We start with a finite dimensional chain complex over $\mathbb{F}\_2$, equipped with a basis. That is, we have finitely many finite dimensional $\mathbb{F}\_2$-vector spaces $C\_0,\dots,C\_k$ with bases $B\_0,\dots,B\_k$, and $\mathbb{F}\_2$-module maps $d\_i\colon C\_i\rightarrow C\_{i-1}$ with $d\_{i-1}d\_i=0$. I want to lift this to a chain complex over $\mathbb{Z}$.


(A cleaner restatement in terms of matrices is: we have matrices $M\_1,\dots,M\_k$ over $\mathbb{F}\_2$ with $M\_{i-1}M\_i=0$, and I want to lift them to matrices over $\mathbb{Z}$ satisfying that property.)


Now, this can always be done (for example, by using the structure theorem for chain complexes over $\mathbb{F}\_2$). I wanted to know if this can always be done subject to the following restrictions.


Restriction 1: $0$ in $\mathbb{F}\_2$ should lift to a $0$ in $\mathbb{Z}$.  

Restriction 2: (Only interested if Restriction 1 can always be achieved) $1$ in $\mathbb{F}\_2$ should lift to $\pm 1$ in $\mathbb{Z}$.


(I believe, for no particular reason, that Restriction 1 can always be achieved, but not Restriction 2 in addition.)


Clarifications in response to comments: $\mathbb{F}\_2$ is the field of two elements (perhaps I should have said $\mathbb{Z}/2\mathbb{Z}$ throughout).  

Explicit restatement: I have matrices $M\_1,\dots,M\_k$ over $\mathbb{F}\_2$ such that for all $i$, the product $M\_{i-1}M\_i$ is defined and is zero. I want to find integer matrices $N\_1,\dots,N\_k$ such that for all $i$, $N\_{i-1}N\_i=0$, and the mod-2 reduction of $N\_i$ is $M\_i$. I want to ensure that if an entry is zero in $M\_i$, then the corresponding entry in $N\_i$ is zero. In addition, I wouldn't mind if I can ensure that if an entry is $1$ in $M\_i$, then the corresponding entry in $N\_i$ is $\pm 1$, but that is perhaps a bit much.



Lift chain complex from $\mathbb{F}_2$ to $\mathbb{Z}$
This is not always possible, even just with condition (1). Consider the complex $\mathbb{F}\_2^7 \to \mathbb{F}\_2^7 \to \mathbb{F}\_2^3$ where the basis of the first vector space is indexed by lines of the [Fano plane](http://en.wikipedia.org/wiki/Fano_plane), the basis of the second vector space is indexed by points of the Fano plane and the third vectors space is just thought of as $\mathbb{F}\_2^3$.


The first map sends a line to the formal sum of the three points on it. The second map embeds the Fano plane as the $7$ nonzero elements of $\mathbb{F}\_2^3$. So the composisition is $0$ because, for any line in the Fano plane, the three points on it add up to $0$ considered as vectors in $\mathbb{F}\_2^3$.


If you had a lift of the sort you describe, you would have $7$ vectors in $\mathbb{Z}^3$ (the image of the basis vectors of the second $\mathbb{Z}^7$) such that each line of the Fano plane gave three vectors with a linear relation between them. In other words, you would have realized the Fano plane in $\mathbb{P}^2(\mathbb{Q})$, which is impossible.


In general, questions about linear algebra while imposing that various matrix entries are zero resemble [Mnev's universality theorem](http://en.wikipedia.org/wiki/Mnev%27s_universality_theorem), so you should expect everything conceivable to go wrong. 


