Rigidity of mutual ordinary embeddings

The composite $f\circ g$ is an injection of the finite set $V(L)$ into itself, hence is surjective. Its image is contained in the image of $f$, so $f$ is surjective; it is injective by hypothesis and therefore bijective.

For each $G\in \{J,L\}$, finiteness of $V(G)$ gives finiteness of $[V(G){]}^2$ and hence of its subset $E(G)$. Define $f_E:E(J)\to E(L)$ by $f_E(\{x,y\})=\{f(x),f(y)\}$, and define $g_E$ analogously. Edge preservation makes these maps well defined. Each is injective: equality of two image pairs and injectivity of the vertex map force equality of the two original pairs. Thus $f_E\circ g_E$ is an injection of the finite set $E(L)$ into itself and is surjective. Its image lies in the image of $f_E$, so $f_E$ is surjective.

If distinct $x,y\in V(J)$ have $\{f(x),f(y)\}\in E(L)$, surjectivity of $f_E$ supplies an edge $\{u,v\}\in E(J)$ with $\{f(u),f(v)\}=\{f(x),f(y)\}$. Injectivity of $f$ gives $\{u,v\}=\{x,y\}$, so $f$ reflects edges. It also preserves edges and is bijective; hence it is an isomorphism. Therefore non-isomorphic finite simple graphs cannot have ordinary embeddings in both directions.

For the induced embedding $h$, its image is a subset of $V(L)$, and $h$ is a bijection onto that image. Cardinality transport gives $\mathrm{\mid }h[V(J)]\mathrm{\mid }=\mathrm{\mid }V(J)\mathrm{\mid }=\mathrm{\mid }V(L)\mathrm{\mid }$; finite-subset rigidity therefore gives $h[V(J)]=V(L)$. Thus $h$ is bijective, and its preservation and reflection of edges make it an isomorphism. This proves the remaining assertion.