Path in GL(V) induces a proper homotopy

Question

Let $V$ be a finite dimensional real vector space, and let $GL(V)$ be the group of its linear automorphisms. Is it true that any homotopy $h\colon V\times I\rightarrow V$ with the property that $h(-, t) \in GL(V)$ for any $t\in I$ (hence corresponding to a path $I\rightarrow GL(V)$) is proper, i.e. for any compact $K\subset V$, $h^{-1}(K)$ is compact?

Answer 1

First give $V$ an inner product so that it is a metric space, and in particular since $V$ is a finite-dimensional real vector space by the Heine-Borel theorem a subset is compact iff it is closed and bounded; this is the criteria we will use for proving compactness. Notice that every element of $GL(V)$ is proper, since they are homeomorphisms (in particular they each have a well-defined and continuous inverse function).

Now give $V \times I$ the product metric (which also has the Heine-Borel property), and consider $h\colon V\times I \to V$ such that $h_t \in GL(V)$ for each $t$. If $K\subset V$ is compact then by continuity $h^{-1} (K)$ is closed; we just need to verify that it is bounded in $V\times I$, but we know already $h_t^{-1}(K)$ is bounded for every $t$.

Suppose for the sake of contradiction that $\tilde{K} = h^{-1}(K)$ is not bounded, and suppose we choose a sequence of points $\{x_i\}\subset \tilde{K}$ such that $|x_i| \to \infty$. By projecting to the $I$ factor this gives a sequence $\{t_i\}\subset I$ which has a convergent subsequence $\{\tilde{t}_i\} \to t_{\omega}\in I$ by Bolzano-Weierstrass and compactness of $I$.

Exercise: show $h_{t_\omega}^{-1}(K)$ is not bounded.

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Edit: note that this phenomenon is not true for an arbitrary homotopy $h\colon X\times I \to Y$ through proper maps, OP provided a link to a counter-example in the comments, which is taken from this paper.