The linkedness of cubical polytopes: The cube

The paper is concerned with the linkedness of the graphs of cubical polytopes. A graph with at least $2k$ vertices is \textit{$k$-linked} if, for every set of $k$ disjoint pairs of vertices, there are $k$ vertex-disjoint paths joining the vertices in the pairs. We say that a polytope is \textit{$k$-linked} if its graph is $k$-linked. We establish that the $d$-dimensional cube is $\lfloor(d+1)/2\rfloor$-linked, for every $d\ne 3$; this is the maximum possible linkedness of a $d$-polytope. This result implies that, for every $d\ge 1$, a cubical $d$-polytope is $\lfloor{d/2}\rfloor$-linked, which answers a question of Wotzlaw \cite{Ron09}. Finally, we introduce the notion of strong linkedness, which is slightly stronger than that of linkedness. A graph $G$ is {\it strongly $k$-linked} if it has at least $2k+1$ vertices and, for every vertex $v$ of $G$, the subgraph $G-v$ is $k$-linked. We show that cubical 4-polytopes are strongly $2$-linked and that, for each $d\ge 1$, $d$-dimensional cubes are strongly $\lfloor{d/2}\rfloor$-linked.


Introduction
A (convex) polytope is the convex hull of a finite set X of points in R d ; the convex hull of X is the smallest convex set containing X . The dimension of a polytope in R d is one less than the maximum number of affinely independent points in the polytope. A polytope of dimension d is referred to as a d-polytope.
A face of a polytope P in R d is P itself, or the intersection of P with a hyperplane in R d that contains P in one of its closed halfspaces. Faces other than P are polytopes of smaller dimension. A face of dimension 0, 1, and d -1 in a d-polytope 2 THE LINKEDNESS OF CUBICAL POLYTOPES: THE CUBE is a vertex, an edge, and a facet, respectively. The graph G(P) of a polytope P is the undirected graph formed by the vertices and edges of the polytope. This paper studies the linkedness of cubical d-polytopes, d-dimensional polytopes with all their facets being cubes. A d-dimensional cube is the convex hull in R d of the 2 d vectors (±1, . . . , ±1). By a cube we mean any polytope that is combinatorially equivalent to a cube; that is, one whose face lattice is isomorphic to the face lattice of a cube. Let G be a graph and X a subset of 2k distinct vertices of G. The elements of X are called terminals. Let Y := {{s 1 , t 1 }, . . . , {s k , t k }} be an arbitrary labelling and (unordered) pairing of all the vertices in X . We say that Y is linked in G if we can find disjoint s i -t i paths for i ∈ [1, k], where [1, k] denotes the interval 1, . . . , k. The set X is linked in G if every such pairing of its vertices is linked in G. Throughout this paper, by a set of disjoint paths, we mean a set of vertex-disjoint paths. If G has at least 2k vertices and every set of exactly 2k vertices is linked in G, we say that G is k-linked. If the graph of a polytope is k-linked we say that the polytope is also k-linked.
Unless otherwise stated, we use the graph theoretical notation and terminology from [4], while the polytope theoretical notation and terminology from [22]. Moreover, when referring to graph-theoretical properties of a polytope such as minimum degree, linkedness and connectivity, we mean properties of its graph.
Linkedness is an attractive property of graphs. Being k-linked imposes a stronger demand on a graph than just being k-connected. Let G be a graph with at least 2k vertices, and let S := {s 1 , . . . , s k } and T := {t 1 , . . . , t k } be two disjoint k-element sets of vertices in G. It follows that, if G is k-connected then the sets S and T can be joined setwise by disjoint paths (namely, by k disjoint S -T paths); this is a consequence of Menger's theorem (Theorem 13). And if G is k-linked then the sets can be joined pointwise by disjoint paths.
From a structural point of view, linkedness guarantees the existence of many subdivisions in a graph. A graph Y is a subdivision of a graph X if it can be obtained from X by subdividing edges of X . The definition of k-linkedness yields that, if a graph is k-linked, then it has a subdivision of every graph on k edges.
From an algorithmic point of view, linkedness is closely related to the classical disjoint paths problem [15]: given a graph G and a set Y := {{s 1 , t 1 }, . . . , {s k , t k }} of k pairs of terminals in G, decide whether or not Y is linked in G. A natural optimisation version of this problem is to find the largest subset of the pairs so that there exist disjoint paths connecting the selected pairs. The disjoint paths problem has found many applications in the field of transportation networks and computer THE LINKEDNESS OF CUBICAL POLYTOPES: THE CUBE 3 science in general [7,8]. It is a special case of a multicommodity flow problem where there exist k different commodities that need to go from the sources s 1 , s 2 , . . . , s k to t 1 , t 2 , . . . , t k ; for information on multicommodity flows consult [10], and for further information on the disjoint paths problem consult [8] and the references therein.
All the 2-linked graphs have been characterised [17,19]. In the context of polytopes, one consequence is that, with the exception of simplicial 3-polytopes, no 3-polytope is 2-linked; a simplicial polytope is one in which every facet is a simplex.
Another consequence is that every 4-polytope is 2-linked. We provide new proofs of these two results: Corollary 5 and Proposition 6.
There is a linear function f (k) such that every f (k)-connected graph is k-linked, which follows from works of Bollobás and Thomason [2]; Kawarabayashi, Kostochka, and Yu [9]; and Thomas and Wollan [18]. In the case of polytopes, Larman Apart from the work of Larman and Mani [11], the study of linkedness in graphs of polytopes has been motivated by a problem in the first edition of the Handbook of Discrete and Computational Geometry [6,Problem 17.2.6]. The problem asked whether or not every d-polytope is d/2 -linked. This question had already been answered in the negative by Gallivan [5] in the 1970s with a construction of a dpolytope that is not 2(d + 4)/5 -linked. A weak positive result however follows from [18]: every d-polytope with minimum degree at least 5d is d/2 -linked.  Theorem 18). We remark that the linkedness of the cube was first established in [12,Prop. 4.4] as part of a study of linkedness in Cartesian products of graphs, but no self-contained proof was available.
In a subsequent paper [3], we prove a stronger result: a cubical d-polytope is (d + 1)/2 -linked, for every d = 3; which is best possible. In anticipation of this result, in Proposition 27 we prove that certain cubical d-polytopes that are embedded in the (d + 1)-cube are (d + 1)/2 -linked, for every d ≥ 3.
Let X be a set of vertices in a graph G. Denote by G[X ] the subgraph of G induced by X , the subgraph of G that contains all the edges of G with vertices Finally, we introduce the notion of strong linkedness, a property marginally stronger than linkedness. We say that a graph G is strongly k-linked if it has at 4 THE LINKEDNESS OF CUBICAL POLYTOPES: THE CUBE least 2k + 1 vertices and, for every vertex v of G, the subgraph G -v is k-linked.
We show that cubical 4-polytopes are strongly 2-linked and that, for each d ≥ 1, d-dimensional cubes are strongly d/2 -linked.

