Even Factors in Edge-Chromatic-Critical Graphs with a Small Number of Divalent Vertices

Abstract

A finite simple connected graph G with maximum degree k is k-critical if it has chromatic index $\chi '(G)=k+1$ and $\chi '(G-e)=k$ for every edge $e\in E(G)$. Bej and the first author [2] raised the question whether every k-critical graph has an even factor. We prove that every k-critical graphs with at most $2k-6$ vertices of degree 2 has an even factor.

We consider finite simple graphs with vertex set V(G) and edge set E(G). The degree of a vertex $v \in V(G)$ is denoted by d(v) and $\Delta (G)$ denotes the maximum degree of a vertex of G. A vertex of degree 2 is also called a divalent vertex. A graph is even if each component is eulerian and different from $K_1$.

For a subset A of V(G), the subgraph induced by $V(G)-A$ is denoted by $G-A$. If no two vertices in A are connected by an edge, then A is stable. For a subset $E'$ of E(G), the graph obtained from G by deleting the edges in $E'$ is denoted by $G-E'$ (if $E'=\{e\}$, then we use the notation $G-e$).

A k-edge-colouring of a graph G is a function $\varphi : E(G) \rightarrow \{1,...,k\}$; the elements of $\{1,...,k\}$ are called colours. A k-edge-colouring $\varphi$ is proper, if no two adjacent edges receive the same colour. For a vertex $v \in V(G)$ and a colour $i \in \{1,..,k\}$, we say colour i is present at v if an edge incident to v is coloured with colour i. Otherwise, colour i is missing at v. The set of colours present at v is denoted by $\varphi (v)$; the set of colours missing at v is denoted by ${\bar{\varphi }}(v)$. If a proper k-edge-colouring of G exists, then G is k-edge-colourable. The chromatic index, denoted $\chi '(G)$, is the smallest integer k such that G is k-edge-colourable. Vizing [10] proved the fundamental result on edge-colouring simple graphs by showing that the chromatic index of a graph G is either $\Delta (G)$ or $\Delta (G)+1$.

An edge $e \in E(G)$ is critical, if $\Delta (G)=k$, $\chi '(G)=k+1$ and $\chi '(G-e)=k$. If G is connected and all edges of G are critical, then G is k-critical. Clearly, every graph H with $\chi '(H) = \Delta (H)+1$ contains a $\Delta (H)$-critical subgraph. There had been several conjectures with regard to the order or to (near) perfect matchings of critical graphs, which all turned out to be false, see [2] for a survey.

The situation changes when we consider 2-factors. In 1965, Vizing [9] conjectured that every critical graph has a 2-factor. This conjecture has been verified for some specific classes of critical graphs as overfull graphs [6] or critical graphs with large maximum degree in relation to their order [3, 8]. Furthermore, some equivalent formulations or reduction to some classes of critical graphs as e.g. critical graphs of even order are proved in [2, 4]. All these approaches have not yet led to significant progress in answering the question whether critical graphs have a 2-factor.

To gain more insight into structural properties of critical graphs it might be useful to investigate slightly easier statements about factors in critical graphs. In [2] Bej and the first author conjectured that (1) every critical graph has a path-factor and (2) every critical graph has an even factor.

The first conjecture is proved in [7]. For the second conjecture note that every bridgeless graph with minimum degree at least 3 has an even factor [5]. Thus, conjecture (2) is reduced to critical graphs with divalent vertices. Our main result is the following:

FormalPara Theorem 1

Let $k \ge 3$ and G be a k-critical graph. If G has at most $2k-6$ divalent vertices, then G has an even factor.

1 Preliminaries and Useful Lemmas

The set of edges with one end in $A_{1}$ and the other in $A_{2}$ is denoted by $E_{G}(A_{1},A_{2})$, where $A_{i}$ is a vertex set, a single vertex (in this case, one end in $A_{i}$ means one end in $\{A_{i}\}$) or a subgraph (in this case, one end in $A_{i}$ means one end in $V(A_{i})$). Furthermore, $e_{G}(A_{1},A_{2})$ is the cardinality of $E_{G}(A_{1},A_{2})$. The set of neighbours of a vertex $v \in V(G)$ is denoted by N(v); for a subset A of V(G) the neighbourhood of A, denoted N(A), is $\bigcup _{v \in A} N(v) -A$. If $G'$ is a subgraph of G, then we write $N(G')$ instead of $N(V(G'))$.

