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Let $0 \rightarrow L \stackrel{\alpha}\rightarrow M\stackrel{\beta}\rightarrow N \rightarrow 0$ be an exact sequence, and $M_1$, $M_2$ be two submodules of $M$; then whether the follwing implications holds or not: $\beta(M_1) = \beta(M_2) $ and $\alpha^{-1}(M_1) = \alpha^{-1}(M_2) \implies M_1 = M_2$

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This implication is not true in general. For an easy counterexample, let's use real vector spaces (modules over $\mathbb R$) so that we can visualize what's going on. Consider an exact sequence, as in the question, with $L=N=\mathbb R$ and $M=\mathbb R^2$, with $\alpha(x)=(x,0)$ and $\beta(x,y)=y$. This gives you a short exact sequence, because $\alpha$ is one-to-one, $\beta$ is surjective, and the kernel of $\beta$ is the $x$-axis, which is the image of $\alpha$. Now let $M_1$ and $M_2$ be two lines of non-zero slope through the origin, say $M_1=\{(x,y):y=x\}$ and $M_2=\{(x,y):y=2x\}$. Then $\beta(M_i)=\mathbb R$ and $\alpha^{-1}(M_i)=\{0\}$ for both values of $i$, even though $M_1\neq M_2$.

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I believe a quick way of doing it is to use the 5-lemma for the sequences, $0\to\alpha^{-1}(M_i) \to M_i \to \beta(M_i) \to 0$

[edit] I took $=$ to mean isomorphic. Not true other wise.

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