Transport, restriction and relabelling of finite colourings

The composite $f\circ h$ is injective: equality of two composite values first gives equality of their $h$-values by injectivity of $f$, then equality of their inputs by injectivity of $h$. An edge of $H$ maps to an edge of $J$ under $h$ and then to an edge of $G$ under $f$. Thus the composite is an ordinary embedding. If its two vertices are images of vertices $x,y$ of $H$, their $c$-colours are $(c\circ f)(h(x))$ and $(c\circ f)(h(y))$, which agree by the given monochromaticity. Its image edges therefore form a monochromatic ordinary copy in $G$.

Since $V(G)$ is finite and $S\subseteq V(G)$, F6 gives that $S$ is finite. Restriction gives a function $S\to [k]$ with the same value as $c$ at each vertex of $S$. Thus $c{\mathrm{\mid }}_S$ is a $k$-colouring of $S$, and all pairwise colour equalities on a subset of $S$ are unchanged. For $x,y\in V(G)$, equality $c(x)=c(y)$ implies $r(c(x))=r(c(y))$, and the converse follows from injectivity of $r$ because both values lie in $P$. Thus injective relabelling preserves and reflects monochromaticity, including for the empty subset. Applied to image vertex sets, these equivalences preserve monochromatic ordinary and induced copies whenever the relevant copy is contained in the restricted set.

If $f$ and $h$ are induced, an edge between the composite images of distinct vertices of $H$ reflects through $f$ to an edge between their $h$-images, and then through $h$ to an edge of $H$. The $h$-images are distinct by injectivity. Together with edge preservation from step 1.1, this makes the composite induced; its induced image is monochromatic by step 1.1.

Put $m=\mathrm{\mid }P\mathrm{\mid }$ and choose a bijection $b:P\to m$. Since $m\le i$, each $b(p)$ belongs to $i$; its successor belongs to $\sigma (i)$ and is nonzero. Hence $r(p)=\sigma (b(p))$ lies in $[i]$. Both $b$ and successor are injective, so $r$ is injective. If $P$ is empty this is the empty injection. Consequently $r\circ d$ is a function $X\to [i]$, and step 1.2's same equality argument shows that it has exactly the monochromatic subsets of $d$. No surjectivity onto $[i]$ is used.