Preliminary results
This section groups a number of results that will be used in later sections of the paper.
Propositions 4 and 6 follow from the characterisation of 2-linked graphs carried out in [17,19]. Both propositions also have proofs stemming from arguments in the form of Lemma 1; for the sake of completeness we give such proofs.
We state Balinski's theorem on the connectivity of polytopes.
A path in the graph is called X -valid if no inner vertex of the path is in X .
The distance between two vertices s and t in a graph G, denoted dist G (s, t), is the length of a shortest path between the vertices.  Proof. Let P be a 3-polytope embedded in R 3 and let X be an arbitrary set of four vertices in G. We first establish the necessary condition by proving the contrapositive. Let F be a 2-face containing the vertices of X and consider a planar THE LINKEDNESS OF CUBICAL POLYTOPES: THE CUBE 5 embedding of G in which F is the outer face. Label the vertices of X so that they appear in the cyclic order s 1 , s 2 , t 1 , t 2 . Then the paths s 1 -t 1 and s 2 -t 2 in G must inevitably intersect, implying that X is not linked.
Assume there is no 2-face of P containing all the vertices of X . Let H be a (linear) hyperplane that contains s 1 , s 2 and t 1 , and let f be a linear function that vanishes on H (this may require a translation of the polytope). Without loss of generality, assume that f (x) > 0 for some x ∈ P and that f (t 2 ) ≥ 0. The subsequent corollary follows at once from Proposition 4.

