On the Connectivity of Token Graphs of Trees

Let $k$ and $n$ be integers such that $1\leq k \leq n-1$, and let $G$ be a simple graph of order $n$. The $k$-token graph $F_k(G)$ of $G$ is the graph whose vertices are the $k$-subsets of $V(G)$, where two vertices are adjacent in $F_k(G)$ whenever their symmetric difference is an edge of $G$. In this paper we show that if $G$ is a tree, then the connectivity of $F_k(G)$ is equal to the minimum degree of $F_k(G)$.


Introduction
Throughout this paper, G is a simple finite graph of order n ≥ 2 and k ∈ {1, . . . , n − 1}. The k-token graph F k (G) of G is the graph whose vertices are all the k-subsets of V (G), where two k-subsets are adjacent whenever their symmetric difference is a pair of adjacent vertices in G. We often write token graph instead of k-token graph. See Figure 1 for an example.
The study of token graphs probably started with Johns (1988) PhD Thesis, in which F k (G) was called the k-subgraph graph of G and some results concerning the diameter of F k (G) were reported. Since then, token graphs have been defined independently at least three more times. Alavi et al. (1991) defined F 2 (G) and call it the double vertex graph of G, and a year later, Zhu et al. (1992) generalized the concept for k ∈ {2, . . . , n − 1} under the name of k-tuple vertex graph of G. In these two papers, the authors studied several combinatorial issues of F k (G) such as Eulerianicity, Hamiltonicity, connectivity, regularity, etc.
This concept was reintroduced for third time by Rudolph (2002), when some connections of F k (G) with quantum mechanics and the graph isomorphism problem were discussed. Regarding the quantum mechanics, Rudolph used F k (G) to model the evolution of a cluster of n interacting qubits (2-level atoms), which must have exactly k qubits in excited state at any time (the n qubits are the vertices of G and their interactions define the edges of G). The use of F k (G) in this direction is still of interest, see Barghi and Ponomarenko (2009); Alzaga et al. (2010); Ouyang (2019). For instance, Ouyang (2019) showed that F k (G) has applications in the Heisenberg model, which is a quantum theory of magnetism. With respect to the graph isomorphism problem, Rudolph (2002) found pairs of cospectral graphs G and H such that F 2 (G) and F 2 (H) are not cospectral, implying that G and H are not isomorphic. Following Rudolph's approach, Audenaert et al. (2007) showed the existence of pairs of non-isomorphic cospectral graphs whose corresponding 2-token graphs are cospectral. A few years later, Barghi and Ponomarenko (2009), and independently, Alzaga et al. (2010), showed that for any k ∈ Z + , there exist infinitely many pairs of non-isomorphic graphs whose corresponding k-token graphs are cospectral. In Rudolph (2002) F k (G) was originally called the k-level matrix of G, but in Audenaert et al. (2007) F k (G) was renamed as the symmetric k-th power of G.
As far as we know, Fabila-Monroy et al. (2012) is the last paper in which F k (G) has been defined, under the name of the k-token graph of G. In that paper, Fabila-Monroy, Flores-Peñaloza, Huemer, Hurtado, Urrutia, and Wood defined F k (G) as "a model in which k indistinguishable tokens move from vertex to vertex along the edges of a graph" and showed several results on the connectivity, diameter, chromatic and clique numbers, and Hamiltonian paths. From this last definition of F k (G), it is not hard to see that the estimation of any parameter involving connectivity or the determination of the distance between vertices of F k (G) can be seen as a reconfiguration problem. We recall that reconfiguration problems are a family of combinatorial problems that ask if there exists a step-by-step transformation between two feasible solutions of a problem such that all intermediate results are also feasible. For two specific examples of theses connections we refer the reader to Ito et al. (2011); Yamanaka et al. (2015).
In 2017 Sloane (i) observed that the problem of determining the maximum size of a binary code of length n and constant weight 2 that can correct a single adjacent transposition is equivalent to determining the packing number of F 2 (P n ), where P n is the path graph of order n. Gómez Soto et al. (2018) solved this problem.
