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