Corollary 5. No nonsimplicial 3-polytope is 2-linked.
The same reasoning employed in the proof of the sufficient condition of Proposition 4 settles Proposition 6. Consider a linear function f that vanishes on a linear hyperplane H passing through X . Consider the two cases in which either H is a supporting hyperplane of a facet F of P or H intersects the interior of P.
Suppose H is a supporting hyperplane of a facet F . First, find an s 1 -t 1 path in the subgraph G(F )-{s 2 , t 2 }, which is connected by Balinski's theorem (Theorem 2).
Second, use Lemma 1 to find an s 2 -t 2 path that touches F only at {s 2 , t 2 }.
If instead H intersects the interior of P then there is a vertex in P with f -value greater than zero and a vertex with f -value less than zero. Use Lemma 1 to find an s 1 -t 1 path in which each inner vertex has negative f -value and an s 2 -t 2 path in which each inner vertex has positive f -value.

d-cube
Consider  Fig. 1(a). We extend this projection to sets of vertices: given a pair {F , F o } of opposite facets and a set X ⊆ V (F ), the projection Let Z be a set of vertices in the graph of a d-cube Q d . If, for some pair of

Definition 7 is exemplified in
Note that an associating pair can associate only one pair of opposite facets. See Fig. 1 The next lemma lies at the core of our methodology. If a direction is present in a cycle C of Q d , then the cycle contains at least two edges from this direction. Indeed, take an edge e = uv on C that belongs to a direction between a pair {F , F o } of opposite facets. After traversing the edge e The relevance of the lemma stems from the fact that a pair of opposite facets {F , F o } not associated with a given set of vertices Z allows each vertex z in Z to have "free projection"; that is, for every

Connectivity of the d-cube
We next unveil some further properties of the cube that will be used in subsequent sections.
Given sets A, B, X of vertices in a graph G, the set X separates A from B if every A -B path in the graph contains a vertex from X . A set X separates two vertices a, b not in X if it separates {a} from {b}. We call the set X a separator of the graph. We will also require the following three assertions.

Proposition 9 ([14, Prop. 1]). Any separator X of cardinality d in Q d consists of the d neighbours of some vertex in the cube.
A set of vertices in a graph is independent if no two of its elements are adjacent.
Since there are no triangles in a d-cube, Proposition 9 gives at once the following corollary.

Corollary 10. A separator of cardinality d in a d-cube is an independent set.
Remark 11. If x and y are vertices of a cube, then they share at most two neighbours. In other words, the complete bipartite graph K 2,3 is not a subgraph of the cube; in fact, it is not an induced subgraph of any simple polytope [13, Cor. 1.12(iii)].

Linkedness of the d-cube
In this section, we establish the linkedness of Q d (Theorem 18). We make heavy use of Menger's theorem [4, Thm. 3.3.1] henceforth, and so we remind the reader of the theorem and one of one of its consequences.  The definition of k-linkedness gives the following lemma at once. Lemma 15. Let ≤ k. Let X * be a set of 2 distinct vertices of a k-linked graph G, let Y * be a labelling and pairing of the vertices in X * , and let Z * be a set of at We require a result on strong linkedness. With Proposition 6 and Lemma 14 at hand, we can verify that cubical 4-polytopes are strongly 2-linked.

