Sheaf of sections vanishing at a point is $\Gamma(E) \otimes I_p$ Notation:


*

*$E$ is a vector bundle with sheaf of sections $\Gamma(E)$.

*$I_p$ is the sheaf of regular functions on the base space vanishing at a point $p$.

*$\Gamma_p(E)$ is the sheaf of sections of $E$ which vanish at a point $p$.


Is it true that $\Gamma_p(E) \simeq \Gamma(E) \otimes I_p$?
How about in the case where the base space is an algebraic curve?

I tried to define a map $I_p\otimes \Gamma(E) \rightarrow \Gamma_p(E)$ by $f\otimes s \mapsto fs$ but it's not clear to me that it's an isomorphism.
I also tried seeing $\Gamma_p(E)$ as the kernel of the evaluation map $\Gamma(E) \rightarrow E$. The resulting quotient is a skyscraper sheaf supported at $p$, right? Does that mean anything?

Context: I'm looking through a proof of Grothendieck's Splitting Theorem, in C. Okonek's book Vector Bundles on Complex Projective Spaces. The base space is $\mathbb{P}^1$, and he has a global section $s$ with $s(p)=0$. He  claims that $s$ "is" in $H^0(\Gamma(E)\otimes I_p)$. Note that in this case, $I_p \simeq \mathcal{O}(-1)$, which maybe makes a difference.

Yes, it is true, and the above defined morphism is indeed an isomorphism.
A morphism of sheaves is an isomorphism if and only if it induces an isomorphism at each stalk, thus one can reduce the discussion to an open covering $\{U_\alpha\}$ such that for each $\alpha$, $\Gamma(E)|_{U_\alpha}$ is a free $\mathcal{O}_x|_{U_\alpha}$ module. But for a free module it is not hard to verify that the defined map is an isomorphism, and we are done.
