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I wonder whether this is true in the categories of groups, monoids, commutative algebras, associative algebras, Lie algebras?

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It is not true in the variety of groups generated by $S_3$. –  Mark Sapir Mar 9 '12 at 17:32
There's a literature on rectracts of polynomial rings, which is referenced in this answer: mathoverflow.net/questions/55931/… –  Charles Rezk Mar 9 '12 at 17:33
So the question basically asks when "free = projective". –  Martin Brandenburg Mar 9 '12 at 20:59
@Charles. One can see from Costa, Douglas L. Retracts of polynomial rings. J. Algebra 44 (1977), no. 2, 492–502. that in 1977 it was unknown whether every retract of $K[X_1,\ldots,X_n]$ is a polynomial ring, where $K$ is a field: The author shows that an affirmative answer to this question would solve the well-known cancellation problem for polynomial rings over fields. Is it still unknown?!! –  Victor Mar 9 '12 at 21:53
It is true both in $Ban_1$ (Banach spaces with morphisms the linear operators of norm at most one) and $Ban$ (Banach spaces with morphisms the bounded linear operators). In both of these analytic categories the free objects are the spaces $\ell_1(S)$ with $S$ any set. –  Bill Johnson Mar 11 '12 at 1:19

3 Answers 3

up vote 9 down vote accepted

A retract of a finitely generated free monoid is free even though submonoids need not be free. I don't know about the infinitely generated case.

Edit: infinitely generated seems ok. The fg case I saw in an automata theory book but I see a general proof.

Added: here is the proof. Let P be a projective monoid (retract of free). Since it is a submonoid of a free monoid it has a unique minimal generating set Y consisting of the elements which are irreducible. Consider the map from the free monoid on Y to P sending generator to generator. Since P is projective it must split. But since elements of Y are irreducible their only preimages are the corresponding generators in the free monoid. Thus the splitting is an inverse to the projection.

Added: It seems to me the above proof works verbatim for free commutative monoids and more generally relatively free monoids in varieties containing all commutative monoids.

Added: Theorem 7 of http://arxiv.org/pdf/math/9711202.pdf seems to imply retracts of free non-associative algebras are free.

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About your last paragraph (free associative case): as the beginning of Section 2.2 says, it is applicable in various non-associative case only, unfortunately. –  Vladimir Dotsenko Mar 10 '12 at 16:51
@Vladimir, thanks! I should have read it more carefully. I will fix the entry. –  Benjamin Steinberg Mar 10 '12 at 18:46

The answer is yes in the category of groups. Suppose that $f: G \to H$ is a retraction with $G$ a free group. Then there is a homomorphism $g: H \to G$ such that $fg = \mathrm{id}_H$. Thus $g$ is injective and hence embeds $H$ isomorphically as a subgroup of $G$. But any subgroup of a free group is free, so $H$ must be free. The same proof works in the category of abelian groups also.

Edit: the same proof will work anytime you have the theorem that a subobject of a free object is free, I think. I don't know if that is true in the other categories that you mention.

Further edit: this property can fail even in very nice categories. For example, let $k$ be a field and consider the matrix algebra $M_n(k)$. In the category of finitely generated modules over $M_n(k)$, $M_n(k)$ itself is a direct sum of $n$ copies of $k^n$, but $k^n$ is not free over $M_n(k)$.

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The same argument works for Lie algebras. –  Mariano Suárez-Alvarez Mar 9 '12 at 17:37
The last example can be generalized as follows: in the category of $R$-modules ($R$ a unital ring), the retracts of free objects are precisely the projective modules. –  Qiaochu Yuan Mar 9 '12 at 20:48

A few months after the last activity on this question, Neena Gupta gave a proof that over a field $k$ of positive characteristic, a retract of a polynomial algebra need not be a polynomial algebra: http://arxiv.org/abs/1208.0483.

In fact, she gives a counterexample to the cancellation problem: there is an algebra $A$ such that $A[t]$ is isomorphic to $k[x_1,x_2,x_3,x_4]$ but $A$ is not isomorphic to $k[y_1,y_2,y_3]$. Composing the isomorphism $k[x_1,x_2,x_3,x_4]\to A[t]$ with the evaluation map $A[t]\to A$ at $t=0$ expresses $A$ as a retract of $k[x_1,x_2,x_3,x_4]$.

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