Theorem 16 (Strong linkedness of cubical 4-polytopes). Every cubical 4-polytope is strongly 2-linked.
Proof. Let G denote the graph of a cubical 4-polytope P embedded in R 4 . Let X be a set of five vertices in G. Arbitrarily pair four vertices of X to obtain Y := {{s 1 , t 1 }, {s 2 , t 2 }}. Let x be the vertex of X not being paired in Y . We aim to find two disjoint paths L 1 := s 1 -t 1 and L 2 := s 2 -t 2 such that each path L i avoids the vertex x. The proof is very similar to that of Propositions 4 and 6.
Consider a linear function f that vanishes on a linear hyperplane H passing through {s 1 , s 2 , t 1 , x}. Assume that f (y) > 0 for some y ∈ P and that f (t 2 ) ≥ 0.
Suppose first that H is a supporting hyperplane of a facet F of P. If t 2 ∈ V (F ) (that is, f (t 2 ) > 0), then find an X -valid L 1 := s 1 -t 1 path in F using the 3connectivity of F (Balinski's theorem). Then use Lemma 1 to find an X -valid Not every 4-polytope is strongly 2-linked. Take a two-fold pyramid P over a quadrangle Q. Then P is a 4-polytope on six vertices, say s 1 , s 2 , t 1 , t 2 , x, y. Let the sequence s 1 , s 2 , t 1 , t 2 appears in Q in cyclic order, and let the vertex x be in V (P) \ V (Q). To see that P is not strongly 2-linked, observe that, for every two paths s 1 -t 1 and s 2 -t 2 in P, they intersect or one of them contains x.
We continue with a simple lemma from [20,Sec. 3]. We first deal with the case of even d ≥ 6. In this setting, d = 2k, and so G is 2k-connected by Balinski's theorem. Furthermore, by the induction hypothesis, the graph G of every facet of Q d , a (d -1)-polytope, is d/2 -linked, namely k-linked.
Lemma 17 now ensures that G is k-linked. As a consequence, for the rest of the proof, we focus on the case of odd d ≥ 5.   From s 1 ∈ F it now follows that π F (s i ) ∈ X , for each i ∈ [2, k]. Besides, since the pair {F , F o } is not associated with X s 1 , we have that It is the case that π F o (t 1 ) = s 1 , otherwise s 1 and t 1 would be adjacent, contradicting the fact that dist(s 1 , t 1 ) = d ≥ 5. Hence π F o (t 1 ) ∈ X . Because k > 2, it is also true that π F o (t i ) = s 1 for some t i ∈ V (F ) with i ∈ [2, k], say π F o (t 2 ) = s 1 . Then π F o (t 2 ) ∈ X . We summarise our discussion below.
Let X := {π F (s 3 ), . . . , π F (s k ), t 3 , . . . , t k }. By the induction hypothesis, F is (k -1)-linked. In the notation of Lemma 15, if we let := k -2, X * := X , , t k }}, and Z * := {t 1 , t 2 }, then we can find k -2 disjoint paths L i in F between π F (s i ) and t i for i ∈ [3, k], with each path avoiding Now we find the paths L 1 and L 2 in F o . The induction hypothesis yields that F o is (k -1)-linked. We also have that π F o (t 1 ) ∈ X and π F o (t 2 ) ∈ X , according to (1). In the notation of Lemma 15, we let := 2, As a consequence, we let L 1 := s 1 L 1 π F o (t 1 )t 1 and L 2 := s 2 L 2 π F o (t 2 )t 2 . In this way, we have found a Y -linkage {L 1 , . . . , L k } with L i joining the pair {s i , t i } for i = 1, . . . , k. This completes the proof of the first scenario.
In the second scenario all vertices in X lie in a facet F of Q d . In this case, Lemma 14 gives an X -valid path L 1 in F joining a pair in Y , say {s 1 , t 1 }.  We define a function ρ : X → X that maps x ∈ X to the terminal with which it is paired in Y : {x, ρ(x)} ∈ Y . Let X F := (X \ {s 1 , t 1 }) ∩ V (F ). We define the 12 THE LINKEDNESS OF CUBICAL POLYTOPES: THE CUBE following sets.

The projection in
We construct the desired Y -linkage {L 1 , . . . , L k } according to the following cases: Case (i) is done, so we focus on Case (ii). We will need to project some terminals from F to F o , namely the ones in X β . However, unlike Scenario 2, it may happen that the projection π F o of a terminal vertex x ∈ X β onto F o is also a terminal vertex, which will cause the problem of having some paths intersect. In order to use the projection π F o , we define an injective map ω : X β → V (F ) so that, for each The motivation for the map ω is to define a path M x = xω(x)π F o (ω(x)), of length at most 2, from each x ∈ X β to F o so that the vertices π F o (ω(x)) and π F o (ω(ρ(x))) can be joined in F o by an X -valid path. Example 23 and Fig. 3 illustrate the function ω. The construction of ω follows a general remark.
Remark 20. Whenever possible we set ω(x) = x. Only when the projection π F o of Lemma 21 shows that the injective map ω exists.

Lemma 21.
There exists an injective map ω : X β → V (F ) such that, for each x ∈ X β , Proof. Let X be the maximal subset of X β such that an injective map ω exists and satisfies Condition (2). We will prove that X = X β by contradiction. Assume that X = X β and let x ∈ X β \ X . Then otherwise setting ω(x) = x would satisfy (2), extending the injection ω to X ∪ {x}.
Let us define the set O x as the subset of vertices v in N F (x) that cannot be selected as v = ω(x), because they violate either Condition (2) or the injectivity of ω.
For a vertex v to violate Condition (2), it must be that either v ∈ X (say v is