Token graphs are also a generalization of Johnson graphs: if G is the complete graph of order n, then F k (G) is isomorphic to the Johnson graph J(n, k). Johnson graphs have been widely studied; the analysis of many of its combinatorial properties is an active area of research (see for instance Alavi (2015); Brouwer and Etzion;Riyono (2007); Etzion and Bitan (1996); Terwilliger (1986)).
For a given graph invariant η, what can be said of η(F k (G)) in terms of G and η(G)? In particular, Fabila-Monroy et al. (2012) gave families of graphs of order n with connectivity exactly t, and whose k-token graphs have connectivity exactly k(t − k + 1), whenever k ≤ t; they also conjectured that if G is t-connected and k ≤ t, then F k (G) is at least k(t − k + 1)-connected. This was proven by Leaños and Trujillo-Negrete (2018). Recently, a similar lower bound was proven for edge-connectivity by Leaños and Ndjatchi (2021); they showed that if G is t-edge-connected and k ≤ t then F k (G) is at least k(t − k + 1)-edge-connected. Infinite families of graphs attaining this lower bound were also given.
In this paper we study the connectivity and edge-connectivity of F k (G) when G is a tree. As usual let κ(G), λ(G), and δ(G) be the connectivity, edge-connectivity, and minimum degree of G, respectively. It is well known that if G is connected then The main result of this paper is the following.
Theorem 1. If G is a tree of order n and 1 ≤ k ≤ n − 1 then We remark that while the hypothesis k ≤ κ(G) has played a central role in both results on κ(F k (G)) stated in Fabila-Monroy et al. (2012); Leaños and Trujillo-Negrete (2018), this hypothesis does not hold when G is a tree; this absence is responsible for the new difficulties in the proof of Theorem 1.
We now recall some standard notation which is used throughout this paper. Let u and v be distinct vertices of G. The distance between u and v in G is denoted by d G (u, v) (we sometimes write d(u, v) when G is understood from the context); we write uv to mean that u and v are adjacent. The neighbourhood of v in G is the set {u ∈ V (G) : uv ∈ E(G)} and it is denoted by N G (v). The degree of v is the number A u − v path of G is starting at u and ending in v. Let U and W be subsets of V (G). We use: G \ W to denote the subgraph of G that results by removing W from G; U \ W to denote set subtraction; and U W to denote symmetric difference. For brevity, if m is a positive integer, then we use [m] to denote {1, . . . , m}. We follow the convention that [m] = ∅ for m = 0.
The rest of the paper is organized as follows. In Section 1.1 we establish several ways to construct paths in F k (G) which come from the concatenation of certain paths of G. These paths of F k (G) play a central role in our constructive proof of Theorem 1. In Section 1.2 we give some basic results on the connectivity structure of F k (G) which help us to simplify significantly the proof of Theorem 1. Finally, in Section 2 we prove Theorem 1.

Constructing Paths of F k (G) from Paths of G
In this section we construct paths in F k (G) using a given set of paths of G. For this purpose, we find it useful to use the following interpretation of F k (G) given by Fabila-Monroy et al. (2012). We consider that there are k indistinguishable tokens placed at the vertices of G (at most one token per vertex). A vertex of F k (G) corresponds to one of this token configurations. Two such configurations are adjacent in F k (G) if and only if one configuration can be reached from the other by moving one token along an edge of G from its current vertex to an unoccupied vertex. These token moves are called admissible moves. Under this interpretation, if A and B are two distinct k-subsets of V (G) then a path in F k (G) with endvertices A and B corresponds to a finite sequence of token configurations that are produced by a corresponding sequence of admissible moves. With this in mind, now we explain how to produce some paths of F k (G) from a certain set of paths of G.
