Prove that $f$ map a neighborhood of $Z$ diffeomorphically on to a neighborhood of $f(Z)$( more detail) Supposed that  the derivative of $f:X\to Y$ is an isomorphism whenever $x$ lies in the sub-manifold $Z \subset X$, and assume that $f$ maps $Z$ diffeomorphically onto a $f(Z)$ . Prove that $f$ map a neighborhood of $Z$ diffeomorphically on to a neighborhood of $f(Z)$
I know that this question has been asked before but get no real answer. I'm struggle at the same point of Diane, so I'm hoping for a detailed answer. My professor show the picture for this problem in class, but it just doesn't make any sense to me.
 A: Assume that the derivative of $f:X→Y$ is an isomorphism whenever $x$ lies in the submanifold $Z⊂X$, and that $f$ maps $Z$ diffeomorphically onto $f(Z)$. Let $U_i$ be the local finite collection of open subset of Y covering $f(Z)$, say $i={1,…,n}$. Since the derivative of $f:X→Y$ is an isomorphism, we can find the local inverse $g_i:U_i→X$.
Let $W=\{y∈U_i:g_i (y)=g_j (y)whenever y∈U_i∩U_j \}$, where $U_i$ is arbitrary. This is the set of $y$ that behave nicely in the intersection of the open covers. Thus we can patch $g_i$ together to define a smooth inverse $g:W→X$ because the elements lying in more than one $U_i$ and have well defined image.
Now we want to show that $W$ contain an open neighborhood of $f(Z)$. So let $x∈Z$, since $f$ maps $Z$ diffeomorphically onto $f(Z)$, $f$ map some neighborhood $V$ of $x$ diffeomorphically onto a neighborhood $U$ of $y=f(x)∈f(Z)$. Since $y∈U_i$ for finitely many $U_i$ as we defined $U_i$ be the local finite collection of open subset of $Y$ covering $f(Z)$,  there is another neighborhood $U'=U∩U_1∩…∩U_n$. By the definition of $W$, $U'⊂W$ and $V'=f^{-1} (U')$ is the neighborhood of $x$. This show that $W$ contain an open neighborhood of $f(Z)$.
