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Let $M$ be a connected open topological $d$-manifold (without boundary).

Whitehead showed that if $M$ has a PL structure, there exists a subcomplex of dimension $\leq d-1$ onto which $M$ deformation retracts.

Can we still find a homotopy equivalent CW complex of dimension $\leq d-1$ when $M$ is not PL?

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    $\begingroup$ I think this is true. If $X$ is homotopy equivalent to a (connected) CW complex and $H^i(X;M)=0$ for all $\pi_1X$-modules $M$ and all $i>n$, then $X$ is a homotopy equivalent to a CW complex of dimension $\leq n$, provided $n\geq3$. The cases $n=1,2$ can be sorted out by hand (although I only know how to do $n=2$ under some basic finiteness assumptions). Bearing in mind that I am quoting 20+ year old literature, so maybe there is no problem at all. $\endgroup$ – Tyrone Jul 11 '20 at 13:25
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    $\begingroup$ @Tyrone Manifolds of dimension $\leq 3$ always have smooth, hence PL structures, so those are certainly okay. $\endgroup$ – Cihan Jul 11 '20 at 13:31
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    $\begingroup$ It is explained in mathoverflow.net/questions/201944/… that a topological $n$-manifold has homotopy type of an $n$-dimensional complex of dimension $\le n$. It remains to exclude dimension $n$ if the manifold is open. $\endgroup$ – Igor Belegradek Jul 11 '20 at 13:32
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    $\begingroup$ The proof for $\le n$ should be generalizable to $<n$, i.e. one has to check that the universal covering of your open $n$-manifold has zero cohomology in dimensions $\ge n$ with any local coefficients. $\endgroup$ – Igor Belegradek Jul 11 '20 at 13:38
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    $\begingroup$ A universal cover has no nontrivial local systems. I think you mean the cohomology of the actual manifold. It holds by Poincaré duality. $\endgroup$ – archipelago Jul 11 '20 at 22:08
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$\DeclareMathOperator{\co}{H}$ $\DeclareMathOperator{\ch}{C}$ $\newcommand{\zz}{\mathbb{Z}}$ $\newcommand{\nn}{\mathbb{N}}$ $\newcommand{\A}{\mathcal{A}}$ $\newcommand{\B}{\mathcal{B}}$ $\DeclareMathOperator{\lf}{lf}$

Let me put together an answer following the pointers in the comments. By Whitehead's result stated in the question and smoothability in lower dimensions we may assume $d \geq 4$. Write $\pi := \pi_1(M)$ for brevity.

As an ANR, M has the homotopy type of a CW-complex, so by a result of Wall it suffices to show that $\co^{j}(M; \A) = 0$ whenever $j \geq d$ and $\A$ is a $\zz \pi$-module. Writing $w$ for the orientation $\zz \pi$-module, by Poincaré duality we have $$\co^j(M;\A) \cong \co^{\lf}_{d-j}(M; \A \otimes_{\zz} w)\,,$$ where $\co^{\lf}_{*}$ denotes the locally finite singular homology (sometimes called the Borel-Moore homology). Therefore the only nontrivial thing to check is the vanishing of the 0-th locally finite homology for every $\zz \pi$-module $\B$. Writing $\tilde{M}$ for the universal cover of $M$, this amounts to showing that the first differential $$\partial_1 \otimes_{\zz\pi} \B \colon \ch^{\lf}_1(\tilde{M}) \otimes_{\zz\pi} \B \rightarrow \ch^{\lf}_0(\tilde{M}) \otimes_{\zz\pi} \B$$ of the locally finite singular chain complex is surjective, for which $\partial_1$ being surjective before tensoring is enough. We can verify $\partial_1$ is surjective by elementary means: Fix a locally finite singular 0-chain $\sigma$; it is necessarily supported on a discrete subset of $\tilde{M}$. Since $\tilde{M}$ is non-compact (and second-countable), we can find a countably infinite discrete subset $$\{x_n : n \in \nn\} \subseteq\tilde{M}$$ which contains the support of $\sigma$. Thus $\sigma$ is a formal sum of the form $$\sigma = \sum_{n \in \nn}a_n x_n$$ with $a_n \in \zz$. Now for each $n \in \nn$ we can find a path $\gamma_n : [0,1] \rightarrow M$ connecting $x_{n+1}$ to $x_{n}$ such that the formal sum $$\tau := \sum_{n \in \nn}b_n \gamma_n$$ with the coefficients $b_n := \sum_{j \leq n} a_j$, is a locally finite $1$-chain with $\partial_1(\tau) = \sigma$.

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