Let P := a 0 a 1 a 2 . . . a m be an a − b path of G (a 0 = a and a m = b); let A, B ∈ V (F k (G)) be such that A B = {a, b}, P ∩ A = {a} and P ∩ B = {b}. A natural way of constructing an A − B path P in F k (G) using P is by moving the token at a along P to b. More precisely, we start at A, then for each i = 0, 1, . . . , m − 1, we move (in this order) the token at a i along the edge a i a i+1 to the vertex a i+1 . We denote this sequence of admissible token moves by Clearly, the first and last configurations of this sequence correspond to the vertices A and B of F k (G), respectively. Moreover, note that if We refer to P as the path of F k (G) induced by P . See Figure 2. Let Q be a path of F k (G) and let {Q 0 , Q 1 , . . . , Q m } be its vertex set. Since each of these Q i 's is a k-set of V (G), then q := k − | ∩ m i=0 Q i | is well defined. We say that Q is a path of Type q. Thus, P and any edge of  The set of red vertices of G defining the left (respectively, right) configuration corresponds to the vertex A (respectively, B) of F k (G). These four configurations together (from left to right) define an A − B path P of F k (G). The path P is induced by P = a0a1a2a3, because the token at a0 is moving along P to a3. Since the remaining k − 1 tokens are fixed on A ∩ B, P is of Type 1.
We now define certain paths of Type 2. Let e 1 = a 1 b 1 and e 2 = a 2 b 2 be independent edges of G, and let A, B ∈ F k (G) be such that A \ B = {a 1 , a 2 } and B \ A = {b 1 , b 2 }. A simple way to construct an A − B path R of Type 2 (and length 2) is by moving the token at a 1 to b 1 along e 1 , and then, by moving the token at a 2 to b 2 along e 2 . We denote this construction by Figure 3). We remark that R can be seen as the concatenation of two paths of Type 1, namely those corresponding to a 1 −→ b 1 and a 2 −→ b 2 . As suggested above, we use a semicolon " ; " to denote the concatenation of paths of Type 1. Now, suppose that A and B are adjacent vertices in F k (G) with A \ B := {a} and B \ A := {b}. Then ab is an edge of G. Let u and v be adjacent vertices of G such that u ∈ A ∩ B and v ∈ V (G) \ (A ∪ B). As we have seen above, a way to produce an A − B path P is simply by moving the token at a to b along the edge ab. Now we use a simple trick, involving the edges uv and ab, to produce a new A − B path P uv of F k (G) that is internally disjoint from P. The path P uv is constructed as follows. First we move the token at u to v along uv, and then we move the token at a to b along ab, and finally we move back the token at v to u along uv. Clearly, each of these moves is admissible and they together define the required P uv path, which we denote by: We say that the vertex v is playing the role of a distractor, which allow us to produce a new path P uv from P and uv. See Figure 4.
We now generalize the above construction. Suppose that P is an A − B path of F k (G) and that uv is an edge of G with u ∈ A ∩ B and v ∈ V (G) \ (A ∪ B). If u ∈ I and v / ∈ I for any internal vertex I of P, then we can get a new A − B path P uv from P and uv as follows. First move the token at u to v along uv. Then, keeping the token at v fixed, move the tokens from the vertices in A \ B to the vertices in B \ A according to P, and finally move back the token at the distractor v to the initial vertex u. Note that at the end we have produced an A − B path P uv with the following property: for each inner vertex J of P uv , we have that v ∈ J and u / ∈ J. This implies that if u v is an edge of G \ {uv} satisfying the same properties as uv with respect to P, then the corresponding path P u v is an A − B path internally disjoint from both P and P uv . The paths produced in this way play an important role in the proof of Theorem 1.

Some basic facts
In this section we prove auxiliary results that are used in the proof of Theorem 1. Proof: The forward implication follows directly from Menger's Theorem. Conversely, let U be a vertex cut of H of minimum order. Let H 1 and H 2 be two distinct components of H − U , and let u ∈ U . Since U is a minimum cut, then u has at least a neighbour v i in H i , for i = 1, 2. Then d H (v 1 , v 2 ) = 2. By hypothesis, H has t pairwise internally disjoint v 1 − v 2 paths. Since each of these t paths intersects U , then we have that |U | ≥ t, as required.
