Let $C$ be a nodal projective curve over an algebraically closed field of genus at least $2$. There are two natural "differential objects" one can consider: The sheaf of differentials $\Omega_C$ and the dualizing sheaf $\omega_C$. Let $\nu: \tilde{C} \to C$ be the normalization and $N$ be the divisor of preimages of nodes under $\nu$. As $\omega_C$ is defined to be the subsheaf $\nu_*\Omega_\tilde{C}(N)$, where the sum of the residues vanishes for each preimage of a node, there is a natural map $\Omega_C \to \omega_C$.
I am interested in the cokernel of the dual of this map. One can show, that the map of duals is injective, so we have a s.e.s. \begin{align*} 0 \to \omega_C^\vee \to \Omega_C^\vee \to A \to 0 \end{align*}
with the cokernel $A$ that I want to understand. As the curve $C$ is stable (and so are the connected components of the normalization after marking the preimage of nodes), they do not have infinitesimal automorphisms and therefore it holds that $H^0(\Omega_C^\vee)=H^0(\omega_C^\vee)=0$.
However, outside of the nodes the map is an isomorphism, so the cokernel sheaf can only be supported at the $0$-dimensional subscheme of the nodes, therefore we have $H^1(A)=0$, so we have a surjection $H^1(\omega_C^\vee) \to H^1(\Omega_C^\vee)$. But the first group parametrizes log-smooth deformations of $C$ while the latter parametrizes "classical" deformations of $C$, so both are vector spaces of dimension $3g-3$ and therefore the map is an isomorphism. But then I can conclude that $H^0(A)=0$ and as its support is discrete it follows $A=0$, which is not true, as another direct calculation shows that $\omega_C^\vee \to \Omega_C^\vee$ is not an isomorphism.
So something is wrong with the argument that I gave and I would be very happy if someone could point out the mistake in my argument.
