Suppose $X$ is a DM stack, and let $E^\bullet$ be a perfect obstruction theory of $X$ such that the $E^{-1}$ term admits a trivial quotient/sub-bundle. Is it true that the virtual fundamental class $[X, E^\bullet]$ is zero?


If $X$ is smooth, then this is true: In such a case, the virtual fundamental class is the top Chern class of the vector bundle $E^{-1}$, which is zero due to the trivial quotient/sub-bundle and the exact sequence
$$0 \to \mathcal{O} \to E^{-1} \to coker \to 0$$
together with the multiplicative nature of $c\_{top}$. Intuitively, you can use the trivial factor to "move" a section of this bundle away from the zero section.


If $X$ is not smooth, then this argument doesn't work and we must use the intrinsic normal cone of Behrend and Fantechi to compute the virtual fundamental class. Instead of bundles, we obtain a cone $C(E^\bullet)$ contained in $(E^{-1})^\vee$, which we intersect with the zero section of $(E^{-1})^\vee$. In comparison with the smooth case, it seems like we should somehow be able to move the cone out of the zero section using the trivial portion of the bundle, but I don't see how to do this.


Is there an easy argument showing that this class is zero? Are there extra conditions required?


**Edit**: The method which I have tried to no avail is to use Proposition 5.10 of Behrend-Fantechi. It states loosely that, given two obstruction theories $F$ and $F'$, and certain compatibility data between them, that
$$
v^![X,F] = [X,F']
$$
However, I was not able to find a clear way to have this yield that my desired virtual class is zero.



Trivial obstructions and virtual fundamental classes
It turns out that this is true. In the paper "Localizing Virtual Cycles by Cosections" by Kiem and Li, they address the case where one has a surjection $Ob \to \mathcal{O}$. In the case of an injection $\mathcal{O} \to Ob$, one can produce via a diagram chase a corresponding surjection, which yields the claim.