Then the following hold: , as required. See Figure 5.
2) Note that |X \ Y | = |Y \ X| = 2 in this case. Since d F k (G) (X, Y ) = 2, there is a way to carry the two tokens at the vertices of X \ Y to the vertices of Y \ X with exactly two admissible token moves. These two token moves corresponds to two independent edges joining vertices of X \ Y with the vertices of Y \ X. See Figure 5 (i).
3) In this case X \ Y and Y \ X each consists of exactly one vertex of G; say x and y, respectively. Since d F k (G) (X, Y ) = 2, then x cannot be adjacent to y in G. On the other hand, d F k (G) (X, Y ) = 2 implies the existence of an X − Y path P produced by exactly 2 admissible token moves. Now note that P necessarily involves two admissible token moves x −→ v and u −→ y. There are two possibilities either x −→ v is applied before u −→ y or u −→ y is applied before x −→ v. Since P is produced by exactly 2 admissible token moves, we have that u = v ∈ N G (x) ∩ N G (y), and xvy is a path of length two in G, as required. The two possibilities are depicted in (ii) and (iii) of Figure 5.
Let X be a vertex of F k (G). From the definition of F k (G) it is not hard to see that the complementary map ψ(X) := V (G) \ X defines an isomorphism between F k (G) and F n−k (G). The next proposition follows from the definition of ψ.
2 Proof of Theorem 1 Throughout this section, T is a tree of order n ≥ 2, and k ∈ {1, 2, . . . , n − 1}. It is sufficient to show that From the definition of F 1 (G) it is straightforward to see that G and F 1 (G) are isomorphic. In this case Theorem 1 holds. We assume that n ≥ 4 and k ∈ {2, . . . , n − 2}. By Proposition 1.1, it suffices to prove the following.
Proof: For brevity of notation, let XY := X ∩ Y , XY := V (T ) \ (X ∪ Y ), and δ := δ(F k (T )). We remark that here XY and XY are subsets of V (T ), but not edges of F k (T ) or F n−k (T ). Informally, the general strategy to show Lemma 2.1 is as follows.
• STEP 1. First, we construct a certain number m of pairwise internally disjoint X − Y paths in F k (T ).

CASE 1:
∈ {x, y}, and P = xvy is a shortest x − y path of T . In view of Proposition 1.3, we can assume without any loss of generality that . We assume that X and Y are as in Figure 5 (ii). Let XY := XY \ {v} and let  Let us define Let m x := min{a, c}, m y := min{b, d}, and m := m x + m y + η + 1.

STEP 1 of CASE 1
We produce the required m X − Y paths by means of four types of constructions.
1. Using the vertex v: 2. Using the edges of E XY,XY . For each z i w j ∈ E XY,XY , let P i,j be the X − Y path defined as follows: Moreover, depending on whether w j = v or w j = v, then A i,j also satisfies the following: We recall that if r = 0, then [r] = ∅.

Using the vertices
, we define the path P i as follows: x }, and at least one of the following holds: 4. Using the vertices w j y ∈ XY (y) and z j y ∈ XY (y). For each j ∈ [m y ], we define the path Q j as follows: y }, and at least one of the following holds: |T 4 | = m y , and m = 1 + η + m x + m y , then in order to finish the STEP 1 of CASE 1, it is enough to show that the paths in T are pairwise internally disjoint.
Proof of Claim 2.2:First we show separately that the paths in T are pairwise internally disjoint for ∈ {2, 3, 4}.
