Exact sequence induced by filtration Let’s suppose that $M_1, M_2, M_3$ are $A$-modules with
$$
  0 \to M_1 \to M_2 \to M_3 \to 0
$$
exact.
Given $I \subset A$, is true that
$$
  0\to M_1 / I^n M_2 \cap M_1 \to M_2 / I^n M_2 \to M_3 / I^n M_3 \to 0
$$
is exact?
In particular, I’ve showed that
$$
  \operatorname{im}(M_1 / I^n M_2 \cap M_1 \to M_2/I^nM_2)
  \subset \ker( M_2/I^nM_2 \to M_3/I^nM_3 ) \,,
$$
but I’m unable to prove the opposite containment. I tried to lift $0$ i.e., an element of $I^n M_3$ and use surjectivity of the above exact sequence to take a preimage, but I’m unable modify it to put it in the $\ker(M_2 \to M_3)$ to continue the standard argument.
Is this true? Any help or solution would be appreciated. I need this lemma in order to proof that the completion of an exact sequence of $A$-modules is exact, and passes from this lemma and inverse limits.
Edit: Is it true that in order to define the map $M_2 / I^n M_2 \to M_3 / I^n M_3$ I need that the image of $I^n M_2$ goes to $I^n M_3$, i.e, that the map respect the filtration? I don’t know whether this is a property that I should require or if it follows from something else.
 A: Notice that the sequence $$M_1 / I^n M_1 \to M_2/I^nM_2 \to M_3/I^nM_3 \to 0$$ is exact, since it's just a tensor product of your first exact sequence with $A/I^n.$ And the kernel of the left map is exactly $I^nM_2\cap M_1$ modulo $I^nM_1$. Taking a quotient by that kernel doesn't change the image of the first map and gives you your second exact sequence.
A: Sergey’s answer is very neat, though I’m posting the answer I was looking for.
I'm calling (from left to right) $\alpha, \beta$ the map of the exact sequence of the $M_i$ and $\tilde{\alpha}, \tilde{\beta}$ the induced map. Note that $\tilde{\beta}$ is well-defined since for every ring morphism $\beta(I^n M_2) \subset I^n M_3$.
Now, taking $[x] \in \ker \tilde{\beta}$ we have that $\tilde{\beta}[x] = \beta(x) + I^n M_3 = 0$, hence $\beta(x) \in I^n M_3$, hence $\beta(x) = \sum_{i=1}^k j_i m_i$ where $j_i \in I^n$ and $m_i \in M_3$. Since ${\beta}$ is surjective we have $m_i = \beta(a_i)$ where $a_i \in M_2$ for each $i$, hence $x - \sum_{i=1}^k j_i a_i \in \ker \beta = \operatorname{im} \alpha$, so that exists $z \in M_1$ such that $\alpha(z) = x - \sum_{i=1}^k j_i a_i$.
At this point should be clear from the commutativity of the diagram induced that, $\tilde{\alpha}$ applied to the image of $z$ in $M_1 / I^nM_2 \cap M_1$ works (by commutativity of the diagram) since $\sum_{i=1}^k j_i a_i \in I^n M_2$, i.e., $x - \sum_{i=1}^k j_i a_i$ is still a lifting of $[x]$.
