Has Pontryagin duality been extended to condensed abelian groups? The obvious approach being to define $\hat M$ as the internal hom to the circle group. Is it true that $\hat{\hat M}=M$ with this definition?



Condensed Pontryagin duality
This cannot be true for all condensed abelian groups. Indeed, in [this answer](https://mathoverflow.net/a/356261/82179) to [Are there (enough) injectives in condensed abelian groups?](https://mathoverflow.net/questions/352448/are-there-enough-injectives-in-condensed-abelian-groups), Scholze explains that there are no nonzero injective condensed abelian groups. So it suffices to prove the following:


**Lemma.** *Let $\mathscr A$ be an abelian category and let $\mathbf D \colon \mathscr A^{\text{op}} \to \mathscr A$ be a left exact functor such that $\mathbf D^2$ is faithful and preserves monomorphisms (e.g. $\mathbf D^2 \cong \mathbf 1\_{\mathscr A}$). Then $\mathbf D$ is exact.*


Indeed, this would prove that $\mathbf T = \mathbf R/\mathbf Z$ is injective if $\mathbf D = \mathbf{Hom}(-,\mathbf T)$ is an auto-duality.


*Proof of Lemma.* If $0 \to A \to B \to C \to 0$ is an exact sequence in $\mathscr A$, then
$$0 \to \mathbf D(C) \to \mathbf D(B) \to \mathbf D(A)$$
is exact by hypothesis. Let $M$ and $N$ be the image and cokernel of $\mathbf D(B) \to \mathbf D(A)$ respectively, so that we get exact sequences
$$\begin{array}{ccccccccc}
0 & \to & \mathbf D(C) & \to & \mathbf D(B) & \to & M & \to & 0,\\
0 & \to & M & \to & \mathbf D(A) & \to & N & \to & 0.
\end{array}$$
Dualising again gives exact sequences
$$\begin{array}{ccccccc}
0 & \to & \mathbf D(M) & \to & \mathbf D^2(B) & \to & \mathbf D^2(C),\\
0 & \to & \mathbf D(N) & \to & \mathbf D^2(A) & \to & \mathbf D(M).
\end{array}$$
The composition $\mathbf D^2(A) \to \mathbf D(M) \to \mathbf D^2(B)$ is injective by hypothesis, so the second sequence gives $\mathbf D(N) = 0$. Faithfulness of $\mathbf D^2$ gives $N = 0$, so $\mathbf D(B) \to \mathbf D(A)$ was surjective to begin with. $\square$


Presumably there is a direct example showing that $\mathbf T$ is not injective. If you want to work with $\mathbf{Cond}\_\kappa$ instead of $\mathbf{Cond}$, then the same argument doesn't work (because $\mathbf{Cond}\_\kappa$ does have enough injectives), and I don't know if $\mathbf T$ is injective in $\mathbf{Cond}\_\kappa$ for any strong limit cardinal $\kappa$. (Surely not, right?)


