The core of the Petersen graph I understand the definition of a core, which is a minimal subgraph that is a homomorphic image, meaning that there exists a homomorphism from the original graph to the core, a homomorphism from the core to the original graph, and the core is minimal with this property.
I also know that the core, as its name suggests, is a simpler graph that resembles the structure of the original graph. However, I am unsure of how to find or validate the core of a particular graph, specifically the Petersen graph.
 A: The Petersen graph is its own core.
In general, determining this for an arbitrary graph is hard for both humans and computers - there is no straightforward algorithm. In this case, we can use some symmetry properties of the Petersen graph to prove it without too much suffering.

A useful fact that is not obvious from the definition is that the core $H$ of a graph $G$ must be a retract of $G$: taking $H$ to be a fixed subgraph of $G$, we may assume that the homomorphism $G \to H$ is the identity on $H$.
(A short proof sketch: the homomorphism $f\colon G \to H$ restricted to the vertices of $H$ gives an endomorphism $f|_H \colon H \to H$. Because $H$ is a core, this must actually be an automorphism. So we can turn $f$ into a retract by undoing that automorphism: in other words, $(f|_H)^{-1} \circ f$ is a retract.)
Now let $G$ be the Petersen graph, and suppose there is a retract $f \colon G \to H$ where $H$ is a smaller subgraph of $G$. Then there must be vertices $v,w$ in $G$ such that $v$ is a vertex of $H$, but $w$ is not. In this case, $f(w)$ must be a different neighbor of $v$. No matter where this happens in the Petersen graph, we can find a $5$-cycle $w-v-f(w)-x-y-w$ n $G$. But then $f(x)$ and $f(y)$ must be adjacent to both each other and $f(w)$, which is impossible because $G$ does not contain a triangle.
This contradiction tells us that no such retract can exist, so $G$ must be its own core.
