Splitting lemma under assumption of the axiom of choice - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T11:59:34Z http://mathoverflow.net/feeds/question/5650 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/5650/splitting-lemma-under-assumption-of-the-axiom-of-choice Splitting lemma under assumption of the axiom of choice Thomas 2009-11-15T22:16:10Z 2012-12-11T19:25:20Z <p>The <a href="http://en.wikipedia.org/wiki/Splitting%5Flemma" rel="nofollow">splitting lemma</a> says:</p> <blockquote> <p>Given a short exact sequence with maps $q$ and $r$:</p> <p>$0 \rightarrow A \overset{q}{\rightarrow} B \overset{r}{\rightarrow} C \rightarrow 0$</p> <p>then the following are equivalent:</p> <ol> <li>...</li> <li>there exists a map $u : C \rightarrow B$ such that $r \circ u = \mathrm{id}_C$</li> <li>$B \cong A \oplus C$</li> </ol> </blockquote> <p>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 $u$.</p> <p>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?</p> http://mathoverflow.net/questions/5650/splitting-lemma-under-assumption-of-the-axiom-of-choice/5651#5651 Answer by Kevin Buzzard for Splitting lemma under assumption of the axiom of choice Kevin Buzzard 2009-11-15T22:29:23Z 2009-11-15T22:29:23Z <p>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 $Z/2Z$.</p> http://mathoverflow.net/questions/5650/splitting-lemma-under-assumption-of-the-axiom-of-choice/5652#5652 Answer by Charles Siegel for Splitting lemma under assumption of the axiom of choice Charles Siegel 2009-11-15T22:30:18Z 2009-11-15T22:30:18Z <p>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. </p> http://mathoverflow.net/questions/5650/splitting-lemma-under-assumption-of-the-axiom-of-choice/5653#5653 Answer by Greg Stevenson for Splitting lemma under assumption of the axiom of choice Greg Stevenson 2009-11-15T22:33:50Z 2009-11-15T22:33:50Z <p>I assume you are working in some fixed abelian category $\mathcal{A}$.</p> <p>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.</p> <p>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.</p> <p>Also if you are not in an abelian category it is not necessarily true that exact sequences display this symmetry. See for example <a href="http://mathoverflow.net/questions/3757/when-does-splits-imply-cosplits" rel="nofollow">this</a> question where the notion is considered in the category of groups.</p> http://mathoverflow.net/questions/5650/splitting-lemma-under-assumption-of-the-axiom-of-choice/5654#5654 Answer by Spinorbundle for Splitting lemma under assumption of the axiom of choice Spinorbundle 2009-11-15T22:39:26Z 2009-11-15T22:59:49Z <p>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.</p> <p>EDIT: In the following I'm talking about groups (vector spaces, vector bundles, presheaves, sheaves,... should also do the job)</p> <p>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.</p> <p>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</p> http://mathoverflow.net/questions/5650/splitting-lemma-under-assumption-of-the-axiom-of-choice/84176#84176 Answer by Daniele for Splitting lemma under assumption of the axiom of choice Daniele 2011-12-23T18:56:41Z 2011-12-24T19:14:08Z <p>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. </p> <p>EDIT - As clarification, in the splitting lemma, the third condition should be:</p> <ol> <li>...</li> <li>...</li> <li>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).</li> </ol> http://mathoverflow.net/questions/5650/splitting-lemma-under-assumption-of-the-axiom-of-choice/98447#98447 Answer by Joe Hannon for Splitting lemma under assumption of the axiom of choice Joe Hannon 2012-05-31T01:35:15Z 2012-12-11T19:25:20Z <p>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 came to the same conclusion myself. To split the sequence on the right, you start by finding a right inverse to a surjection, which requires AC. I asked about the situation on math.SE, with a negative answer. Basically, you can construct the splitting morphism without first choosing an arbitrary section of the surjection.</p> <p><a href="http://math.stackexchange.com/questions/151438/does-the-splitting-lemma-hold-without-the-axiom-of-choice/" rel="nofollow">http://math.stackexchange.com/questions/151438/does-the-splitting-lemma-hold-without-the-axiom-of-choice/</a></p>