Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $G$ and $H$ be two graphs. It is known that if there is a homomorphism from $G$ to $H$, then $\omega(G) \leq \omega(H)$ where $\omega(G)$ is the clique number of $G$.

When does the converse hold and when does it fail?

share|cite|improve this question
A trivial example of the converse's failure would be to take $G$ to be a complete graph and $H$ to be an edgeless graph. – Austin Mohr Jan 25 '13 at 4:02
It should be $\omega(G) \leq \omega(H)$, not $\geq$. – polkjh Jan 25 '13 at 7:38
@polkjh Look at observation $2.6$ here ( – Turbo Jan 25 '13 at 7:57
I think that is a mistake. Suppose there is a homomorphism from $G$ to $H$. Adding edges and vertices to $H$ retains the homomorphism from $G$ to $H$. So we can increase $\omega(H)$ indefinitely and still have a homomorphism from $G$ to $H$, which contradicts that $\omega(H)$ cannot be more than $\omega(G)$. – polkjh Jan 25 '13 at 8:07

2 Answers 2

up vote 1 down vote accepted

If $H$ is a subgraph of $G$, then there is a homomorphism $H\to G$. But clearly $\omega(H)$ can even be zero, while $\omega(H)$ could be as large as $|V(G)|$.

For the converse, there is no hope. As an example consider triangle-free graphs, with $\omega=2$. For each positive integer $m$, there are triangle-free graphs with chromatic number greater than $m$, and for such graphs there is no homomorphism to $K_m$.

share|cite|improve this answer

Let $H=K_n$. Then existence of homomorphism from $G$ to $H$ implies $G$ is $n$-colorable. But $\omega(G) \leq n$ does not imply that $G$ is $n$-colorable (consider an odd cycle and $n=2$). So atleast for complete graphs $H$, the converse holds only if $\omega(H) \geq \chi(G)$ (chromatic number). This is not necessary for general graphs $H$. But I doubt there is any simple characterization of when the converse holds for generel $H$.

share|cite|improve this answer
I think if a homomorphism from $G$ to $H = K_{n}$ exists, then $\omega(G) \geq n$ ( observation $2.6$) – Turbo Jan 25 '13 at 8:07
I think that might be a typo in the paper. Look at my comment above. – polkjh Jan 25 '13 at 8:09
It might help if you try to prove the statement yourself (homomorphism from $G$ to $H$ implies $\omega(G) \leq \omega(H)$). It is not too difficult. – polkjh Jan 25 '13 at 8:16

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.