THE LINKEDNESS OF CUBICAL POLYTOPES: THE CUBE 13
For a vertex v to violate the injectivity of ω, there must exist z ∈ X such that v = ω(z); we remark that a vertex of type 1 or 2 could violate the injectivity of ω.
As a consequence, we say that v is of type 3 if it violates the injectivity but it is not of type 1 or 2.
Therefore the set O x can be defined as the subset of N F (x) that violate Condition (2) or the injectivity of ω.
implying that the vertices z, x, and ω(z) would all be pairwise neighbours but there are no triangles in Q d .
. Finally, suppose that v / ∈ X and v = ω(z) for some z ∈ X . Further assume that v is not of type 2 (namely, v = z p F for any z ∈ X \{ρ(x)}), since this case was already considered. Then v is of type 3. Because v / ∈ X , we have that z = ω(z). From z = ω(z) it follows that z p F o ∈ X (Remark 20). If z = ρ(x), then map v to z, else map v to z p F o . We prove in Claim 1 that the map ψ x is indeed injective.
Then ψ x is injective. Figure 3 depicts the different types of neighbours of the vertex s 2 and the injec- Proof. For the proof of the claim, we say that a v is of type 3(a) when it satisfies the third line of the definition of ψ x , namely v / ∈ X , v = ω(z) for some z ∈ X , z = ρ(x), and v is not of type 2. And we say that v is of type 3(b) if it satisfies the fourth line of the definition of ψ x , namely if v / ∈ X , v = ω(z) for some z ∈ X , z = ρ(x), and v is not of type 2. First note the following: • If v is of type 1, then ψ

THE LINKEDNESS OF CUBICAL POLYTOPES: THE CUBE
• if v is of type 3(a), then ψ x (v) ∈ X F \ N F (x) (see Remark 22).
Suppose that γ ∈ V (F ). If γ ∈ N F (x), then both v 1 and v 2 must be of type 1.
In this case, from the definition of ψ x we conclude that v  (ρ(x)). Finally suppose that v 1 is of type 2 and v 2 is of type 3(b).
Since v 1 is of type 2, we get that v 1 = γ p F , and since v 2 is of type 3(b) we get that ρ(x) = γ p F . This implies that ρ(x) = v 1 ∈ N F (x), which in turn implies that {x, ρ(x)} ∈ Y α and x ∈ X α , contradicting the assumption that x / ∈ X α . Therefore, for every v 1 , v 2 ∈ O x , the equality ψ x (v 1 ) = ψ x (v 2 ) implies that v 1 = v 2 , and the map ψ x is injective. injection ω can be extended to X ∪ {x}.This contradicts the maximality of X and concludes the proof of the lemma.
For every x ∈ X β , we define the path M x = xω(x)π F o (ω(x)), of length at most 2, from X β to F o . The injectivity of ω and the injectivity of the restriction of π F o to V (F ) ensure that the paths M x are pairwise disjoint. For every x ∈ X ∩ V (F o ) we set M x := x. Because ω satisfies Condition (2), the only case when the paths M x and M y intersect is when y = ρ(x), which is not a problem.
We now finalise this third scenario. Applying Lemma 21 to X β , we get the paths M x from all the terminals in X β to F o . We also consider the paths Let Y o be the corresponding pairing of the vertices in The induction hypothesis ensures that F o is (k -1)-linked. As a consequence, because of (4), It only remains to show the existence of a path L 1 := s 1 -t 1 in F disjoint from the paths L i for i ∈ [2, k] (Case (iii)). Suppose that we cannot find a path L 1 disjoint from the other paths L i with i ∈ [2, k]. Then there would be a set S in V (F ) separating s 1 from t 1 . The set S would consist of terminal vertices in X F and nonterminal vertices in ω(X β ) (see Lemma 21). Since x = ω(x) implies that By the (d -1)-connectivity of F , which follows from Balinski's theorem (Theorem 2), the set S would have cardinality d -1, which implies that every terminal in X F and every nonterminal ω(x) in F are in S. By Proposition 9, the set S would consist of the neighbours of s 1 or t 1 , say of s 1 , and therefore the vertices in S are pairwise nonadjacent (as there are no triangles in Q d ), that is, ω(x) = x for each x ∈ X β (ω(x) is either x or a neighbour of x). This implies that S ⊂ X , and so The proof of the theorem is now complete. We next illustrate Scenario 3 of Theorem 18. Lemma 21 gives us an injection ω from (X \{s 1 , t 1 })∩V (F ) to V (F ) to find X -valid paths from the d -1 terminals in (X \ {s 1 , t 1 }) ∩ V (F ) to V (F o ). Then we use the 2-linkedness of F o to find the paths L 2 , L 3 . We follow the notation of the proof of Theorem 18. For each terminal x ∈ (X \ {s 1 , t 1 }) ∩ V (F ), we define the path First look at Fig. 3(a). Then ω( say M s 2 = s 2 v 2 t 2 , M t 2 := t 2 , M s 3 := s 3 , and M t 3 := t 3 . It follow that We have paths L p 2 := t 2 and L p 3 := s 3 -t 3 in F o . A path L 1 := s 1 -t 1 in F should avoid only s 2 and ω(s 2 ) = v 2 , and so it exists by the 4-connectivity of F . As a result, the Y -linkage in this setting is given by L 1 , L 2 := s 2 M s 2 t 2 , and L 3 := s 3 L p 3 t 3 . Now look at Fig. 3 We are now in a position to answer Wotzlaw . Let X p := π F (X x ); that is, the set X p comprises the vertices in X x ∩ V (F ) plus the projections of X x ∩ V (F o ) onto F . Denote by Y p the corresponding pairing of the vertices in X p ; that is,

