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Given a finite dimensional finite $CW$ complex $X$ of dimension $d$, I want to build a compact manifold $M$ (with least dimension possible) with boundary with the property that,

  1. $M$ has the same homotopy type as $X$.
  2. $M$ is inductively built and each stage is a compact manifold with boundary.
  3. When we go to the next stage from the previous one, there is exactly one critical point of index $\leq d$.

Point (2) can be done using a theorem related to elementary cobordism which I found in "Lectures on the h-Cobordism theorem" (theorem 3.12) by Milnor. But (1), I don't know how to start the induction process and wish to understand the mechanism.

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    $\begingroup$ What does $M$ have to do with the CW complex? $\endgroup$
    – Bma
    Commented Dec 13, 2021 at 3:44
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    $\begingroup$ Assuming you want the manifold to have the homotopy-type of the CW-complex, you would do it by performing handle attachments corresponding to your cellular attachments, and building the corresponding Morse function that has that handle attachment. $\endgroup$
    – Ryan Budney
    Commented Dec 13, 2021 at 5:12
  • $\begingroup$ @RyanBudney while constructing the handle decomposition, corresponding to each skeleton $X_0, X_1, X_2$ (suppose the CW has only three skeleton) how to start the induction? (For an example take $D^2$ attached to $S^1$ via degree 2 map) If I call the corresponding manifolds $M_0, M_1, M_2$ so that $M=\cup_{i=0}^2 M_i$. can you elaborate for this example? $\endgroup$
    – piper1967
    Commented Dec 13, 2021 at 5:58
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    $\begingroup$ I'm a little confused by what you mean by "start the induction". Isn't every handle attachment an inductive step? What do you want to induct on? $\endgroup$
    – Ryan Budney
    Commented Dec 13, 2021 at 6:06
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    $\begingroup$ Not sure but for that example, if you embed $D^2$ into $\mathbb R^4$ in such a way that $S^1$ is its intersection with a hyperplane $\mathbb R^3\subset\mathbb R^4$, then you can take $M_1$ a tubular neighborhood of $S^1$ in $R^3$ which is a solid torus $T$, and $M^2$ a tubular neighborhood of $D^2$ in $\mathbb R^4$ whose intersection with $\mathbb R^3$ is $T$. $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Dec 13, 2021 at 6:10

1 Answer 1

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There is a more 'geometric' construction that does not use induction and gives you the correct dimension bound. A rough sketch of how this works follows.

If $\text{dim}(X) = d$, then there exists a local embedding $X \rightarrow \mathbb R^{2d}$ with only finitely many self-intersection points all of which lie in top cells of $X$. Taking a 'tubular neighbourhood' (one has to make sense of this appropriately) of this embedding gives you a $2d$-manifold with boundary $M$ that is homotopy equivalent to $X$. Now pick any self-indexing Morse $\phi$ function on $M$ and consider the stages $\varphi^{-1}(i)$, $0 \leq i \leq 2d-1$ of building $M$ using $\varphi$.

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    $\begingroup$ @JensReinhold by tubular neighbourhood in the above answer you meant the closed tubular neighbourhood right? $\endgroup$
    – piper1967
    Commented Dec 14, 2021 at 2:33

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