Onto extensions of free groups

An extension of subgroups $H\leqslant K\leqslant F_A$ of the free group of rank $|A|=r\geqslant 2$ is called onto when, for every ambient free basis $A'$, the Stallings graph $\Gamma_{A'}(K)$ is a quotient of $\Gamma_{A'}(H)$. Algebraic extensions are onto and the converse implication was conjectured by Miasnikov-Ventura-Weil, and resolved in the negative, first by Parzanchevski-Puder for rank $r=2$, and recently by Kolodner for general rank. In this note we study properties of this new type of extension among free groups (as well as the fully onto variant), and investigate their corresponding closure operators. Interestingly, the natural attempt for a dual notion -- into extensions -- becomes trivial, making a Takahasi type theorem not possible in this setting.


Context and background
Let A = {a 1 , . . . , a r } be an alphabet of r letters, let A ± = {a 1 , . . . , a r , a −1 1 , . . . , a −1 r } be its formal involutive closure, and let F A be the free group on A (formally, the free monoid on A ± modulo the equivalence relation generated by elementary reduction, a i a −1 i ∼ a −1 i a i ∼ 1). In 1983, elaborating on previous ideas by several authors, Stallings [11] established the notion of so-called Stallings A-automata: oriented graphs (allowing loops and parallel edges) with labels from A ± at the edges, being involutive (i.e., for every edge e from p to q with label a, there is another one e −1 from q to p and labelled a −1 ; e and e −1 are said to be inverse to each other), with a selected vertex called the basepoint (denoted ), and being connected, deterministic (no two different edges with the same label from, or into, the same vertex) and trim (every vertex appears in some reduced closed path at the basepoint). Here, a path is called reduced if it has no backtracking, i.e., no crossings of an edge immediately followed by its inverse (equivalently in the deterministic case, a path is reduced if and only if its label is a reduced word).
With this notion, Stallings [11] established a bijection between the set of (free) subgroups of F A and the set of isomorphism classes of Stallings A-automata. Further, finitely generated subgroups correspond to finite Stallings A-automata and, when restricted to these subsets, the Stallings bijection is algorithmic friendly, i.e., there are fast algorithms for both directions: given a set of words h 1 , . . . , h n on A, one can compute the Stallings graph Γ A (H) corresponding to the subgroup they generate, H = h 1 , . . . , h n F A . And, given a Stallings A-automaton Γ, one can compute a basis for its language subgroup L(Γ) F A (the set of labels of reduced closed paths at the basepoint in Γ). We assume the reader is familiar with this theory; see [1,5,11] for details. Theorem 1.1 (Stallings,[11]). The following is a bijection: From now on, we will mostly consider involutive A-automata and will describe and draw them just mentioning their positive part (i.e., those edges labelled by letters in A); next to each positive edge we implicitly assume its inverse is also there (even if we do not mention it), ready to be used by paths around the automaton. The positive subautomaton of Γ is denoted Γ + (and it is obviouly not involutive, unless edgeless).
For later use, we briefly explain how the above mentioned algorithms work. Suppose we are given a finite set of reduced words, W = {h 1 , . . . , h n } ⊂ F A . Draw the so-called flower automaton F(W ): for each i = 1, . . . , n, consider a circular graph with A-labels at the edges (usually called a petal ) in such a way that, when travelled around, it spells the word h i = h i (a 1 , . . . , a r ) (or its inverse if travelled in the opposite direction), and glue all of them together along their basepoints . Note that F(W ) is trim, deterministic except maybe at the basepoint, and satisfies L(F(W )) = H, where H = W F A . To obtain a deterministic automaton, successively apply elementary foldings: whenever we have two edges with the same label from, or into, the same vertex, identify them into a single one (as well as their terminal, or initial, vertices). This process always terminates in a finite number of steps, and it can be proven that the final A-automaton obtained in this way, denoted , is deterministic, trim, reads the same language L(Γ A (H)) = L(F(W )) = H, and, more significantly, is independent from the specific sequence of foldings applied, and even from the set of generators of H we started with: the Stallings graph Γ A (H) only depends on the subgroup H F A (and on the ambient basis A chosen to work with). Conversely, given a finite Stallings A-automaton Γ, we can choose a spanning tree T (more precisely, the involutive closure of a spanning tree of Γ + ), and get the basis of H = L(Γ) F A given by {h e | e ∈ EΓ + \ ET }, where h e = lab(T [ , ιe]eT [τ e, ]) and T [p, q] stands for the unique reduced path in T from vertex p to vertex q, i.e., the T -geodesic from p to q. By degree of a vertex p in an involutive A-automaton Γ we mean the out-degree of p. Note that this equals the in-degree of p in Γ, and also the total (in-plus out-) degree of p in its positive part Γ + .
