Given a manifold $X$ and short exact sequence of abelian groups
$$
1\rightarrow A\_1\overset{\iota}{\rightarrow} A\_2\overset{\pi}{\rightarrow} A\_3\rightarrow 1
$$
we get the Bockstein map in cohomology $\beta : H^p(X,A\_3)\rightarrow H^{p+1}(X,A\_1)$. On cochains this is defined as follows. Choose a section $s:A\_3\rightarrow A\_2$, then given $f\in Z^p(X,A\_3)$, $s(f)\in C^{p}(X,A\_2)$ is not closed, but $\pi(\delta (s(f)))=0$ and thus there exists $\beta(f)\in Z^{p+1}(X,A\_1)$ such that $\iota(\beta(f))=\delta s(f)$. In a similar way one can define the Bockstein map in homology $b : H\_k(X,A\_3)\rightarrow H\_{k-1}(X,A\_1)$. On a chain $c\in Z\_k(X,A\_3)$ this is defined such that $\iota(b(c))=\partial s(c)$. In the last equation for any map $m : A\rightarrow B$ (even not a homomorphism) of abelian groups and $c\in C\_k(X,A)$ I denoted by $m(c)\in C\_k(X,B)$ the chains obtained by acting with $m$ on the coefficients.


On the other hand, if $X$ is orientable, by using Poincarè duality $\Phi\_A :H\_p(X,A)\rightarrow H^{d-p}(X,A)$ for homology/cohomology we induce a map in homology $b': H\_p(X,A\_3)\rightarrow H\_{p-1}(X,A\_1)$ as follows. Given $c\in H\_p(X,A\_3)$, construct $\Phi \_{A\_3}(c)\in H^{d-p}(X,A\_3)$, and then by using the Bockstein $\beta(\Phi\_{A\_3}(c))\in H^{d-p+1}(X,A\_1)$. Then define
$$
b'(c)=\Phi\_{A\_1}^{-1}\left(\beta(\Phi\_{A\_3}(c))\right)
$$


It is natural to expect that $b'=b$ but this does not seem to me totally obvious. I tried to prove it but at some point I need to use $\Phi$ on non-closed cycles (which I don't know how much is legal) and I would need to prove that this action commutes with the section $s:A\_3\rightarrow A\_2$, and I don't know how to do this...



Does the cohomology Bockstein homomorphism map to the homology Bockstein homomorphism under Poincarè duality?
I just found another proof you might prefer as Lemma 2.4 of this paper:
<https://arxiv.org/pdf/1708.03754.pdf>.
It's stated only for one case of groups, but I don't see why it wouldn't extend more generally.
Note that the Poincar'e duality isomorphism is given by the cap product with a fundamental class. That's the inverse of the map you denote $\Phi$.


