Perfect complexes of plane nodal cubic curve

Question

Let $C\subset\mathbb{P}^2$ be a plane nodal cubic curve with a unique singular point $O$ at the origin. Then I consider its normalization, denoted by $\widetilde{C}$ and let $\pi:\widetilde{C}\rightarrow C$ be the normalization map. Then I consider the associated functor $\pi^*:\mathrm{Perf}(C)\rightarrow\mathrm{Perf}(\widetilde{C})$, where $\mathrm{Perf}$ is the derived category of perfect complexes. I was wondering how I can describe the category $\mathrm{Ker}\pi^*$, I think this category is nonzero and the objects in $\mathrm{Ker}\pi^*$ should support on the nodal singularity $O$.

Answer 1

The kernel is zero.

Indeed, on the one hand, since $\pi$ is an isomorphism over the complement of $O$, any object in $\operatorname{Ker}(\pi^*)$ is supported at $O$.

On the other hand, if $\tilde{O}$ is the fiber of $\pi$ over the node, $i \colon O \to C$ and $\tilde{i} \colon \tilde{O} \to \tilde{C}$ are the embeddings, and $\pi_O \colon \tilde{O} \to O$ is the restriction of $\pi$ then $$ \tilde{i}^* \circ \pi^* \cong \pi_O^* \circ i^*. $$ The functor $\pi_O^*$ is obviously fully faithful, hence any object in the kernel of $\pi^*$ must lie in the kernel of $i^*$, but then it must be supported away from $O$.

Thus, the kernel of $\pi^*$ is zero.