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The statement of the problem (Aluffi III.7.17) is as follows:

Consider a (possibly infinite) commutative diagram of $R$-modules in which the central column is a complex and every row is exact. Prove that the left and right columns are also complexes. Prove that if any two of the columns are exact, so is the third.

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I have no problem proving the first statement, it is the second in bold that is causing me some difficulty. What remains is showing 3 cases. Here is what I have attempted so far:

Case 1: If column 1 and 2 are exact then column 3 is exact. Need to show $\ker\varepsilon_i\subseteq\text{im }\varepsilon_{i+1}$.

Let $n_i\in\ker\varepsilon_i$. Then $\varepsilon_i(n_i)=0$. Since $\beta_i$ is surjective $\exists m_i\in M_i$ such that $\beta_i(m_i)=n_i$. So $\varepsilon_i\circ\beta_i(m_i)=0$. By commutativity of the diagram, $\beta_{i-1}\circ\delta_i(m_i)=0$. Then $\delta_i(m_i)\in\ker\beta_{i-1}=\text{im }\alpha_{i-1}$ so $\exists l_{i-1}\in L_{i-1}$ such that $\alpha_{i-1}(l_{i-1})=\delta(m_i)$. (Not sure where to go from here, I suspect injectivity of $\alpha_{i-1}$ may need to be used but I’m not exactly sure how).

Case 2: If column 1 and 3 are exact then column 2 is exact. Need to show $\ker\delta_i\subseteq\text{im }\delta_{i+1}$.

Let $m_i\in\ker\delta_i$. Then $\delta_i(m_i) = 0$ and thus $\beta_{i-1}\circ\delta_i(m_i)=0$. By commutativity of the diagram, $\varepsilon_i\circ\beta_i(m_i)=0$. Then $\beta_i(m_i)\in\ker\varepsilon_i=\text{im }\varepsilon_{i+1}$. So $\exists n_{i+1}\in N_{i+1}$ such that $\varepsilon_{i+1}(n_{i+1}) = \beta_i(m_i)$. $\beta_{i+1}$ is surjective so $\exists m_{i+1}\in M_{i+1}$ such that $\beta_{i+1}(m_{i+1})=n_{i+1}$. So $\varepsilon_{i+1}\circ\beta_{i+1}(m_{i+1})=\beta_i(m_i)$. By commutativity of the diagram once more, $\beta_i\circ\delta_{i+1}(m_{i+1})=\beta_i(m_i)$. So $\beta_i(\delta_{i+1}(m_{i+1})-m_i)=0$. Then $\delta_{i+1}(m_{i+1})-m_i\in\text{im }(\alpha_i)$ so $\exists l_i\in L_i$ such that $\alpha_i(l_i)=\delta_{i+1}(m_{i+1})-m_i$. Separately, $\delta_i\circ\alpha_i(l_i)=\alpha_{i-1}\circ\gamma_i(l_i)$ and thus $\gamma_i(l_i)=0$. So $l_i\in \text{im }\gamma_{i+1}$ so $\exists l_{i+1}\in L_{i+1}$ such that $\gamma_{i+1}(l_{i+1})=l_i$. So $\alpha_i\circ\gamma_{i+1}(l_{i+1})=\delta_{i+1}(m_{i+1})-m_i$ and by commutativity it follows that $\delta_{i+1}(m_{i+1}-\alpha_{i+1}(l_{i+1}))=m_i$, i.e. $\ker\delta_i\subseteq\text{im }\delta_{i+1}$.

Case 3: If column 2 and 3 are exact then column 1 is exact. Need to show $\ker\gamma_i\subseteq\text{im }\gamma_{i+1}$.

Let $l_i\in\ker\gamma_i$. Then $\gamma_i(l_i)=0$ so $\alpha_{i-1}\circ\gamma_i(l_i) = 0$. Then $\delta_i\circ\alpha_i(l_i)=0$. By exactness of the middle column there exists $m_{i+1}\in M_{i+1}$ such that $\delta_{i+1}(m_{i+1})=\alpha_i(l_i)$. (Similarly to case 1 I get stuck here and suspect I need to use injectivity of $\alpha_i$).

Any pointers in the right direction would be very much appreciated. Thanks in advance!

iceberg56
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    Search for the ninr lemma or 3x3 lemma. This is almost certainly a duplicate. – Lukas Heger May 25 '25 at 21:59
  • @LukasHeger nine lemma seems to require top / bottom rows being all zeroes which is not the case here – iceberg56 May 25 '25 at 22:08
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    I don't have Aluffi at hand, but does he cover the long exact sequence in cohomology induced by a short exact sequence of complexes? If you apply this, it's pretty straightforward. A complex is exact iff it has zero cohomology. – Lukas Heger May 25 '25 at 22:12
  • @LukasHeger he does make a comment to this effect: The first part is straightforward. The second part will take you a couple of minutes now due to the needed diagram chases, and a couple of seconds later, once you learn about the long exact (co)homology sequence in §IX.3.3. Since this is in a very introductory section to homology where we really only have the snake lemma I would like to try and prove the two cases with a diagram chase instead of resorting to the more advanced methods that will show up later in the book – iceberg56 May 25 '25 at 22:17
  • Usually this exercise is with zero rows on top and bottom, see here. Then the Snake Lemma resp. Salamander Lemma can be used. But diagram chasing also works. I suppose you have now enough ideas to proceed. – Dietrich Burde May 26 '25 at 09:18
  • @DietrichBurde do you have any advice on how I might be able to proceed with a diagram chase given the work I have shown above? – iceberg56 May 26 '25 at 17:37

1 Answers1

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The question in case 1: Let $m_{i-1}$ denote $\delta_i (m_i)$. Note that, because the central column is a complex, $\delta_{i-1}(m_{i-1}) =0$. By commutativity this means that $\alpha_{i-2}(\gamma_{i-1}(l_{i-1}) =0 $, and because this map is injective we conclude that $\gamma_{i-1}(l_{i-1}) = 0$. Thus it lifts to some $l_{i}$, now consider $m_i - \alpha_i(l_i)$. On the one hand this maps to the original $n_i$ through $\beta_i$, and on the other hand it is mapped to $0$ by $\delta_i$. Hence it lifts to some $m_{i+1}$ and $\beta_{i+1}(m_{i+1})$ is your desired lift.

The question in case 3: Let $m_i$ denote $\alpha_i(l_i)$. As you mentioned this lifts to some $m_{i+1}$. Let $n_{i+1}$ denote $\beta_{i+1}(m_{i+1})$. Observe that, $\varepsilon_{i+1}(n_{i+1})=0$ because $\beta_i(m_i)=0$, hence $n_{i+1}$ lifts to some $n_{i+2}$. By exactness of the rows there is some $m_{i+2}$ such that $\beta_{i+2}(m_{i+2})=n_{i+2}$. Consider $m_{i+1} - \delta_{i+2}(m_{i+2})$. On the one hand this maps to $0$ through $\beta_{i+1}$, and on the other hand it is mapped to $m_i$ by $\delta_{i+1}$. Now, by exactness of the $(i+1)$'th row there is some $l_{i+1}\in L_{i+1}$ such that $\alpha_{i+1}(l_{i+1}) = m_{i+1} - \delta_{i+2}(m_{i+2})$. This is the desired lift of $l_i$ because $\gamma_{i+1}(l_{i+1}) - l_i$ gets mapped to $0$ by $\alpha_i$ because of the commuting square, the claim then follows from the injectivity of $\alpha_i$.

Hidde Bakker
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