Lift chain complex from $\mathbb{F}_2$ to $\mathbb{Z}$ 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.
 A: 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, 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, so you should expect everything conceivable to go wrong. 
