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.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.