Let $R$ be a ring and $M,N$ be $R-$modules. Let $\pi: M \to N$ be a module homomorphism. Let $K$ be a $R$-submodule of $N$. Then $\pi^{-1}(K)$ is a submodule of $M$

Surprisingly, I couldn't find this result on google.

Is my proof correct?

Proof: Since $\pi (0) = 0 \in K$, $0 \in \pi^{-1}(K)$ and $\pi^{-1}(K) \neq \emptyset$

Let $x,y \in \pi^{-1}(K).$ Let $r,s \in R$. Then:

$\pi(x), \pi(y) \in K$ and hence:

$$\pi(rx + sy) = r\pi(x) + s\pi(y) \in K$$

because $K$ is a submodule of $N$.

From this, it follows that $rx + sy \in \pi^{-1}(K)$ and we are done.

  • $\begingroup$ Yes, it's true. $\endgroup$ – egreg Dec 8 '17 at 17:05
  • $\begingroup$ Thanks! Is my proof correct? $\endgroup$ – user370967 Dec 8 '17 at 17:06
  • $\begingroup$ Looks solid to me. $\endgroup$ – stressed out Dec 8 '17 at 17:13

Yes, your statement and proof are correct.

This fact is important in the sometimes called “correspondence theorem”: if $f\colon M\to N$ is a homomorphism, the mappings $H\mapsto f(H)$ and $K\mapsto f^{-1}(K)$ define bijections, inverse of each other, between the set of submodules of $M$ containing $\ker f$ and the set of submodules of $N$ contained in $\operatorname{im}f$.

Moreover, for $H$ in the former set, we have $$ H/\ker f\cong f(H) $$ and, for $K$ in the latter set, we have $$ f^{-1}(K)/\ker f\cong K $$

You can also prove that, for every submodule $H$ of $M$ and $K$ of $N$, $$ f^{-1}(f(H))=H+\ker f \qquad f(f^{-1}(K))=K\cap\operatorname{im}f $$


You can also note that $\pi^{-1}(K)$ is the kernel of the $R$-module homomorphism $M\xrightarrow{\pi} N\longrightarrow N/K$, where $N\longrightarrow N/K$ is the canonical projection onto the factor module. Since kernels are submodules this proves again that $\pi^{-1}(K)$ is an $R$-module.


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