# Regular value of $g \circ f$ is a regular value of $g$

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?

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Do you know the chain rule? –  Ryan Budney Dec 2 '12 at 21:59
Yeah, 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 $g$. –  JJJ Dec 2 '12 at 22:16
So 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? –  Ryan Budney Dec 2 '12 at 22:55
So, 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 $x \in X$ –  JJJ Dec 2 '12 at 23:00
Correct. So now that you have this insight, what are you going to do with it? It's everything you need to answer your question. –  Ryan Budney Dec 2 '12 at 23:03
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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 $g'(5)=0$.