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(cross-post from stack exchange after not receiving any answers)

I'm wondering the following: if we have a finite-index subgroup $H\subset G$, and a cocycle $[c]\in H^1(H,\mathbb{Z})$, is there any way to get an explicit cochain representing its image under the isomorphism $H^1(H,\mathbb{Z})\to H^1(G, \hom(G/H,\mathbb{Z}))$? Maybe if we pick an explicit set of representatives for $G/H$ there's some canonical choice?

Something similar was asked previously here, but didn't receive an answer, and the given reference to Neukirch didn't explain how "picking a transversal" worked.

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    $\begingroup$ I think if c is a homomorphism from H to Z and T is a transversal then the corresponding cocycle takes g in G to the mapping sending tH to c(h) where gt=t'h with t,t' in the transversal and h in H. But possibly it is c(h^{-1}) $\endgroup$
    – Benjamin Steinberg
    Commented Dec 1, 2022 at 20:42
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    $\begingroup$ The linked question provides and answer "See Cohomology of Number Fields, (1.6.4). " See also : mathoverflow.net/questions/259123/… $\endgroup$
    – Chris Wuthrich
    Commented Dec 2, 2022 at 9:33

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