Let $\sigma = \{E\}$ be a signature with a binary relation symbol $E$ and let $\mathcal{G}$ be a $\sigma$-structure given by the following graph:
I'm wondering what the size of the smallest image of an endomorphism $g : \mathcal{G} \rightarrow_{hom} \mathcal{G}$ could be. My initial thought was "cutting" it in half i.e. $$ g(v_0) := v_0 \quad g(v_1) := v_1 \quad g(v_2) := v_2 \quad g(v_3) := v_1 \quad g(v_4) := v_4 \quad g(v_5) := v_5 \quad g(v_6) := v_6 \quad g(v_7) := v_5 $$ until it resulted in $$ g(v_0) := v_5 \quad g(v_1) := v_1 \quad g(v_2) := v_5 \quad g(v_3) := v_1 \quad g(v_4) := v_1 \quad g(v_5) := v_5 \quad g(v_6) := v_1 \quad g(v_7) := v_5 $$ Since I (guess I) can't project everything on one vertex the smallest image would be $|\{v_1, v_5\}| = 2$, right? Is it still an endomorphism?
A simpler way to see this is that $\mathcal G$ is bipartite as the subgraphs induced by $P_1 = \{v_1, v_3, v_4, v_6\}$ and $P_2 = \{v_0, v_2, v_5, v_7\}$ are independent sets and hence, the map $h$ defined as $$h \colon \mathcal G \to \mathcal G, v \mapsto \begin{cases} v_1, \text{ if } v \in P_1 \\ v_5, \text{ if } v \in P_2 \end{cases} $$ is an endomorphism of $\mathcal G$ of minimal image size.