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 $f( G)=S$.



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> 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?
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Define a homomorphism of a set of graphs to its power set
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*.


