Question about image of proper smooth map of constant rank. (Undergraduate) I have to find a proof of the following theorem
Given a smooth function f between smooth manifolds X and Y that:


*

*has constant rank

*is proper

*the preimage of every point in f(X) is connected and simply connected


Then f(X) is a smooth submanifold of Y.
I know that locally we have a submanifold structure by the constant rank theorem. I see that we want the preimage to be connected to avoid self-intersections and properness is also plausible. How to extend this globally?
I read this theorem in "The moment map revisited" by Guillemin and Sternberg
[ http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.210.4537&rep=rep1&type=pdf#page26 ] 
page 149
and I have also read the reference, but I have not understood how that helps (it stays in a local setting)
Does anyone have an idea on what a good strategy would be? I do not see how to use the simply connected assumption. I have looked up in many books, but I have found no mention of anything similar. Any good reference?
I feel like this is an easy exercise, but I am missing something.
Thank you very much in advance.
 A: It seems that the simply connected assumption is not needed. Or maybe there is a mistake in the proof below.
By Hormander's local statement, we know that for every $x \in X$ with $y=f(x)$ there exists an open $U_x \subset X$ and an open $V_y \subset Y$ such that $f(U_x)\subset V_y$ is a submanifold of $V_y$. We would like to prove that there is an open $W_y \subset V_y$ such that $f(X) \cap W_y= f(U_x)\cap W_y$, thus making $f(X)$ a submanifold of $Y$.
First, let's prove that any point $x$ in the preimage of $y$ ''locally detects the local image of $f$ near $y$". Namely, consider the equivalence relation on points of $f^{-1}(y)$ given by
$x \sim x'$ if and only if for any open $O_x$ and $O_{x'}$ there exist open neighborhoods $U_x \subset O_x , U_{x'} \subset O_{x'} $ s.t. $ f(U_x)=f(U_x')$. 
By local structure qiven in Hormander, each equivelence class is open in  $f^{-1}(y)$. The classes cover $f^{-1}(y)$. But it is compact. So we have a finite open subcover. But then the classes are closed. By connectedness, there is only one class. We restate this for future use as follows:
For any $x, x' \in f^{-1} (y)$ for any open $O_x$ and $O_{x'}$ there exist open neighborhoods $U_x \subset O(x) , U_{x'} \subset O(x') $s.t. $ f(U_x)=f(U_x')$. 
Now, let's go back to our main proof. Suppose no such $W_y$ exists. Then we get a sequence $y_k$ in $Y$ converging to $y$ with $y_k \in f(X)$ but with $y_k \notin f(U_x)$. Pick arbitrary preimages $x_k$ of $y_k$. Since $f$ is proper, they lie in a compact subset of $X$, hence have convergent subsequence, converging to $x'$ which is forced to lie in $f^{-1} (y)$. By the observation above, there is $U_{x'}$ such that $f(U_{x'})=f(U'_x)$ for some open $U'_x \subset U_x$ around $x$. For large $k$ the points $x_k$ are in $U_{x'}$ so that $y_k \in f(U_{x'})=f(U'_x) \subset f(U_x)$. This is a contradiction which proves that we indeed have a $W_y$ and that $f(X)$ is a submanifold.  
