Three marked points stabilize the constant sphere and give the point-target invariant

For the two-marked constant map, every automorphism $z\mapsto az$ with $a\in {\mathbb{C}}^{\times }$ fixes $0$ and $\mathrm{\infty }$ and preserves the constant map. This gives an infinite automorphism group, so the two-marked map is not stable by [F2].

A Möbius transformation fixing the three distinct points $0,1,\mathrm{\infty }$ is the identity. Thus the three-marked constant map has trivial automorphism group, and its domain is stable by [F2]. The unique map to a point satisfies the $J$-holomorphic equation because both sides are the zero tangent map; its homology class is $\beta =0$.

Let $C$ be any genus-zero nodal curve with three ordered markings; write $k$ for its number of irreducible components, $e$ for its number of nodes, and $h_v$ for the genus of the normalization of component $v$. At each node the local ring $\mathbb{C}\{x,y\}\mathrm{/}(xy)$ embeds in the normalization $\mathbb{C}\{x\}\oplus \mathbb{C}\{y\}$ as the pairs with equal constant terms, whose quotient is $\mathbb{C}$; away from nodes the normalization is an isomorphism. Hence the normalization sequence is $0\to {\mathcal{O}}_C\to {\nu }_{\ast }{\mathcal{O}}_{\tilde{C}}\to {⨁}_{q\text{ a node}}{\mathbb{C}}_q\to 0$, so $\chi ({\mathcal{O}}_C)={\sum }_v(1-h_v)-e=k-{\sum }_v{h}_v-e$ and $g(C)={\sum }_v{h}_v+e-k+1$. A connected graph with $k$ vertices and $e$ edges has first Betti number $e-k+1$: the boundaries of a spanning tree generate the rank-$(k-1)$ augmentation-zero subgroup of its vertex group. Thus all terms in the genus formula are nonnegative, so every $h_v=0$ and the dual graph is a tree with $e=k-1$. Counting special points over all components counts each marking once and each node twice, giving $3+2e=2k+1$. Stability of every rational component requires at least three special points, so $2k+1\ge 3k$ and $k\le 1$. Since $k\ge 1$, $k=1$. Therefore the domain is a smooth projective line; an ordered triple of distinct points is carried to $(0,1,\mathrm{\infty })$ by a unique Möbius transformation. The unique map to the point makes the stable-map moduli one point. Its domain tangent space and target-map deformation/obstruction spaces are zero; the linearized section $0\to 0$ is surjective of index zero. Hence $Z=\{\ast \}$, $E=0$, $s=0$ is a regular zero-dimensional chart. Orient it by the positive generator of $H_0(\{\ast \};\mathbb{Z})$.

By [F3], this oriented regular atlas has virtual class the ordinary fundamental class $[\ast ]$. The stabilization/evaluation map is the identity of the one-point space, so its pushforward remains $[\ast ]$. On the cohomology-dual convention this is the functional $a\mapsto a(\ast )$ on ${\check{H}}^0(\{\ast \};\mathbb{Q})=\mathbb{Q}$: the singleton cover is cofinal, with degree-zero cochains $\mathbb{Q}$ and no higher cochains. Identifying that functional with its value at $1$ gives the ordinary point class. By [F4] with $G=\mathbb{Q}$, this is the generator $1\in H_0(\{\ast \};\mathbb{Q})\cong \mathbb{Q}$. This is a mathematical model calculation, not empirical evidence.