mukherjee Oct 27 '16 at 17:28

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