exact sequence and modules proposition. I have problems to prove the following proposition:

Let's consider $$0 \rightarrow L \stackrel{\alpha}{\rightarrow} M
\stackrel{\beta}{\rightarrow} N \rightarrow 0$$ an exact sequence of modules
  and $M_1$, $M_2$ are submodules of $M$.
Prove the following statement in case it's true or give a counterexample if not:
  $$\beta(M_1)=\beta(M_2) \mbox{ and } \alpha^{-1}(M_1)=\alpha^{-1}(M_2) \Rightarrow M_1=M_2.$$

My attemp: I've been trying to prove it's true, because I haven't found a counterexample.
In the given conditions, for submodules $M_a \subset M_b \subset M$, I've proved that: 
$$[\beta(M_a)=\beta(M_b) \mbox{ and } L \cap M_a=L \cap M_b] \Rightarrow M_a=M_b$$
and I've been trying to use that by doing $M_a=M_1$ and $M_b=M_1 \cup M_2$ and applying this but I don't get what I want.
Any help would be appreciate. Thanks in advance.
 A: Let $$0 \to L \xrightarrow \alpha M \xrightarrow \beta N \to 0$$ be a short exact sequence of $A$-modules.
You had the following result at your disposal (except in your notation you identify $\alpha(L)$ with $L$):
Lemma: For submodules $M_1 \subseteq M_2 \subseteq M$, $\alpha(L) \cap M_1 = \alpha(L) \cap M_2$ and $\beta(M_1) = \beta(M_2)$ implies $M_1 = M_2$.
Claim: If we assume that $M_1 \subseteq M_2 \subseteq M$, then $\alpha^{-1}(M_1) = \alpha^{-1}(M_2)$ and $\beta(M_1) = \beta(M_2)$ implies $M_1 = M_2$. (Notice that this is the result you're trying to prove with the added hypothesis that $M_1 \subseteq M_2 \subseteq M$.)
This follows immediately from the following fact:
Claim: $\alpha(L) \cap M_1 = \alpha(L) \cap M_2$ if and only if $\alpha^{-1}(M_1) = \alpha^{-1}(M_2)$.
Proof: Suppose $\alpha(L) \cap M_1 = \alpha(L) \cap M_2$. Let $l_1 \in \alpha^{-1}(M_1)$, then $\alpha(l_1) \in M_1$ and $\alpha(l_1) \in \alpha(L)$ trivially, so we have $\alpha(l_1) \in \alpha(L) \cap M_1 = \alpha(L) \cap M_2$ which means $\alpha(l_1) \in M_2$ and hence $l_1 \in \alpha^{-1}(M_2)$. Thus we've shown that $\alpha^{-1}(M_1) \subseteq \alpha^{-1}(M_2)$. The reverse containment follows from symmetric reasoning.
Conversely, suppose that $\alpha^{-1}(M_1) = \alpha^{-1}(M_2)$. Let $m_1 \in \alpha(L) \cap M_1$, then $m_1 \in \alpha(L)$ so there exists $l_1 \in L$ such that $m_1 = \alpha(l_1) \in M_1$. This means that $l_1 \in \alpha^{-1}(M_1) = \alpha^{-1}(M_2)$ so $m_1 = \alpha(l_1) \in M_2$. Hence $m_1 \in \alpha(L) \cap M_2$. Thus we've shown that $\alpha(L) \cap M_1 \subseteq \alpha(L) \cap M_2$. The reverse containment follows from symmetric reasoning.

Now, if we remove the hypothesis that $M_1 \subseteq M_2 \subseteq M$, then the statement doesn't hold. The example given in the comments is perfect to show this:
Define $L := \mathbb Z$, $M := \mathbb Z \times \mathbb Z$, $N := \mathbb Z$. 
Define $\alpha : L \to M$ by $\alpha(x) = (x, 0)$. Define $\beta : M \to N$ by $\beta(x, y) = x$. Notice that $\alpha$ is clearly injective, $\beta$ is clearly surjective, and $\operatorname{im} \alpha = \ker \beta$, so this in fact defines a short exact sequence.
Define $M_1 := \{(x, x) : x \in \mathbb Z\}$ and $M_2 := \{(-x, x) : x \in \mathbb Z\}$. (Notice that neither is contained in the other!)
Observe that $\beta(M_1) = \beta(M_2) = \mathbb Z$, so that hypothesis has been satisfied. Observe that $\alpha^{-1}(M_1) = \alpha^{-1}(M_2) = \{0\} = 0$, so that hypothesis has been satisfied. But clearly $M_1 \neq M_2$!
