Let us consider a category $C$ formed by topological spaces and continuous functions (or by smooth manifolds and smooth functions). We consider the morphism category $C_{2}$. An object of $C_{2}$ is a morphism $\rho \colon A \to X$ in $C$. A morphism between $\rho \colon A \to X$ and $\eta \colon B \to Y$ is a pair of continuous functions $(f \colon X \to Y, g \colon A \to B)$ such that $f \circ \rho = \eta \circ g$.

Moreover, given two morphisms $(f_{0}, g_{0}), (f_{1}, g_{1}) \colon \rho \to \eta$, a homotopy between them is defined in the following way. Let us consider the object $\rho_{I} := \rho \times id_{I} \colon A \times I \to X \times I$. A morphism $(F, G) \colon \rho_{I} \to \eta$ is a homotopy between $(f_{0}, g_{0})$ and $(f_{1}, g_{1})$ if $F\vert_{X \times \{i\}} = f_{i}$ and $G\vert_{A \times \{i\}} = g_{i}$ for $i = 0, 1$. Finally, a homotopy equivalence between $\rho \colon A \to X$ and $\eta \colon B \to Y$ is a morphism $(f, g) \colon \rho \to \eta$ such that there exists $(f', g') \colon \eta \to \rho$ verifying $(f', g') \circ (f, g) \simeq id_{\rho}$ and $(f, g) \circ (f', g') \simeq id_{\eta}$.

My question is the following. Let us consider the category $M$ of smooth manifolds and the category $CW$ of spaces having the same homotopy type of a countable finite-dimensional CW-complex. It is known that, given an object $X$ of $CW$, there exist an object $\hat{X}$ of $M$ and a homotopy equivalence $\xi \colon X \to \hat{X}$ (a proof is shown in Kreck, Singhof, "Homology and cohomology theories on manifolds", section 2, fact 2). Does the same result hold about the categories $M_{2}$ and $CW_{2}$? Explicitly, given a continuous function $\rho \colon A \to X$ between objects of $CW$, can we find a smooth function $\hat{\rho} \colon \hat{A} \to \hat{X}$ between manifolds and a homotopy equivalence $(f, g) \colon \rho \to \hat{\rho}$ in $CW_{2}$?

  • $\begingroup$ That's interesting, but when you take a homotoy pull-back the diagram commutes up to homotopy, thus it is not a morphism in CW2. I don't see how to reach a morphism. $\endgroup$ – Fabio Jul 16 '16 at 20:52
  • $\begingroup$ Oh sorry I was thinking to something else... I have removed the comment $\endgroup$ – Simon Henry Jul 17 '16 at 13:09
  • $\begingroup$ Does there (for a given object $\rho$) always exist a morphism $\rho\to\hat{\rho}$ to a "smooth object" $\hat{\rho}$ such that $A\to\hat{A}$ and $X\to\hat{X}$ are homotopy equivalences? Same question for the other direction? $\endgroup$ – user83633 Jul 18 '16 at 7:16

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