Let $\varphi$ be a proper k-edge-colouring of a graph G and $v \in V(G)$. For two different colours $i,j \in \{1,...,k\}$, the subgraph induced by the edges that are coloured i or j is denoted by K(i, j). Its components are called (i, j)-Kempe chains or sometimes just Kempe chains. Clearly, a Kempe chain is a path or a circuit. If $\{i,j\} \cap \varphi (v) \ne \emptyset$, then the unique component of K(i, j) that contains v is denoted by $P_{v}^{\varphi }(i,j)$. We will omit the upper index if this does not cause any ambiguity. A new proper k-edge-colouring, denoted by $\varphi /P_{v}(i,j)$, can be obtained from $\varphi$ by interchanging colours i and j in $P_{v}(i,j)$.

In the proofs of Lemma 2 and 3 we will use the following basic observation without reference: Let G be a graph with a critical edge vw and let $\varphi$ be a proper $\Delta (G)$-edge-colouring of $G-vw$. If colour i is missing at v and j is missing at w, then colour i is present at w, colour j is present at v and $P_{v}^{\varphi }(i,j)$ is a v, w-path.

Lemma 2

Let G be a graph with $\Delta (G)=k$, $\chi '(G)=k+1$, and let $A \subseteq V(G)$ be a set of vertices such that

• $e_{G}(A,v)=1$ for every $v \in N(A)$, and

• $N(A)=\{x,y,w_{1},...,w_{l}\}$ with $l\ge 1$, $d(y)\le d(x)<k$ and $d(w_{i})=2$ for every $i \in \{1,...,l\}$.

If at least one edge in $E_{G}(A,\{w_{1},...,w_{l}\})$ is critical, then $l > k(k-d(y))-d(x)+1$.

Proof

Let $w \in \{w_{1},...,w_{l}\}$ be a divalent vertex, let $w'$ be the unique neighbour of w that belongs to A, and let the edge $w'w$ be critical.

Claim 1: There is a proper k-edge-colouring $\varphi$ of $G-w'w$ such that ${\bar{\varphi }}(w')=\varphi (w)=\{1\}$ and $1 \in {\bar{\varphi }}(x)$.

Proof

Since $w'w$ is critical there is a proper k-edge-colouring $\varphi '$ of $G-w'w$. Furthermore, ${\bar{\varphi }}'(w')=\varphi '(w)=\{i\}$ for a colour $i \in \{1,...,k\}$. Since $d(x)<k$, there is a colour j that is missing at x. If $i=j \ne 1$, then we obtain a colouring with the desired properties by interchanging colours i and 1. If $i \ne j$, then $P_{w'}(i,j)$ is a $w',w$-path and thus does not contain x. Therefore, the colouring $\varphi ''$, defined by $\varphi ''=\varphi '/P_{w'}(i,j)$, satisfies ${\bar{\varphi }}''(w')=\varphi ''(w)=\{j\}$ and $j \in {\bar{\varphi }}''(x)$. Again, if $j \not = 1$, then a colouring with the desired properties can be obtained by interchanging colours j and 1. Thus, Claim 1 is proved.

Now fix a proper k-edge-colouring $\varphi$ of $G-w'w$ with the properties stated in Claim 1. Define a set M as follows:

$$\begin{aligned} M=\{(h,z,h'): \text { }&z \in N(A)-\{w\}, \{h,h'\} \subseteq \varphi (z), h \ne h',\\ {}&\varphi (e)=h \text {, where } e \text { is the unique edge in } E_{G}(A,z)\}. \end{aligned}$$

We prove a lower bound for the number of triples in M, which will be used to obtain the lower bound for l.

For each triple $(h,z,h')$ of M there is a unique Kempe chain P, that contains the two edges incident with z that are coloured h and $h'$. In this case we say P contains $(h,z,h')$. Furthermore, if P is a path and v is an end vertex of P, then we can interpret P as a vertex-list starting with v. This gives an order of the vertices of P and thus an order of the triples contained in P. We define the first and the last triple contained in P (starting with v) in the natural way. An example is given in Fig. 1.

