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If $G_1, G_2, G_3$ are abelian groups and $0 \to G_1 \xrightarrow{\varphi_1} G_2 \xrightarrow{\varphi_2} G_3 \to 0$ is exact, then $G_2 \simeq \ker(\varphi_2) \oplus \text{im}(\varphi_2) \simeq G_1 \oplus G_3$

The above is the conclusion of Example 12.2 in Nonlinear Analysis and Semilinear Elliptic Problems, by Ambrosetti and Malchiodi.

I worked out as follows:

By exactness, ${0} = \ker(\varphi_1)$, so $G_1 \simeq \text{im}(\varphi_1) = \ker(\varphi_2) \lhd G_2$, so it makes sense to consider $G_2/G_1$. On the other hand, again by exactness, $\text{im}(\varphi_2) = G_3$. By the First Isomorphism Theorem, $$ G_2/ \ker(\varphi_2) \simeq G_2/G_1 \simeq G_3. $$

What one would like to do now is to "multiply both sides by $G_1$ and cancel out in the left-hand side". My question is, how to do it in a rigorous way?

I tried to write $G_2/G_1$ explicitly, but it was a dead end. I also tried to follow the hint by Najib Idrissi in this question, but failed.

Thanks in advace.

Danilo Gregorin Afonso
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    This doesn't work: $0\to\mathbb{Z}\xrightarrow{2}\mathbb{Z}\to\mathbb{Z}/2\to 0$. – user10354138 Feb 25 '21 at 13:07
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    As stated, it is false. Counterexample: $0 \to n\mathbf Z\to\mathbf Z\to\mathbf Z/n\mathbf Z\to 0$ is exact, but your assertion would imply the ideal $n\mathbf Z$ is generated by an idempotent, which is impossible as $\mathbf Z$ is an integral domain. I guess you're for getting a hypothesis. – Bernard Feb 25 '21 at 13:08
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    The claim is false. See split exact sequence. –  Feb 25 '21 at 13:08
  • Thank you all for the examples. I think that I didn't forget any details, maybe it was a distraction by the authors. The "Fundamental Theorem of Homomorphism" is the First Isomorphism Theorem, isn't it? – Danilo Gregorin Afonso Feb 25 '21 at 13:10
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    they are closely related, see: https://en.wikipedia.org/wiki/Fundamental_theorem_on_homomorphisms – Zest Feb 25 '21 at 13:14
  • I am a bit confused by the example of @user10354138. If I understand correctly, 2 is the map that multiply by 2, right? So its image are the even numbers. But when we take this to $\mathbb Z/2$, don't we map everybody to $[0]$? I'm sorry if I'm saying insanities – Danilo Gregorin Afonso Feb 25 '21 at 13:21
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    The last map sends an integer onto its congruence class modulo $2$, so even numbers indeed map to $[0]$ and odd numbers map to $[1]$. Actually, the decomposition of $G_2$ as a direct sum is correct if you can prove $G_3$ is a projective abelian group, for instance. – Bernard Feb 25 '21 at 13:24
  • @Bernard indeed was an insanity, I'm sorry. And thank you! – Danilo Gregorin Afonso Feb 25 '21 at 13:25
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    minor correction: You have written a $\varphi_3$ in the first line. Is that supposed to be $\varphi_2$? – Aryaman Maithani Feb 25 '21 at 13:33
  • @AryamanMaithani yes! Thanks for pointing out – Danilo Gregorin Afonso Feb 25 '21 at 13:38
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    The usual counterexample with finite groups is $0\to\mathbb Z/2\mathbb Z\to \mathbb Z/4\mathbb Z\to\mathbb Z/2\mathbb Z\to 0$, where the second map is reduction mod $2$ and the first map send $0, 1\mapsto 0, 2$. – Mathmo123 Feb 25 '21 at 13:54
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    The authors of that book are just wrong, there is no way around this. I checked the book, that's verbatim what they wrote, and it's simply false. – Najib Idrissi Feb 25 '21 at 14:28
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    @Najib: I don't have a copy of the book. Is it possible that the authors only intended free abelian groups? E.g., do they then append a proof of the statement which indicates they are only thinking about free abelian groups? – Jason DeVito - on hiatus Feb 25 '21 at 15:22
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    @JasonDeVito No, I'm pretty sure it's an oversight on their part. Here is a screenshot of the relevant part: https://imgur.com/AM9nZzb Afterwards, they apply it to e.g. long exact sequences in homology, which can certainly involve non-free groups. – Najib Idrissi Feb 25 '21 at 15:30
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    @NajibIdrissi: Thanks for the screenshot. That is incredibly weird because the counterexamples are so well known and easy to construct. I must say, I'm very curious as to what the "fundamental theorem of the homomorphism" is. It sounds like the first isomorphism theorem, but it obviously is being misused if so. – Jason DeVito - on hiatus Feb 25 '21 at 17:39
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    @JasonDeVito: I don't understand why the "First Isomorphism Theorem" is a misuse here? we have a surjective homomorphism and its kernel is $G_1$. What is the problem? – C.F.G Feb 25 '21 at 19:02
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    @C.F.G: The first isomorphism theorem asserts an isomorphism between $G_3$ and $G_2/\phi_1(G_1)$ (which is correct), but does not allow to to conclude that we can "multiply" by $\phi_1(G_1)$ to obtain $G_2\cong G_3\times \phi_1(G_1)\cong G_3\times G_1$. – Jason DeVito - on hiatus Feb 25 '21 at 19:09
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    @DaniloGregorinAfonso: Thanks for bringing the book name to me, I like its style as Idrissi screenshot shows, but simple mistakes like this post make me hesitant to read it. – C.F.G Feb 25 '21 at 19:17
  • @C.F.G Not very bad, I actually like it and think its worth reading. But they make simple computations a bit messy at times, and some proofs are rather sketchy. If you need references for the subjects the book is about, please send me an email. – Danilo Gregorin Afonso Feb 25 '21 at 19:21
  • @JasonDeVito: I had seen this fact about "splitting" over and over like this common false beliefs but I never understand why it means really until your last comment about "multiply" end it. Thanks a lot. – C.F.G Feb 26 '21 at 12:57

1 Answers1

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This is unfortunately false. Consider the standard counterexample

$$0 \to \mathbb{Z} \xrightarrow{\varphi} \mathbb{Z} \xrightarrow{\pi} \mathbb{Z}/2\mathbb{Z} \to 0$$

where $\varphi(x)=2x$ and $\pi(x)=x+2\mathbb{Z}$ is the quotient map. The sequence is exact but clearly $\mathbb{Z}$ is not a direct sum $\mathbb{Z}\oplus(\mathbb{Z}/2\mathbb{Z})$.

For that to hold we need the $0 \to G_1 \xrightarrow{\varphi_1} G_2 \xrightarrow{\varphi_2} G_3 \to 0$ sequence to be a split exact sequence (see also: splitting lemma). This is for example true whenever $G_3$ is free abelian, i.e. $G_3\simeq \bigoplus\mathbb{Z}$.

freakish
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