Given smooth maps $f: X \to Y, g: Y \to Z$, where $X, Y, Z$ are boundaryless, compact manifolds of dimension $n$, is the statement in the title true?
Regular value of $g \circ f$ is a regular value of $g$
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0So is the proof of problem 4 incorrect here, since it assumes that $z$ is a regular value of $g$ http://www.stanford.edu/~ronen/math147/midtermSol.pdf – 2012-12-02
1 Answers
For the definition of regular value that I know ("a point in the complement of the set of critical values"), the claim is clearly false. Consider maps $\mathbb R\to\mathbb R$ defined by $f(x)\equiv 1$ and $g(x)=x^2$. Then $0$ is a regular value of $g\circ f$ (vacuously, since it's not attained at all), but not of $g$. One can modify this example to satisfy the "boundaryless, compact" part, but I won't bother to: it's clear that the matter is local.
Even if we are to take the more restrictive definition of a regular value, requiring it to be attained, the claim is still false. Again on $\mathbb R$, take $f(x)=\tan^{-1} x$ and $g(x)=x(x-5)^2$. Now $0$ is a regular value of $g\circ f$, since $g(f(x))=0$ only when $x=0$, and the derivative of $g\circ f$ at $0$ does not vanish. But $0$ is not a regular value of $g$, because $g(5)=0$ and $g'(5)=0$.
The claim is true, and easy to prove, for surjective maps.