Fig. 1

$P_{v}(1,2)$ contains (1, y, 2), $(2,z',1)$ and (2, z, 1). (1, y, 2) is the first and (2, z, 1) the last triple contained in $P_{v}(1,2)$ (starting with v)

If $i \in \{2,...,k\}$, then the Kempe chain $P_{w'}(1,i)$ is a path with end vertices $w'$ and w and thus, it contains at least one triple of M. Therefore, we can define a subset $M_1$ of M as follows: For every $i \in \{2,...k\}$ let $(i_{1},z_{i},i_{2})$ be the last triple contained in $P_{w'}(1,i)$ (starting with $w'$). Let $M_1=\{(i_{1},z_{i},i_{2}):i \in \{2,...k\}\}$. Figure 2 shows an example. We note, that $\{i_1,i_2\}=\{1,i\}$ for every $i \in \{2,...,k\}$ and in particular, x is not in a triple of $M_1$, since colour 1 is missing at x.

Fig. 2

The triple $(3,z_3,1)$ is the last triple contained in $P_{w'}(1,3)$; (1, y, 2) is the last triple contained in $P_{w'}(1,2)$(starting with $w'$). Thus, $(3,z_3,1), (1,z_2,2) \in M_1$ where $z_2=y$

$\underline{\hbox {Claim 2: }}\vert M_{1} \vert = k-1$.

Proof

Let $i, i'$ be two different colours of $\{2,...,k\}$. Then, $\{i_1,i_2\}=\{1,i\}\ne \{1,i'\}=\{i'_1,i'_2\}$, and hence $(i_{1},z_{i},i_{2}) \ne (i'_{1},z_{i'},i'_{2})$.

$\underline{\hbox {Claim 3:}}$ Let $i \in \{2,...,k\}$ and $j \in {\bar{\varphi }}(z_{i})$. Then $P_{z_{i}}(i_{1},j)$ contains a triple of M.

Proof

Suppose, $P_{z_{i}}(i_{1},j)$ does not contain a triple of M. Then, the colouring $\varphi '$, defined by $\varphi '=\varphi /P_{z_{i}}(i_{1},j)$, satisfies ${\bar{\varphi }}'(w')=\varphi '(w)=\{1\}$. Since $(i_{1},z_{i},i_{2})$ is the last triple contained in $P_{w'}(1,i)$ (starting with $w'$), the Kempe chain $P_{w}^{\varphi '}(i_{1},i_{2})$ has end vertices w and $z_{i}$. In particular $P_{w}^{\varphi '}(i_{1},i_{2})$ is not a $w',w$-path. We have either $i_{1}=1$ or $i_{2}=1$, a contradiction. See Fig. 3 for an example.

Fig. 3

The triple $(3,z_3,1)$ is in $M_1$. Colour 2 is missing at $z_3$. The Kempe chain $P_{z_3}(3,2)$ does not contain a triple of M. Interchanging colours 3 and 2 in $P_{z_3}(3,2)$ produces a contradiction, since $P_{w}(1,3)$ is no longer a $w',w$-path

Let $M_{2}=\{(h,z,h'): i \in \{2,...,k\}, j \in {\bar{\varphi }}(z_{i}),(h,z,h') \text { is the first triple contained in } P_{z_{i}}(i_{1},j) \text { (starting with } z_{i})\}.$ An example is given in Fig. 4.

Fig. 4

The triple $(1,z_2,2)$ is in $M_1$, where $z_2=y$. Colour 3 is missing at y. The triple (3, z, 1) is the first triple contained in $P_{y}(1,3)$ (starting with y) and thus in $M_2$

$\underline{\hbox {Claim 4: }} M_{1} \cap M_{2}=\emptyset$.

Proof

We have $z_{i} \notin \{w',w\}$ for every $i \in \{2,...,k\}$. Hence, every triple of $M_2$ is contained in a path with an end vertex that is neither $w'$ nor w, whereas every triple of $M_1$ is contained in a $w',w$-path.

$\underline{\hbox {Claim 5: }} \vert M_{2} \vert = \vert \{(j,z_{i}): i \in \{2,...,k\}, j \in {\bar{\varphi }}(z_{i})\} \vert$.

Proof

Let $i,i' \in \{2,...,k\}$, $j \in {\bar{\varphi }}(z_{i})$ and $j' \in {\bar{\varphi }}(z_{i'})$ such that $(j,z_i)\ne (j',z_{i'})$. Then, the paths $P_{z_{i}}(i_{1},j)$ and $P_{z_{i'}}(i'_{1},j')$ have at least one different colour or a different starting vertex (interpreted as a vertex-list starting with $z_{i}$ or $z_{i'}$ respectively). Therefore, these two paths have different first triples (in the case $z_{i} \ne z_{i'}$ we use the fact that both triples are first triples).

