Let $K$ be a number field, let $A$ be an abelian variety over $K$, and let $H$ be a torus over $K$. For a prime $l$, we have the natural map $$\mathrm{Ext}^1(A, H) \otimes_{\mathbb{Z}} \mathbb{Z}_l \rightarrow \mathrm{Ext}^1_{\mathrm{Gal}(\overline{K}/K)}(T_l A, T_l H),$$ where the first $\mathrm{Ext}$ is in the category of commutative algebraic groups over $K$ and the second one is in the category of continuous representations of $\mathrm{Gal}(\overline{K}/K)$ on finite free $\mathbb{Z}_l$modules. Is the displayed map injective and, if so, why?

$\begingroup$ Did you try studying this using the exact sequence $0\to A_n\to A\to A\to 0$ arising from multiplication by $l^n$ on $A$, and then the similar sequence for $H$? $\endgroup$ – anon Jan 10 '15 at 4:54
It is affirmative (up to a compatibility check at the end of the argument given here). If $K'/K$ is a finite Galois splitting field of $H$ and $T' = H_{K'}$ is the associated split $K'$torus then we have an exact sequence of $K$tori $$0 \rightarrow H \rightarrow {\rm{R}}_{K'/K}(T') \rightarrow T \rightarrow 0.$$ But ${\rm{Hom}}(A,T)=0$ since $T$ is affine and ${\rm{Hom}}_K(T_{\ell}(A),T_{\ell}(T))=0$ by consideration of Frobenius eigenvalues. Thus, the natural maps $${\rm{Ext}}^1_K(A,H) \rightarrow {\rm{Ext}}^1_K(A,{\rm{R}}_{K'/K}(T')),\,\,\, {\rm{Ext}}^1_K(T_{\ell}(A),T_{\ell}(H)) \rightarrow {\rm{Ext}}^1_K(T_{\ell}(A),T_{\ell}({\rm{R}}_{K'/K}(T'))$$ are injective. It therefore suffices to treat ${\rm{R}}_{K'/K}(T')$ in place of $H$.
Using pushout along ${\rm{R}}_{K'/K}(T')_{K'} \rightarrow T'$ and pullback along $A \rightarrow {\rm{R}}_{K'/K}(A_{K'})$, the Ext's are compatibly identified with ${\rm{Ext}}^1_{K'}(A_{K'},T')$ and ${\rm{Ext}}^1_{K'}(T_{\ell}(A_{K'}),T_{\ell}(T'))$, so this reduces us (upon renaming $K'$ as $K$) to treating the case when $H$ is a split torus, and then even $H = {\rm{GL}}_1$ by direct sum compatibility in $H$. So far, so good.
But in the special case when $H={\rm{GL}}_1$ we naturally have $${\rm{Ext}}^1_K(A,{\rm{GL}}_1) \simeq A^{\vee}(K)$$ for the dual abelian variety $A^{\vee}$ (Barsotti's formula, valid over any field), and likewise the Galois Ext is ${\rm{H}}^1(K,T_{\ell}(A^{\vee}))$. The $\ell^n$power Kummer sequence of $A^{\vee}$ defines an injective map $$\delta:A^{\vee}(K)/(\ell^n) \rightarrow {\rm{H}}^1(K,A^{\vee}[\ell^n]).$$ Letting $S$ be a nonempty finite set of places containing the archimedean places and the bad places of $A$ over $K$ and the $\ell$adic places, $\delta$ is seen to factor through ${\rm{H}}^1(K_S/K,A^{\vee}[\ell^n])$ where $K_S/K$ is the maximal extension unramified outside $S$.
Since ${\rm{Gal}}(K_S/K)$ has good cohomological finiteness properties, it is legitimate to identified the injective map between inverse limits over $n$ with a map $$A^{\vee}(K) \otimes_{\mathbf{Z}} \mathbf{Z}_{\ell} \rightarrow {\rm{H}}^1(K_S/K,T_{\ell}(A^{\vee})).$$ The correctness of the left side uses that $A^{\vee}(K)$ is finitely generated (so its $\ell$adic completion is given by tensoring against $\mathbf{Z}_{\ell}$). Composing with the injective inflation map then defines an injection $${\rm{Ext}}^1_K(A,{\rm{GL}}_1) \otimes_{\mathbf{Z}} \mathbf{Z}_{\ell} \rightarrow {\rm{Ext}}^1_K(T_{\ell}(A), T_{\ell}({\rm{GL}}_1)).$$ The final step is to check that this injection is actually the original map of interest (in the special case $H = {\rm{GL}}_1$); this is left to the reader to sort out (maybe it is correct up to a harmless sign or something like that).

1$\begingroup$ Thank you for your answer. I don't understand the compatible Ext identification part, but I think one may reduce to the $H = \mathbb{G}_m$ case simply due to the injectivity of $\mathrm{Ext}^1(A, H) \rightarrow \mathrm{Ext}^1(A_{K'}, H_{K'})$ (and likewise for the $l$adic Tate modules), which results from the uniqueness of splittings over $K'$. Also, it seems that there is no need to work with $K_S$ instead of $K$, since continuous cohomology vs. $\varprojlim H^1$ identification results from mere finiteness of the $H^0(K, A[l^n])$. As for the final compatibility check... $\endgroup$ – Lisa S. Jan 10 '15 at 19:04

1$\begingroup$ ...I think one may check it after passage to mod $l^n$ (for varying $n$), and then it results from the "standard" compatibility of $A^\vee(K) \rightarrow H^1(K, A^\vee[l^n]) \rightarrow \mathrm{Ext}^1(A[l^n], \mathbb{G}_m)$ and $\mathrm{Ext}^1(A, \mathbb{G}_m) \rightarrow \mathrm{Ext}^1(A[l^n], \mathbb{G}_m)$, which, with a little bit of good luck, may even be in Oort's book. $\endgroup$ – Lisa S. Jan 10 '15 at 19:09

1$\begingroup$ @LisaS.: I agree, it was overkill to bring in Weil restriction, and with $K_S/K$ I was just being overly cautious with the Rlim stuff. As for the final compatibility, I don't remember it being in Oort's book, but let me know if you find a proof in the literature. (It surely cannot be deduced from the sheafified version, since the sheafified analogue of the target in this torsionlevel compatibility vanishes.) $\endgroup$ – user74230 Jan 10 '15 at 21:43