Proof:

We shall prove that $\eta$ is injective, leaving surjectivity as an exercise. Assume $\eta(a_3) = 0$ for some $a_3 \in A_3$ so that $\beta_3\circ \eta(a_3) = 0$. The commutativity of the third square now gives $\phi_1\circ \alpha_3(a_3) = 0$. Using injectivity of $\phi_1$ and exactness of the top row we arrive at

$$a_3 = \alpha_2(a_2) \eqno(38.1)$$

for some $a_2 \in A_2$. Again, $0 = \eta(a_3) = \eta\circ\alpha_2(a_2) = \beta_2\circ \psi_2(a_2)$ showing that $\psi_2(a_2) \in\;$im$\;\beta_1$. Thus $\psi_2(a_2) = \beta_1(b_1)$ for some $b_1 \in B_1$ and using the surjectivity of $\psi_1$ we get for some $a_1 \in A_1$,

$$\psi_2(a_2) = \beta_1\circ\psi_1(a_1) = \psi_2\circ \alpha_1(a_1).$$

Injectivity of $\psi_2$ now gives $a_2 - \alpha_1(a_1) = 0$. Substituting into (38.1) we conclude $a_3 = 0$.