The A1 Nakajima varieties and the Hecke correspondence are points

When $v=0$, both maps have zero source or target and hence are zero. There is one solution, stability is vacuous, and $G_0$ is trivial, so $\mathfrak{M}(0,1)$ is a point. Its invariant coordinate ring is $\mathbb{C}$, whose only prime ideal is $(0)$; hence ${\mathfrak{M}}_0(0,1)$ and the fibre $\mathcal{L}(0,1)$ are points by [F2]–[F3].

When $v=1$, write $I(w)=iw$ and $J(z)=jz$ for $i,j\in \mathbb{C}$. The moment equation is $ij=0$ and stability is $j\mathrm{\ne }0$, so $i=0$ on the stable locus. Since $G_1={\mathbb{C}}^{\times }$ sends $j$ to $j{g}^{-1}$ transitively on ${\mathbb{C}}^{\times }$, $\mathfrak{M}(1,1)$ is one point.

A monomial basis of $A_1=\mathbb{C}[i,j]\mathrm{/}(ij)$ is $1$, $i^a$ ($a\ge 1$), and $j^b$ ($b\ge 1$). Under the point action $(i,j)\mapsto (gi,j{g}^{-1})$, evaluation of these monomials scales by $1$, $g^a$, and $g^{-b}$, respectively. The nonconstant basis monomials have pairwise distinct nonzero characters $g^a$ and $g^{-b}$; their linear independence therefore implies that an invariant polynomial has only its constant term; hence $A_1^{G_1}=\mathbb{C}$. Its spectrum has one point by [F3], and the image of the stable orbit is that point, so ${\mathfrak{M}}_0(1,1)$ and $\mathcal{L}(1,1)$ are points.

The product $\mathfrak{M}(0,1)\times \mathfrak{M}(1,1)$ is a point by steps 1.1 and 1.2. It lies in $P_{0,1}$: use $V^{\mathrm{\prime }}=0$ and the stable middle datum $(V_1,I=0,J=1)$; the inclusion is injective and the quotient has one-dimensional vertex space, zero framing space and zero induced maps, hence is $S$. Thus $P_{0,1}$ is a nonempty subset of a point, so it is a point; both projections are the unique identity map.