Constructing a topological space from a chain complex I know that given a topological space $X$, I can construct(functorially) the singular chain-complex $C_{\bullet}(X,R)$ (where $R$ is a commutative ring with unit).
I wondered if it was possibile to reverse this construction:
"Starting from a chain complex of $R$-modules $C_{\bullet}$(concentrated at non-negative degrees), can I determine a topological space such that its singular chain complex is isomorphic to $C_{\bullet}$?"
I found out by myself a counterexample:
$$\mathbb{Z}_3 \leftarrow 0  \leftarrow 0  \leftarrow ...$$
Notice that it should be $| \mathbb{Z}_3|=|\mathbb{Z}_3 X|$ that is to say $X$ is a point. But the singular chain complex of a point is different from the one we started with. So there cannot be a topological space that induces the chain complex in the beginning.
So this is not always possible. But I would like to know when this is possible:
Question 1
"When does a chain complex (concentrated at non-negative degrees) arise from a topological space? If so, is there a canonical way of constructing this topological space?"
A question that is different, but strictly linked in someway to this one is:
Question 2
"Given a succession of $R$-modules $(M_n)_{n\in \mathbb{N}_0}$, when is it possible to find a topological space $X$ such that its $n$-th homology is isomorphic to $M_n$? If so, is there a canonical way of constructing this topological space?"
I'm interested also in partial(non-trivial) results.
Thank you.
 A: Question 1: This question really can't be interpreted exactly as you ask it since the singular chain complex is so large. It turns out that to turn this into an interesting question with a positive, nontrivial answer, we must restrict coefficients to $\mathbb{Q}$ and also consider the cup product structure on cohomology (or if you really want to work with homology you have to use what is called a coalgebra structure on homology). We must also replace isomorphism with quasi-isomorphism, i.e. a chain map, respecting multiplicative structure (technical difficulties here that I will ignore) that induces isomorphisms on cohomology.
Question 1 is a very classic question and it is answered by rational homotopy theory. The answer is very strong, it is that every rational cochain complex with an algebra structure occurs (up to quasiisomorphism) as the singular chains (correctly interpreted) of some space. In fact, this space is unique up to (rational) homotopy equivalence and it can be picked in a functorial way.
Question 2: One can equip this sequence with trivial differential and trivial multiplication and apply rational homotopy theory to produce such a space. Explicitly, you will get a wedge of Moore spaces (these are spaces with only one nontrivial homology group).
For either of these questions, it becomes much much more difficult if $\mathbb{Q}$ is replace by anything besides another characteristic 0 field. Classifying what cohomology rings occur are classic and presumably still open problems.
