Universal covering of abelian variety Let $A$ be an abelian variety over a field $K$. It is shown that its  $p$-adic Tate module $T_p(A)= \varprojlim_{n} A[p^n](\overline{K}) \cong Hom(\mathbb{Q}_p/\mathbb{Z}_p,A(\overline{K})= \varprojlim Hom (\mathbb{Z}/p^n \mathbb{Z}, A(\overline{K}))$.
Now I consider the following projective system $[p]: A(\overline{K}) \rightarrow A(\overline{K}) $ and I want to compute
$B(A)= \varprojlim \left(A(\overline{K}) \overset{[p]}{\leftarrow} A(\overline{K}) \leftarrow \cdots \leftarrow A(\overline{K}) \leftarrow \cdots \right)$
We have the exact sequence $0 \rightarrow T_p A \rightarrow B(A) \rightarrow A(\overline{K}) \rightarrow 0$
My question is can I have any similar result that is $B(A) \cong Hom(?,A(\overline{K}))$ ?
 A: I think the group $(\mathbf{Q}_p/\mathbf{Z}_p)\oplus\mathbf{Z}[\tfrac1p]$ does the job, i.e.
$$
\mathrm{Hom}((\mathbf{Q}_p/\mathbf{Z}_p)\oplus\mathbf{Z}[\tfrac1p],A(\bar K))\cong  B(A).
$$
Indeed, let $A(\bar K)_{\mathrm{tor}}$ be the torsion subgroup of $A(\bar K)$. We have a short exact sequence
$$
0\rightarrow A(\bar K)_{\mathrm{tor}}\rightarrow A(\bar K)\rightarrow A(\bar K)/A(\bar K)_{\mathrm{tor}}\rightarrow 0.
$$
Since $A(\bar K)$ is a divisible group, the quotient group $A(\bar K)/A(\bar K)_{\mathrm{tor}}$ is a uniquely divisible abelian group, i.e., a
$\mathbf{Q}$-vector space $V$. Since $A(\bar K)_{\mathrm{tor}}$ is also divisible, the sequence splits and one has an isomorphism
$$
A(\bar K)\cong A(\bar K)_{\mathrm{tor}}\oplus V.
$$
Denoting by $A(\bar K)[p^\infty]$ the subgroup of $A(\bar K)$ of all elements whose order is a power of $p$, and by
$A(\bar K)[\not p]$ the subgroup of $A(\bar K)$ of all elements of order prime to $p$, one has
$$
A(\bar K)_{\mathrm{tor}}=A(\bar K)[p^\infty]\oplus A(\bar K)[\not p].
$$
Hence,
$$
A(\bar K)\cong A(\bar K)[p^\infty]\oplus W
$$
with $W=A(\bar K)[\not p]\oplus V$ an abelian group for which multiplication-by-$p$ is a bijection. It follows that
$$
B(A)\cong T_pA\oplus W.
$$
Now,
$$
\mathrm{Hom}(\mathbf{Z}[\tfrac1p],A(\bar K)[p^\infty])=0
$$
since all elements of $A(\bar K)[p^\infty]$ are of $p$-power torsion. On the other hand
$$
\mathrm{Hom}(\mathbf{Z}[\tfrac1p],W)\cong W
$$
since $W$ is uniquely $p$-divisible. It follows that
$$
\mathrm{Hom}((\mathbf{Q}_p/\mathbf{Z}_p)\oplus\mathbf{Z}[\tfrac1p],A(\bar K))\cong T_pA\oplus W\cong B(A).
$$
