If $H$ is an induced subgraph $G$ and there is a homomorphism $G\to H$, then $H$ is a retract of $G$?

Question

If $H$ is an induced subgraph of $G$ and there is a homomorphism $G\to H$, then $H$ is a retract of $G$?

I ask because:

A subgraph $H$ of $G$ is called a core of $G$ if there is a homomorphism $G \to H$ but no homomorphism $G \to H'$ for any proper subgraph $H'$ of $H$.

$H$ is a core of $G$ if and only if it is a retract of $G$ and, among retracts of $G$, it is minimal with respect to inclusion.

Answer 1

Let me address your concern about cores. I will start by some definition of core of graph and show the equivalences of definitions one by one. Throughout all graphs are finite.

Def: A core is a graph $C$ such that any homomorphism $C\to C$ is a bijection.

Now that if $H\subset G$ is a proper subgraph, then if there exists a homomorphism $G\to H$, composing with inclusion $\imath: H\to G$, we get a homomorphism $G\to G$ which is not a bijection. As a result $G$ is a core then there are no homomorphisms $G\to H$ with $H\subsetneq G$.

Conversely suppose for all $H\subsetneq G$ no homomorphism $G\to H$ exists. Take any homomorphism $\phi: G\to G$. Then the image of this homomorphism $I=\mathrm{Im}\: \phi$ is a subgraph of $G$. So unless $I=G$, we have found a homomorphism to a subgraph. So if there are no homomorphisms to a subgraph, then $G$ is a core.

Def: Given a graph $G$, $H$ is a core of $G$ if $H$ is a core and there is a homomorphism $G\to H$. One can also show that $H$ is isomorphic to an induced subgraph of $G$, but let us assume for simplicity $H$ is a subgraph.

Suppose $\phi: G\to H$ is a homomorphism where $H$ is the core of $G$. Then, looking at the restriction homomorphism $\phi|H: H\to H$, since $H$ is a core, we find that $\phi|H$ is a bijection. As a result by definition, $H$ is a retract of $G$. Moreover, $H$ cannot be retracted any further either, since there is no homomorphism from $H$ to any subgraphs of $H$.

Conversely, if $H$ is a minimal retract of $G$, then the retraction homomorphism $r:G\to H$ shows that $H$ is a core of $G$.

Answer 2

Counterexample:

  1---2---6---7---9
       \ /     \ /
        3       8

where $H$ is the subgraph induced by $\{6,7,8,9\}$.

There is a (surjective) homomorphism $G\to H$, namely $$\begin{array}{c|cc}v & 1 &2 & 3&6&7&8&9 \\ \hline f(v) & 6 & 7 & 8 & 9 & 7 & 8 & 9 \end{array} $$ But $H$ is not a retract of $G$, because $6$ is part of a triangle in $G$ but not in $H$.