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Let's state this question in cokernels first. The case of kernels are easily solved since free modules are projective.

Let $0 \to P \to N \to M \to 0$ be a short exact sequence of modules over a ring $R$. Given free resolutions $0 \to L_n \to L_{n-1} \to \cdots \to L_0 \to N \to 0$ and $0 \to K_m \to K_{m-1} \to \cdots \to K_0 \to P \to 0$, how to construct a free resolution of $M$? The idea is to fit the given information into an exact diagram, but I cannot prove the canonical injections $K_n \to L_n$ assured by projectivity of free modules split. Would this work with some extra care, or do I need to follow a different approach?

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  • $\begingroup$ What are the objects here exactly? $\endgroup$ Mar 30 at 11:43
  • $\begingroup$ @AntonOdina Apologies for not specifying- they ought to be modules over a ring $R$. $\endgroup$ Mar 31 at 3:15

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The map $P \to N$ induces a map of complexes $f \colon K_\bullet \to L_\bullet$. The mapping cone of $f$ gives a free resolution for $M$. https://en.wikipedia.org/wiki/Mapping_cone_(homological_algebra)

This follows by considering the long exact sequence on homology of $0 \to L_\bullet \to cone(f) \to K_\bullet[1] \to 0$ (where $[1]$ means homological shift), and the existence of this short exact sequence follows from the definition of mapping cone.

I think you are mistaken about the kernel case being easy. If you have free resolutions $L_\bullet \to N$ and $F_\bullet \to M$ instead, then you get an induced map $g \colon L_\bullet \to F_\bullet$ between their free resolutions, but those maps don't have to be surjective, so they don't necessarily split. Again you can take the mapping cone of $g$, call it $C_\bullet$. But $H_1(C_\bullet) = P$ (and all other homology vanishes, can be proven from what I said above) so you take the truncation: $\cdots \to C_n \to \cdots \to C_2 \to C'_1$ where $C'_1 = \ker(C_1\to C_0)$.

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I am not sure if I understood you correctly. But if I did, the problem is that the inclusions do not split in general.

Take the usual example of a non-split SES $$0\rightarrow \Bbb Z \overset{\cdot 2}\rightarrow \Bbb Z \rightarrow \Bbb Z/2 \rightarrow 0$$ and note that the first two factors are already free. Hence their free resolutions turn out to be just $\Bbb Z \overset{=}\rightarrow \Bbb Z$. But then the natural inclusion is again the map $\Bbb Z \overset{\cdot 2}\rightarrow\Bbb Z$, which does not split. This is to say, in general you won't be able to construct a free resolution of the quotient as the quotient of the free resolutions (which I think you were trying to do).

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  • $\begingroup$ Yeah, I am aware of this.. Are there alternative methods though? $\endgroup$ Mar 31 at 3:17
  • $\begingroup$ As I said, I was not 100% sure how to read your question. I don‘t know about a way to get at a free resolution of the quotient. Based on my counterexample I feel like the obstruction is that the free resolution of the extension might be too small to admit a quotient, which is free again. I guess this is why one usually resolves kernel and quotient and adds them up to resolve the extension. But maybe someone with more experience in homological algebra should elaborate on this… $\endgroup$ Mar 31 at 10:11

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