How do you show the existence and well definedness of a map between two smooth manifolds?
For instance, if we have some submanifold $\Sigma$ with dimesion $n$ of $M\times N$, where both $M$ and $N$ are smooth manifolds with dimension $n$, then we have the projection map defined by $\pi_M :M\times N \rightarrow M$ and $\pi_N :M\times N \rightarrow N$ and the inclusion map $i_{\Sigma}:\Sigma\rightarrow M\times N.$ If $\pi_M\circ i_{\Sigma}$ is bijective, then how can we show that there exists a smooth well defined map $\Psi:M\rightarrow N$, such that $\Sigma = \{(p,\Psi(p))\in M\times N; p\in M\}.$
(Further note, there exists some $n$-form $\omega$ on $M$, such that $i_{\Sigma}^*\pi_M^*\omega$ is nowhere zero).
Drawing the map, it seems obvious that there should be a map $\Psi$, but not sure how to show it is well defined.
This is a typical problem in my book, and I am struggling to show formally why this is true. I would appreciate the help.
Let $f$ be the inverse of $\pi_M\circ i$. The following diagram is commutative. $\require{AMScd}$ \begin{CD} \Sigma @>{i}>> U\times N\\ @A{f}AA @VV{\pi_M}V \\ U @<{id}<< U \end{CD} But $f$ is a bijection and $\dim M = \dim\Sigma $, so $U = M$ (edit: This is is wrong, and it is the crucial point. Is $\pi_M\circ i$ a bijection into the image of $i$, or into $M$? The latter would solve the problem.). Then, the diagram can be extended to \begin{CD} \Sigma @>{i}>> M\times N@>{\pi_N}>>N\\ @A{f}AA @VV{\pi_M}V \\ M @<{id}<< M \end{CD} Choosing $\Psi = \pi_N\circ i\circ f$ does the trick.
Being well defined means that $\Psi(p)$ is actually a point of $N$. It clearly is.