This is a question a friend of mine asked me some time ago. I suspect the answer is "no" but can't prove it.


Every free complex of abelian groups is isomorphic to the reduced cellular complex of some finite CW-complex; in fact, one can take a wedge of balls and Moore spaces. The question is whether there is a similar result for complexes equipped with a basis.


Namely, let $C\_{\*}$ be a finite chain complex (i.e., the differential decreases the degree) of free abelian groups. Suppose a basis $B\_i$ of each $C\_i$ is chosen.


1. Is it true that $(C\_{\*},(B\_i))$ is isomorphic as a complex with a basis to the (possibly shifted) reduced cellular complex of some CW-complex with the basis given by the cells?
2. What happens if one works with complexes of finite-dimensional vector spaces over $\mathbb{Z}/2$ instead?



Realizing complexes with bases as cellular complexes
Here is a sketch of an argument to show that all based chain complexes are realizable. (This might end up being pretty similar to Tyler's argument.)


First one gives an algebraic argument that by a change of basis the chain complex can be put in a standard "diagonal" form. Moreover, the change of basis can be achieved by a sequence of elementary operations, as in linear algebra, but now over the integers rather than a field, using the fact that the group $GL(n,Z)$ is generated by elementary matrices, including signed permutations. The most important of the elementary operations is to add plus or minus one basis element to another. Doing such an operation in $C\_i$ changes the boundary maps to and from $C\_i$ by multiplication by an elementary matrix and its inverse.


The "diagonalized" chain complex can easily be realized geometrically, so it remains to see that the elementary basis change operations can be realized geometrically. In the special case of top-dimensional cells, one can slide a part of one such cell over another to achieve the elementary operation of adding plus or minus one column of the outgoing boundary matrix to another. For lower-dimensional cells one wants to do the same thing and then extend the deformation over the higher cells. It should be possible to do this directly without great difficulty. The slide gives a way of attaching a product $\{cell\}\times I$, and this product deformation retracts onto either end, so one can use the deformation retraction to change how the higher-dimensional cells attach.


The argument should work for 1-cells as well as for higher-dimensional cells, so it shouldn't be necessary to assume that $C\_1$ is trivial.


An alternative approach would be to temporarily thicken the cell complex into a handle structure on a smooth compact manifold-with-boundary of sufficiently large dimension, with one i-handle for each i-cell. Sliding an i-cell then corresponds to sliding an i-handle, and there is well-established machinery on how to do this sort of thing, as one sees in the proof of the h-cobordism theorem for example. Or one can use the language of morse functions and gradient-like vector fields as in Milnor's book on the h-cobordism theorem. Either way, after all the elementary basis changes have been realized by handle slides, one can collapse the handles back down to their core cells to get the desired based cellular chain complex.


There are plenty of details to fill in here in either the cell or handle approach. I don't recall seeing this result in the classical literature, but I wouldn't be too surprised if it were there somewhere, maybe in some paper or book on J.H.C.Whitehead's simple homotopy theory where elementary row and column operations play a big role.


