# isotopy equivalence (topological meaning) between $CW$-complexes

Let $M$ and $N$ be $CW$-complexes.

Definition. (different from the isotopy notion in geometry of submanifolds). A (topological) isotopy is a fibre-wise continuous map $$F: M\times [0,1]\longrightarrow N\times [0,1]$$ such that $F$ maps the fibre $M\times t$ homeomorphically onto a subset of the fibre $N\times t$ for each $t\in [0,1]$. Two injective continuous maps $$f_1,f_2: M\longrightarrow N$$ are said isotopy equivalent, if there exists an isotopy $$F: M\times [0,1]\longrightarrow N\times [0,1]$$ such that $$f_1=F(\cdot,0),\\ f_2=F(\cdot,1).$$ The $CW$-complexes $M$, $N$ are said isotopy equivalent, if there exist injective continuous maps $$f: M\longrightarrow N,\\ g: N\longrightarrow M$$ such that their compositions $fg$ is isotopy equivalent to $Id_N$ and $gf$ is isotopy equivalent to $Id_M$.

Question. Suppose $M$ is a compact manifold without boundary. Does there exist any non-compact $CW$-complex $N$ such that $M$ is isotopy equivalent to $N$? I cannot figure out any such example...

Since $M$ is a closed manifold and $gf$ is homotopic to the identity, it must be surjective, otherwise it wouldn't preserve the fundamental class mod 2. In particular $g$ is surjective. Your isotopy condition implies that $fg$ is a homeomorphism onto the image, which coincides with the image of $f$ since $g$ is onto. The image of $f$ is compact since $M$ is compact, so $N$ should be compact too.
The isotopy condition also implies that $gf$ is injective, so $f$ is injective too. Since $M$ is compact and $N$ is Hausdorff, $f$ is closed and hence a homeomorphism onto the image. we have seen that the image of $f$ coincides with the image of $fg$. Therefore we conclude that $N$ and $M$ are homeomorphic.