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I'm reading the proof of Poitou-Tate duality in the book Galois Cohomology and Class Field Theory by David Harari.

After some arguments, we get a exact sequence

$$ \mathbf{P}^1_S(k,M^{'})^* \longrightarrow H^1(G_S,M^{'})^* \longrightarrow \text{Ш}_S^2(k,M) \longrightarrow0 $$ and the author said it shows that $Ш^2_S(k,M)$ is the dual of $Ш^1_S(k,M^{'})$, which I can't understand.

Since $Ш^1_S(k,M^{'})$ is defined as the kernel of $ H^1(G_S,M^{'})\to \mathbf{P}^1_S(k,M^{'})$, that is, we have $$0 \to Ш^1_S(k,M^{'})\to H^1(G_S,M^{'})\to \mathbf{P}^1_S(k,M^{'})$$

After taking Pontryagin duality, we have

$$ \mathbf{P}^1_S(k,M^{'})^* \longrightarrow H^1(G_S,M^{'})^* \longrightarrow \text{Ш}_S^1(k,M^{'})^* \longrightarrow 0 $$

If we could show this complex is also exact, then we can conclude $Ш^2_S(k,M) \cong Ш^1_S(k,M^{'})^*$ as cokernel of a same morphism. Now the exactness at $Ш^1_S(k,M^{'})$ is because $\mathbf{Q}/\mathbf{Z}$ is injective module, but I can't show the exactness at the middle term.

( Pontryagin dual in this book is defined as $G^*:=\mathrm{Hom}_{contiuous}(G,\mathbf{Q}/\mathbf{Z})$, but $\mathbf{Q}/\mathbf{Z}$ is viewed as a discrete group in the Notations and Conventions at beginning of the book, which confuse me again because the Euclidean topology is more natrual to me. I hope the conclusion is still valid if $\mathbf{Q}/\mathbf{Z}$ is given Euclidean topology, but I don't know if it's true)

Thanks for any help!

Edit: All the $\mathbf{P}^1_S(k,M)$ should be $\mathbf{P}^1_S(k,M^{'})$, I corrected them and sorry for my mistakes.

Edit2：Here $M$ is assumed to be finite, but $S$ could be infinite.

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    $\begingroup$ Is $M$ finite? Is $S$ finite? If so then all groups in these sequences are finite and Pontyagin duality is exact. Why it does not matter for these cohomology groups, if we take $\mathbb{Q}/\mathbb{Z}$ with the discrete topology, look at the remark after Theorem 1.1.8 in "Cohomology of number fields". $\endgroup$
    – Chris Wuthrich
    Commented Dec 1, 2021 at 9:21
  • $\begingroup$ @ChrisWuthrich $M$ is finite, but $S$ is not assumed to be finite. $\endgroup$
    – Mugenen
    Commented Dec 1, 2021 at 9:41
  • $\begingroup$ It is true for finite $S$ and then take the limit. See the Theorem 1.1.8. mentioned above which states Pontryagin duality with all details. $\endgroup$
    – Chris Wuthrich
    Commented Dec 1, 2021 at 14:47

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