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Suppose $X$ is a CW-complex. The monoid of homotopy self-equivalences $M = hAut(X)$ is the subspace of $Map(X,X)$ consisting of those maps with a homotopy inverse. It is a union of path components. It obviously acts on $X$, and the homotopy type only depends on the homotopy type of $X$.

It is known that we can find maps $G \leftarrow M' \to M$ of topological monoids, all homotopy equivalences, with $G$ a topological group.

It is also known that we can use this to find a $G$-space $Y$ and maps of $M'$-spaces $Y \leftarrow X' \to X$ which are all homotopy equivalence. In other words, this rigidifies the action of $hAut(X)$ to an honest action of a topological group.

However, even in this situation we have a composite map $G \to Aut(Y) \to hAut(Y)$ that we know is a homotopy equivalence, but it is unlikely to be the case that $Aut(Y)$ is homotopy equivalent to $hAut(Y)$.

Can we rigidify this and find an $Y$ whose automorphism group is equivalent to its homotopy automorphism group? Or does there exist a space for which the map $Aut(Y) \to hAut(Y)$ is never a homotopy equivalence for any space homotopy equivalent to $X$?

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Just a naive remark: on the "yes" side one has it in special cases: if $X$ is an oriented surface for example. Furthermore, rigidity theorems such as the Mostow rigidity give it on $\pi_0$ for $X$ in the class of examples for which the rigidity theorem holds (such as hyperbolic manifolds). – John Klein Jan 23 '11 at 6:27
This is related to an aside from one of JK's drafts: if $Q$ is the Hilbert cube and $X$ is finite, then the homeomorphism type of $X\times Q$ is the same information as the simple homotopy type of $X$. If $X$ is simply connected, then $Aut(X\times Q)\to hAut(X)$ is 1-connected, but far from an equivalence. If there is torsion in the fundamental group, there may be self-equivalences with non-trivial Whitehead torsion, which thus never lift to homeomorphisms of compact models. I suspect that if you also cross by $R^n$, there are related obstructions for higher $\pi_k$. Maybe $R^\infty$ helps. – Ben Wieland Jan 25 '11 at 4:12
@Ben: Another interesting sequence associated with the finite complex $Y$ is gotten by choosing a homotopy equivalence $Y \simeq M$, where the latter is a codimension zero open submanifold of some Euclidean space. Then in a range we have a fibration sequence $F(M,\text{Top}_n) \to \text{TOP}(M) \to G(M)$. The range tends to infinity when we stabilize $M$ by taking iterated products with the reals. – John Klein Jan 26 '11 at 1:25
up vote 4 down vote accepted

This is a substantial revision of my original post. It shows that if we replace the "equivalence" Tyler is asking for by a "retract" then the answer is yes.

  1. Given a CW space $Y$, we can take $G(Y) =$ the topological monoid of homotopy automorphisms of $Y$. The Borel construction $$ EG(Y) \times_{G(Y)} Y \to BG(Y) $$ is then a quasifibration. Let $U \to BG(Y)$ be the effect of converting it into a fibration.

  2. Let $G$ be a topological group with a chosen homotopy equivalence $$BG\simeq BG(Y). $$ For example, we can do what Tyler does, or we can simply take $\Omega BG(Y)$, where this means the realized Kan loop of the total singular complex of $BG(Y)$.

  3. Let $EG \to BG$ be a universal $G$-principal bundle, and set $$ Z \quad := \quad \text{pullback}(EG \to BG \simeq BG(Y) \leftarrow U) $$ Then $Z \subset EG \times U$ inherits a $G$-action and its underlying homotopy type is that of $Y$. Then the Borel construction $$ EG\times_G Z \to BG $$ is a fiber bundle which is weak fiber homotopy equivalent to $U \to BG(Y)$.

  4. Step 3 implies that $BG(Y)$ is a retract up to homotopy of $B\text{homeo}(Z)$. This will imply that $G(Y)$ is a homotopy retract of $\text{homeo}(Z)$ in the $A_\infty$ sense, with $Z \simeq Y$.

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OK. So it seems that the hard question is really whether one can make alterations to the "kernel" group of homeomorphisms equipped with homotopies to the identity, even for finite CW objects. – Tyler Lawson Jan 25 '11 at 3:23

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