Let $G,H$ be infinite simple undirected graphs with the property that for any graph $X$ we have $|\text{Hom}(X,G)| = |\text{Hom}(X,H)|$. Does this imply that $G$ is isomorphic to a subgraph of $H$, and vice versa?

(Note that in the finite case the condition above implies $G\cong H$, see here.)

  • $\begingroup$ I assume that you mean "edge preserving" by homomorphism (not necessarily non-edge preserving). I also assume that you really mean subgraph, not "induced subgraph". Correct? $\endgroup$
    – Goldstern
    May 29, 2015 at 12:12
  • $\begingroup$ Thanks for asking for clarification. You're right: I mean homomorphisms to be edge-preserving, and I mean subgraphs that are not necessarily induced. $\endgroup$ May 29, 2015 at 12:28

1 Answer 1


Did you want both G to be a subgraph of H and H to be a subgraph of G? Then I think the answer is negative. For example if G is infinite and have no isolated vertices and H is G together with an isolated vertex. If X have no isolated vertices the Hom-sets are the same. The set of maps from the set of isolated vertices of X to G or H should be of the same size as the cardinality of the vertex sets of G and H should be the same.

Edit: And if G for example is the integers with edges $\{i,i+1\}$ then I don't think H can be a subgraph.


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