Recall that a chain complex $(C\_\*,d)$ of abelian groups is contractible if it is homotopic to the zero map. Or equivalently: there exists a degree 1 map $F: C\_\* \to C\_\*$ such that $\operatorname{Id}= dF+ Fd$. 


Question: does there exist a topological space $X$ which is not contractible (in the sense of topology), but whose complex of singular chain $C\_\*(X)$ is contractible?


More generally, one can ask whether the functor $X \mapsto C\_\*(X)$ from the category of topological spaces to the homotopy category of chain complexes of abelian groups forgets any information. I assume the answer is "yes", but I can't seem to come up with a counterexample.



Contractible chain complex from non-contractible space
These are known as [*acyclic spaces*](https://en.wikipedia.org/wiki/Acyclic_space) (note that since $\tilde C\_\*(X)$ is a bounded below complex of projectives, it being contractible is equivalent to its homology being trivial).


There's an extensive literature about them, starting with this [Emanuel Farjoun's paper](http://www.math.huji.ac.il/~farjoun/acyclic%20spaces.pdf).


In general, yes $C\_\*(X)$ does forget some information about $X$: first of all it factors through the stable homotopy type $\Sigma^\infty\_+X$ of $X$, and it forgets further information from there (the slogan is that it remembers only the "$\mathbb{Z}$-linear" information contained in $\Sigma^\infty\_+X$). For example it is unable to distinguish between $\mathbb{CP}^2$ and $S^2\vee S^4$ (since the attaching map $\eta:S^3\to S^2$ is sent to a map homotopic to 0 by $C\_\*(-)$).


You can get much closer to reconstructing the full homotopy type of $X$, by remembering the coalgebra structure on $C\_\*(X)$ induced by the diagonal, although in general that is still not enough.


