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In part of the proof of the splitting lemma (a left-split short exact sequence of abelian groups is right-split) it seems necessary to invoke the axiom of choice. That is, if $0\to A\overset{f}{\to} B\overset{g}{\to} C\to 0$ is exact and there is a retraction $B\overset{r}{\to}A$, then we can find a section $C\overset{s}{\to}B$ by choosing any right inverse of $g$ and removing the part in the kernel of $f$, which gives a well-defined morphism independent of the choice.

Is this invocation of the axiom of choice essential? I thought I had an example that showed it was: $0\to\mathbb{Q}\to\mathbb{R}\to\mathbb{R}/\mathbb{Q}\to0$ splits on the right if you can choose a basis for $\mathbb{R}$ as a vector space over $\mathbb{Q}$. But actually now I think it doesn't split on the left without a basis either.

The map $C\to B$ is supposed to be a canonical injection map. Can something be "canonical" if it requires choice?

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    The sequence splits if and only if $\mathbb R\cong\mathbb R\oplus\mathbb Q$. I know this follows from a Hamel basis of $\mathbb R$ over $\mathbb Q$ but I'm not sure it implies such basis exists.2012-05-30
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    I deleted an earlier answer because I missed the second part of the question. The first part can be done without choice: Consider the morphism $h = 1_B-fr\colon B \to B$. Then $hf = f -frf = 0$, so $h$ factors as $h = sg$ for a unique morphism $s\colon C \to B$ (note that $s(b+A) = h(b)$ is well-defined as a map $s: B/A \to B$). Now $gsg = gh =g-gfr = g = 1_C g$, so $gs = 1_C$ because $g$ is onto, so $s$ is a section of $g$.2012-05-30
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    @t.b.: It might just be the hangover, but either I answered the "hard" part, or I missed something. Have I missed something?2012-05-30
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    @Asaf: I don't understand why the map $\mathbb{Q} \to \mathbb{R} \oplus \mathbb{Q}$ given by composing $\mathbb{Q} \to \mathbb{R} \to \mathbb{R} \oplus \mathbb{Q}$ should have a left inverse. There is *some* map having a left inverse, namely the inclusion into the second summand, but why is this the same as the map obtained from composing the natural inclusion with the isomorphism $\mathbb{R} \to \mathbb{R} \oplus \mathbb{Q}$ you assume to exist?2012-05-30
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    To be clear, the splitting lemma does not require choice. Or even the law of excluded middle. It's completely constructive.2012-05-30
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    @t.b. Ah, okay. I *am* missing something (I have to admit that I cannot say that this is all the hangover's fault! :-)) I will remove my answer until I figure this out.2012-05-30
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    @AsafKaragila I guess you're right that Hamel basis is not equivalent to splitting. I edited the question to change "iff" to "if".2012-05-31

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There's no choice involved in the following argument showing that $f$ has a left inverse (if and) only if $g$ has a right inverse:

Consider the morphism $h = 1_B-fr\colon B \to B$. Then $hf = f -frf = 0$, so $h$ factors as $h = sg$ for a unique morphism $s\colon C \to B$ (note that $s(b+A) = h(b)$ is well-defined as a map $s: B/A \to B$). Now $gsg = gh =g-gfr = g = 1_C g$, so $gs = 1_C$ because $g$ is onto, so $s$ is a section of $g$.


We also have $frsg = frh = fr-frfr = 0$, so $frs = 0$ because $g$ is onto, so $rs = 0$ because $f$ is injective and by construction of $s$ we have $fr + sg = fr + h = 1_B$. In particular, we have an isomorphism $$ \begin{bmatrix} f & s \end{bmatrix}\colon A \oplus C \longleftrightarrow {B :} \begin{bmatrix} r \\ g \end{bmatrix}. $$ Note that $s$ is uniquely determined by $r$ and similarly one shows that if $s$ is a right inverse of $g$ then there is a uniquely determined right inverse $r$ of $g$ such that $fr = 1-sg$.


Concerning the existence of a retraction of the inclusion $\mathbb{Q} \to \mathbb{R}$: assuming such a retraction exists, the previous part gives us a section $s\colon\mathbb{R/Q} \to \mathbb{R}$ of $g\colon \mathbb{R} \to \mathbb{R/Q}$. Modifiying this section by setting $t(x) = s(x) - \lfloor s(x) \rfloor$, we get a Vitali set $t(\mathbb{R}/\mathbb{Q}) \subset [0,1]$, whose existence we cannot prove from ZF alone, so we cannot prove from ZF that there is a left inverse of the inclusion $\mathbb{Q} \to \mathbb{R}$.


A simple example of something “canonical” that requires choice in order to be non-trivial would be a product of an arbitrary collection of sets.

My naïve way of thinking of the axiom of choice is that it is first and foremost an axiom ensuring the existence of things. In my experience it is quite often the case that things can be defined and shown to be unique (hence “canonical”) if they exist (or non-uniqueness is controlled in some tractable way), but their existence requires additional assumptions.

