A proper singular curve with weight-zero cohomology

Use the reduced Cech nerve $C_p=({\mathbb{P}}^1{\times }_C\cdots {\times }_C{\mathbb{P}}^1{)}_{\mathrm{r}\mathrm{e}\mathrm{d}}$, with $p+1$ factors and maps deleting or repeating entries. Reduction is functorial: a map from a reduced scheme kills nilpotents and hence factors through the target reduction. The underlying points off the node form the diagonal; over the node the tuples are in $\{a,b{\}}^{p+1}$. The diagonal is a closed immersion, and its complement consists of the $2^{p+1}-2$ mixed tuples, an open finite set of closed complex points. This set is also closed, so these points and the diagonal are open-and-closed components. The mixed components are reduced points. The reduced diagonal component equals ${\mathbb{P}}^1$: its closed immersion has the whole underlying support, so its ideal lies in the zero nilradical of that reduced component. Thus $C_p={\mathbb{P}}_d^1\bigsqcup T_p$ is smooth projective. Every compatible boundary tuple of geometric points has exactly the $p+1$ compatible vertices of the Cech nerve, so it lifts to $C_p$; this proves surjectivity of every matching map, even if its target has nilpotents. These maps are proper by [F4], since their sources are projective and the finite matching limits are separated. The augmentation is proper surjective by [F1]. The reduced nerve is also split: repeated adjacent entries are degeneracies, and compressing successive equal entries gives the unique alternating nondegenerate tuple; for $p=0$ the nondegenerate part is ${\mathbb{P}}^1$. Apply [F2]. Its row $E_1^{p,0}$ has one copy of $\mathbb{Q}$ for the diagonal and one per mixed tuple; by [F3] the positive rows vanish except for the single diagonal copy in degree two.

Compute the degree-zero row in low degrees. On $C_0={\mathbb{P}}^1$ one has $E_1^{0,0}=\mathbb{Q}$, and on $C_1={\mathbb{P}}_d^1\bigsqcup \{ab\}\bigsqcup \{ba\}$ one has $E_1^{1,0}={\mathbb{Q}}_d\oplus {\mathbb{Q}}_{ab}\oplus {\mathbb{Q}}_{ba}$; the alternating sum of the two face pullbacks ${\delta }^0$ sends a constant to the differences of the two diagonal values, hence is zero. On $C_2$ the six nonconstant tuples contribute the coordinates for $(aab),(aba),(abb),(baa),(bab),(bba)$, and evaluating the alternating sum ${\delta }^1(c)$ on these components and on the diagonal gives $c_d$ on the diagonal and on $(aab),(abb),(baa),(bba)$, and $c_{ab}+c_{ba}-c_d$ on $(aba),(bab)$, where $c=(c_d,c_{ab},c_{ba})$. Hence ${\delta }^1(c)=0$ exactly when $c_d=0$ and $c_{ba}=-c_{ab}$, so $E_2^{1,0}=\ker {\delta }^1\mathrm{/}\mathrm{im}{\delta }^0$ is one-dimensional, generated by $(0,1,-1)$.

The other total-degree-one entry is $E_1^{0,1}=H^1({\mathbb{P}}^1,\mathbb{Q})=0$ by [F3], so it contributes nothing; the spectral sequence is first-quadrant with finite diagonals and degenerates at $E_2$ by [F2]. Therefore $H^1(C,\mathbb{Q})\cong E_2^{1,0}$ is one-dimensional, and by the weight identification of [F2] its graded pieces are ${\mathrm{Gr}}_0^W{H}^1=E_2^{1,0}=\mathbb{Q}$ and ${\mathrm{Gr}}_1^W{H}^1=E_2^{0,1}=0$. The first-quadrant weight identification also makes all graded pieces of negative weight zero; together with the upper bound $1$ and finiteness of the weight filtration, this gives $W_{-1}{H}^1=0$ and $W_0{H}^1=H^1$: the whole one-dimensional space is pure of weight $0$ and type $(0,0)$ (Rational mixed Hodge structures), as asserted.