Let $X = \mathbb{A}_{\mathbb{C}}^{2}$ and let $Y$ be the blowup of $X$ at the origin. Let $E \cong \mathbb{P}^{1}$ the exceptional divisor.
I think that we have a canonical inclusion $\mathcal{O}_{Y} \hookrightarrow \mathcal{O}_{Y}(E)$. To my understanding, when we restrict this map to $E$ we should get an inclusion $\mathcal{O}_{E} \hookrightarrow (\mathcal{O}_{Y}(E))\vert_E \cong \mathcal{O}_{E}(-1)$, where the last isomorphism follows from the fact that the exceptional divisor has self-intersection $-1$. But such an inclusion cannot exist, since the bundle $\mathcal{O}_{E}(-1)$ has no global sections, whereas $\mathcal{O}_{E}$ clearly does.
Where is my mistake? What does the restriction to $E$ of the similar canonical inclusion $\mathcal{O}_{Y}(E) \hookrightarrow \mathcal{O}_{Y}(2E)$ look like?
The ideal sheaf of $E$ is $\mathcal{O}_Y(-E)$ and one has $0\to \mathcal{O}_Y(-E)\to \mathcal{O}_Y\to \mathcal{O}_E\to 0$ coming from multiplying by a local equations for $E$. One can twist this by $\mathcal{O}_Y(E)$ to get $0\to \mathcal{O}_Y\to \mathcal{O}_Y(E)$ as you suggest. The error is that restriction is right exact rather than left exact. So, if $j:E\hookrightarrow Y$ is the inclusion of the exceptional curve then $0\to j^*\mathcal{O}_Y\to j^*\mathcal{O}_Y(E)$ need not be exact.