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This is not a complete answer, but a conditional one. If you assume that $ \bigcap_{N\in\mathbb N} \pi^N\left(L_2/\pi^n L_1\right)=0, $ or, what amounts to the same: $\bigcap_N(\pi^NL_2+ L_1)=L_1$, then the map is injective. It suffices to show that the map $\hat L_1\to \hat L_2$ is injective, where $L_j=\lim_{\leftarrow}L_j/\pi^n$. So let $l=(l_n)$ be in the kernel of that map, i.e., $l_n\in\pi^nL_2$, where $l_n\in L_1$ is only determined up to $\pi^nL_1$. Further $l_{n}\equiv l_{n+1}\mod \pi^{n}L_1$, so that $$ l_{n}=l_{n+1}+\pi^{n}\tilde l_1=\pi^{n+1}\tilde l_2+\pi^{n}\tilde l_1 $$ for some $\tilde l_1\in L_1$ and $\tilde l_2\in L_2$. Now we can modify $l_{n}$ so that $\tilde l_1$ is zero and thus get $l_{n}\in\pi^{n+1} L_2$, i.e., we have increased the power of $\pi$ by one. We can iterate this step to see that $l_n$ lies in the image of $L_1\cap\pi^NL_2$ in $L_1/\pi^n L_1$ for every $N\in\mathbb N$. As an element of $L_2/\pi^nL_1$, the element $l_n$ lies in $\pi^N(L_2/\pi^nL_1)$ for every $N$, so by our assumption, it must be zero.