For an A-automaton Γ, define its core, denoted c(Γ), as its largest trim subautomaton, i.e., that determined by the vertices and edges appearing in some reduced closed path at ; so, Γ is trim if and only if c(Γ) = Γ. It is easy to see that, when Γ is finite and connected, this is equivalent to saying that no vertex in Γ has degree 1 except maybe ; in this case, one can get c(Γ) from Γ by applying finitely many times the trim operation: remove a vertex of degree one different from (together with the corresponding edge).
An A-automata homomorphism (A-homomorphism, for short) from Γ to Γ is a pair of maps θ = (θ V , θ E ), θ V : V Γ → V Γ and θ E : EΓ → EΓ (where V Γ and EΓ denote the sets of vertices and edges of Γ, respectively), such that θ = and, for every a-labelled edge e from p to q in Γ, eθ E is an a-labelled edge from pθ V to qθ V in Γ ; we shall abuse language and write θ = θ V = θ E . Such a θ is called onto (resp. into) if both θ V and θ E are onto (resp. into). For an onto A-homomorphism we will use the notation θ : Γ Γ . Observe that, for Γ connected and Γ deterministic, there exists at most one A-homomorphism from Γ to Γ .
Stallings bijection behaves well with respect to inclusions in the sense that, for two subgroups H, K F A , H K if and only if there is an A-homomorphism from Γ A (H) to Γ A (K) which, in this case, is unique and will be denoted θ H,K : Γ A (H) → Γ A (K).
Since 1983, the Stallings bijection became central for the modern understanding and study of the lattice of subgroups of a free group. With the development of these graphical techniques, many new results have been obtained about free groups and their subgroups. Also, most of the results known before Stallings [11] have been reproven using graphical techniques, usually with conceptually simpler and more transparent proofs. Takahasi theorem is a typical example illustrating this fact.
An extension of subgroups H K F A is called free (we also say that H is a free factor of K), denoted H ff K, whenever some (so, any) basis of H can be extended to a basis for K; for example, it is easy to see that, for H K F A , if Γ A (H) is a subautomaton of Γ A (K) then H ff K (just extend a spanning tree for Γ A (H) to a spanning tree for Γ A (K)), while the converse is far from true. The notion of free factor is the non-abelian version of the notion of direct summand from commutative algebra. In a vector space, every pair of subspaces E F is in direct sum position, E ⊕ F , i.e., F = E ⊕E for some complementary subspace E . When we consider, for example, free abelian groups (i.e., free modules over Z) the exact same result is not true, but it works if we admit a bit of flexibility: every subgroup H Z m is of finite index only in finitely many subgroups H = H 0 fi H 1 , . . . , H n Z m and, for every K Z m containing H, there exists a unique i such that H fi H i ⊕ K Z m . Of course, the situation in the free group seems much wilder, starting from the well known fact that H K F A does not even imply rk(H) rk(K) (to the extreme that F ℵ 0 can be viewed as a subgroup of F 2 ). However, back in the 1950's, Takahasi [12] proved that, 2:4

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Vol. 13:1 again, the same result adapts to the free group case, after admitting a little bit more of degeneration: we will have to restrict ourselves to finitely generated subgroups, and we will lose the finite index condition. Takahasi theorem was proved 70 years ago using purely combinatorial and algebraic techniques. However, in more recent years, it was rediscovered independently, by Ventura [13] in 1997, by Margolis-Sapir-Weil [4] in 2001, and by Kapovich-Miasnikov [1] in 2002, in slightly different contexts; see also the subsequent paper Miasnikov-Ventura-Weil [5] joining the three points of view. These authors, independently, gave their own proofs of Takahasi's theorem, and they happened to be essentially the same proof; we would say, the "natural" proof of this result using Stallings graphs. Let us sketch it here, since it will play a central role along the rest of the paper.  