In Zieschang's "Theory of Association Schemes", in section 5.2, he remarks that the composition of homomorphisms is not always a homomorphism. I've been struggling to find an example of that claim. I've found certain papers which reference that fact, but I'm unable to find any example.
What would be a simple counterexample? By the lemma 5.2.1, the first homomorphism has to be non-surjective, so I've tried to find some simple examples for 2 or 3 element sets, but couldn't get it.
You can find an example using the Hamming square, Hamming cube, and complete graph on four vertices, which I think may be one of the simplest example. Two and three element sets are too small, as the only schemes are equivalent to directed/undirected complete graphs.
Scheme Morphisms and Homomorphisms. Just to review, here are the definitions of a scheme morphism and homomorphism in graph terms. Notationally, this is different from the book.
Let $(X, S)$ and $(\tilde{X}, \tilde{S})$ be association schemes, so $(X, s)$ and $(\tilde{X}, \tilde{s})$ are directed graphs for each $s \in S$ and $\tilde{s} \in \tilde{S}$. A scheme morphism is a pair of functions $f:X \to \tilde{X}$ and $g:S \to \tilde{S}$ such that $f:(X,s) \to (\tilde{X},g(s))$ is a graph homomorphism for each $s$, meaning for all $x, y \in X$, $(x,y) \in s$ implies $(f(x), f(y)) \in g(s)$.
The scheme homomorphism condition is like a converse or surjective condition to this. $f$ and $g$ form a scheme homomorphism if for each $s$, $f:(X,s) \to (f(X), g(s))$ maps edges surjectively, meaning for all $\tilde{x}, \tilde{y} \in f(X)$, $(\tilde{x}, \tilde{y}) \in g(s)$ implies there exists $(x, y) \in s$ such that $f(x) = \tilde{x}$ and $f(y) = \tilde{y}$.
Hamming Schemes. Let $X_n = \{0,1\}^n$ be the set of binary vectors of length $n$. $X_n$ is a metric space with respect to the Hamming metric, $d$ where $d(x, y)$ equals the number of components $x$ and $y$ differ. For $0 \leq i \leq n$, let $$s_i = \{(x, y) \in X_n \times X_n \mid d(x, y) = i\}$$ and $S_n = \{s_0,\ldots,s_n\}$. $(X_n, S_n)$ forms a scheme called the Hamming scheme. For $n = 2$, we have that $X_2$ contains four points, $s_1$ forms a square with the four points, and $s_2$ forms the diagonals of the square. For $n = 3$, we have a cube where $s_1$ forms the edges of the cube, $s_2$ forms the diagonals of the faces, and $s_3$ forms the diagonals of the cube.
Now, let $\tilde{X} = \{1,2,3,4\}$ and $\tilde{S}$ contain two classes, $\tilde{s_0}$ and $\tilde{s_1}$ so $(\tilde{X}, \tilde{S})$ is the complete graph scheme. We define scheme homomorphisms $(X_2, S_2) \to (X_3, S_2)$ and $(X_3,S_3) \to (\tilde{X}, \tilde{S})$. For the first one, we map $(X_2, S_2)$ onto a face of $(X_3, S_3)$. For example, take: $$f(00) = 000, f(01) = 001, f(10) = 010, f(11) = 011$$ and $$g(s_0) = s_0 \in S_3, g(s_1) = s_1 \in S_3, g(s_2) = s_2 \in S_3.$$ From the description, it's relatively easy to verify that this is a scheme homomorphism. For the second one, we map the same face to each of the vertices of the complete graph: $$f(000) = 1, f(001) = 2, f(010) = 3, f(011) = 4$$ Note that $s_3 \in S_3$ pairs opposite corners of the cube, so $f$ must map each pair to the same point to even be a scheme morphism: $$f(111) = 1, f(110) = 2, f(101) = 3, f(100) = 4$$ $g$ must then be $$g(s_0) = g(s_3) = \tilde{s_0}, g(s_1) = g(s_2) = \tilde{s_1}.$$
$s_0$ and $s_3$ trivially satisfy the edge-surjectivity property. For $s_1$, you can see that you get at least the square connecting $1$, $2$, $3$, $4$. For the diagonals of the square, you see that for example $f(000) = 1$, $f(100) = 4$, and $(000,100) \in s_1$ so the diagonals have a preimage in $s_1$. $s_2$ contains the edges of two different complete graphs on four vertices, and you can similarly see every edge in $g(s_2)$ has a preimage in $s_2$.
Now, the composition is not a homomorphism. Basically, the composition maps the vertices bijectively, but $s_1 \in S_2$ has $4$ edges, $s_2 \in S_2$ has $2$ edges, and $\tilde{s_1}$ has $6$ edges, so $s_1$ and $s_2$ separately do not give preimages for every edge in $\tilde{s_1}$.
