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A proper edge $k$-coloring of a graph is an assignment of $k$ colors to the edges of the graph so that no two adjacent edges have the same color. The smallest integer $k$ such that $G$ has a proper edge $k$-coloring is the chromatic index of $G$.

Giving a simple graph $G$, the well-known Vizing's theorem tells us that the chromatic index of any simple graph $G$ is either $\Delta(G)$ or $\Delta(G)+1$. In particular, if the chromatic index of a graph $G$ is $\Delta(G)+1$, then we say that $G$ is of class two.

Suppose that $G$ is a graph of class two and $\varphi$ is a proper edge coloring using $\Delta(G)+1$ colors. I wonder whether there always exists an edge $uv$ such that $\varphi(uv)\cup \{\varphi(e)~|~e~is~an~edge~adjacent~to~uv\}$ covers all of the $\Delta(G)+1$ colors.

I guess the answer is yes, however, cannot find any source supporting this. If anyone knows some relative references or can prove or disprove this, please reply me. Thanks in advance.

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    $\begingroup$ This is a homework level question, please try math.stackexchange.com, as this site is for researchers. $\endgroup$
    – domotorp
    Sep 26, 2021 at 15:21

1 Answer 1

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Yes!

Let $uv$ be an edge colored by the last color, say $\Delta+1$. If $uv$ is incidence with all colors, then it is the required edge. So the only case is that every edge colored with $\Delta+1$ is not incident with some color in $[\Delta]$, and thus it can be recolored by that color. This results an edge $\Delta$-coloring, a contradiciton.

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