Smooth homotopy of proper maps

Question

Suppose $M$ is a manifold. We call a map $F:M \rightarrow \mathbb{R}^m$ proper if for every compact subset $K \subset \mathbb{R}^m$ we have that $F^{-1}(K) \subset M$ is compact.
Suppose now that $F: M \times [0,T) \rightarrow \mathbb{R}^m$ is a smooth family of immersions such that $F(\cdot ,0)$ is proper. Is it true that $\forall t \in [0,T): F(\cdot, t)$ is proper?

Answer 1

This is false as stated. For an (analytic) counterexample, take $M = \mathbb{R}$ and define $F:\mathbb{R} \times [0,\pi] \to \mathbb{R}$ via

$$F(x,t):= (1-\sin(t))x + \sin(t) \arctan(x).$$

For any fixed $t\in [0,\pi]$, $F(\cdot,t)$ is an immersion because the maximum slope of $\arctan$ is $1$. Clearly $F(\cdot,0)$ is proper. But $F(\cdot,\pi/2) = \arctan$, which is bounded and therefore not proper.

The conjecture is still false if $m > \text{dim}(M)$. As a counterexample, take $M= \mathbb{R}$ again, let $F$ be as above and define the map $G: \mathbb{R}\times [0,\pi]\to\mathbb{R}^m$ via

$$ G(x,t):= (F(x,t),\ldots,F(x,t)).$$

$G(\cdot,t)$ is an immersion for any $t \in [0,\pi]$ because $F(\cdot,t)$ is, and $G(\cdot,0)$ is proper, but $G(\cdot,\pi/2)$ is not proper because it is given by the product of $\arctan$ functions.