Sketch of proof. Given H fg F A , consider its (finite) Stallings graph Γ A (H). Identifying certain sets of vertices, we get a new A-automaton which may very well be not deterministic; apply then a sequence of foldings until obtaining a deterministic one, say Γ 1 , which will correspond to a finitely generated overgroup of H, say H L(Γ 1 ) F A . Repeating this operation for every possible partition on the set of vertices V Γ A (H) (and possibly getting the same result for different initial partitions), we obtain a finite number of overgroups of H, say O A (H) = {H = H 0 , H 1 , . . . , H n }, called the A-fringe of H in [13], which satisfies the statement of Takahasi theorem. In fact, let K be a (non-necessarily finitely generated) subgroup with H K F A , and look at the corresponding A-homomorphism θ H,K : Γ A (H) → Γ A (K). Looking at the image Im(θ H,K ) as a (finite and deterministic) subautomaton of Γ A (K), we see that: (1) the A-homomorphism θ H,K : Γ A (H) Im(θ H,K ) is onto and so, L(Im(θ H,K )) ∈ O A (H); and (2) Im(θ H,K ) is a subautomaton of Γ A (K) and so, L(Im(θ H,K )) ff K. This proves that the A-fringe of H is a finite family of finitely generated overgroups of H satisfying the property stated in Takahasi theorem (a Takahasi family, for short).
As done in Kapovich-Miasnikov [1] and in Miasnikov-Ventura-Weil [5], it is possible to clean up the fringe of H, O A (H), in order to obtain a minimal Takahasi family, and gain uniqueness of the middle subgroup H i in Theorem 1.2; moreover, this minimal Takahasi family will have a clear algebraic meaning as follows. A subgroup extension H K F A is called algebraic, denoted H alg K, if H is not contained in any proper free factor of K; denote by AE(H) the set of algebraic extensions of H (within F A ). By Takahasi's theorem, AE(H) ⊆ O A (H) and so, |AE(H)| < ∞ for every finitely generated H. Furthermore it can be seen that, applying the following cleaning process to O A (H), one obtains precisely AE(H): for each pair of distinct subgroups H i , H j ∈ O A (H), if H i ff H j then delete H j from the list; see [5] for details.
Using the language of algebraic extensions, one can deduce the following variant (and slight improvement) of Takahasi's theorem: free factor of K containing H, and with the largest algebraic extension of H contained in K.
In Theorem 1.3, the words smallest and largest make sense because of the following basic properties of free and algebraic extensions: For later use, we collect some more properties of free and algebraic extensions, highlighting the duality of these two notions.
Let us insist on the computability part in Theorem 1.3. If we start with a finitely generated subgroup H fg F A (given by a finite set of generators), we can effectively compute: (1) its Stallings graph Γ A (H); (2) all its (finitely many) quotients, resulting from identifying vertices in Γ A (H) in all possible ways followed by folding (i.e., we can effectively compute bases for all the members of the fringe O A (H)); and (3) the cleaning process until getting the set of (bases of all the) algebraic extensions of H. This last step requires an algorithm deciding whether a given extension H K is free or not; this can be done using the classical Whitehead techniques (see Roig-Ventura-Weil [9] for a significant improvement on the time complexity, from exponential to polynomial), or using more modern algorithms based on Stallings automata (see Silva-Weil [10], and Puder [8]). Therefore, for H fg F A , the set AE(H) is finite and computable.
