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The splitting lemma says:

Given a short exact sequence with maps $q$ and $r$:

$0 \rightarrow A \overset{q}{\rightarrow} B \overset{r}{\rightarrow} C \rightarrow 0$

then the following are equivalent:

  1. ...
  2. there exists a map $u : C \rightarrow B$ such that $r \circ u = \mathrm{id}_C$
  3. $B \cong A \oplus C$

Now I figured, since $r(B) = \ker 0$, $r$ is surjective. Hence, for every $c \in C$ we have some $b \in B$ such that $r(b) = c$. Simply set $u(c) = b$ for one of these $b$s and you have your map $u$.

However, it turns out that this construction assumes the axiom of choice; it chooses one element from each of an infinite number of sets. So my question is: assuming the axiom of choice, is condition (2) always satisfied? Because this would imply that (3) holds for any such short exact sequence. Or am I making some mistake here?

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    $\begingroup$ You aren't guaranteed that u is linear. $\endgroup$ – Qiaochu Yuan Nov 15 '09 at 22:27
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I assume you are working in some fixed abelian category $\mathcal{A}$.

It is not true in general that every short exact sequence in $\mathcal{A}$ will split. The problem is that although you can pick a preimage for every 'element' $c\in C$ there is no guarantee that you can assemble this into a morphism in $\mathcal{A}$. It is true in the category of sets that every surjection splits if one assumes the axiom of choice but this is only a set map.

For instance in the category of abelian groups $$0 \to \mathbb{Z}/2\mathbb{Z} \stackrel{2}{\to} \mathbb{Z}/4\mathbb{Z} \to \mathbb{Z}/2\mathbb{Z} \to 0$$ is exact but $\mathbb{Z}/4\mathbb{Z}$ is indecomposable.

Also if you are not in an abelian category it is not necessarily true that exact sequences display this symmetry. See for example this question where the notion is considered in the category of groups.

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  • $\begingroup$ So it should read "morphism" instead of "map" in the formulation of the lemma. My book (Miles Reid, Undergrad. Comm. Algebra) also calls it a "map". I'm not really familiar with category theory, but in algebraic geometry (homo)morphisms are always explicitly called so, and are not the same as maps in general. Hence the confusion; thank you all for the answers. $\endgroup$ – Thomas Nov 16 '09 at 11:39
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    $\begingroup$ In category theory, "map" and "morphism" (and "arrow") are synonymous. $\endgroup$ – Tom Leinster May 31 '12 at 10:19
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The mistake you're making is that your map $u$ is not a homomorphism of (whatever $A$, $B$, $C$ are---possibly groups or modules), it's just going to be a map of sets, if you define it the way you defined it. Read the proof of the lemma to see why this isn't good enough. Or think of $Z/2Z$ living in $Z/4Z$ with quotient $Z/2Z$.

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For the record, the axiom of choice is not required to prove the splitting lemma. I can see why you might think it might; I once came to the same conclusion myself. To split the sequence on the right, you could start by finding a section of the surjection, invoking AC. Then you can project out the part of your section which doesn't lie in the kernel, which ensures the map is a homomorphism.

I asked about for clarification on math.SE, with a negative answer: no choice necessary. Basically, instead of starting by splitting the surjection, you recall that a retraction of the injection was part of your given data, and use it to construct the splitting of the surjection, without making any arbitrary choices.

https://math.stackexchange.com/questions/151438/does-the-splitting-lemma-hold-without-the-axiom-of-choice/

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The axiom of choice may give you a map $u:C\to B$ as SETS. However, it may not be a morphism in the category of $R$-modules.

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This is right: a map $f:A \rightarrow B$ is surjectiv $\Longleftrightarrow$ $f$ has a right-inverse. The proof needs the axiom of choice, as you pointed out correctly. But this is just a map of sets.

EDIT: In the following I'm talking about groups (vector spaces, vector bundles, presheaves, sheaves,... should also do the job)

Every short exact seqeunce can be seen as a sequence $$0 \overset{inc_0}\rightarrow A \overset{inc}{\rightarrow} B \overset{\pi}{\rightarrow} B/A \rightarrow 0$$ where $inc$ denotes the inclusion and $\pi$ the projection.

But (as for example Charles Siegel pointed out) surjectivity gives you just a rightinverse $u:C \rightarrow B$ as map of sets. So if you have further structures (let's say a group structure, vector space structure, etc.), this doesn't mean, that the map $u$ is an inverse with respect to these structures

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It is not true: consider the exact sequence of abelian groups $$0 \longrightarrow \mathbb{Z} \xrightarrow{x\mapsto 3x} \mathbb{Z} \xrightarrow{x\mapsto [x]_3} \mathbb{Z}/3 \longrightarrow 0 \quad$$ and note that the only homomorphism $\mathbb{Z}/3 \rightarrow \mathbb{Z}$ is the zero one. So the sequence does not split.

EDIT - As clarification, in the splitting lemma, the third condition should be:

  1. ...
  2. ...
  3. The sequence $0\rightarrow A \xrightarrow{q} B \xrightarrow{r} C\rightarrow 0$ and the canonical one $0\rightarrow A \xrightarrow{i} A \oplus C \xrightarrow{\pi} C\rightarrow 0$ are isomorphic (with the identities on A and on C).
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