Can you tell me what's wrong in my proof ?


Let f be a surjective R-module homomorphism from M to N.

For every $x\in N$ we let:

$A_{x}=\lbrace y \in M \mid f(y)=x \rbrace$, then $A_{x}$ is non-empty.

So $\prod A_{ x }$, $x\in M$ is non-empty, that mean there is an element $(y_{ x })_{ x\in M }$ (Choice' axiom)

Now, we define $f_{1}$ from N to M as follows:

for every $x \in M$ let $f_{1}(x)=y_{x}$.

Then $f\circ f_{1}=Id_{N}$

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    $\begingroup$ It's interesting because we can define $P$ to be a projective $R$-module provided any surjective $\alpha : M \to P$ has a right inverse $\beta : P \to M$ (i.e., $\alpha \beta = 1_P$). This suggests that if it's not projective, the result you want might not hold. $\endgroup$ – Robert Cardona Sep 19 '15 at 9:16
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    $\begingroup$ oh I have known my wrong !!!!!!!!!! $f_{1}$ which is establish above, maybe not R-modules homomorphism. $\endgroup$ – PHU CUONG LE VAN Sep 19 '15 at 9:17
  • $\begingroup$ @Robert Carndona: That's right, I understand what your mean. Thanks!!!! $\endgroup$ – PHU CUONG LE VAN Sep 19 '15 at 9:20
  • $\begingroup$ I tried to edit $x\in M$ after the definition of $A_x$ into $x\in N$ but it did not work, could someone do it? $\endgroup$ – Noix07 Sep 30 '16 at 10:35

This community wiki solution is intended to clear the question from the unanswered queue.

You have used the Axiom of Choice to show that there exists a function $f_1 : N \to M$. You have not shown it is a module homomorphism, which may not be the case unless you add more hypothesis.


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