# Given a commutative diagram, show that $\delta$ is isomorphism

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

Attempt:

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.

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

$$\DeclareMathOperator{\im}{Im}$$ 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.