Given a finite graph $\Gamma$, one has the 
 [right-angled Artin group](http://en.wikipedia.org/wiki/Right-angled_Artin_group#Right-angled_Artin_groups) $A(\Gamma )$. 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?



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


