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](https://mathoverflow.net/questions/120351/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. 



Homotopy equivalence from contractibility of fiber
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. 