Another important observation is the following. The fringe of H strongly depends on the ambient basis A (reflected in the notation with the subscript A in O A (H)), while the set of algebraic extensions AE(H) does not, and is canonically associated to the subgroup H, since it is defined completely in algebraic terms. To illustrate this fact, see Example 2.5 from Miasnikov-Ventura-Weil [5], where the fringe of H = ab, acba

S. Mijares and E. Ventura
Vol. 13:1 We interpret the above fact by thinking that AE(H) is what really carries relevant algebraic information about the subgroup H and its relative position within the lattice of subgroups of F A . And O A (H) is the same set locally distorted with some accidental new members depending on the ambient basis used to draw and work with the graphs. From this point of view, Miasnikov-Ventura-Weil launched in [5] the following natural conjecture: the common subgroups in O A (H), when A runs over every ambient basis might be, precisely, the algebraic extensions: Conjecture 1.6 (Miasnikov-Ventura-Weil, [5]). Let A be a finite alphabet, and F A be the free group on A. Then, for every H fg F A , In [5] it was mentioned that this is clearly true in the two extremal situations H fi F A and H ff F A , but nothing else was known at that time. Seven years later, in 2014, the paper Parzanchevski-Puder [7] appeared showing that the conjecture is not true as stated: They also proposed two possible reformulations making the conjecture more plausible. On one hand, the authors recognize that their counterexample exploits many idiosyncrasies of the (automorphism group of the) free group of rank two, and it could be that Miasnikov-Ventura-Weil conjecture holds true for ambient free groups of rank three or more (i.e., Aut(F 2 ) is much "smaller" and easier in structure than Aut(F r ) for r 3 and so, the intersection of fringes with respect to all ambient basis is "too lax" in the case of rank two). On the other hand, they made the elementary but clever observation that the independence of AE(H) from the ambient basis (i.e., the reason for the obvious inclusion AE(H) ⊆ A O A (H)) has an even more restrictive consequence: adding new letters, extend A to a bigger (possibly infinite) new alphabet A ⊆ B, and consider the free extension F A ff F B ; viewed as subgroups of F B , it is still true that H alg K if and only if Hϕ alg Kϕ, for every ϕ ∈ Aut(F B ). So, the same reasoning gives us the stronger inclusion AE(H) ⊆ B O B (H), where B runs now over all the ambient bases of all free extensions F B , B ⊇ A. Parzanchevski and Puder finished their paper [7] by reformulating Conjecture 1.6 into the following two variations: [7]). Let A be a finite alphabet with |A| 3, and F A be the free group on A. Then, for every H fg F A , They also observed that their counterexample to Conjecture 1.6 does not serve as a counterexample for either Conjecture 1.8 or Conjecture 1.9. In fact, H = a 2 b 2 and K = a 2 b 2 , ab live inside the free group of rank two F A , with A = {a, b}; but if we extend this ambient free group with a third letter, say B = {a, b, c}, then the new ambient basis B = {x, y, z}, with x = a, y = cb −1 , z = cbc −1 , breaks the counterexample since H = x 2 y −1 z 2 y , K = x 2 y −1 z 2 y, xy −1 zy and K ∈ O B (H). We shall further exploit this example below. The last step in this story is the recent 2020 preprint Kolodner [2], where the author definitively disproves the conjecture in all its mentioned forms. In fact, he shows the following stronger result:

Closing the MVW conjecture
In this section, we want to elaborate more on Parzanchevski-Puder's idea about possible natural modifications of the original Miasnikov-Ventura-Weil conjecture, which could make it true.
