# 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?

-
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
show 1 more comment