Let $k$ be an algebraically closed field of characteristic $0$. Consider the closed subvariety $\mathbb{A}^1_{k}$ of $\mathbb{A}^3_{k}$.
By the local cohomology theory, we have the cohomology sequence \begin{equation} \begin{split} 0 &\rightarrow H^0_{\mathbb{A}^1}(\mathbb{A}^3, \mathcal{O}) \rightarrow H^0(\mathbb{A}^3, \mathcal{O}) \rightarrow H^0(\mathbb{A}^3 \setminus \mathbb{A}^1, \mathcal{O}) \\ &\rightarrow H^1_{\mathbb{A}^1}(\mathbb{A}^3, \mathcal{O}) \rightarrow H^1(\mathbb{A}^3, \mathcal{O}). \end{split} \end{equation}
It is well known that $H^1(\mathbb{A}^3, \mathcal{O}) = 0$, and by Hartogs theorem, we have $H^0(\mathbb{A}^3, \mathcal{O}) \simeq H^0(\mathbb{A}^3 \setminus \mathbb{A}^1, \mathcal{O})$. Therefore, we must have $$ H^1_{\mathbb{A}^1}(\mathbb{A}^3,\mathcal{O}) = 0.$$
On the other hand, there is the short exact sequence $$0 \rightarrow I_{\mathbb{A}^1}^n \rightarrow \mathcal{O}_{\mathbb{A}^3} \rightarrow \mathcal{O}_{\mathbb{A}^3}/I_{\mathbb{A}^1}^n \rightarrow 0$$ which induces the cohomology sequence \begin{equation} \begin{split} 0 &\rightarrow \mathcal{H}om(\mathcal{O}_{\mathbb{A}^3}/I_{\mathbb{A}^1}^n, \mathcal{O}_{\mathbb{A}^3}) \rightarrow \mathcal{H}om(\mathcal{O}_{\mathbb{A}^3}, \mathcal{O}_{\mathbb{A}^3}) \rightarrow \mathcal{H}om(I_{\mathbb{A}^1}^n, \mathcal{O}_{\mathbb{A}^3})\\ &\rightarrow \mathcal{E}xt^1(\mathcal{O}_{\mathbb{A}^3}/I_{\mathbb{A}^1}^n, \mathcal{O}_{\mathbb{A}^3}) \rightarrow 0. \end{split} \end{equation}
Since $\mathcal{O}_{\mathbb{A}^3}$ is locally free, $\mathcal{H}om(\mathcal{O}_{\mathbb{A}^3}/I_{\mathbb{A}^1}^n, \mathcal{O}_{\mathbb{A}^3}) = 0$.
By local cohomology theory again, there exists an isomorphism $$\varinjlim Ext^i_{\mathcal{O}}(\mathcal{O}_{\mathbb{A}^3}/I_{\mathbb{A}^1}^n, \mathcal{O}) \xrightarrow{\sim} H^i_{\mathbb{A}^1}(\mathbb{A^3, \mathcal{O}}).$$
Therefore,we have $\mathcal{E}xt^1(\mathcal{O}_{\mathbb{A}^3}/I_{\mathbb{A}^1}^n, \mathcal{O}_{\mathbb{A}^3})=0$, and the morphism $\mathcal{H}om(\mathcal{O}_{\mathbb{A}^3}, \mathcal{O}_{\mathbb{A}^3}) \rightarrow \mathcal{H}om(I_{\mathbb{A}^1}^n, \mathcal{O}_{\mathbb{A}^3})$ should be isomorphism for sufficiently large $n$.
My question is
Can we prove this isomorphism $\mathcal{H}om(\mathcal{O}_{\mathbb{A}^3}, \mathcal{O}_{\mathbb{A}^3}) \rightarrow \mathcal{H}om(I_{\mathbb{A}^1}^n, \mathcal{O}_{\mathbb{A}^3})$ directly?
Is the argument so far correct?
The motivation of this question is that I heard that there exists an isomorphism $$ \varinjlim H^{i-1}(Y, N_{nY}) \xrightarrow{\sim} H^i_Y(X, \mathcal{O})$$ for closed subvariety $Y \subset X$. However, applying this isomorphism to the above discussion, we have $$ Ext^1(\mathcal{O}_{\mathbb{A}^3}/I_{\mathbb{A}^1}^n, \mathcal{O}_{\mathbb{A}^3}) \simeq H^0(\mathbb{A}^1, N_{n\mathbb{A}^1}) \neq 0$$ which is a contradiction.
So, I want to understand the relation between $H^{i-1}(Y, N_{nY})$ and $H^i_Y(X, \mathcal{O})$ more clearly.