Suppose that = 2, and let P i,j and P s,t be distinct paths in T 2 . Let A i,j and A s,t be inner vertices of P i,j and P s,t , respectively. Since (i, j) = (s, t), then z i = z s or w j = w t . If Suppose that = 3, and let P s and P t be distinct paths in T 3 . For r ∈ {s, t}, let A r be an inner vertex of P r . From the last assertion of (C3) we know that Suppose that = 4. This case can be handled in a totally analogous manner as previous case. Let A 0 , A i,j , A s , and A t be inner vertices of P 0 ∈ T 1 , P i,j ∈ T 2 , P s ∈ T 3 , and Q t ∈ T 4 , respectively. It remains to show that P 0 , P i,j , P s , and Q t are pairwise internally disjoint. We analyze separately each pair.
{A 0 , A s }: By (C1) we know that A 0 ∩ XY = XY and that A 0 ∩ XY = ∅. Similarly, by the last assertion of (C3), we know that either {A 0 , A t }: As in previous case, the last assertion of (C4) implies that either x , as otherwise the vertex set {x, z i , v} forms a cycle, contradicting that T is a tree. Since A i,j ∩ XY = XY \ {z i }, and either A s ∩ XY = XY or Then we can assume that z s x = z i . This implies that w s x = w j , as otherwise {z i , x, w j } forms a cycle. By (C2.1) we know that A i,j ∩ XY = {w j }, and by (C3) we have that either {A i,j , A t }: Again, this case can be handled in a totally analogous manner as previous case.
{A s , A t }: Since XY (x), XY (y), XY (x), and XY (y) are pairwise disjoint, then z s x = z t y and w s x = w t y . From these inequalities and (C3)-(C4) we have that either A s ∩ XY = A t ∩ XY or This completes the proof of Claim 2.2.
2.1.2 STEP 2 of CASE 1 We start by showing that δ − m ≤ 2.
This completes the proof of Claim 2.3. Claim 2.3 shows that almost all X − Y paths claimed by Lemma 2.1 are provided by T, when |XY | = k − 1. We finish the proof of CASE 1 with the construction of the remaining δ − m X − Y paths.
Proof of Claim 2.4: We have already constructed m X − Y pairwise internally disjoint paths, namely the elements of T. Then, it remains to show the existence of δ − m additional X − Y paths with similar properties. Since if δ ≤ m then there is nothing to prove, we assume that δ > m. From this and Claim 2.3 it follows that b ≥ d + 1. Moreover, since a + b ≤ c + d, then c ≥ a + 1. Hence, a = min{a, c} and d = min{b, d}.
Suppose first that b = d + 1. By Claim 2.3 we have that δ ≤ m + 1. Thus, it is enough to construct a new X − Y path internally disjoint to each path in T. Since b = d + 1 > d = min{b, d} and c ≥ a + 1 > a = min{a, c}, then the vertices z b y and z c x were not used in the construction of the paths of T 3 ∪ T 4 . We construct the required path P as follows: Let A be an inner vertex of P. From the definition of P it follows that Now we show that P is internally disjoint to any path in T. Let A 0 , A i,j , A s , and A t be inner vertices of P 0 ∈ T 1 , P i,j ∈ T 2 , P s ∈ T 3 , and Q t ∈ T 4 , respectively.
We analyze these cases separately.

as otherwise T has a cycle. Then, by (C2) and (C5) we have that
Then, (C3) and (C5) implies that A s ∩ XY = A ∩ XY , and so A = A s . {A t , A}: We proceed as in previous case. Since t ≤ d < b, then z t y = z b y , and z t y = z c x because XY (x) ∩ XY (y) = ∅. Then, (C4) and (C5) implies that A t ∩ XY = A ∩ XY , and so A = A t .
Finally, suppose that b ≥ d + 2. By Claim 2.3 we have that δ ≤ m + 2. Thus, it is enough to construct two X − Y paths, say P and P , such that {P, P } ∪ T is a set of pairwise internally disjoint paths.
Since b ≥ d + 2 and a + b ≤ c + d, then c ≥ a + 2. Now we use z b y , z b−1 y , z c x , and z c−1 x to construct P and P as follows.
