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Suppose we have $L \xrightarrow{f} M \xrightarrow{g} N$ where $L,M,N$ are $R$-modules and $f,g$ are homomorphisms. A pseudo-isomorphism of $R$-modules is a homomorphism with finite kernel and cokernel. Is it the case that if 2 out of the 3 maps $f,g,gf$ are pseudo-isomorphisms, then the third is as well?

Here's what I haven't been able to show:

  1. If $f,gf$ are pseudo-isomorphisms, then $g$ has finite kernel.
  2. If $g,gf$ are pseudo-isomorphisms, then $f$ has finite cokernel.

If those two statements can be shown, I believe the answer to my question is yes. I'd be grateful to see either a proof of the two statements above or a counter example. If the 2 of 3 property doesn't hold for $R$-modules generally, are there "nice" restrictions one can place on $L,M,N$ or $R$, so that it does hold? E.g. $L,M,N$ are finitely generated, torsion, etc.

Austin Maison
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2 Answers2

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Let me give a concrete and an abstract answer.

Concretely, let's consider only case 1 you struggled with: If $f$ and $gf$ are pseudo-isomorphisms, then $g$ has finite kernel: Consider $\text{ker}(gf)\stackrel{f}{\longrightarrow}\text{ker}(g)$. Its domain is finite by assumption, and its cokernel $\text{ker}(g)/f(\text{ker}(fg))$ embeds into $\text{coker}(f)$, which is also finite. Hence, $\text{ker}(g)$ is finite.

Abstractly, you may want to look up the notion of a Serre subcategory: The class of finite $R$-modules is a Serre subcategory of the category of all $R$-modules, and the pseudo-isomorphisms are those morphisms mapping to isomorphisms under the corresponding quotient functor. As such, they satisfy the 2-out-of-3 property without further calculation.

Hanno
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  • This is really helpful, and it's the first I've heard of Serre subcategories. Thanks. If you have any hints for statement 2, I'd be very grateful. – Austin Maison Apr 08 '23 at 22:17
  • I'm just now rereading your answer, and I realize I misunderstood it. Your claim is that pseudo-isomorphisms of $R$-modules (not just the finite ones) do satisfy the 2-out-of-3 property. This is great news. I suppose I'd still like to see the explicit calculation of statement 2 if anyone can provide it. – Austin Maison Apr 09 '23 at 01:41
  • I'm hoping you could clarify your answer. The corresponding quotient functor you mention is an exact functor, and so I can see that it must send pseudo-isomorphisms to isomorphisms. Certainly isomorphisms satisfy the 2-out-of-3 property, but I don't see why that implies that the pseudo-isomorphisms must also. Could you provide more details? – Austin Maison Apr 12 '23 at 16:54
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@Hanno has posted the idea of a categorical answer to my question which sounds promising, but since I can't quite follow it to the end, I wanted to post a concrete way to see why the answer to my question is yes.

It follows from an answer to this post without much extra work. The idea is that, by applying the snake lemma to the correct diagram, we get the exact sequence $$0 \to \operatorname{ker}(f) \to \operatorname{ker}(gf) \to \operatorname{ker}(g) \to \operatorname{coker}(f) \to \operatorname{coker}(gf) \to \operatorname{coker}(g) \to 0$$

We'll also use the following easy-to-check fact:

Lemma: If $A \to B \to C$ is an exact sequence of $R$-modules and $A$ and $C$ are finite, then $B$ must also be finite.

The proof of the 2-out-of-3 property for pseudo-isomorphisms of $R$-modules goes as follows.

Suppose $f$ and $g$ are pseudo-isomorphisms. Then $\operatorname{ker}(f), \operatorname{ker}(g), \operatorname{coker}(f), \operatorname{coker}(g)$ are all finite. Applying our lemma to the above exact sequence immediately gives us that $\operatorname{ker}(gf)$ and $\operatorname{coker}(gf)$ must be finite, and hence $gf$ is a pseudo-isomorphism. The other two cases follow for the same reasons.

Austin Maison
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