Is a smooth surjective map $f:X\to Y$ with connected, compact fibers always closed? Let $X$ and $Y$ be smooth manifolds and let $f:X\to Y$ be a smooth surjective submersion, such that the fibers of $f$ are connected and compact. Is it true that $f$ is necessarily closed?
I am trying to find a proof of this fact, but I am not even sure if it is true. I need it because I want to check that my map is proper by only checking that the fibers are compact.
 A: Note that $X, Y$ are Hausdorff and locally compact. Thus, verifying properness of $f$ is equivalent to verifying closeness of $f$ (provided $f$ has compact fibers, as you are assuming). I will prove that $f$ is proper under a weaker set of assumptions than in your question:

*

*$X, Y$ are metrizable, $X$ is locally compact; in particular, every compact $K\subset X$ is contained in the interior of another compact $K'\subset X$.


*$f: X\to Y$ is surjective, continuous, with connected and compact fibers (point-preimages).


*$f$ is open, i.e. sends open sets to open sets. (This is a consequence of the assumption  that your map is a submersion.) Open maps $f: X\to Y$ of metrizable spaces have the following useful property: For every $x\in X, y=f(x)$, and a sequence $y_n\in Y$ converging to $y$, there exists a sequence $x_n\in X$ converging to $x$ such that $f(x_n)=y_n$.
Theorem. Under the assumptions 1, 2, 3 above, the map $f$ is proper.
Proof. Suppose that $f$ is not proper. Then there is a sequence $y_n\in Y$ converging to some $y\in Y$ and a sequence $x_n\in f^{-1}(y_n)$ such that $(x_n)$ diverges to infinity in the sense that for every compact $K\subset X$ all but finitely many members of the sequence $(x_n)$ do not belong to $K$.
Since $f$ is surjective, there exists $x\in f^{-1}(y)$. By the openness of $f$, there exists $N\in {\mathbb N}$ and a sequence $x'_n\in f^{-1}(y_n), n\ge N$, converging to $x$.  Let $K$ be a compact whose interior contains
$f^{-1}(y)\cup \{x'_n: n\ge N\}$. (I am using compactness of $f^{-1}(y)$ and local compactness of $X$ here.) By taking a larger $N$, we can assume that $x_n\notin K$ for all $n\ge N$. Since $f^{-1}(x_n)$ is connected, for each $n$ there exists $z_n\in f^{-1}(y_n)\cap \partial K$. Since $\partial K$ is compact, after passing to a subsequence, we obtain $z_n\to z\in \partial K$. By continuity of $f$, $f(z)=y$, i.e. $z\in \partial K\cap f^{-1}(y)$. This is a contradiction since $K$ was chosen so that $f^{-1}(y)\subset int(K)$. qed
Lastly, the condition that fibers of $f$ are connected is clearly needed in your question, otherwise, for instance, the map
$$
f: (0, 4\pi)\to S^1, f(t)=\exp(it)
$$
is a counter-example.
