Show that $\phi$ is surjective in commutative staircase diagram

Question

I'm having some trouble with the following problem involving a sort of staircase diagram and I would really appreciate some help.

Assume the following commutative diagram of modules and module morphisms has exact rows and columns:

Show that $\phi$ is surjective.

---

Here is what I have so far:

• Let some $y\in H_4$ be given. Since the diagram has exact rows, $\psi_3$ is surjective so there exists some $u\in L_3$ with $\psi_3(u)=\alpha_4(y)$.
• Since the diagram is commutative and since $\text{im}(\alpha_4)=\text{ker}(\beta_4)$, $$\chi_3(\beta_3(u)) = \beta_4(\psi_3(u)) = \beta_4(\alpha_4(u)) = 0$$ so $\beta_3(u) \in \text{ker}(\chi_3) = \text{im}(\chi_2)$. Let $v\in M_2$ be such that $\chi_2(v)=\beta_3(u)$.
• Again we have $$\eta_2(\gamma_2(v)) = \gamma_3(\chi_2(v))=\gamma_3(\beta_3(u))=0$$ so $\gamma_2(v)\in \text{ker}(\eta_2) = \text{im}(\eta_1)$. Let $w\in N_1$ be such that $\eta_1(w) = \gamma_2(v)$
• Since $\gamma_1$ is surjective, there exists some $a\in M_1$ with $\gamma_1(a) = w$.

This is where I get stuck. I think I somehow need to use all this to swim back up the diagram but I can't figure out how. I also had a thought that if $\phi(x)=y$ then $$\alpha_4(y)=\alpha_4(\phi(x))=\psi_3(\alpha_3(x))$$ so it would be sufficient to find some $x$ such that $\psi_3(\alpha_3(x))=\alpha_4(y)$ since $\alpha_4$ is injective, however, I haven't figured out if this is useful.

Answer 1

Following your proof (with different notation) we have the following elements:

• $h_4\in H_4$
• $\ell_4\in L_4$ such that $\ell_4=\alpha_4(h_4)$
• $\ell_3\in L_3$ such that $\ell_4=\psi_3(\ell_3)$
• $m_3\in M_3$ such that $m_3=\beta_3(\ell_3)$
• $m_2\in M_2$ such that $m_3=\chi_2(m_2)$
• $n_2\in N_2$ such that $n_2=\gamma_2(m_2)$
• $n_1\in N_1$ such that $n_2=\eta_1(n_1)$
• $m_1\in M_1$ such that $n_1=\gamma_1(n_1)$
• $\ell_2\in L_2$ such that $m_2-\chi_1(m_1)=\beta_2(\ell_2)$
• $h_3\in H_3$ such that $\ell_3-\psi_2(\ell_2)=\alpha_3(h_3)$

Then we have: \begin{align} (\alpha_4\circ\phi)(h_3) &=(\psi_3\circ\alpha_3)(h_3)\\ &=\psi_3(\ell_3-\psi_2(\ell_2))\\ &=\ell_4-(\psi_3\circ\psi_2)(\ell_2)\\ &=\ell_4\\ &=\alpha_4(h_4) \end{align} from which $\phi(h_3)=h_4$.