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All spaces below will be assumed simply connected. A continuous map is a rational equivalence if it induces an isomorphism of the rational homology groups. Two spaces are rationally equivalent if they can be connected by a sequence of rational equivalences (not necessarily all going in the same direction). Now suppose $X$ and $Y$ are rationally equivalent smooth compact manifolds without boundary. Can one choose a sequence of rational equivalences from $X$ to $Y$ that only passes through smooth compact manifolds of dimension $\dim X=\dim Y$?

Note: if $X$ and $Y$ are CW-complexes, or more generally, compactly generated Hausdorff spaces, then $X$ is rationally equivalent to $Y$ if and only if the $\mathbb{Q}$-localizations $X_\mathbb{Q}$ and $Y_\mathbb{Q}$ are homotopy equivalent, see e.g. Felix, Halperin, Thomas, Rational homotopy theory, Proposition 9.8. If this is the case, then there is a roof $X\to X_\mathbb{Q}\gets Y$ of rational equivalences. The middle of the roof is of course an infinite complex (except when it is a point).

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    $\begingroup$ I think there's no way to connect $S^2 \times S^3$ with the unique other 5-mfd with $H_2 = \Bbb Z$ (double of nontrivial $D^3$-bundle over $S^2$) through degree 1 maps because one has trivial $w_2$ and other one does not. They are rationally equivalent, since they are both formal. $\endgroup$
    – Denis T
    Commented Aug 30, 2022 at 0:05
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    $\begingroup$ Denis: this is a good example, but rational equivalences may have degree $\neq 1$. For example, any map $S^n\to S^n$ of degree $\neq 0$ is a rational equivalence. $\endgroup$
    – algori
    Commented Aug 30, 2022 at 0:19
  • $\begingroup$ .. also, the tangent Stiefel-Whitney classes are preserved under genuine homotopy equivalences of manifolds, but not in general under rational ones. $\endgroup$
    – algori
    Commented Aug 30, 2022 at 0:30
  • $\begingroup$ Well, yeah, that's why I'm writing it as a comment and not an answer, but I think my non-example can be promoted. By methods of Manuel Amann I think one can produce "sturdy" PD rational types, i. e. ones without self-maps of degree $\neq 1$ and retaining that property under connected sums. Then one can use some secondary obstructions (i. e. homology classes not realizable as manifolds) to distinguish two realisations of that type. Sturdiness will prevent positive degree things from killing our obstructions, which will forbid any nonzero degree maps in either direction. $\endgroup$
    – Denis T
    Commented Aug 30, 2022 at 1:37
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    $\begingroup$ To add to Denis T's comment, if the rational homotopy type in question is formal, then $X$ admits endomorphisms of arbitrarily divisible degree (see Infinitesimal Computations in Topology Theorem 12.2, and F. Manin's "Positive weights and self-maps" for a detailed discussion, Corollary 1.1 arxiv.org/abs/2108.02173v3) and so, upon precomposing with an appropriate endomorphism one can lift the map $X \to X_{Q}$ through $Y \to X_{Q}$ to get a direct rational equivalence $X \to Y$. $\endgroup$ Commented Aug 30, 2022 at 6:02

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