$\underline{\hbox {Claim 6: }} \vert \{(j,z_{i}): i \in \{2,...,k\}, j \in {\bar{\varphi }}(z_{i})\} \vert \ge (k-d(y))(k-1)$.

Proof

Since colour 1 is missing at x, there is no $i \in \{2,...,k\}$ with $z_{i}=x$. If $z_i=z_{i'}$ for two different integers $i,i'$ of $\{2,...,k\}$, then $z_i=z_{i'}=y$, since every vertex in $N(A)\setminus \{x,y\}$ is divalent. Furthermore, the number of indices $i \in \{2,...,k\}$ with $z_{i}=y$ is at most $d(y)-1$. Vertex y misses $k-d(y)$ colours whereas all other vertices in $N(A)-\{w,x\}$ miss $k-2$ colours. In conclusion:

$$\begin{aligned}&\vert \{(j,z_{i}): i \in \{2,...,k\}, j \in {\bar{\varphi }}(z_{i})\} \vert \\ \ge&(k-d(y))+(k-1-(d(y)-1))(k-2)\\ =&(k-d(y))(k-1). \end{aligned}$$

We now prove that $l \ge k(k-d(y))-d(x)+1$. Since $e_{G}(A,v)=1$ for every $v \in N(A)$ and all vertices in $N(A)-\{x,y\}$ are divalent, the inequality $l >\vert M \vert - (d(x)-1)-(d(y)-1)$ holds. By Claims 2, 4, 5 and 6, we have:

$$\begin{aligned} l&>\vert M \vert - (d(x)-1)-(d(y)-1)\\&\ge \vert M_{1} \vert +\vert M_{2} \vert - (d(x)-1)-(d(y)-1)\\&\ge k-1+(k-d(y))(k-1)-(d(x)-1)-(d(y)-1)\\&= k(k-d(y))-d(x)+1. \end{aligned}$$

$\square$

Lemma 3

Let $k > 3$ and let G be a graph with $\Delta (G)=k$ and $\chi '(G)=k+1$. If $E'\subseteq E(G)$ is an inclusion-minimal edge-cut consisting of three critical edges, then no edge in $E'$ is incident to a divalent vertex.

Proof

Let G be a graph with $\Delta (G)=k$ and $\chi '(G)=k+1$. Furthermore, let $E'\subseteq E(G)$ be an inclusion-minimal edge-cut consisting of three critical edges $e_{1},e_{2}$ and $e_{3}$; let A and B be the components of $G-E'$, and let $e_{i}=x_{i}y_{i}$, where $x_{i}$ belongs to A and $y_{i}$ to B. Let $G_A$ be the subgraph induced by $V(A) \cup \{y_{1},y_{2}, y_{3}\}$, and let $G_B$ be the subgraph induced by $V(B) \cup \{x_{1},x_{2}, x_{3}\}$. We say a k-edge-colouring $\varphi$ of $G_A$ or $G_B$ is of

• type 1, if $\varphi (e_{1})=\varphi (e_{2})=\varphi (e_{3})$,

• type 2, if $\varphi (e_{1})=\varphi (e_{2})\ne \varphi (e_{3})$,

• type 3, if $\varphi (e_{1})=\varphi (e_{3})\ne \varphi (e_{2})$,

• type 4, if $\varphi (e_{2})=\varphi (e_{3})\ne \varphi (e_{1})$,

• type 5, if $\varphi (e_{1})\ne \varphi (e_{2})$, $\varphi (e_{1}) \ne \varphi (e_{3})$, $\varphi (e_{2})\ne \varphi (e_{3})$.

Suppose to the contrary that there is an edge of $E'$ that is incident to a divalent vertex. We will show that there is a proper k-edge-colouring $\varphi _{A}$ of $G_A$ and a proper k-edge-colouring $\varphi _{B}$ of $G_B$ such that $\varphi _{A}$ and $\varphi _{B}$ can be combined to a proper k-edge-colouring of G (by possibly relabeling the colours in one of the colourings), a contradiction.

In order to label the appearing colourings properly, we use the following definition: For an edge $e \in E'$, a k-edge-colouring $\varphi$ of $G-e$ and a colour $i \in \{1,...,k\}$, let $\varphi _{i}$ denote the k-edge-colouring of G obtained from $\varphi$ by colouring e with i.

