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Let $M$ and $N$ be projective $A$-modules. If we know that $f: M \to N$ is surjective and $g: N \to M$ is surjective, can we conclude that $M$ is isomorphic to $N$? More generally, if $M$ and $N$ are not necessarily projective, can we conclude that $M$ is isomorphic to $N$?

I am asking this question because it seems that in the first four lines of page 30 of the book elements of representation theory of associative algebras volumn 1 (see the picture below), it is said that g is a surjective map $P'$ to $P(M)$ and g' is a surjective map from $P(M)$ to $P'$ imply that g is a bijection. Thank you very much.

enter image description here

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Please revise your title to reflect the content of your question. –  Alexander Gruber Mar 22 '13 at 9:58
For a non-projective counterexample, take $A = \mathbb{Z}$, $M = \mathbb{Z}^{\oplus \mathbb{N}} \oplus \mathbb{Z} / 2 \mathbb{Z}$, $N = \mathbb{Z}^{\oplus \mathbb{N}}$. –  Zhen Lin Mar 22 '13 at 10:23
@Alexander, thank you very much. I have revised the title. –  LJR Mar 25 '13 at 10:15
@JianrongLi Thanks! –  Alexander Gruber Mar 25 '13 at 19:45

2 Answers 2

up vote 2 down vote accepted

As noted in the comments and the answer by Bob, this does not hold in general. What holds in general is that if you have a surjective map (i.e. epimorphism) $f:M\to P$. Then $P$ is a direct summand of $M$.

However, there are some general assumptions in the book you are reading. One that is crucial here is that $A$ is finite dimensional and the modules, they are talking about are finite dimensional. And this implies that if you have surjective maps $f:M\to N$ and $g:N\to M$, then $\dim M=\dim N$. And furthermore $$\dim M=\dim \operatorname{ker} f + \dim\operatorname{im} f=\dim \operatorname{ker} f+ \dim M.$$ Hence $f$ is also injective and hence an isomorphism.

This can be a made a bit more general: It holds whenever $M$ and $N$ are of finite length.

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No, only surjectivity doesnt imply isomorphism.

$g: N \to M$ is surjective doesn't imply that $f: M \to N$ is injective, what might be what your thinking.

The 3 necessary conditions are:

$\bullet$ the mapping is linear

$\bullet$ the mapping is injective

$\bullet$ the mapping is surjective

For less abstract cases showing there's an inverse might be time saving.

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