If a graph homomorphism induces an isomorphism, is it a homotopy equivalence?

Question

I am taking graphs to potentially have loops and multiple edges. That is, they look like 1-dimensional CW complexes.

If $f\colon\Gamma\to\Delta$ is a graph homomorphism (a.k.a. graph map, map of graphs) where $f_*\colon\pi_1(\Gamma,v)\to\pi_1(\Delta,f(v))$ is an isomorphism then is the same map on the realisation of $\Gamma$ and $\Delta$ as CW complexes a homotopy equivalence? The converse is clearly true.

Answer 1

I guess you define $\pi_1(\Gamma,v)$ as the fundamental group of the CW-realization $\lvert \Gamma \rvert$ of $\Gamma$.

For non-connected graphs your claim is true only if

(1) $f(\Gamma)$ meets all connected components of $\Delta$.

(2) $f_*$ is an isomorphism for all $v \in \Gamma$.

The second requirement can be weakened as follows: For each connected component $\Gamma'$ of $\Gamma$, $f_*$ is an isomorphism for some $v \in \Gamma'$.

So let us confine to connected graphs.

Choose maximal trees $T_\Gamma, T_\Delta$ in $\Gamma$ and $\Delta$ such that $v \in T_\Gamma, f(v) \in T_\Delta$. It is well-known that the quotient maps $p_\Gamma : \lvert \Gamma \rvert \to \lvert \Gamma \rvert / \lvert T_\Gamma \rvert$ and $p_\Delta : \lvert \Delta \rvert \to \lvert \Delta \rvert / \lvert T_\Delta \rvert$ are pointed homotopy equivalences. Let $q : (\lvert \Gamma \rvert / \lvert T_\Gamma \rvert,\ast) \to (\lvert \Gamma \rvert ,v)$ be a pointed homotopy inverse for $p_\Gamma$. Then $F = p_\Delta f q : (\lvert \Gamma \rvert / \lvert T_\Gamma \rvert,\ast) \to (\lvert \Delta \rvert / \lvert T_\Delta \rvert,\ast)$ is a map which induces an isomorphism on fundamental groups. We have $(\lvert \Gamma \rvert / \lvert T_\Gamma \rvert,\ast) \approx \bigvee_{i \in I} S^1_i$ and $(\lvert \Delta \rvert / \lvert T_\Delta \rvert,\ast) \approx \bigvee_{j \in J} S^1_j$, where $I,J$ are suitable index sets, $S^1_i, S^1_j$ are copies of $S^1$ and $\bigvee$ denotes wedge (= one-point-union). It is well-known that all higher homotopy groups of wegdes of circles are trivial. See for example Higher homotopy groups of wedge of circles. Therefore $F$ is a weak homotopy equivalence (i.e. induces isomorphisms on all homotopy groups $\pi_n$). Since domain and range of $F$ are CW-complexes, $F$ is a homotopy equivalence. This implies that $f$ is a homotopy equivalence.