# Explicit description of dualizing sheaf of nodal curve

Let $$C$$ be a nodal curve with one single node $$p$$ and $$f: N \to C$$ it's normalization. Let $$r,s$$ preimages of $$p$$. In Geometry of Algebraic Curves (p 91) is stated without proof that the dualizing sheaf $$\omega_C$$ can be explicitly described as invertible subsheaf $$\omega_C \subset f_*(\Omega_N(r+s))$$ with sections $$\varphi \in \omega_C (C)$$ beeing sections of $$\Omega_N(r+s)$$ of satisfying the condition

$$\text{Res}_r(\varphi) +\text{Res}_s(\varphi)=0$$

where $$\text{Res}_x:\Omega_{\text{Frac}(C)} \to \kappa(x)$$ is the residue map (details eg in Hartshorne, p 248).

I'm seeking for a source discussion in detail why this gives exactly the dualizing sheaf of the nodal curve.

Let $$X$$ be a nodal curve, say with a single node $$q$$. Let $$\nu :\tilde{X} \to X$$ be its normalization with $$\nu^{-1}(p)=\{p_1,p_2\}$$. Let $$\imath : \{q\} \hookrightarrow X$$ and $$\jmath : X \backslash q \hookrightarrow X$$. Similarly $$\tilde{\imath} : \{p_1,p_2\} \hookrightarrow \tilde{X}$$ and $$\tilde{\jmath} : \tilde{X} \backslash \{p_1,p_2\} \hookrightarrow \tilde{X}$$. We get an exact triangle
$$\tilde{\imath}_*\tilde{\imath}^!\omega_{\tilde{X}} \to \omega_{\tilde{X}} \to \tilde{\jmath}_*\tilde{\jmath}^!\omega_{\tilde{X}}$$ or $$\tilde{\imath}_*\omega_{\{p_1,p_2\}} \to \omega_{\tilde{X}} \to \tilde{\jmath}_*\omega_{\tilde{X} \backslash \{p_1,p_2\}}$$ and similarly for $$X$$. $$\nu$$ induces a map between the sequences, which is an isomorphism on the last term and a sum for the first one. This, together with the fact that the first term is concentrated in degree $$0$$ and the last two in degree $$1$$ implies that $$\omega_X$$ is the kernel of $$\nu_* \tilde{\jmath}_*\omega_{\tilde{X} \backslash \{p_1,p_2\}} \xrightarrow{\text{res}} \nu_*\tilde{\imath}_*\omega_{\{p_1,p_2\}}[1] \xrightarrow{\text{sum}}\imath_*\omega_{q}[1]$$ i.e. exactly elements of $$\omega_{\tilde{X}}(\infty\{p_1,p_2\})$$ with residue summing to zero and any higher order poles vanishing.