Let H K F A . Firstly, we can look above F A , not just through free extensions but using all possible extensions. That is, consider a new free group F B and an arbitrary injective homomorphism F A → F B , not necessarily with A ⊆ B, i.e., with the image not necessarily being a free factor of F B . In this situation, H alg K still implies that, for every ϕ ∈ Aut(F B ), Hϕ alg Kϕ F B ; or, in other words, H alg K implies that, for every basis B of F B , the B -homomorphism θ H,K : Γ B (H) Γ B (K) is onto. Secondly, we can look downwards instead of upwards: H alg K F A also implies that, for every subgroup K L F A and every automorphism ϕ ∈ Aut(L), Hϕ alg Kϕ L F A ; or, in other words, H alg K implies that, for every K L F A and every basis C of L, the C -homomorphism θ H,K : Γ C (H) Γ C (K) is onto. Note that, in general, there are plenty of automorphisms of L which do not extend to automorphisms of F A ; furthermore, L may very well be not finitely generated.
Thirdly, we can combine the two effects upwards/downwards: H alg K also implies that, for every free group inclusion F A → F B , every subgroup L F B containing K, and every automorphism ϕ ∈ Aut(L), Hϕ alg Kϕ L F B (note that this is more general than before since L is now not necessarily a subgroup of F A ); equivalently, H alg K implies that, for every free group inclusion F A → F B , every subgroup L F B containing K, and every basis C of L, the C -homomorphism θ H,K : Γ C (H) Γ C (K) is still onto. As we show in the following result, all these generalizations of Parzanchevski-Puder's idea are, in fact, (tautologically) equivalent to the algebraicity of the initial extension H K. For the relevant one, (e)⇒(a), consider a free decomposition K = K 1 * K 2 with H K 1 ; applying the hypothesis to a basis of K of the form C = C 1 C 2 , where C i is a basis of K i , i = 1, 2, we have that θ H,K : Γ C (H) Γ C (K) is onto. But, by construction, Γ C (H) contains only edges labelled by letters from C 1 , while Γ C (K) has a single vertex, and edges in bijection with C . Therefore, C 2 must be empty and K 2 = 1. This proves that H alg K.

Onto extensions
Interestingly, the counterexample given by Kolodner in Theorem 1.10 opens up a possible new line of research considering and studying two new types of subgroup extensions within the lattice of subgroups of a free group (which do not coincide, in general, with algebraic extensions).

That is, AE(H) ⊆ f Ω(H) ⊆ Ω(H) ⊆ O A (H).
Proof. We have already seen the three implications in the previous section. As for counterexamples, take A = {a, b}, and observe that Kolodner example  Let us investigate now the properties of these two new types of extensions among free groups. To do this, we need to use an idea, which is not explicitly written in Kolodner [2] but it is reminiscent in the arguments there.
Let Γ 0 and ∆ 0 be connected A-automata (neither necessarily deterministic, nor trim) and let θ 0 : Γ 0 → ∆ 0 be an A-homomorphism; let H = L(Γ 0 ) and K = L(∆ 0 ). We want to fold and trim both Γ 0 and ∆ 0 until obtaining the Stallings graphs Γ A (H) and Γ A (K), respectively, but in a synchronized way so that θ 0 keeps inducing A-homomorphisms down the tower of (1) For every pair of edges, e 1 , e 2 , violating determinism in Γ 0 , fold them in Γ 0 and simultaneously fold their images e 1 θ 0 and e 2 θ 0 in ∆ 0 ; there is the possibility that e 1 θ 0 and e 2 θ 0 are already equal in ∆ 0 , in which case we do nothing on the right hand side. Observe that after this (or these) folding operation(s), the A-homomorphism θ 0 determines naturally an A-homomorphism among the resulting A-automata. Repeat this process until having no more foldings to do at the left hand side, and denote the result by θ 1 : Γ 1 → ∆ 1 . By construction, Γ 1 is deterministic. Note also that if θ 0 is onto then θ 1 is also onto. (2) Now