$f:[0,1]\times X\to X$ continuous is proper if $f(t,\cdot)$ is homeomorphism for any $t\in[0,1] $ Let $f:[0,1]\times X\to X$ a continuous map  such that for any $t\in [0,1]$, $f(t,\cdot):X\to X$ is an homeomorphism.

Is $f$ a proper map ? I recall that proper means that  for any $K\subset X$, $f^{-1}(K)$ is compact.

Notice that  if instead of $[0,1]$ we consider $[0,+\infty)$, then the answer is no, as we can use as counterexample $f:[0,\infty)\times \mathbb{R} \to \mathbb{R}$, $f(t,x) = x-t$ so that the fiber is $f^{-1}\{0\} = \{(t,t)\in [0,\infty)\times \mathbb{R}\}$ which is obviously non compact.
 A: As requested, here is a proof in the manifold case. The key is that if $Y, Z$ are $n$-dimensional manifolds with boundary and $h: (Y,\partial Y)\to (Z,\partial Z)$ is a continuous bijection, then $h$ is open, hence, a homeomorphism. (This is just the invariance of domain theorem for manifolds with boundary.)
Now, set $I:= [0,1]$. Define a map
$$
g: Y=I\times X\to Y
$$
by $g(t,x)=(t, f(x))$. Clearly, $g$ is again continuous and bijective. Since $X$ is a manifold, $Y$ is a manifold with boundary. The map $g$ sends boundary to boundary. Hence, $g$ is a homeomorphism. For each compact $K\subset X$, $L=I\times K$. Then
$g^{-1}(L)=f^{-1}(K)$ is compact in $Y$ since $g^{-1}$ is continuous.
The same proof works if $X$ is a manifold with boundary, you just have to realize that $g$ still sends boundary to boundary since
$$
\partial Y= \partial I\times X \cup I\times \partial X. 
$$
Since each $f(t, \cdot)$ is a homeomorphism, it sends $\partial X$ to $\partial X$. From this, you see that $g$ sends boundary to boundary.
