Let $M$ be a simplicial complex and $Π$ be a finite abelian group. In the simplest case $Π = \mathbb{Z}_r$ finite group with $r$ even, the Pontryagin square is a cohomological operation which maps an element $f ∈ H^p(M,\mathbb Z_r)$, $p$ even, to an element $\in H^{2p}(M,\mathbb Z_{2r})$.
Questions:
If the homology group $H_{p−1}(M,\mathbb Z)$ is torsion-free. Then how can we show that every $p$-cocycle modulo $r$ can be lifted to an integral $p$-cocycle. If $\tilde f$ is a lift of $f$, we define $Pf = \tilde f ∪ \tilde f \mod 2r$. How do we prove that this is well-defined (i.e. independent of the choice of the lift)?
If the homology group $H_{p−1}(M,\mathbb Z)$ is NOT torsion-free. Can we still define the $$ f ∈ H^p(M,\mathbb Z_r)?$$ And its lifting to an integral $p$-cocycle?
Say $p=2$, then how to show that the Pontryagin square can be thought of as a distinguished element in $$H^4 (B^2 \mathbb Z_r, \mathbb Z_{2r}) \simeq \text{Hom}(\mathbb Z_{2r},\mathbb Z_{2r}),$$ is this true? In fact, does it corresponds to the identity element in Hom$(\mathbb Z_{2r},\mathbb Z_{2r})$? Does the latter also provide a quadratic refinement of twice the cup product?