One-to-one correspondence between n-edge-colourings of $K_{n,n}$ and Latin squares

Question

I need help with this. I don't know how to do this (especially case b) ).

a) An $n \times n$ array $A = (a_{ij})$, whose entries are taken from some set $S$ of $n$ symbols, is called a Latin square of order $n$ if each symbol appears precisely once in each row and precisely once in each column of $A$. Show that there is a one-to-one correspondence between $n$-edge-colourings of $K_{n,n}$ in colours $1, 2, \dots , n$ and Latin squares of order $n$ in symbols $1, 2, \dots, n$.

b) Deduce from the (below) Theorem the following assertion, a special case of a conjecture due to J. Dinitz.

For $1 \leq i\leq j \leq n$, let $S_{ij}$ be a set of $n$ elements. Then there exists a Latin square $A = (a_{ij})$ of order $n$ using a set $S$ of $n$ symbols such that $a_{ij} \in S_{ij} \cap S$, $1 \leq i \leq j \leq n$.

Theorem: Every simple bipartite graph $G$ is $\Delta$-list-edge-colourable.

Answer 1

Presumably your answer to (a) formalized the bijection illustrated below:

Basically, the symbol $s$ in cell $(a,b)$ maps to the edge $\text{row}_a \text{col}_b$, say, of color $s$.

For the second part, it's the same idea with "symbol"/"color" replaced by "list".

The list $L$ assigned to cell $(a,b)$ maps to the edge $\text{row}_a \text{col}_b$ which is also assigned the list $L$. The theorem implies a $\Delta$-list-edge-coloring exists, which we convert back to give a $n \times n$ matrix which we can show has no repeated symbols in the rows and columns.

As bof points out, however, the statement of the Dinitz Conjecture Theorem is incorrect. I.e., it doesn't guarantee a Latin square, just no repeats in rows/columns (a generalization of a Latin square).