# Contractible chain complex from non-contractible space

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.

• I've decided to interpret the question as asking whether the reduced singular complex is trivial, or equivalently if the canonical map $C_*(X)→\mathbb{Z}$ is a homotopy equivalence, since that's the property that's analogue to the contractibility of $X$ (i.e. $X\to \ast$ is a homotopy equivalence). – Denis Nardin Apr 5 at 8:20

These are known as acyclic spaces (note that since $$\tilde C_*(X)$$ is a bounded below complex of projectives, it being contractible is equivalent to its homology being trivial).
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.
• For more information relating to the last paragraph, one can look up Mandell's theorem that gives some conditions for $C^*(X)$ to lose no information – Maxime Ramzi Apr 5 at 8:27