Graph Homomorphism is a well-known NP-complete problem. Given graph $G$ and $H$, $G$ is said to be homomorphic to $H$ if there is a mapping $f: V(G) \mapsto V(H)$ such that $(u,v) \in E(G) \implies (f(u), f(v)) \in E(H)$.

The mapping in above is unrestricted -- and hence, multiple nodes of $G$ can map to a single node in $H$. If we restrict the mapping to be injective, then only one node of $G$ can map to a node in $H$. Thus, we have the injective-homomorphism variant:

Given graph $G$ and $H$, $G$ is said to be injective-homomorphic to $H$ if there is an injective mapping $f: V(G) \mapsto V(H)$ such that $(u,v) \in E(G) \implies (f(u), f(v)) \in E(H)$.

To me -- the injective homomorphic is a very natural variant of Graph Homomorphism problem, but I am unable to find a single research paper on it (the closest is "locally injective" homomorphisms). Why has there been zero interest in Injective-Homomorphism? Could I be missing something? BTW, its easy to see that injective-homomorphism decision problem is also NP-complete, by a simple reduction from the Homomorphism problem.

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    $\begingroup$ This is also related to the subgraph isomorphism problem. In particular, if there is such an injection, then $(f(V(G)),f(E(G)))$ (a subgraph of $H$) is similar to $G$, except that it might have additional edges. Your problem might be related to subgraph monomorphism but I'm not certain about that. $\endgroup$
    – D.W.
    Commented May 10, 2022 at 18:22
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    $\begingroup$ Yup - but does monomorphism allow for additional edges? I have never understand that part of this problem space so I might be talking nonsense - I apologize if this is not helpful. $\endgroup$
    – D.W.
    Commented May 10, 2022 at 21:08
  • $\begingroup$ You are right. Injective-homomorphism is same as subgraph isomorphism (which allows for extra edges in H; induced subgraph isomorphism doesn't). $\endgroup$
    – Brian
    Commented May 10, 2022 at 23:35

2 Answers 2


As D.W. mentioned, $G$ is injective-homomorphic to $H$ if and only if $G$ is isomorphic to a subgraph of $H$. That is why you are "unable to find a single research paper on 'injective-homomorphism'". There are plenty of paper about subgraphs or about subgraph isomorphism, of course.

More specifically, suppose we are given graph $G$ and $H$ and an injective-homomorphism from $G$ to $H$ $f: G\to H$, i.e., $f|_{V(G}:V(G) \to V(H)$ is injective and, $(u,v) \in E(G) \implies (f(u), f(v)) \in E(H)$. Then $f(G):=(f(V(G)),f(E(G)))$ is a subgraph of $H$. Moreover, the map $f: G\to f(G)$ is a graph isomorphism.

Conversely, if there is a graph isomorphism $f: G\to S$ where $S$ is a subgraph of $H$, then the composition of $f$ with the natural inclusion of $S$ into $H$ is an "injective-homomorphism" from $G$ to $H$.


If the homomorphism from a graph $G$ to a graph $H$ is injective, then $G$ is isomorphic to a subgraph of $H$. An injective homomorphism is called a monomorphism. Perhaps you can search for research papers on specific types of injective homomorphisms, such as co-retractions or isometric embeddings.

Given any homomorphism $f$ from $G$ to $H$, we can construct an injective homomorphism $\psi$ from the quotient $G/P$ to $H$ and a map $\pi_P$ such that $f = \psi \cdot \pi_P$. Here, $P$ is the partition of $V(G)$ obtained from the equivalence relation $x \sim y$ if and only if $f(x)=f(y)$, the quotient $G/P$ is defined to have vertex set $P$ and two vertices $V_i$ and $V_j$ are adjacent in this quotient whenever they each contain vertices that are adjacent in $G$, and $\pi_P: V(G) \rightarrow V(G/P)$ maps each vertex of $G$ to the part it belongs to.

A map $f:V(G) \rightarrow V(H)$ is an isometric embedding if it preserves distances in the sense that $d_H(f(u),f(v)) = d_G(u,v)$ for any two vertices $u,v \in V(G)$. Thus, an isometric embedding is a special type of injective homomorphism; it arises in the context of routing problems.

See (Hahn and Tardif, "Graph Homomorphisms: structure and symmetry") for properties of isometric embeddings and co-retractions.


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