Suppose to the contrary that $d(y_{1})=2$. First of all we use the fact that $e_{1}$ is critical. Let $\varphi$ be a proper k-edge-colouring of $G-e_{1}$ such that w.l.o.g. ${\bar{\varphi }}(x_{1})=\varphi (y_{1})=\{1\}$ holds. Thus, for every $i \in \{2,...,k\}$ the Kempe chain $P_{x_{1}}(1,i)$ is an $x_{1},y_{1}$-path. Since $k>3$, at least two of these paths, say $P_{x_{1}}(1,2)$ and $P_{x_{1}}(1,3)$, contain w.l.o.g. $e_{2}$, which means $\varphi (e_{2})=1$. We first prove that the colouring $\varphi$ can be used to obtain a proper type 1 and a proper type 2 k-edge-colouring of $G_A$ and a proper type 3, a proper type 4 and a proper type 5 k-edge-colouring of $G_B$, no matter which colour the edge $e_{3}$ has received.

$\underline{\hbox {Case 1: } \varphi (e_{3})=1}$

In this case, the colouring $\varphi _{1}\mid _{E(G_A)}$ is a proper type 1 k-edge-colouring of $G_A$. On the other hand, $\varphi _{2}\mid _{E(G_B)}$ is a proper type 4 k-edge-colouring of $G_B$. Furthermore, the colouring $\varphi '$, defined by $\varphi '=\varphi /P_{x_{3}}(1,2)$, satisfies $\varphi '(e_{3})=2$, while ${\bar{\varphi }}'(x_{1})=\varphi '(y_{1})=\{1\}$ and $\varphi '(e_{2})=1$ still hold. Therefore, $\varphi '_{1}\mid _{E(G_A)}$ is a proper type 2 k-edge-colouring of $G_A$, the colouring $\varphi '_{2}\mid _{E(G_B)}$ is a proper type 3 k-edge-colouring of $G_B$, and $\varphi '_{3}\mid _{E(G_B)}$ is a proper type 5 k-edge-colouring of $G_B$.

$\underline{\hbox {Case 2: } \varphi (e_{3}) \ne 1}$

If $e_{3} \in P_{x_{1}}(1,\varphi (e_{3}))$, then the colouring $\varphi '$, defined by $\varphi '=\varphi /P_{x_{1}}(1,\varphi (e_{3}))$, satisfies ${\bar{\varphi }}'(x_{1})=\varphi '(y_{1})=\{\varphi (e_{3})\}$ and $\varphi '(e_{2})=\varphi '(e_{3})=1$. Hence, for any $i \in \{2,...,k\}\setminus \{\varphi (e_{3})\}$ the Kempe chain $P_{x_{1}}^{\varphi '}(i,\varphi (e_{3}))$ is not an $x_{1},y_{1}$-path, a contradiction. Therefore, we may assume $e_{3} \notin P_{x_{1}}(1,\varphi (e_{3}))$, which implies $e_{2} \in P_{x_{1}}(1,\varphi (e_{3}))$. As a consequence, the colouring $\varphi '$, defined by $\varphi '=\varphi /P_{x_{3}}(1,\varphi (e_{3}))$, satisfies ${\bar{\varphi }}'(x_{1})=\varphi '(y_{1})=\{1\}$ and $\varphi '(e_{2})=\varphi '(e_{3})=1$. Since $P_{x_{1}}^{\varphi '}(1,2)$ and $P_{x_{1}}^{\varphi '}(1,3)$ still contain $e_{2}$, Case 1 applies.

In both cases there is a proper type 1 and a proper type 2 k-edge-colouring of $G_A$, and a proper type 3, a proper type 4 and a proper type 5 k-edge-colouring of $G_B$.

We now use the fact that the edge $e_{3}$ is critical as well. Let $\varphi '$ be a proper k-edge-colouring of $G-e_{3}$ with $i \in {\bar{\varphi }}'(x_{3})$ and $j \in {\bar{\varphi }}'(y_{3})$. If $e_{1}$ and $e_{2}$ are coloured with the same colour, then $\varphi '_{j}\mid _{E(G_B)}$ is a proper k-edge-colouring of $G_B$ that is of type 1 or 2. On the other hand, if $\varphi '(e_{1}) \ne \varphi '(e_{2})$, then $\varphi '_{i}\mid _{E(G_A)}$ is a proper type 3, type 4 or type 5 k-edge-colouring of $G_A$.