perform all possible foldings remaining to be done at the right hand side (and nothing on the left hand side): the homomorphism θ 1 naturally transfers to the new situation, and we get θ 2 : Γ 2 → ∆ 2 , where Γ 2 = Γ 1 , and now both Γ 2 and ∆ 2 are deterministic (and not trim in general). Note that, again, if θ 1 is onto then θ 2 is also onto. (3) At this point, observe that the edges in the core of Γ 2 must map through θ 2 to edges in the core of ∆ 2 (this is because both Γ 2 and ∆ 2 are deterministic and θ 2 is an Ahomomorphism). So, edges outside the core of ∆ 2 can only be images of edges from outside the core of Γ 2 . Hence, trimming all the edges from outside the core of ∆ 2 , and trimming simultaneously all their θ 2 -preimages in Γ 2 , we obtain θ 3 : Γ 3 → ∆ 3 , where ∆ 3 = c(∆ 2 ) is deterministic and trim, and Γ 3 is deterministic (and not yet trim, in general). Again, θ 2 onto implies θ 3 onto. Moreover, observe also that L(∆ 3 ) = L(∆ 2 ) = L(∆ 1 ) = L(∆ 0 ) = K and so, ∆ 3 = Γ A (K). (4) Finally, let us finish trimming Γ 3 (and do nothing on the right hand side) to obtain θ 4 : Γ 4 → ∆ 4 , where ∆ 4 = ∆ 3 = Γ A (K), Γ 4 = Γ A (H) since it is already a Stallings A-automaton with L(Γ 4 ) = H, θ 4 = θ H,K , and we are done. It is crucial to note, however, that in this critical last step we may very well lose surjectivity: in fact, even with θ 3 being onto, removing edges from Γ 3 may result into some edges from ∆ 4 = ∆ 3 having no θ 4 -preimages in Γ 4 . This synchronized folding process is crucial for the proof of the next proposition. Observe that Propositions 3.4 and 3.6 express the fact that onto and fully onto extensions satisfy the same properties we already know for algebraic extensions so, they behave very similarly to them (compare with Propositions 1.4 and 1.5). Fix a basis A of F (A). The hypothesis tells us that, for i ∈ I, θ H i ,K i : Γ A (H i ) Γ A (K i ) is an onto A -homomorphism. Glue together all the Γ A (H i ) (resp., Γ A (K i )), for i ∈ I, along their basepoints, to get Γ 0 (resp., ∆ 0 ) and the natural onto A -homomorphism θ 0 : Γ 0 ∆ 0 induced by the θ H i ,K i 's. Observe that neither Γ 0 nor ∆ 0 is deterministic, in general, and that L(Γ 0 ) = H and L(∆ 0 ) = K, where H = H i , i ∈ I and K = K i , i ∈ I . Now let us apply the synchronized folding process described above, starting with the onto A -homomorphism θ 0 : Γ 0 ∆ 0 , until obtaining θ H,K : Γ A (H) → Γ A (K). We will 2:10

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Vol. 13:1 deduce that this last A -homomorphism is onto, after arguing that all the left hand side A -automata Γ 0 , Γ 1 , Γ 2 , along the process (as well as the right hand side ones) are trim and so, neither step (3) nor the critical step (4) take place. By construction, θ 0 : Γ 0 ∆ 0 is onto, and Γ 0 and ∆ 0 are both trim, and not necessarily deterministic. In general, along the individual folding processes applied to Γ 0 and ∆ 0 in steps (1) and (2), the fact of being trim can be lost, since some vertices along the process decrease their degrees and could eventually become degree one vertices. However, we claim that this will not be the case, neither for Γ 0 , nor for ∆ 0 . Observe that these two A -automata are trim in a stronger way: for every vertex p = , not only the degree is bigger than one, |{e ∈ E | ιe = p}| > 1, but also its label-degree, |{a ∈ (A ) ± | ∃ e ∈ E s.t. ιe = p, lab(e) = a}| > 1. And, meanwhile the degree of a vertex could decrease along the folding process, its label-degree stays constant or increases. Therefore, at the end of step (2), the label-degree of all vertices p = in Γ 2 and ∆ 2 are bigger than one and hence, so are their degrees too. This means that steps (3) and (4) are empty and Γ 2 = Γ 3 = Γ 4 = Γ A (H), Since this is valid for every initially fixed basis A of F (A), we deduce that H ont K.
