Proof of Kőnig's line coloring theorem ($\chi'(G) = \Delta(G)$) I'm trying to find a proof of Kőnig's line coloring theorem, i.e.:

The chromatic index of any bipartite graph equals its maximum degree

But to my surprise, I've only* been able to find two questions touching the subject:

*

*Edge-coloring of bipartite graphs

*Edge coloring of a bipartite graph with a maximum degree of D requires only D colors
As graphs are my Achilles' heel, I'm incapable to use the information contained in the above to prove $\chi'(G) = \Delta(G)$ myself.

* I've found many papers referring to it, but none proving, except for page 4 of CH6.pdf from first question, but I don't think it's sufficient.
 A: I will try and give an overview of the prerequisite knowledge, and include sources at each step so that you can sequentially understand. If you don't understand certain parts (like the construction at the end) I recommend you work some small examples.
Let's first introduce Hall's Theorem:

Theorem: (Hall's Theorem) Let $G$ be a bipartite graph with parts $A$ and $B$. Then $G$ has a matching (independent edge set) saturating $A$ (every vertex of $A$ is the endpoint of some edge in the matching) if and only if for every $X \subseteq A$ we have $|X| \le |N(X)|$.

The two sources I recommend for a good view of Hall's Theorem is Diestel's Graph Theory (which, if I recall, gives four proofs) and West's Introduction to Graph Theory.
The significance of Hall's Theorem here is that for $k$-regular bipartite graphs, we can find a perfect matching. This comes from two things:

*

*A $k$-regular bipartite graph is balanced.

*A $k$-regular bipartite graph satisfies Hall's condition.

So now we can prove the following:

Lemma: If $G$ is a $k$-regular bipartite graph, then $\chi'(G) = k$.

We can use induction on $k$. By Hall's Theorem, $G$ has a perfect matching $M$. Consider $G-M$, which is $k-1$-regular (why?). By the induction hypothesis, $\chi'(G) = k-1$, and so we can add $M$ back as a new color, hence extending a proper $k-1$-edge-coloring from $G-M$ to a proper $k$-edge-coloring on $G$.
If you're not familiar with induction, here's a different description: Removing a perfect matching from a $k$-regular bipartite graph gives a $k-1$-regular graph, which must also have a perfect matching... Iterate this process $k$ times.
Now for the finish line. We want to prove the result for any bipartite graph $G$.

Result: If $G$ is a bipartite graph, then $\chi'(G) = \Delta(G)$.

If $G$ is regular, then we are done by the Lemma. Otherwise, there is at least one vertex $v$ in $G$ with $\deg(v) < \Delta(G)$. We can construct a graph $R$ such that

*

*$R$ is bipartite.

*$R$ is $\Delta(G)$-regular.

*$G \subseteq R$.

One construction is as follows. We have $G$ bipartite with parts $A$ and $B$. Take a copy of $G$, say $G'$ with parts $A'$ and $B'$. Then for each vertex $v$ not of degree $\Delta(G)$ in $G$, we add an edge between $v$ and it's copy $v' \in G'$. This newly obtained graph is bipartite with parts $A \cup B'$ and $B \cup A'$. Repeat this process as necessary. You'll notice that at each iteration the gap between the minimum degree and maximum degree decreases, so we must terminate with a $\Delta(G)$-regular graph $R$ as desired. You'll find this construction is the one given by Jon Noel's comment here.
Using the Lemma, $\chi'(R) = \Delta(G)$, and thus there is a proper $\Delta(G)$-edge-coloring of $R$. Since $G \subseteq R$, this proper coloring works for $G$. I.e. $\chi'(G) = \Delta(G)$.

Some notes.
Note we used the general fact that $\chi'(H) \le \chi'(G)$ for $H \subseteq G$ at the end.
One thing I did glance over is if we are allowing multiple edges, but things still work that way. If we allow multiple edges, can you see why the way we've constructed $R$ takes exactly $1$ iteration? I don't believe there's any real reason to exclude use of multiple edges.
One key take away is to think of color classes in an edge coloring as what they are: matchings.
