Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $\alpha^{-1}(M_1) = \alpha^{-1}(M_2) \implies M_1 = M_2$

share|cite|improve this question
up vote 2 down vote accepted

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 $M_1\neq M_2$.

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.