Let $G$ be a simple graph and $S$ be the set of all sub graphs of $G$. Define two operations on $S$ as: $union$ of two graphs $ G_1$ and $G_2$ is, $G_1\cup G_2$

$=\langle V(G_1)\cup V(G_2), (E(G_1)\cup E(G_2)\rangle$

and the graphs $intersection$ is, $G_1\cap G_2$

$=\langle V(G_1)\cap V(G_2), (E(G_1)\cap E(G_2)\rangle$.

Let $P(S)$ be the power set of $ S$ and for any two subsets of $S$ namely, $ A, B\in P(S)$, define:

$A\sqcup B=\lbrace G_i\cup G_j:~G_i\in A, G_j\in B\rbrace$ and

$A\sqcap B=\lbrace G_i\cap G_j:~G_i\in A, G_j\in B\rbrace$. Also, consider a mapping

$ f: S\rightarrow P(S)$ such that $f(empty~ graph)= emptyset $ and $f( G)=S$.

So, how should $f$ be defined so that it is a homomorphism keeping in mind that every element of $ S$ is a graph while every element of $P(S)$ is a subset containing graphs?

  • $\begingroup$ I hope it is not a stupid question. What do you consider a homomorphism in this case? $\endgroup$ Commented Sep 19, 2019 at 21:44
  • $\begingroup$ @user2679290 Even breakthrough results were borned out of stupid questions only. So, please bear. $\endgroup$
    – gete
    Commented Sep 20, 2019 at 1:27

1 Answer 1


A possible answer is the following, if you are willing to relax the definitions (in a very minor way) of union and intersection of two graphs.

For clarity, let me introduce an additional notation. For any subgraph $H$ of $G$, considered as a graph with vertex set $V(G)$ (this is trivially possible by adding those vertices of $G$ into $H$, which do not participate in the edge set of $H$), let $Q_G(H)$ be the set of all subgraphs of $H$, where again the vertex set of each graph in $Q_G(H)$ is $V(G)$ (hence the subscript $G$ in $Q_G$).

Define the union and intersection of two graphs in $Q_G(G)$ as you have defined for the edge set, but now over vertex set $V(G)$. Now the map $f:Q_G(G)\to P(Q_G(G))$ defined by $f(H)=Q_G(H)$ should give you the desired homomorphism.

  • $\begingroup$ In that case $Q_G(G)\subseteq S$. Right? $\endgroup$
    – gete
    Commented Sep 20, 2019 at 13:29
  • $\begingroup$ They'll be essentially equal. Every graph H in S will be in Q_G(G), however,the vertex set of H will not be V(H) but instead V(G). $\endgroup$
    Commented Sep 20, 2019 at 13:54
  • $\begingroup$ Noted. Thanks... $\endgroup$
    – gete
    Commented Sep 20, 2019 at 15:02

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