Note that a similar argument to the one used above (for the case b = d + 1) can be applied to show that P and P are internally disjoint of each path in T. Hence all that remains to be checked is that P and P are internally disjoint.
Let A and A be inner vertices of P and P , respectively. From the definition of P (respectively, P ) we know that either then in all the arising cases, we always have A = A , as required. This completes the proof of Claim 2.4, and hence the proof of CASE 1.

STEP 1 of CASE 2
We proceed similarly as in CASE 1. In particular, we often use slight adaptation of many arguments given in CASE 1. We start by producing m X − Y paths by means of six types of constructions.
1. Let us define P x1 and P x2 as follows: Let L 1 := {P x1 , P x2 }. Let P ∈ L 1 , and let A be an inner vertex of P. Then (D1) A ∩ XY = XY and A ∩ XY = ∅.

For each edge
Let L 2 := {P i,j : z i w j ∈ E XY,XY }. Let P i,j ∈ L 2 , and let A i,j be an inner vertex of P i,j . Then 3. For each i ∈ [m x1 ], we define the path P i as follows: Let P i ∈ L 3 , and let A i be an inner vertex of P i . Then and at least one of the following holds: 4. For each j ∈ [m y1 ], we define the path Q j as follows: y1 }, and at least one of the following holds:

5.
For each i ∈ [m x2 ], we define P * i as follows: Let P * i ∈ L * 3 , and let A * i be an inner vertex of P * i . Then and at least one of the following holds: 6. For each j ∈ [m y2 ], we define Q * j as follows: and at least one of the following holds: , and m = 2 + η + m x1 + m y1 + m x2 + m y2 , then in order to finish the STEP 1 of CASE 2, it is enough to show that the paths in L are pairwise internally disjoint.
Claim 2.6. The X − Y paths in L are pairwise internally disjoint.
Proof of Claim 2.6: We start by noting that, in some sense, the four ways in which the paths of T were constructed in STEP 1 of CASE 1 have been "repeated" in the construction of the paths of L. This close relationship between T and L is the main ingredient in the proof of Claim 2.6.
Before moving on any further, let us verify that the two paths of L 1 are internally disjoint. Let A 1 and A 2 be the inner vertices of P x1 and P x2 , respectively. Then The analogies between the paths of T and L are given by the interactions that the inner vertices of the X − Y paths have with XY and XY in the CASE 1 and with XY and XY in the CASE 2. More formally, let T ∈ T and L ∈ L. We say that T and L are analogous, if A ∩ XY = B ∩ XY and A ∩ XY = B ∩ XY , for any A and B inner vertices of T and L, respectively. For T ⊆ T and L ⊆ L we write T ∼ L to mean that any path of T is analogous to any path of L . For instance, note that T 1 ∼ L 1 . Indeed, let P 0 ∈ T 1 and P xi ∈ L 1 , and let A 0 and A be inner vertices of P 0 and P xi , respectively. From (C1) we know that A 0 ∩ XY = XY , and from (D1) we have that A ∩ XY = XY . Similarly, from (C1) it follows that A 0 ∩ XY = ∅, and from (D1) that A ∩ XY = ∅. Analogously, we can verify that: • (C1) and (D1) imply that T 1 ∼ L 1 . For completeness of this list, we include this case here again.
We recall that the strategy in the proof of Claim 2.2 was the following. Given two inner vertices A and B belonging to distinct paths of T, we always conclude that A = B by showing that at least one of A ∩ XY = B ∩ XY or A ∩ XY = B ∩ XY holds. From this fact, the definition of ∼, and the above list, it is not hard to see that analogous arguments as those used in the proof of Claim 2.2 imply that the X − Y paths belonging to L 1 ∪ L 2 ∪ L 3 ∪ L 4 (resp. L 1 ∪ L 2 ∪ L * 3 ∪ L * 4 ) are pairwise internally disjoint. Thus, it remains to show that the paths in L 3 (resp. L 4 ) are pairwise internally disjoint from the paths in L * 3 ∪ L * 4 . Let A i , A j , A * s , and A * t be inner vertices of P i ∈ L 3 , Q j ∈ L 4 , P * s ∈ L * 3 , and Q * t ∈ L * 4 , respectively. We analyze these cases separately.
