Theorem on Formal Function Hartshorne

Question

Let consider the Thm on Formal Functions [see Hartshorne, III.11.1]. We have with $f:X \to Y$ a projective morphism of noetherian schemes, $\mathcal{F}$ be a coherent sheaf on $X$ and let $y\in Y$. Then the natural map $$ R^i f_\ast (\mathcal{F})_y^\hat{} \to \varprojlim H^i(X_n, \mathcal{F}_n) $$ is an isomorphism for all $i \geq 0$.

My question is from what $n$ does the inverse limit start? From $n=1$ or $n=0$?

The background of my question is that all $\mathcal{F}_n $ come from the exact sequence $$\displaystyle 0 \rightarrow \mathcal{I}^n \mathcal{F} \rightarrow \mathcal{F} \rightarrow \mathcal{F}_n \rightarrow 0, $$

And by $\mathcal{I}^0\mathcal{F}$ it would imply that $\mathcal{F}_0 =\mathcal{F} $ and therefore the limit would map surjectively on global sections $H^i(X_n, \mathcal{F})$ what would be absurd for example if we consider the structure sheaf $\mathcal{F}= \mathcal{O}_X$.

Answer 1

The answer is in Hartshorne, right at the bottom of p. 276: $n \geq 1$.

(You want $\mathrm{Spec}\,\mathcal O_y/\mathfrak m^n_y$ to be topologically just a point, and this happens starting from $n = 1$.)

Answer 2

It doesn't matter; you can restrict to any infinite subset of natural numbers and get the same limit. (see final functor for the general notion)

In general, the canonical 'projections' $$ \left(\varprojlim F_\bullet \right) \to F_n $$ need not be epimorphisms. So, if you find yourself in a situation where it would be absurd for one of these maps to be an epimorphism... that simply isn't a problem at all.

As an example, consider the family of subsets of $\mathbb{R}$ defined by $S_n = (-\infty, \frac{1}{n})$, where the transition maps are all inclusions. The inverse limit of these sets is simply their intersection: $$ \left(\varprojlim S_\bullet\right) = (-\infty, 0] $$ and the canonical projections are simply inclusion. And clearly, none of the inclusions $(-\infty, 0] \to (-\infty, \frac{1}{n})$ are surjective.