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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differential-topology
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2Do you know the chain rule? – 2012-12-02
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0Yeah, so for any regular value $z$ of $g \circ f$, if $x \in (g \circ f)^{-1}(z)$, then $D(g \circ f)_x = Dg_y \circ Df_x$, where $y = f(x)$ So $Dg_y$ is an isomorphism, but that doesn't seem sufficient to show that $z$ is a regular value for $g$. – 2012-12-02
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0So how could it fail to be a regular value? In your argument you have $Dg$ at $f(x)$ is onto. What more do you need? – 2012-12-02
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0So, an element $z \in Z$ is a regular value if for every $y \in g^{-1}(z)$, $Dg_y$ is onto. The chain rule only seems to show that $Dg_y$ is onto only when $y = f(x)$ for some $x \in X$ – 2012-12-02
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0Correct. So now that you have this insight, what are you going to do with it? It's everything you need to answer your question. – 2012-12-02
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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