ScalingStacks

Proof: Since we already know that $X_{1},X_{2}$ and $X_{3}$ are graded subcomplexes of $\overline{C}(D_{1}),$ it suffices to check (112) on the level of underlying abelian groups. We have $\alpha=\beta\psi_{3}$ and, therefore, for $z\in\overline{C}(D_{1}(\ast 01))[-1]\{-1\}$

$$\alpha z=\beta\psi_{3}z\in X_{3}$$

(113)

Subcomplex $X_{1}$ consists of elements $z+\alpha z$ and we know that $\alpha z\in X_{3}.$ We are thus reduced to proving the following direct sum splitting of abelian groups

$$\overline{C}(D_{1})=\overline{C}(D_{1}(\ast 01))[-1]\{-1\}\oplus X_{2}\oplus X_{3}$$

(114)

Next recall that $X_{2}$ consists of elements $z+dw$ for $z,w\in\overline{C}(D_{1}(\ast 00)).$ The differential $dw$ reads

$$dw=d_{00}w+[-1]\psi_{1}w+[-1]\psi_{2}w$$

(115)

Note that $[-1]\psi_{2}(w)\in\overline{C}(D_{1}(\ast 01))[-1]\{-1\}$ and $d_{00}w\in\overline{C}(D_{1}(\ast 00)).$ Let $X^{\prime}_{2}$ be the subgroup of $\overline{C}(D_{1})$ given by

$$X^{\prime}_{2}=\{z+[-1]\psi_{1}w|z,w\in\overline{C}(D_{1}(\ast 00))\}$$

(116)

Then it is enough to verify that $\overline{C}(D_{1})$ is a direct sum of its subgroups $\overline{C}(D_{1}(\ast 01))[-1]\{-1\},X^{\prime}_{2}$ and $X_{3}$:

$$\overline{C}(D_{1})=\overline{C}(D_{1}(\ast 01))[-1]\{-1\}\oplus X^{\prime}_{2}\oplus X_{3}.$$

(117)

Note that $X_{3}$ contains $\overline{C}(D_{1}(\ast 11))[-2]\{-2\}$ and $X^{\prime}_{2}$ contains $\overline{C}(D_{1}(\ast 00)).$ Recall the direct sum decomposition

	$\displaystyle\overline{C}(D_{1})$	$\displaystyle=$	$\displaystyle\overline{C}(D_{1}(\ast 00))\oplus\overline{C}(D_{1}(\ast 01))[-1]\{-1\}$
		$\displaystyle\oplus$	$\displaystyle\overline{C}(D_{1}(\ast 10))[-1]\{-1\}\oplus\overline{C}(D_{1}(\ast 11))[-2]\{-2\}$

of $\overline{C}(D_{1}).$ Let $X^{\prime\prime}_{2}$ and $X^{\prime}_{3}$ be the following abelian subgroups of $\overline{C}(D_{1}(\ast 10))[-1]\{-1\}$:

	$\displaystyle X^{\prime\prime}_{2}$	$\displaystyle=$	$\displaystyle\{[-1]\psi_{1}(w)|w\in\overline{C}(D_{1}(\ast 00))\}$
	$\displaystyle X^{\prime}_{3}$	$\displaystyle=$	$\displaystyle\{\beta(w)|w\in\overline{C}(D_{1}(\ast 11))[-2]\{-2\}\}$

Now we are reduced to proving the direct sum decomposition

$$\overline{C}(D_{1}(\ast 10))[-1]\{-1\}=X^{\prime\prime}_{2}\oplus X^{\prime}_{3}$$

(118)

in the category of abelian groups. As an abelian group, $\overline{C}(D_{1}(\ast 10))[-1]\{-1\}$ is a direct sum of $F(D_{1}(\mathcal{L}a))$ over all possible resolutions of the $(n-2)$-double points of $D_{1}.$ Similar direct sum splittings can be formed for $X^{\prime\prime}_{2}$ and $X^{\prime}_{3}$ and one sees then that it suffices to check (118) when $D_{1}$ has only two double points. There are two such $D_{1}$’s: