Following the proof outlined in here, I wanted to show that when

$$\begin{alignedat}{3} 0 \longrightarrow \mathord{} & K_j & \mathord{} \longrightarrow \mathord{} & A_j & \mathord{} \longrightarrow \mathord{} & B_j \\ & \downarrow && \downarrow && \downarrow \\ 0 \longrightarrow \mathord{} & K & \mathord{} \longrightarrow \mathord{} & A & \mathord{} \longrightarrow \mathord{} & B \\ \end{alignedat}$$

commutes, the rightmost arrow being mono and both rows being exact, then the left square is a pullback. However, I only needed the bottom row to be exact at $ A $, which threw me off a bit:

If you have $$ \require{AMScd} \begin{CD} U@>>>A_j\\ @VVV @VVV\\ K @>>> A \end{CD} $$

Then by exactness at $ A $ and commutativity, you have $$\begin{CD} U @>>> A_j @>>> B_j\\ @.@.@VVV\\ @.@.B \end{CD}$$ equal to $ 0 $, and then because $ B_j \longrightarrow B $ is mono, $ U \longrightarrow A_j $ factors uniquely through $ K_j \longrightarrow A_j $. The other leg of the pullback diagram commutes by uniqueness of the kernel factorisation through $ K \longrightarrow A $.

Does that proof seem correct?


You have the right idea, but you really need the bottom row to be exact at $K$ and not necessarily at $A$.

Indeed, to get the factorisation of $U\to A_j$ through $K_j$ you need to prove that the composition $U\to A_j\to B_j$ is zero, and for this you only need to know that $K\to A\to B$ is zero, which is weaker than exactness at $A$.

On the other hand, to prove that the second leg of the pullback commutes, you use the uniqueness of the factorisation through $K$, or in other words, the fact that $K\to A$ is a monomorphism, which is equivalent to the exactness of $0\to K\to A$ at $K$.

  • $\begingroup$ Thank you very much, it's a shame I didn't notice the missing assumption... $\endgroup$ – FreeSalad May 21 at 10:14

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