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The lemma is as follows

Let $f:X\to N$ be a smooth mapping, where $X$ is $m$ dimensional and $N$ is $n$ dimensional. Moreover, $m\geq n$. If $y\in N$ is a regular value, both for $f$ and 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 $\partial X$.

1) In the proof of the lemma he says 'Since we have to prove a local property, it suffices to consider the special case of a map $f:\mathbb H^m\rightarrow \mathbb R^n$'. I don't see how this is true. Also what exactly is meant by local property?

2)'$f^{-1}(y)$ is a smooth manifold in the neighbourhood of $x$'. What does it mean to be smooth manifold in the "neighbourhood" of $x$ ?

3)A point $x$ was taken from boundary. Another smooth map $g:U \to \mathbb{R}^n$ was constructed on a neighborhood $U$ of ${x}$ which agrees with $f$ on $U \cap H^m$. Then it was stated that we can replace $U$ by a sufficiently small neighborhood around $\bar{x}$ on which $g$ will have no critical point. Hence $g^{-1}(y)$ will be a smooth $(m-n)$-manifold. Why do we require a neighbourhood on which there is no critical point?

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1) Smoothness is a local property, that's why you can always restrict to a chart (or local coordinates). Moreover, if you're not yet talking about the boundary then everything else is an open set i.e it can be mapped to the upper half space. Here that's $\mathbb{H}^m$.

2) Think of a point x in the plane, let's say the center of an open disk. Now take a smaller disk in this region, say $B_y(x) = \{y: |x-y|<\epsilon\}$. This is a smooth manifold in this neighborhood (the disk) of the point x.

3) If $y$ is a regular value, then by definition each point $p \in f^{-1}(y)$ is regular i.e $J_f(p)$ has full rank which implies no point is critical.

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  • $\begingroup$ In 3 why do we require an absence of a critical point in the neighbourhood of each $p$ $\endgroup$ – mathemather Jul 10 '18 at 6:00
  • $\begingroup$ In 1, we want to prove $f^{-1}(y) $ is a smooth manifold and not that the neighbourhood of $x \in f^{-1}(y)$ is, so I still don't get how it works. $\endgroup$ – mathemather Jul 10 '18 at 12:45
  • $\begingroup$ I'm sorry for being so late, but f^{-1}(y) is a sub-manifold i.e it has a tangent plane at each point i.e no point can be critical since this would imply the non-existence of a tangent plane. I've answered you question about (1), re-read it. $\endgroup$ – Faraad Armwood Jul 17 '18 at 3:00

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