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Suppose $X$ and $Y$ are two $CW$ complexes and $f:X\rightarrow Y$ is a continuous surjection such that fiber of each point (i.e. $f^{-1}(y)$ for each $y\in Y$) is contractible. Does it implies that $X$ and $Y$ are homotopy equivalent.

PS-1:By Whitehead's Theorem it will be enough to show that $f$ induces an isomorphism between all homotopy groups.

PS-2:In question Equivariant Cohomology for actions with finite stabilizers there are some discussion regarding the above question but in terms of homology. If anybody thinks that my question can be a consequence of this discussion please explain the connection.

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    $\begingroup$ Let $Y=[0,1]$, $X=[0,1]^\delta$, the same set with the discrete topology, and $f$ be the identity function. This data satisfies the hypotheses, but is not a homotopy equivalence. $\endgroup$
    – Oscar Randal-Williams
    Commented Apr 3, 2013 at 19:02
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    $\begingroup$ @Anton: A discrete topological space can be made (easily) into a CW-complex, with only $0$-cells. So maybe one wants to add to the question that $f$ is cellular, so that the two CW-structures have some relationship via $f$. Is the Leray spectral sequence relevant? I need to look it up! $\endgroup$
    – Ronnie Brown
    Commented Apr 3, 2013 at 20:48
  • $\begingroup$ In many cases f happens to be a quasifibration and then contractibility of the fiber implies f is a weak homotopy equivalence. Conditions under which f is a quasifibration are to be found in Dold-Thom "Quasifaserungen und Symmetrische Produkte". $\endgroup$
    – ThiKu
    Commented Apr 7, 2013 at 15:20

1 Answer 1

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In his paper

MR0087106 (19,302f) Smale, Stephen A Vietoris mapping theorem for homotopy. Proc. Amer. Math. Soc. 8 (1957), 604–610.

Smale proved the following theorem:

Theorem : Let $X$ and $Y$ be connected, locally compact separable metric spaces. Assume also that $X$ is locally contractible. Consider a proper surjective continuous map $f : X \rightarrow Y$. Assume that for all $y \in Y$, the space $f^{-1}(y)$ is contractible and locally contractible. Then $f$ is a weak homotopy equivalence.

To see how this fits into your situation, remember that (for instance) finite CW complexes are locally compact and locally contractible. So you need to impose conditions on the fibers to ensure that they are also locally contractible.

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  • $\begingroup$ I just want to add a small, tangential refinement. In Andy's conclusion for finite dimensional complexes, we can actually do a bit better. If the fibres of a map between finite CW-complexes are all contractible and locally contractible, then the map is actually a "cell-like map", and thus a simple homotopy equivalence. $\endgroup$
    – Ricardo Andrade
    Commented Apr 3, 2013 at 22:13
  • $\begingroup$ Andy: this is how I have seen (and stated) this theorem also, but does one really need to say "surjective"? The empty set does not have the homotopy type of a point, does it? $\endgroup$
    – Vidit Nanda
    Commented Apr 4, 2013 at 3:02
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    $\begingroup$ @Vel Nias : It depends on your conventions. All right-thinking people consider the empty set non-connected (and certainly not contractible), but just in case the OP has other ideas I thought I'd avoid this issue by assuming surjectivity. $\endgroup$
    – Andy Putman
    Commented Apr 4, 2013 at 3:31
  • $\begingroup$ @Andy: Thanks. I am a little curious about the property that the map $f$ should proper. How important this condition really is? What I mean is can I replace this condition by a more simple condition like taking $f$ to be cellular or something of that sort. $\endgroup$
    – Cusp
    Commented Apr 4, 2013 at 3:47
  • $\begingroup$ Properness is certainly needed; it is not hard to construct counterexamples without it. I do not know whether assuming that $f$ is cellular is good enough. $\endgroup$
    – Andy Putman
    Commented Apr 4, 2013 at 4:01

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