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In a previous post Lift chain complex from $\mathbb{F}_2$ to $\mathbb{Z}$ the body of the question mentions that this (lifting a chain complex from $\mathbb Z/2\mathbb Z$ to $\mathbb Z$) is always possible :

"Now, this can always be done (for example, by using the structure theorem for chain complexes over $\mathbb F_2$)."

What would be a constructive way of doing this?

(Unlike the original post, I'm removing the other two restrictions.)

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    $\begingroup$ A chain complex over $\mathbb{F}_2$ is just a direct sum of copies of $0\to\mathbb{F}_2\to 0$ and $0 \to \mathbb{F}_2\xrightarrow{\sim}\mathbb{F}_2 \to 0$ and shifts of these. Each of these has an obvious lift. $\endgroup$ Commented Apr 30, 2023 at 17:54
  • $\begingroup$ @DaveBenson excuse me if I'm being dense but I'm not sure how to use this. I have a list of matrices $\partial_n : \mathbb{F}_2^{a_n} \to \mathbb{F}_2^{a_{n+1}}$ with $\partial_n \partial_{n+1}=0$; how would I transform these into integer matrices? $\endgroup$
    – unknown
    Commented Apr 30, 2023 at 18:35
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    $\begingroup$ @unknown Change basis as suggested by Dave's comment. Split each $C_n$ as $H_n \oplus dC_{n+1} \oplus B_n$, where $d: B_n \to dC_n$ is a bijection. Choose a basis $B_n = \Bbb F_2^{b_n}$ and $H_n = \Bbb F_2^{h_n}$ and lift this to $C'_n = \Bbb Z^{h_n} \oplus \Bbb Z_{b_{n+1}} \oplus \Bbb Z_{b_n}$; the differential sends the last factor in $C'_n$ identically onto the second factor of $C'_{n-1}$, and is otherwise zero. All that remains is to show that you can lift the change of basis: every invertible matrix over $\Bbb F_2$ lifts to an invertible matrix over $\Bbb Z$ (lift elementary matrices) $\endgroup$
    – mme
    Commented Apr 30, 2023 at 18:51
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    $\begingroup$ This is "the structure theorem for chain complexes over F2". The hard part if you're trying to be concrete is to find such a splitting. $\endgroup$
    – mme
    Commented Apr 30, 2023 at 18:53

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I presume “constructive” means a computational algorithm is desired.

Assuming the chain complex $F$ over $\def\Z{{\bf Z}}\Z/2$ is bounded from below, we are going to construct by induction on $n$ a basis $E_n=A_n⊔B_n⊔C_n$ of $F_n$ with the following properties:

  • $A_n$ is a basis of exact elements in $F_n$;
  • $A_n⊔B_n$ is a basis of closed elements in $F_n$.

For sufficiently small $n$ we have $F_n=0$, which gives $A_n=B_n=C_n=∅$.

If $A_{n-1}$, $B_{n-1}$, $C_{n-1}$ have already been constructed, take $A_n$ to be the image of $C_{n-1}$ under the differential. Then extend $A_n$ to a basis $A_n⊔B_n$ of closed elements in $F_n$ using Gaussian elimination over $\Z/2$. Finally, extend $A_n⊔B_n⊔C_n$ to a basis of $F_n$ using Gaussian elimination again.

Now it is easy to lift $F$ to a chain complex over $\Z$: take the same basis elements and construct the corresponding differentials over $\Z$ as matrices with respect to these bases, where all entries are zero except for the entries corresponding to elements of $C_{n-1}$ mapping to elements of $A_n$, which we take to be 1.

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