Let $p \subset \mathbb{P}^1$ a closed point of $\mathbb{P}^1$ and $i: p \to \mathbb{P}^1$ the canonical closed imbedding. I want to see how to explicitely prove $i^*\mathcal{O}_{\mathbb{P}^1}(1) = \mathcal{O}_p $.
I tried to start using the general definition of the inverse image presheaf by $U \to i^{-1} \mathcal{O}_{\mathbb{P}^1}(1)(U) \otimes _{i^{-1}\mathcal{O}_{\mathbb{P}^1}(U)} \mathcal{O}_{\mathbb{P}^1}(U) $
Considering $i^{-1} \mathcal{O}_{\mathbb{P}^1}(1)(V)= \varinjlim _{p \in V \subset \mathbb{P}^1} \mathcal{O}_{\mathbb{P}^1}(1)(V)= \mathcal{O}_{\mathbb{P}^1}(1)_{\mathbb{P}^1, p}$,
as well as $i^{-1}\mathcal{O}_{\mathbb{P}^1}(V)= \varinjlim _{p \in V \subset \mathbb{P}^1} \mathcal{O}_{\mathbb{P}^1}(V) = \mathcal{O}_{\mathbb{P}^1, p}$.
so I get $i^{-1} \mathcal{O}_{\mathbb{P}^1}(1)(U) \otimes _{i^{-1}\mathcal{O}_{\mathbb{P}^1}(U)} \mathcal{O}_{P}(U) = \mathcal{O}_{\mathbb{P}^1}(1)_{\mathbb{P}^1, p} \otimes _{\mathcal{O}_{\mathbb{P}^1, p}} \mathcal{O}_{\mathbb{P}^1}(U)$ as presheaf.
But I don't see the why this simplification imply $i^*\mathcal{O}_{\mathbb{P}^1}(1) = \mathcal{O}_p $.
The background of the question is here: Image of Morphism $C \to \mathbb{P}^1$ not a Closed Point