"Remove a vertex" map for right-angled Artin groups

Question

Given a finite graph $\Gamma$, one has the right-angled Artin group $A(\Gamma )$. Its generators $s_1, \dots s_n$ bijectively correspond to vertices of $\Gamma$ and the relators are $s_is_j=s_js_i$ provided the corresponding vertices are joined by an edge.

Let $A_i(\Gamma)$ be the group obtained from $A(\Gamma)$ by setting $s_i=1$; this corresponds to removing the vertex $i$ from $\Gamma$.

I know very little of these matters but it seems plausible that any nontrivial element of $A(\Gamma)$ projects to a nontrivial element of $A_i(\Gamma)$ for some $i$; is this correct?

Answer 1

No. Let $\Gamma$ be the graph with two vertices and no edges - the non-abelian free group of rank two - and let $g$ be the commutator of the two generators $s_1$ and $s_2$. Then $g$ is certainly non-trivial, but $g$ dies whenever you kill $s_1$ or $s_2$.

UPDATE:

For an example with a connected graph, let's take $\Gamma$ to be the straight-line graph with four vertices $a,b,c,d$ (so $[a,b]=[b,c]=[c,d]=1$). Now consider $g=[[c,a],[b,d]]$. Clearly this dies when you kill any generator. On the other hand,

$g=cac^{-1}a^{-1}bdb^{-1}d^{-1}aca^{-1}c^{-1}dbd^{-1}b^{-1}$

and a well-known solution to the word problem in right-angled Artin groups tells you that $g$ is non-trivial.