Our strategy is as follows. In any of the analyzed cases we show that F k (G) has a vertex X 1 "close to" X whose degree is at most m + 1. Recall that we need to consider only the cases (2), (4), (6), (7), (8) and (16).
Then a 1 > 0 and a 2 > 0. Moreover, our suppositions and (2) imply that d 2 > b 2 . Let U := XY ∪ {x 1 , x 2 , y 1 , y 2 }. From a 1 > 0, a 2 > 0, and d 2 > 0 it follows that x 1 , x 2 , and y 2 have degree at least 2 in T (2.1) If y 1 is not adjacent to neither x 2 nor y 2 , then (2.2) If y 1 is adjacent to some of x 2 or y 2 , then it is adjacent to exactly one of them, because T has no cycles. Hence, in this case and so (J1) holds.
(4) a 1 > c 1 , a 2 > c 2 , b 1 ≤ d 1 and b 2 ≤ d 2 . If x 2 , y 1 , y 2 } and i ∈ {1, 2}. These and the fact that T [U ] is a forest imply the existence of two vertices We now suppose d 1 > 0 and d 2 > 0 does not hold. Then d 1 = 0 or d 2 = 0. By symmetry, we may assume that d 1 = 0. Then b 1 = 0, and d 2 > 0 by (2). Then for U : has no cycles, then y 1 is adjacent to at most one of x 2 or y 2 . From this fact, b 1 = d 1 = 0, and Again, these and the fact that T [U ] is a forest imply the existence of two distinct vertices u 1 , From (2) and these inequalities it follows that at least one of c 2 > a 2 or d 2 > b 2 holds. Let Since T has no cycles, then it contains at most one of x 1 x 2 or y 1 y 2 .
(6.1) Suppose that none of x 1 x 2 or y 1 y 2 is in T . Then (6.2) Suppose that exactly one of x 1 x 2 or y 1 y 2 is in T . Then and so (J2) holds.
Let X 1 := (X \ {x 1 }) ∪ {y 2 }. Again, since T has no cycles, then there is at most one edge in T with one endvertex in {x 1 , y 1 } and the other endvertex in {x 2 , y 2 }. Then As we have mentioned above, T has at most one edge with one end in {x 1 , y 1 } and the other end in {x 2 , y 2 }.
(8.2) Suppose that d 2 ≥ b 2 + 2. Then a 1 > 0 and d 2 ≥ 2, and hence x 1 and y 2 have degree at least 2 in T [U ], for U := XY ∪ {x 1 , y 1 , y 2 }. Since T [U ] is a forest, then there is a vertex (8.2.1) Suppose that x 2 is not adjacent to neither x 1 nor y 1 . Then, (8.2.2) Suppose that x 2 is adjacent to some of x 1 or y 1 . Since there is at most one edge with one end in {x 1 , y 1 } and the other end in {x 2 , y 2 }, then x 2 is adjacent to exactly one of x 1 or y 1 . Then, implying that (J3) holds.
Since there is at most one edge with one end in {x 1 , y 1 } and the other end in {x 2 , y 2 }, then T contains at most one of x 1 y 2 or x 2 y 1 .
(16.1) Suppose that neither x 1 y 2 nor x 2 y 1 is in T . Then, (16.2) Suppose that some of x 1 y 2 or x 2 y 1 is in T . Then exactly one of x 1 y 2 or x 2 y 1 belongs to T . By symmetry, we may assume that x 1 is adjacent to y 2 . Let X 1 := (X \ {x 1 }) ∪ {y 2 }. Then, and so (J4) holds.