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    It is worth mentioning that for a particular sequence of split (e.g. the one mentioned about $\mathbb{Q\to R}$) one can always generate models in which the axiom of choice holds "high enough" to ensure the splitting; but later on fails in the worst possible ways.2012-05-30
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    So my proof used the axiom of choice, but it didn't need to. If you use the fact that abelian groups are an abelian category, and use the universal properties of kernels and cokernels, you won't need AC. I had come to suspect that must be the case. Thanks.2012-05-30
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    But wait... in your argument, you're using the fact that $f$ is the kernel of $g$, I guess? Because $f$ is mono, it must be the kernel of something, and that something must be $g$. But just to be sure, we're not going to need the axiom of choice to prove that the category of abelian groups is an abelian category, right? That every surjection of abelian groups is a cokernel is constructive as well?2012-05-30
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    @ziggurism: I use nothing fancy and wouldn't express it this way (however, the fact that abelian groups and more generally $R$-modules form an abelian category doesn't need any choice either). Here I use that $g$ induces an isomorphism $B/\ker{g} \to C$ (it is surjective by exactness and the induced map is injective); $h$ factors over $B/\ker{g}$. This works entirely in ZF, you can just write it down: $h$ is constant on the classes $b + \ker{g}$, so it gives a well-defined map on the quotient. Similarly, $f$ is injective and gives an isomorphism from $A$ to the kernel of $g$ by exactness.2012-05-30
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    @ziggurism: Another side remark on this discussion is that Andreas Blass proved that the assertions "Every abelian group is projective" and "Every divisible abelian group is injective" are equivalent to the axiom of choice (each of those on its own).2012-05-30
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    @t.b. The statement "if $hf=0$ then there is a unique $s$ with $h=sg$" is the universal property for a cokernel, so it just says "$g$ is the cokernel of $f$", right? Which we know since $f$ is monic, it is the kernel of something, say $f=\ker x$, so $\im f=\ker \coker \ker x=\ker x=\ker g.$ Then since $g$ is epi, $g=\coker \ker g=\coker \im f=\coker f.$ So what you said follows if abelian groups are an abelian cat. I can also see that this is equivalent to the statement in group theory that a map which vanishes on $\ker g$ is well-defined on $B/\ker g$, i.e. factors through $B\to B/\ker g$.2012-05-30
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    @ziggurism: yes, right: the factorization $h = sg$ follows from the statement that $g$ is the cokernel of $f$. Yes, if we know that abelian groups are an abelian category then all what you say is correct. But proving this reduces these statements to the basic abstract algebra statement I used, so I preferred that route: an injective homomorphism between abelian groups is a kernel and a surjective homomorphism between abelian groups is a cokernel. We don't disagree, we simply phrase differently :)2012-05-30
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    @t.b. So I also asked a colleague, who gave yet another way of looking at it. Although there is a step in my proof where I choose an element $b\in g^{-1}(c)$, the next step is to define $s(c)=b-fr(b)$ and then show that this value is independent of the choice of $b$. In other words, there's a canonical choice in each fiber $g^{-1}(c)$, so the axiom of choice isn't actually required to construct this function. It took some convincing for me to see. So we have three ways to see it: the "choice" of "b" is actually canonical, $(1-fr)f=0$ so $1-fr$ factors over $g$, and $g$ is the cokernel of $f$.2012-05-31
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    And of course all three ways are just different ways of saying the same thing, as you point out. Good! This helped me a lot! Thank you, @t.b. It is a thorough answer.2012-05-31
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    @AsafKaragila I think this result, or one like it, that the axiom of choice is equivalent to the assertion that category of abelian groups has enough injectives, is what I was thinking of when I wondered whether AC was required to show that abelian groups constitute an abelian category. Thanks for mentioning that, it helped me straighten out my confusions.2012-05-31
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    @Asaf: "every abelian group is projective" sounds funny. Perhaps you meant "every free abelian group is projective"? That result is neat! Does it also work for modules over an arbitrary ring (not necessarily $\mathbb{Z}$)?2012-05-31
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    @ziggurism: No, enough projectives/injectives only imply that the axiom of choice can be restored through forcing (with a set of conditions that is). It also means that in models that choice cannot be restored in such way there are simply not enough projective groups. Furthermore, to your comment about independence of representatives: you **are** using choice. You show that your definition is good regardless to the choice, but you are choosing arbitrarily from the fibers which is the definition of choice.2012-05-31
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    @Bruno: Well, that is what the paper says. I will be glad to fully understand what does the term suggests (I suppose it might be buried in the text of the paper, which I keep not reading due to time budgets). [Here is a link to the paper](http://www.ams.org/journals/tran/1979-255-00/S0002-9947-1979-0542870-6/home.html).2012-05-31
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    @AsafKaragila So we can't, without invoking AC, do some set theoretic gymnastics where we observe that the fiber $p^{-1}(c)$ is nonempty, therefore it has an element $b$, and therefore it has a canonical element $b-fr(b)$. And once we know that there is a canonical element in each fiber, we define a function. This doesn't work? It does seem suspicious. "I don't need to use choice if I just choose one at a time, infinitely often."2012-05-31
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    @ziggurism: Not in the general case, no.2012-05-31