Let $\Delta : X \rightarrow X \times X$ denote the diagonal embedding where $X$ is a $\mathbb{C}$-variety. In a paper I am reading it asserts that the pull-back of $X$ diagonally embedded in $X \times X$ by $\Delta$ is equal to $-\omega_{X}$, where $\omega_{X}$ is the canonical bundle of $X$. Could anyone explain why this is the case or point me in the right direction of where to look in order to prove this claim?
It is worth mentioning that $X$ is a disjoint union of modular curves in the paper I am reading. So, in particular, $X$ is smooth and projective.
Here's what's (probably) happening. You're pulling back $\mathcal{O}_{X\times X}(\Delta(X))=\mathcal{I}_{\Delta(X)}^{-1}$ along $\Delta:X\to X\times X$, which gives $\mathcal{I}_{\Delta(X)}^{-1}\otimes_{\mathcal{O}_{X\times X}} \mathcal{O}_{X\times X}/\mathcal{I}_{\Delta(X)}$, or $(\mathcal{I}_{\Delta(X)}/\mathcal{I}_{\Delta(X)}^2)^{-1}$. As $\mathcal{I}_{\Delta(X)}/\mathcal{I}_{\Delta(X)}^2\cong \Omega_X$ and $\omega_X\cong\bigwedge^1\Omega_X\cong \Omega_X$, we see that the pullback of $\mathcal{O}_{X\times X}(\Delta(X))$ along $\Delta$ is $\omega_X^{-1}$.