Finite colour-class capacity and prescribed partitions

Let $c:X\to [k]$ be a $k$-colouring and suppose $\mathrm{\mid }X\mathrm{\mid }>k(n-1)$. Applying the counting form of the pigeonhole principle to $c$ with $t=n-1$ gives a fibre of size greater than $n-1$, hence at least $n$.

Put $m=\mathrm{\mid }X\mathrm{\mid }$ and choose a bijection $e:X\to \{0,\dots ,m-1\}$. If $m\le kn$, divide each $e(x)$ uniquely as $qn+r$ with $0\le r<n$. Since $0\le e(x)<kn$, necessarily $0\le q<k$; define $c(x)=q+1$. This is a $k$-colouring, including the empty function when $m=0$.

Suppose $m=k(n-1)+1$. In the enumeration given by $e$, assign indices $0,\dots ,n-1$ colour $1$ and, for each $i=2,\dots ,k$, assign the next block of $n-1$ consecutive indices colour $i$. These disjoint blocks cover exactly $n+(k-1)(n-1)=m$ indices, and their sizes are $n,n-1,\dots ,n-1$. Pulling them back along $e$ gives the stated colour classes. For $n=1$ the later blocks are empty; for $k=1$ there are no later blocks.

For each $i\in [k]$, the map sending $x$ in the $i$th fibre to its remainder $r$ is injective: equal remainders and the common quotient $i-1$ give equal $e(x)$, hence equal $x$. By definition this map is surjective onto its image and hence is a bijection from the fibre to that image. Its image is a subset of $\{0,\dots ,n-1\}$, so the fibre has at most $n$ elements by cardinality transport and the finite-subset bound.

Steps 1.1, 2.1 and 1.3 establish the three assertions.