(ii). Fix a free extension F A ff F B , B ⊇ A, and a basis B for F B . The hypothesis tells us that, each B -homomorphism θ H i ,K i : Γ B (H i ) Γ B (K i ), for i ∈ I, is onto. Glue them together along their basepoints and apply the synchronized folding process described above. The exact same argument as in (i) tells us that θ H,K : Since this is valid for every basis B , we deduce that H f.ont K.
Remark 3.5. The argument in the proof of Proposition 3.4 is not technically correct for the case |I| = ∞, since the processes of folding edges, pair by pair, in steps (1) and (2) could very well be infinitely long. This is not a conceptual obstacle, but only a matter of expression: one should do all these foldings in a single step (losing, of course, the algorithmic aspect of the proof, valid only when |I| < ∞). In a non-deterministic (possibly infinite) A-automata Γ, one can define the equivalence relation among vertices given by p ∼ q ⇔ there is a path γ in Γ satisfying ιγ = p, τ γ = q, and lab(γ) = 1 ∈ F A . It is straightforward to see that Γ/ ∼ is automatically deterministic and has the same language L(Γ/ ∼) = L(Γ); further, for the case when Γ is finite, Γ/ ∼ equals the final output of the sequence of foldings. Similarly, when Γ is infinite, the trim process cannot be done edge by edge (there could even be no vertex of degree 1, and infinitely many edges to be trimmed out). Instead, one should delete, in a single step, all the edges not visited by any reduced closed path at .
is onto as well. Hence, H ont K. The proof of (i') is analogous.
(ii)-(ii'). Let A be a basis for F A . Since, by hypothesis, the A -homomorphism θ H,K : Γ A (H) Γ A (K) is onto, and the inclusions H is onto as well. Hence, M 1 ont K. The proof of (ii') is analogous.
For a counterexample to the other assertion, consider a 2 b 2 alg a, b = F {a,b} , which is an algebraic extension because a 2 b 2 is neither a proper power, nor a primitive element in a, b ; consequently, a 2 b 2 f.ont a, b and a 2 b 2 ont a, b but, clearly,  F {a,b,c} , while it is a free extension so it is not algebraic. However, we have to be careful with a similar attempt to use Proposition 3.4(i): a 2 b 2 ont a 2 b 2 , ab as subgroups of F {a,b} , and a 2 c 2 ont a 2 c 2 , ac as subgroups of F {a,c} , but we cannot conclude that a 2 b 2 , a 2 c 2 ont a 2 b 2 , ab, a 2 c 2 , ac as subgroups of F {a,b,c} since, in the ambient basis A = {x, y, z}, with x = a, y = cb −1 , z = cbc −1 , the corresponding A -homomorphism is not onto (the problem being that neither of the two initial extensions is onto when viewed as subgroups of F {a,b,c} ).
As a second example, consider a 2 b 2 ont a 2 b 2 , ab and a −2 b −2 a 2 b 2 ont a 2 , b 2 , both as subgroups of F {a,b} (the first one is Parzanchevski-Puder example, and the second one is algebraic). Then, by Proposition 3.4(i), a 2 b 2 , b 2 a 2 ont ab, a 2 , b 2 , while a 2 b 2 , b 2 a 2 while H ff K. Hence, Cl K (H) = H, f OCl K (H) = K, and OCl K (H) = K. On the other hand, we also know that H = a 2 b 2 and K = a 2 b 2 , ab satisfy H ont K Proof. The chain of inclusions is clear by construction. Also, by construction, we have H alg Cl K (H), H f.ont f OCl K (H) and H ont OCl K (H) therefore, by Proposition 3.6(ii)(ii'), To see the free factors, observe that Cl K (H) ff K and so, by Proposition 1.5(ii'), Cl K (H) ff f OCl K (H) and Cl K (H) ff OCl K (H). The remaining two free factors come from transitivity of onto and fully onto extensions (see Proposition 3.6(i)(i')) and from applying Theorem 1.3 to the corresponding extensions: there exists a subgroup L such that Proof. It is enough to see the inclusion OCl F A (H) Cl F A (H). In fact, Cl F A (H) ff F A and we consider a basis {a 1 , . . . , a r } for Cl F A (H) and an extension of it to a basis A = {a 1 , . . . , a r , a r+1 , . . . , a n } of F A . By construction H ont OCl F A (H) so, in particular, the A -homomorphism θ H,OCl F A (H) : Γ A (H) Γ A (OCl F A (H)) is onto. But H Cl F A (H) = a 1 , . . . , a r so Γ A (H), and hence Γ A (OCl F A (H)), has no edges labelled by a r+1 , . . . , a n . Therefore, OCl F A (H) a 1 , . . . , a r = Cl F A (H).
