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I am trying to prove the existence part of the comparison theorem for injective resolutions. Every reference I find says that it is just the dualization of the proof for projective resolutions, but I am not seeing how to dualize it. The issue is that in the proof of the projective case, they assume inductively that a chain map has been constructed at degree $n$. Then they use the commutativity of the square to show that $n$-cycles get taken to $n$-cycles.

My idea to dualize this was to use the fact that boundaries get taken to boundaries. However, there's no way to do this because in the dual case, you don't yet have the chain map $f_{n+1}$ that can take those boundaries to boundaries.

Has anyone got a link to a proof for the injective case, or is someone able to tell me how to dualize the standard proof for the projective case?

Daniela
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  • I think your issue would be more clearer if your post actually contained the diagram (or a complete description of it) that you are referring to. –  Sep 17 '17 at 11:44
  • As a quick sanity check, make sure you've actually reversed all of the arrows. And while in the projective case the induction goes in the opposite direction of of the boundary maps, in the injective case the induction goes in the same direction as the boundary maps. (i.e. both cases take the end of the resolution as the base case and proceed along the resolution from there) –  Sep 17 '17 at 11:45

1 Answers1

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Suppose that $f : M \to N$ is a map of modules, and that $M\to I^*$ and $N\to J^*$ are injective resolutions of $M$ and $N$ respectively. Begin by noting that we have an incomplete diagram of the form $$ \require{AMScd} \begin{CD} M @>>> I^0\\ @VfVV \\ N @>>> J^0 \end{CD}$$

where both horizontal maps are injective. Now consider the map $M\to J^0$ obtained by composition. Since $J^0$ is injective, this can be extended to a map $I^0\to J^0$, rendering the diagram commutative. This starts the induction. In the projective case you would now consider the following diagram

$$ \require{AMScd} \begin{CD} && P^k\\ &&@VVV \\ Q^k @>>> \operatorname{im} d@>>> 0 \end{CD}$$

to complete the induction. Can you imagine what to do in this dual case? You should try to obtain a diagram of the form

$$\require{AMScd} \begin{CD} 0 @>>> ?? @>>> I^k\\ &&@VVV \\ && J^k \end{CD}$$

and use that $J^k$ is injective.

Pedro
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  • (Old thread, I know:) My approach was to replace $I^{k-1}$ with $I^{k-1} / \ker(d_{k-1})$ to induce an injective map to $I^k$, but I have not been able to make this work as I would also need to replace $J^k$ with a factor module, for which I don't know why it should be injective. Also, this thread suggests that this is not easily doable. Do you still know what your ?? was supposed to be? – NiklasvMoers Sep 15 '20 at 18:29
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    @NiklasvMoers That's the right approach, but you just want to observe that, by exactness that, say at the beginning of the proof, the quotient $I^0/\ker^0$ is isomorphic to the image of $\varepsilon$. Hence, you do have a map to $J^1$. This should get you going. – Pedro Sep 15 '20 at 20:01