Let $R$ be a commutative ring with a unit and let the following picture be a commutative diagram of $R$-modules with exact rows. ive Suppose that $\alpha$ and $\beta$ are isomorphisms. Show that the natural homomorphism $\bar{\delta}:\ker(\varphi)\to\ker(\psi)$ is an isomorphism.


According to the snake' lemma, there is a natural homomorphism $\bar{\delta}:\ker(\varphi)\to\ker(\psi)$.

$\bar{\delta}$ is surjective:

Let $n\in\ker{\psi}$. Then $\psi(n)=0\Rightarrow \beta(\eta(n))=\eta'(\psi(n))=0$. Because that $\beta$ is injective then $\eta(n)=0\Rightarrow n\in\ker\eta=\operatorname{Im}\delta\Rightarrow\exists m\in M,\delta(m)=n.$

We know that $\delta'(\varphi(m))=\psi(\delta(m))=0\Rightarrow \varphi(m)\in\ker\delta'=\operatorname{Im}\epsilon'\Rightarrow\exists k'\in K', \epsilon'(k')=\varphi(m).$

$\alpha$ is surjective. Hence, $$ \exists k\in K, \alpha(k)=k' $$Here I stuck. If I succeed to show that $\varphi(m)=0$, (or that $\epsilon'=0$) I'll finish the proof.

$\bar{\delta}$ is injective:

In this part I had no problem:

Let $m\in\ker\bar\delta$. Then $0=\bar\delta(m)=\delta(m)\Rightarrow m\in\ker\delta=\operatorname{Im}\epsilon\Rightarrow\exists k\in K,\epsilon(k)=m$.

We know that $\varphi(m)=0\Rightarrow0=\varphi(\epsilon(k))=\epsilon'(\alpha(k))$. Hence $\alpha(k)\in\ker\epsilon'=\operatorname{Im}(0\to K')\Rightarrow \alpha(k)=0$. But $\alpha$ is injective, then $k=0\Rightarrow m=\epsilon(k)=0$. Thus $\bar\delta$ is injective.

  • $\begingroup$ Do you know the snake lemma? $\endgroup$
    – Bernard
    Sep 10, 2019 at 14:29
  • $\begingroup$ Yes, but I don't see how it can help. $\endgroup$
    – J. Doe
    Sep 10, 2019 at 14:33
  • $\begingroup$ Are the rows exact? $\endgroup$ Sep 10, 2019 at 14:34
  • $\begingroup$ If the rows are exact (I suppose so) the beginning of the $6$ terms exact sequence in the snake lemma makes it obvious: $\ker\alpha=0=\operatorname{coker}\alpha$. $\endgroup$
    – Bernard
    Sep 10, 2019 at 14:38
  • 2
    $\begingroup$ Diagram chasing is a nice way of passing the time. $\endgroup$
    – Wuestenfux
    Sep 10, 2019 at 15:36

1 Answer 1


$\DeclareMathOperator{\im}{Im}$ Let's do some grunt diagram chasing.

First, we prove $\overline{\delta}$ is injective. Suppose $x\in\ker\overline{\delta}$. Then $x\in\ker(\delta)=\im(\varepsilon)$, so $x=\varepsilon(k)$ for some $k\in K$. But also, $x\in\ker\varphi$, so $$0=\varphi(x)=\varphi(\varepsilon(k))=\varepsilon'(\alpha(k))$$ But $\epsilon'$ is injective, so $\alpha(k)=0$. Since $\alpha$ is an isomorphism $k=0$, so $x=\varepsilon(k)=\varepsilon(0)=0$.

Therefore, $\overline{\delta}$ is injective.

Now let us prove $\overline{\delta}$ is surjective. Let $y\in\ker\psi$. Then $\psi(y)=0$, so $$0=\eta'(\psi(y))=\beta(\eta(y))$$ Since $\beta$ is an isomorphism, $\eta(y)=0$, so $y\in\ker(\eta)=\im(\delta)$. Write $y=\delta(x)$ for some $x\in M$.

We have $\delta'(\varphi(x))=\psi(\delta(x))=\psi(y)=0$, so $\varphi(x)\in\ker\delta'=\im(\varepsilon')$. Let $k'\in K'$ such that $\varphi(x)=\varepsilon'(k')$. Since $\alpha$ is an isomorphism, there is $k\in K$ such that $k'=\alpha(k)$. Then $$\varphi(\varepsilon(k))=\varepsilon'(\alpha(k))=\varepsilon'(k')=\varphi(x)$$ So now we use the $R$-module structure: Let $z=x-\varepsilon(k)$. Then the above means that $z\in\ker\varphi$. We prove that $\overline{\delta}(z)=y$: $$\overline{\delta}(z)=\delta(x)-\delta\varepsilon(k)=\delta(x)=y$$ because $\delta\varepsilon=0$, since the first row is exact.


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