Let $R$ be a ring and $M,N,S$ be $R$-modules. Let $\varphi_1 : M\to N$ and $\varphi_2 : N\to S$ be homomorphisms. Then $\operatorname{im}(\varphi_2\circ\varphi_1) \subseteq \operatorname{im}(\varphi_2)$ and $\ker(\varphi_1) \subseteq \ker(\varphi_2\circ\varphi_1)$, so we may form the quotients $\operatorname{im}(\varphi_2)/\operatorname{im}(\varphi_2\circ\varphi_1)$ and $\ker(\varphi_2\circ\varphi_1)/\ker(\varphi_1)$.

For the kernels, the application of the first isomorphism theorem to the homomorphism $\varphi_1|_{\ker(\varphi_2\circ \varphi_1)} : \ker(\varphi_2\circ \varphi_1) \to N$ yields the nice description $$ \ker(\varphi_2\circ\varphi_1)/\ker(\varphi_1) \cong \operatorname{im}(\varphi_1) \cap \ker(\varphi_2). $$

Now my question is:

Is there a similar description for the quotient $$\operatorname{im}(\varphi_2)/\operatorname{im}(\varphi_2\circ\varphi_1)?$$

I didn't succeed in finding a suitable application of the first isomorphism theorem in this case.

We may observe that the above reasoning for the kernels works exactly the same for homomorphisms of groups. However for the quotient of the images to make sense for groups, we need the additional property $\operatorname{im}(\varphi_2\circ\varphi_1)$ is a normal subgroup of $\operatorname{im}(\varphi_2)$.

This indicates that the situation might be a bit different for the quotient of the images. If needed, please assume some additional property on the ring $R$ to get a meaningful answer.

  • $\begingroup$ In the general case, why would this quotient be a submodule of $S$ at all? And if that can be false, what sort of nice description are you looking for? $\endgroup$
    – Asaf Karagila
    Dec 1, 2013 at 11:26
  • $\begingroup$ @AsafKaragila: Thanks for your comment! I'm looking for a description similar to that for the kernels. Note that in general, the quotient of the kernels isn't a submodule of $M$, $N$ or $S$, and still it admits that nice description. $\endgroup$
    – azimut
    Dec 1, 2013 at 12:03
  • 1
    $\begingroup$ I guess $\operatorname{im}\varphi_2/\operatorname{im}(\varphi_2\circ\varphi_1) \cong N/(\operatorname{im}\varphi_1 + \ker \varphi_2)$ doesn't count as "nice"? $\endgroup$ Dec 1, 2013 at 16:23
  • $\begingroup$ @DanielFischer: Thanks! This is "nice", in my opinion. Would you mind making an answer out of this, including a proof? $\endgroup$
    – azimut
    Dec 1, 2013 at 22:55

1 Answer 1


For ease of notation, let $A = \operatorname{im}\varphi_1 \subset N$, $T = \operatorname{im}\varphi_2 \subset S$, and $Q = T/\operatorname{im} (\varphi_2\circ\varphi_1) = T/\varphi_2(A)$. Consider the map $\psi = \pi \circ \varphi_2 \colon N \to Q$, where $\pi \colon T \to Q$ is the canonical projection.

The kernel of $\psi$ is

$$\ker \psi = \psi^{-1}(0) = \varphi_2^{-1}(\pi^{-1}(0)) = \varphi_2^{-1}(\varphi_2(A)) = A + \ker\varphi_2,$$

and $\psi$ is surjective because $\varphi_2$ and $\pi$ are ($\varphi_2$ considered as a homomorphism to $T$). By the homomorphism theorem, we have an induced isomorphism

$$\overline{\psi} \colon \frac{N}{\operatorname{im}\varphi_1 + \ker \varphi_2} \to Q = \frac{\operatorname{im}\varphi_2}{\operatorname{im}(\varphi_2\circ\varphi_1)}.$$

  • $\begingroup$ Many thanks for this answer! This is exactly the kind of representation I was looking for, but somehow I wasn't able to find it. $\endgroup$
    – azimut
    Dec 2, 2013 at 11:49

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