Claim 2.7 shows that almost all X − Y paths claimed by Lemma 2.1 are provided by L, when |XY | = k − 2. We finish the proof of CASE 2 with the construction of the remaining δ − m X − Y paths.
Proof of Claim 2.8: Consider the m X − Y paths of L. Clearly, if m ≥ δ, then we are done. Then by Claim 2.7 we can assume that m + 1 = δ, and that one of (J1), (J2), (J3) or (J4) holds.
In view of these facts, it is enough to exhibit a new X − Y path P / ∈ L with P internally disjoint from any path in L. We note that in any of these four cases, T has one edge e with an endvertex in {x 1 , y 1 } and the other endvertex in {x 2 , y 2 }. Since T has no cycles, then e is the only edge of T with this property. Then XY (x 1 ), XY (x 2 ), XY (y 1 ), XY (y 2 ), XY (x 1 ), XY (x 2 ), XY (y 1 ), and XY (y 2 ) are pairwise disjoint, as otherwise T has a cycle.
Our strategy is as follows. First we define a set P = {P 1 , P 2 , P 3 , P 4 } consisting of four new X − Y paths of F k (T ). Then we show that for each of the four cases mentioned in previous paragraph, there is a path in P which is internally disjoint from any path of L, providing the additional required path.
1. If a 1 > c 1 and d 2 > b 2 , then we define the X − Y path P 1 as follows: From the definition of P 1 it follows that if A 1 is an inner vertex of P 1 , then 2. If a 1 > c 1 and c 2 > a 2 , then we define the X − Y path P 2 as follows: From the definition of P 2 it follows that if A 2 is an inner vertex of P 2 , then (E2) Either A 2 ∩ XY = XY or A 2 ∩ XY = XY \ {z c2 x2 }, and either A 2 ∩ XY = ∅ or A 2 ∩ XY = {w a1 x1 }, and at least one of the following holds: 3. If c 1 > a 1 and x 1 y 2 ∈ E(T ), then we define the X − Y path P 3 as follows: From the definition of P 3 it follows that if A 3 is an inner vertex of P 3 , then (E3) A 3 ∩ XY = ∅, and A 3 ∩ XY ∈ XY, XY \ {z c1 x1 } .
4. If d 2 > b 2 and x 1 y 2 ∈ E(T ), then we define the X − Y path P 4 as follows: From the definition of P 4 it follows that if A 4 is an inner vertex of P 4 , then We now proceed to show that for i ∈ {1, 2, 3, 4}, the X − Y paths in {P i } ∪ L are internally disjoint. For this, let us assume that A 1 , A 2 , A 3 , A 4 , A, A i,j , A i , A j , A * s , and A * t are inner vertices of P 1 , P 2 , P 3 , P 4 , P ∈ L 1 , P i,j ∈ L 2 , P i ∈ L 3 , Q j ∈ L 4 , P * s ∈ L * 3 , and Q * t ∈ L * 4 , respectively.
Clearly, the proof of Claim 2.8 finishes the proof of Lemma 2.1, which implies Theorem 1.

Concluding remarks
The trees and the complete graphs are two families of graphs which are extremely distinct from the point of view of the connectivity. Here we have shown that if G is a tree, then κ(F k (G)) = λ(F k (G)) = δ(F k (G)). Surprisingly, these same equalities hold for the case of the complete graph. More precisely, from Leaños and Trujillo-Negrete (2018) and Leaños and Ndjatchi (2021) we know that the connectivity and the edge-connectivity of F k (K n ) are equal to δ(F k (K n )) the minimum degree of F k (K n ). However, these equalities do not hold in general. For instance, it is not hard to see that for the graph H of Figure 7 we have κ(F 2 (H)) = m − 1 = λ(F 2 (H)) and δ(F 2 (H)) = 2(m − 2). On the other hand, based on computational experimentation and on some analytic approaches we have the following conjecture.