Arriving at this point, it seems natural to ask which could be the possible dual notions to the concepts of onto and fully onto, in the same way that free extensions are dual to algebraic extensions. Fulfilling this duality, Theorem 1.3 states that the K-algebraic closure of H K can alternatively be reached by taking the biggest algebraic extension of H contained in K, or the smallest free factor of K containing H. Is there a dual notion for onto (resp., fully onto) extensions in this sense? i.e., is there a property P of extensions for which OCl K (H) (resp., f OCl K (H)) is the smallest P-subgroup of K containing H?
Since being onto is less restrictive than being algebraic, the dual notion should be a more restrictive notion than being a free factor. We presume the reader can easily imagine a very natural candidate for this possible dual notion: we could say that an extension H K F A is into if θ H,K : Γ A (H) → Γ A (K) is injective for every basis A of F A . And similarly, for fully into.
However, somehow against intuition, this dualization project fails dramatically. After studying these two concepts, and proving some promising properties (very similar to those of free factors), we realized that the only into extension H K F A is the equality H = K, 2:14

S. Mijares and E. Ventura
Vol. 13:1 by first proving that H into K implies that Γ A (H) must be a full subgraph of Γ A (K) for every ambient basis A , and then seeing that this situation forces equality H = K. So, these notions of into and fully into are trivial and, by no means, can they provide a Theorem similar to 1.3. At the time of writing the present paper, a subsequent preprint Kolodner [3] appeared providing an easier proof for this same fact. We redirect the reader there.

Open questions
We conclude with a list of interesting related questions, which remain open as far as we know.
Question 5.1. Is there an algorithm to decide whether a given extension H fg K fg F A is onto? and fully onto? Is there an algorithm to compute onto and fully onto closures of given extensions H K of finitely generated subgroups?
Remark 5.2. In the situation H fg K fg F A , we can compute O A (H) and keep all those overgroups of H contained in K. This provides a finite list of subgroups containing both OCl K (H) and f OCl K (H). However, to finish deciding who they are among the candidates in the list, we would need an algorithm deciding whether a given extension H fg L fg F A is onto (resp., fully onto) or not. A procedure for this was designed by Kolodner in [2], which may stop and give the correct answer, or may work forever (the example in Theorem 1.10 was obtained, precisely, as a stopping instance for this procedure). To make it into a true algorithm we would need a proof that it always stops, or an additional criterium to kill the process at certain point, and get the answer in finitely many steps.
Question 5.3. What is the algebraic meaning of an extension being onto, or fully onto? Is it possible to characterize the facts H ont K F A and H f.ont K F A without refereing to the bases of F A ?
Remark 5.4. These two definitions are canonical in the sense that they do not depend on any prefixed ambient basis. It would be interesting, then, to characterize them in algebraic terms with respect to H and K, but not talking about ambient bases (like the definition of algebraic extension). It seems, however, that free factors will probably not help in this, since there exist onto, and even fully onto, extensions being simultaneously free factors. A tricky detail to take into account here is that fully onto transfers through free extensions while onto does not, i.e., H f.ont K as subgroups of F A implies H f.ont K as subgroups of F B for any B ⊇ A, while the same is not true in general for onto extensions.
Question 5.5. Is there a notion dual to onto, or dual to fully onto, which may lead to a Theorem similar to 1.3? Vol