I am studying differential geometry, and I am having problems with the following exercise:
Let $M,N,W$ be smooth manifolds, $F:M\to N$ a smooth map and $i:W\to N$ a smooth embedding such that there exists a map $G$ such that $F=i\circ G$. Prove that $G$ is smooth.
Since $i$ is an embedding it is a homeomorphism in the image, so there exists $i^{-1}$ and it is continuous. Hence $G=i^{-1}\circ F$, which is continuous becuase it is the composition of two continuous maps. However, I don't know how to prove that $G$ is smooth, the only way I know to find two charts $\phi, \psi$ such that $\phi\circ G\circ\psi^{-1}$ is smooth for every $p\in G$. So my idea was to use that $i$ and $F$ are smooth to find those charts, but this has no sense, since this aproach does not use the fact that $i$ is an embedding. Another idea I had was to take diferentials to use the fact that $i$ is an immersion , but this aproach fails too, since we can't take the diferential of $G$, because it requires smoothness, which is exactly what we are looking for.
I am looking for an elementary approach (which can be done with the tools given in Lee ''introduction to smooth manifolds'' in chapters 1-4) Thanks for your help.
Best Answer
The question is purely local. Choose $x_0\in M$ and let $y_0=f(x_0)\in W\subset N$. Choose local coordinates $y^1,\dots,y^m$ around $y_0\in N$ so that $W = \{y^{n+1}=y^{n+2}=\dots=y^m=0\}$. That is, $i(y^1,\dots,y^n) = (y^1,\dots,y^n,0,\dots,0)$ gives the inclusion $W\hookrightarrow N$.
Write $F(x) = (f^1(x),\dots,f^m(x)) = (f^1(x),\dots,f^n(x),0,\dots,0)$, since the image of $F$ is contained in $W$. Since projection $\pi\colon\Bbb R^m\to\Bbb R^n$, being linear, is smooth, we see that $G=\pi\circ F$ is smooth.