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...



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