Linkedness inside the cube
The boundary complex of a polytope P is the set of faces of P other than P itself.
And the link of a vertex v in a polytope P, denoted link(v, P), is the set of faces of P that do not contain v but lie in a facet of P that contains v (Fig. 4). According to [22,Ex. 8.6], the link of a vertex in a d-polytope is combinatorially equivalent to the boundary complex of a (d -1)-polytope; in particular, for d ≥ 3 the graph of the link is isomorphic to the graph of a (d -1)-polytope. It follows that the link of a   (Fig. 4).
We verify that, for every d ≥ 2 such that d = 3, the link of a vertex in a (d + 1)-cube is (d + 1)/2 -linked. Since |X | -1 ≤ d and there are d + 1 pairs of opposite facets in Q d+1 , from Lemma 8 there exists a pair {F , F o } of opposite facets of Q d+1 that is not associated with X . This means that, for every x ∈ X ∩ V (F ), its projection π Q d+1 F o (x) ∈ X , and that, for every x ∈ X ∩ V (F o ), its projection π Q d+1 F (x) ∈ X . Henceforth we write π F rather than π Q d+1 F . Assume that v ∈ F and v o ∈ F o . We consider two cases based on the number of terminals in the facet F .
In what follows, we implicitly use the d-connectivity of F or F o for d ≥ 5. Since F is a d-cube, it is (d + 1)/2 -linked by Theorem 18, and hence, we can find k pairwise disjoint pathsL 1 , . . . ,L k in F between s i and t i for each i ∈ [1, k].
If no path L i passes through v, we let L i :=L i for i ∈ [1, k], and so {L 1 , . . . , L k } is the desired Y -linkage. So suppose one of those paths, sayL 1 , passes through v; there can be only one such path.
In this case, we consider the two neighbours w 1 and w 2 of v onL 1 so that v / ∈ w 1L1 s 1 and v / ∈ w 2L1 t 1 . Since dist Q d+1 (v, v o ) = d + 1 ≥ 3, we have that we let L i :=L i for i ∈ [2, k], and so {L 1 , . . . , L k } is the desired Y -linkage.
By symmetry, the proposition also holds if |X ∩ V (F o )| = d + 1.
It follows that |X ∩ V (F o )| ≤ (d + 1)/2 . If some terminal x in F o is adjacent to v: π F (x) = v, without loss of generality, assume that it is t 1 . In any case, there is at least one terminal in F o , and we may assume that it is t 1 . The facet F is a d-cube, and so it is (d + 1)/2 -linked by Theorem 18. Let w ∈ V (F ) be a neighbour of v on M 1 . This neighbour exists, since π F (s 1 ) = v, and so the length of M 1 is at least 1. The projection π F o (w) of w onto F o is not in X \ {s 1 } because w / ∈ X p \ {π F (s 1 )} and the path M 1 is disjoint from the paths L i , i ∈ [2, k]. By the d-connectivity of F o we can find a pathL 1 in F o from π F o (w) to t 1 that avoids S. Hence the path L 1 then becomes s 1 π F (s 1 )M 1 wπ F o (w)L 1 t 1 .
This completes the proof of the case and of the proposition.