In every case there are two proper k-edge-colourings, one of $G_A$ and one of $G_B$, that are of the same type. This contradicts the fact that G is not k-edge-colourable. $\square$

A graph without an even factor can be characterized as follows (see Theorem 6.2 (p. 221) in [1]).

Theorem 4

If G is a graph, then G has no even factor, if and only if there is an $X \subset V(G)$ with

$$\begin{aligned} \sum \limits _{v \in X} (d(v)-2) - q(G;X) < 0, \end{aligned}$$

(1)

where q(G; X) denotes the number of components D of $G-X$ such that $e_{G}(D,X) \equiv 1 \text { (mod 2)}$.

We will give a more detailed formulation of Theorem 4 with regard to a minimal set X that satisfies inequality (1).

Theorem 5

If G is a connected graph, then G has no even factor, if and only if there is an $X \subset V(G)$ with the following properties: Let $D_1, \dots , D_n$ be the components of $G-X$.

• (a) $\sum \limits _{v \in X} (d(v)-2) - q(G;X) < 0$,

• (b) $e_{G}(D_{i},v) \le 1$ for every $v \in X$ and every $i \in \{1,...,n\}$,

• (c) X is stable,

• (d) $e_{G}(D_{i},X) \equiv 1 \text { (mod 2)}$ for every $i \in \{1,...,n\}$,

• (e) $\sum \limits _{\begin{array}{c} v\in X \\ d(v)\ne 2 \end{array}} (d(v)-3) + \frac{1}{2} \sum \limits _{i=1}^{n} (e_{G}(D_{i},X)-3) < \vert \{ v\in X : d(v)=2\}\vert$.

Proof

By Theorem 4 it suffices to prove one direction. Let G be a connected graph without an even factor. By Theorem 4, there is a set that satisfies inequality (1). Let $X \subset V(G)$ be the smallest set with $\sum _{v \in X}(d(v)-2)-q(G;X)<0$. We show that X satisfies (b)–(e).

For each $v \in X$ let c(v) be the number of components D of $G-X$ with $e_{G}(D,X) \equiv 1 \text { (mod 2)}$ and $e_{G}(D,v) \ge 1$. We first prove $c(v)=d(v)$ for every $v \in X$, which implies properties (b)–(d), since G is connected.

Let $x \in X$ and $X'=X\setminus \{x\}$. By the choice of X, the set $X'$ does not satisfy inequality (a). Furthermore, we observe that $-2 \vert X\vert + \sum _{v \in X} d(v) - q(G;X)$ is even. As a consequence,

$$\begin{aligned} 0&\le \sum \limits _{v \in X'} (d(v)-2) - q(G;X')\\&\le -2\vert X \vert + 2 + \sum \limits _{v \in X} d(v) - d(x) - (q(G;X)-c(x))\\&= -2\vert X \vert + \sum \limits _{v \in X} d(v) - q(G;X) + 2 - d(x) + c(x)\\&\le -2 +2 - d(x) + c(x). \end{aligned}$$

Thus, $d(x) \le c(x)$, which implies $d(x) = c(x)$. Therefore, the set X satisfies (b)–(d).

Next, by using (c) and (d) we can transform (a) to (e) as follows:

$$\begin{aligned}&\sum \limits _{v \in X} (d(v)-2) - q(G;X)< 0\\ {\underset{{\textit{(d)}}}{\Leftrightarrow }} \quad&\sum \limits _{v \in X} \left( d(v)-2\right)< n\\ \Leftrightarrow \quad&\frac{1}{2} \sum \limits _{v \in X} \left( d(v)-2\right) +\sum \limits _{v \in X} \left( d(v)-2\right)< \frac{3}{2} \sum \limits _{i=1}^{n} 1\\ {\underset{{\textit{(c)}}}{\Leftrightarrow }} \quad&\frac{1}{2}\sum \limits _{i=1}^{n} \left( e_{G}(D_{i},X)\right) - \vert X \vert +\sum \limits _{v \in X} \left( d(v)-2\right)< \sum \limits _{i=1}^{n} \frac{3}{2}\\ \Leftrightarrow \quad&\frac{1}{2}\sum \limits _{i=1}^{n} \left( e_{G}(D_{i},X)-3\right) +\sum \limits _{v \in X} \left( d(v)-3\right)< 0\\ \Leftrightarrow \quad&\frac{1}{2}\sum \limits _{i=1}^{n} \left( e_{G}(D_{i},X)-3\right) +\sum \limits _{\begin{array}{c} v\in X \\ d(v)\ne 2 \end{array}} \left( d(v)-3\right) - \vert \{v\in X: d(v)= 2\} \vert< 0\\ \Leftrightarrow \quad&\sum \limits _{\begin{array}{c} v\in X \\ d(v)\ne 2 \end{array}} \left( d(v)-3\right) + \frac{1}{2} \sum \limits _{i=1}^{n} \left( e_{G}(D_{i},X)-3\right) < \vert \{ v\in X: d(v)=2\}\vert . \end{aligned}$$

