If two proper maps are homotopic, then they are properly homotopic

Question

Let $M$ and $N$ be smooth manifolds, and let $f$ and $g$ be proper smooth maps $M\to N$ such that there is a smooth map $F:\mathbb{R}\times M \to N$ with $F_0=f$ and $F_1=g$. Is there a proper smooth map $G:\mathbb{R}\times M \to N$ with $G_0=f$ and $G_1=g$?

If needed, we may assume $M$ and $N$ oriented and without boundary. If this result is much more general, please tell me.

I know for instance that, in the category of topological spaces and continuous maps, $F_t$ proper for all $t$ does not imply $F$ proper, but here we are considering the existence of a proper homotopy.

I need this result in order to know if I can pullback a compactly supported form along an homotopy $F$ and still get a compactly supported form, for instance in arguments using Stokes' Theorem.

Answer 1

Here is another example.

Let $M = N = \mathbb R$ and $f(x) = x, g(x)=-x$. Then

$$F : \mathbb R \times \mathbb R \to \mathbb R, F(t,x) = (1-2t)x$$ is a smooth homotopy from $f$ to $g$. However, there is no proper map $H : [0,1] \times \mathbb R \to \mathbb R$ such that $H_0 = f$ and $H_1 = g$, and this prevents the existence of a proper smooth homotopy $G : \mathbb R \times \mathbb R \to \mathbb R$ from $f$ to $g$.

To see this, assume there is a proper map $H : [0,1] \times \mathbb R \to \mathbb R$ such that $H_0 = f$ and $H_1 = g$, This map has a unique continuous extension $\bar H : X \to Y$ from the one-point compactification $X$ of $[0,1] \times \mathbb R$ to the one-point compactification $Y$ of $\mathbb R$, where $\bar H(\infty) = \infty$. See for example Is one point compactification functorial?

The map $$p : [0,1] \times Y \to X, p(t,\xi) = \begin{cases} \xi & \xi \in \mathbb R \\ \infty & \xi = \infty & \end{cases} $$ is easily seen to be continuous. Hence $\bar H \circ p : [0,1] \times Y \to Y$ is a homotopy.

Stereographic projection $s : S^1 \setminus \{(1,0)\} \to \mathbb R$ from $(1,0)$ to the $y$-axis is a homeomorphism, and it extends to a homeomorphism $\bar s : S^1 \to Y$ such that $\bar s(1,0) = \infty$. This provides a homotopy $$\phi : [0,1] \times S^1 \xrightarrow{id_{[0,1]} \times \bar s} [0,1] \times Y \xrightarrow{\bar H \circ p} Y \xrightarrow{\bar s^{-1}} S^1$$ such that $\phi_0(a,b) = (a,b)$ and $\phi_1(a,b) = (a,-b)$. But $\phi_0 = id$ has degree $1$ and $\phi_1$ has degree $-1$, thus they cannot be homotopic.

Answer 2

This is not true, take $M=\mathbb{R}$, $N=\mathbb{R}^2 \setminus (\mathbb{Z}\times \{0\})$, $f(x)=(x,-1)$, $g(x)=(x,1)$. Of course there is an homotopy between $f$ and $g$ since $\mathbb{R}$ is contractible, and it can always be upgraded to a smooth homotopy. For any homotopy $G:[0,1]\times \mathbb{R} \to N$, there must be a gate $I$ in $(\mathbb{R}\setminus \mathbb{Z})\times\{0\}$ such that $G([0,1]\times\{x\})$ pass through it for all $x\in \mathbb{R}$ (by considering the fundamental group of $\mathbb{R}^2$ minus some points). Take a compact annulus $K$ in $N$ that goes around the gate $I$ and does not intersect the images of $f$ and $g$. Therefore $K$ intersects $G([0,1]\times\{x\})$ for all $x \in \mathbb{R}$, and so $G^{-1}(K)$ is not compact (since it is not bounded in $[0,1]\times \mathbb{R}$).