Virtual fundamental class

Remarks

The class does not depend on the choice of $s'$. Two choices are joined by a generic homotopy whose zero set is an oriented cobordism between the two zero sets inside $U$. The classes for smaller neighbourhoods map to the classes for larger ones. If $s$ itself is transverse to the zero section, then $X$ is a compact oriented manifold of dimension $d$ and $[X]^{\mathrm{vir}}$ is its fundamental class. For a general Kuranishi structure the same construction is done chart by chart with multivalued perturbations, which is why rational coefficients are used.

Example: let $V = S^2$, $E = T S^2$ and $s = 0$, so $X = S^2$ and $d = 0$. A generic vector field on $S^2$ has isolated zeros whose signed count is $2$, the value of the Euler class of $T S^2$ on the fundamental class of $S^2$. So $[X]^{\mathrm{vir}} = 2 \, [\mathrm{pt}] \in H_0(S^2; \Q)$, although $X$ has dimension $2$.

Non-example: for $V = \R$, $E = \R \times \R$ and $s = 0$, the zero set $X = \R$ is not compact. The perturbations $s' = \varepsilon$ and $s' = \varepsilon x$ have $0$ and $1$ zeros, so no well-defined class arises; this is why compactness is required.