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Let $M$, $N$ and $P$ be smooth manifolds and $f:M\longrightarrow N$ and $g:N\longrightarrow P$ be two smooth maps. Let us write $h:=g\circ f:M\longrightarrow P$.

Suppose $h$ is a surjective submersion and $f$ is surjective. Does this imply $g$ is a surjective submersion?

Thanks.

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    $\begingroup$ Compositions of surjections are surjections. This applies to manifolds and to tangent spaces. $\endgroup$ – Angina Seng Aug 26 '17 at 19:47
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Since $h$ is surjective, $g$ is surjective. This is true for any maps of sets satisfying the hypotheses. By the chain rule, $$ dh_x = dg_{f(x)}\circ df_x, $$ and $dh_x$ is surjective. Thus we glean that $dg_{f(x)}$ is surjective, so that $g$ is a submersion at $f(x)$. Since any point $y\in N$ may be written $f(x)$ for some $x\in M$, $dg_y$ is surjective for all $y\in N$, so $g$ is a submersion. Putting it all together, $g$ is a surjective submersion.

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