Let $\phi: M \to M'$ be a homomorphism of left R-modules, and let $N' \subset M'$ be a submodule. Construct a natural module homomorphism:

$\tilde{\phi}: M/\phi^{-1}(N') \to M'/N'$

and show that $\tilde{\phi}$ is injective. Also, show that $\tilde{\phi}$ is an isomorphism if $\phi$ is surjective.

Thanks in advance.


Hint: simply set $\widetilde\phi(m+\phi^{-1}(N'))= \phi(m)+N'$ and check it is well defined, and it is an injective module homorphism.

For the surjectivity if $\phi$ is surjective, it comes from a general fact about mappings of sets.

  • $\begingroup$ Okay, that helped to understand it better, thanks. So this homomorphism takes the equivalent classes of $m, n \in M$ with the equivalence relation $m$ ~ $ n <=> m + \phi^{-1} = n + \phi^{-1}$, and points them to an $m' \in M' mod N'$? So the check for them being well-defined is, I guess, to take two elements from the same equivalence class and check that they're pointed to the same element of $M'/N'$? $\endgroup$
    – moran
    Apr 9 '15 at 16:40
  • $\begingroup$ Exactly. You might also apply to the composed homomorphism: $M\to M'/N $the well-known factorisation through the cokernel to the composed homomorphism $\,M\to M'\to M'/N'$. $\endgroup$
    – Bernard
    Apr 9 '15 at 16:45

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