Let $h_t:X\to Y$ be a homotopy, we assume both spaces are locally compact and hausdorff, and each $h_t$ is proper, when can we lift it to a homotopy of one point compactification $\bar X\to \bar Y$? I think this is true when $h$ is a isotopy, or at least when $h_t$ is a homeomorphism at each $t$?
Besides, if $X,Y$ are smooth manifolds, do we have different story in smooth category?
Remark: the motivation of this question is isotopy class of knots in $\mathbb R^3$ is same as in 3 sphere.
Let $Z$ be a locally compact Hausdorff topological space. Recall that the 1-point compactification of $Z$, denoted $Z^+$ or $Z\cup \infty$ is the disjoint union of $Z$ and a singleton $\{\infty\}$, where the neighborhoods of $\infty$ are complements to compact subsets in $Z$, while $Z$ is topologically embedded in $Z^+$. A sequence in $Z$ is said to be divergent if it converges to $\infty$ in $Z^+$.
A (continuous) map of two Hausdorff spaces is called proper if the preimage of every compact is again compact. A continuous map $f: X\to Y$ of two locally compact Hausdorff spaces is proper if and only if it extends to a continuous map $f: X^+\to Y^+$ by sending $X^+\setminus X$ to $Y^+\setminus Y$. A homotopy $H: X\times I\to Y$ is said to be proper if it is a proper map. Suppose that $X, Y$ are manifolds. Then it is easy to see that a homotopy $H: X\times I\to Y$ is proper if and only if the following holds:
For every sequence $t_n\to t_0\in I=[0,1]$ and every divergent sequence $x_n\in X$, the sequence $H(x_n, t_n)$ in $Y$ is also divergent.
Lemma. Suppose that $X, Y$ are topological manifolds and $H: X\times I\to Y$ is an isotopy, i.e. is a homotopy such that for each $t\in I$, $H(\cdot, t): X\to Y$ is a homeomorphism. Then $H$ is a proper isotopy.
Proof. Suppose that $H$ is not proper. Then, there exists a sequence $t_n\to t_0\in I=[0,1]$ and a divergent sequence $x_n\in X$ such that the sequence $y_n=H(x_n, t_n)$ converges to some $y\in Y$. Since $h= H(\cdot, t_0): X\to Y$ is a homeomorphism, there exists $x\in X$ such that $h(x)=y$. Let $B\subset X$ be a closed ball whose interior contains $x$. Then $h(int B)$ contains $y$. Let $W\subset h(int B)$ be a compact whose interior contains $y$. Now, I will need some Degree Theory. We get, $deg(h, B, w)=1$ for all $w\in W$. By the degree theory, for all sufficiently large $n$ and all $w\in W$, $deg(h_{t_n}, B, w)=1$ as well, where $h_{t_n}= H(\cdot, t_n)$. Hence, for all large $n$ there exists $z_n\in B$ such that $h_{t_n}(z_n)=y_n$. But we also have $h_n(x_n)=y_n$. Since the sequence $(x_n)$ diverges while the sequence $(z_n)$ is in the compact $B$, we get that $x_n\ne z_n$ for all large $n$. This contradicts injectivity of $h_{t_n}$. qed
Note that if we have a homotopy $H: {\mathbb R}^n\times I\to {\mathbb R}^n$ through proper maps $H(\cdot, t)$, it does not follow that $H$ is proper, see the example given in this answer.
Lastly, for diffeomorphisms $f: {\mathbb R}^n\to {\mathbb R}^n, f(0)=0$, there is an easy construction of a proper smooth isotopy of $f$ to a linear map (through diffeomorphisms): $$ H(x,t)= \frac{1}{t} f(tx), t\in (0,1], $$ $$ H(x,0)=L(x), $$ where $L$ is the differential of $f$ at $0$. In particular, if $f$ is orientation-preserving, then $f$ is smoothly properly isotopic to the identity map. As noted above, this isotopy extends to a topological isotopy $S^n\times I\to I$ to the identity map. In general, there is no smooth isotopy $S^n\times I\to I$ of an orientation-preserving diffeomorphism $f$ to the identity map (for $n\ge 6$). These are Milnor's famous examples.