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If we have a short exact sequence $0 \to A \to B \to C \to 0$, then the map $A \to B$ is injective and the map $B \to C$ is surjective.

Therefore, there always exists a left inverse for $i$ and a right inverse for $j$. So, every short exact sequence splits?

user5826
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  • How about $0\to \mathbb Z\to\mathbb Z\to\mathbb Z/7\mathbb Z\to 0$ in abelian groups? There is no non-zero homomorphism $\mathbb Z/7\mathbb Z\to\mathbb Z.$ – Thomas Andrews Mar 28 '19 at 19:44
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    As functions, onto maps always have inverses, but those inverses might not work as algebraic homomorphisms. – Thomas Andrews Mar 28 '19 at 19:46
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    @ThomasAndrews That's not even a hint. That's a full answer. Please don't put those in comments. – Arthur Mar 28 '19 at 19:47
  • It's not true, for example, if a map $f:X\to Y$ is continuous on topological spaces, $X,Y$ and onto, it might not be true tha the inverse is continuous. Consider the standard continuuous function $[0,1]\to S^1.$ There is no continous inverse. – Thomas Andrews Mar 28 '19 at 19:48
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    As the excerpt notes, if $0\to A\to B\to C\to 0$ is exact and splits, then you get $B\cong A\oplus C$. So the question you should ask yourself is, "If $B$ is an abelian group and $A$ is a subgroup, can I always write $B$ as a direct sum with $A$ as one summand? – Arturo Magidin Mar 28 '19 at 19:48
  • Ahh I see now. I just glossed over the homomorphism part. I got used to all maps being structure preserving. – user5826 Mar 28 '19 at 20:24
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    You should look up the terms "projective module" and "injective module". If you have not seen modules, a quick explanation is that general R-modules have the same notion of exact sequences as abelian groups. Then, $C$ is projective, if and only if, any exact sequence $ 0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0$ splits. You might be able to guess the dual equivalence for injective modules. A great exercise would be to use this as a definition for projective abelian group and to try to characterize them. – Connor Malin Mar 28 '19 at 20:46
  • @ConnorMalin Modules are abelian groups with extra structure. Isn't this result, which is only about abelian groups, stronger? – user5826 Mar 29 '19 at 01:06
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    Two things: you are correct that modules are abelian groups with extra structure, but that doesn't mean this result is stronger because the result for modules gives decompositions into direct sums of modules. The result for abelian groups does not imply it for modules (imply being used informally). In fact, the opposite is true. The result for modules implies it for abelian groups because abelian groups are $\mathbb{Z}$ modules. – Connor Malin Mar 29 '19 at 03:36

1 Answers1

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The question's already been answered in the comments, but for posterity there should be an official one.

The answer is No and a nice counterexample is

$$ 0 \to \mathbb{Z} \to \mathbb{Z} \to \mathbb{Z}/2 \to 0$$

We know this can't split because the Splitting Lemma would then imply $\mathbb{Z} \cong \mathbb{Z} \oplus \mathbb{Z}/2$, which is a contradiction because $\mathbb{Z}$ has no torsion.

It is true that in the category of Sets every surjective function has a right inverse and every injective function has a left inverse, but this is not true in the category of Abelian Groups. Indeed a right inverse $\mathbb{Z}/2 \to \mathbb{Z}$ of the quotient map would be any function sending $0$ to an even number and $1$ to an odd number, but this can never be a homomorphism, again because $\mathbb{Z}$ has no torsion.

William
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  • i just stumbled upon your answer and i am slightly confused about the splitting lemma. According to the splitting lemma the existence of a left split $\mathbb{Z} \leftarrow \mathbb{Z}$ in the SES $$0 \to \mathbb{Z} \to \mathbb{Z} \to \mathbb{Z}_2 \to 0$$ would imply $\mathbb{Z} \cong \mathbb{Z}\oplus \mathbb{Z}_2$. What if we choose the identity as the left split? That would lead to the same contradiction you mentioned, so either the identity does not qualify as a left split or i am missing something. What is it? – Zest Jun 09 '20 at 05:23
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    Yes, the problem is that the identity is not a left split. The first map in the sequence is multiplication by $2$, so a left split $p\colon \mathbb{Z} \to \mathbb{Z}$ would have to satisfy $p(2n) = n$. – William Jun 09 '20 at 06:10
  • Ah, of course. I missed that. Thanks William! – Zest Jun 09 '20 at 06:14