$\square$

2 Proof of Theorem 1

For $k=3$ there is nothing to prove. Let $k > 3$. Let G be a k-critical graph without an even factor. Hence, there is a subset $X \subset V(G)$ that satisfies conditions (a)–(e) of Theorem 5. We show that X contains more than $2k-6$ divalent vertices.

Let $D_{1},...,D_{n}$ be the components of $G-X$ and $g: \{D_{1},...,D_{n}\} \rightarrow {\mathbb {R}}$ with

$$\begin{aligned} g(D_{i}):=\sum \limits _{v \in N(D_{i})} \frac{d(v)-2}{d(v)} \hbox {for} i \in \{1,...,n\}. \end{aligned}$$

Properties (a)–(d) imply

$$\begin{aligned} \sum \limits _{i=1}^{n} g(D_{i}) \quad = \quad \sum \limits _{i=1}^{n} \sum \limits _{v \in N(D_{i})} \frac{d(v)-2}{d(v)} \quad {\underset{{\textit{(b)},\textit{(c)}}}{=}} \quad \sum \limits _{v \in X} d(v)-2 \quad {\underset{{\textit{(a)}}}{<}} \quad q(G;X) \quad {\underset{{\textit{(d)}}}{=}} \quad n. \end{aligned}$$

Thus, there is at least one component $D\in \{D_{1},...,D_{n}\}$ with $g(D) < 1$. Every critical graph does not contain a vertex of degree 1. Therefore, there are at most two vertices in N(D) that are not divalent. Moreover, if N(D) contains two vertices of degree at least 3, then one of them is of degree 3 and the other is of degree at most 5. Furthermore, since every critical graph is bridgeless, the component D has at least three neighbours in X by (b) and (d). In conclusion, we can assume $N(D)=\{x,y,w_{1},...,w_{l}\}$, where $l \ge 1$, the vertices $w_{1},...,w_{l}$ are divalent and either $d(y)=2$, or $d(y)=3$ and $d(x)\le 5$. We consider the following two cases:

$\underline{\hbox {Case 1}: d(x)<k}$

By condition (b) and Lemma 2 it follows that

$$\begin{aligned} l > k(k-d(y))-d(x)+1 \ge k(k-3)-k+2 = k(k-4)+2 \ge 2k-6. \end{aligned}$$

$\underline{\hbox {Case} 2: d(x)=k}$

If there are three components adjacent to x such that each has exactly three edges to X, then none of this components is adjacent with a divalent vertex by Lemma 3. In conclusion, we obtain with property (b)

$$\begin{aligned} \sum _{\begin{array}{c} i \in \{1,...,n\} \\ x \in N(D_{i}) \end{array}} g(D_{i}) \ge d(x)-2+6\left( \frac{1}{3}\right) =d(x). \end{aligned}$$

Since $\sum _{i=1}^{n} g(D_{i})<n$, there is another component $D' \in \{D_{1},...,D_{n}\}$ with $g(D')<1$, but $x \notin N(D')$. If $D'$ is not adjacent to a vertex of degree k, then $N(D')$ contains at least $2k-6$ divalent vertices since Case 1 applies. Otherwise X contains at least two vertices of degree k. Since G is bridgeless, property (d) implies that $e_{G}(D_{i},X) \ge 3$ for every $i \in \{1,...,n\}$. In conclusion, property (e) implies that

$$\begin{aligned} \vert \{ v\in X: d(v)=2\}\vert > 2(k-3)=2k-6. \end{aligned}$$

If at most two components adjacent to x have exactly three edges to X, then by properties (b)–(d) there are at least $d(x)-2$ components such that each has at least five edges to X. Therefore, property (e) implies that

$$\begin{aligned} \vert \{ v\in X: d(v)=2\}\vert >d(x)-3+(d(x)-2)\frac{1}{2}(5-3)=2k-5, \end{aligned}$$

and the proof is completed.