The Theorem on Formal Funtions - Harthshorne Theorem 11.1 Let $f:X \rightarrow Y$ a projective morphism of noetherian schemes, let $\mathcal{F}$ be a coherent sheaf on $X$, and let $y\in Y$ be a point. For each $n\geq 1$ we define 
$X_n=X \times_Y Spec (\mathcal O_y/{m_y}^n) $
$\require{AMScd}$
\begin{CD}
    X_n @>v>> X\\
    @V f' V V\Box @VV f V\\
    Spec \mathcal O_y/{m_y}^n @>>v'> Y
    \end{CD}
1.For n=1, we get $X_n=X_y$
2.For $n\geq 2$, we get a scheme with nilpotent elements having the same underlying space as $X_y$. It is kind of "thickened fibre" of $X$ over the point $y$.


*Let $\mathcal{F_n}=v^{*}\mathcal {F}$, Then we have natural maps, for each $n$
$R^{i}f_{*}\mathcal{F}\otimes \mathcal{O_y}/m_y^n\rightarrow R^{i}f'_{*}(\mathcal{F_n})$
Question- How does we get this natural map? 
My Attempt: We have a map 
$v'^{*}({R^if_{*}(\mathcal F)})\rightarrow R^i f'_{*}{v^*{\mathcal{F}}}=R^i f'_{*}{\mathcal {F_n}}$.  ---(1)
I don't understand why is $v'^{*}({R^if_{*}(\mathcal F)})=R^if_{*}(\mathcal F)\otimes\mathcal O_y/{m_y}^n $.


*Since $Spec(\mathcal O_y/{m_y}^n)$ is an affine scheme concentrated at a point the right hand side of the equation is $H^i(X_n, \mathcal{F_n})^\sim =H^i(X_n, \mathcal{F_n})$ (Since the scheme is one-pointed) 


*As n varies, both sides form inverse systems. 



Question: How does it forms an inverse system? What are the maps  $H^i(X_n, \mathcal{F_n})\rightarrow H^i(X_m, \mathcal{F_m})$ and $R^{i}f_{*}\mathcal F\otimes \mathcal{O_y}/m_y^n \rightarrow R^{i}f_{*}\mathcal F\otimes \mathcal{O_y}/m_y^m$ which makes them an inverse system?
 A: The natural map $H^i(X_n, \mathcal F_n) \to H^i(X_m, \mathcal F_m)$ when $n \ge m$ can be established as follows:
Since $f:\ X \to Y$ is projective by hypothesis, $f$ is separated. Therefore from the canonical morphism: 
$$ X_n = X \times_Y \mathrm{Spec} \mathcal O_y/m_y^n \to \mathrm{Spec} \mathcal O_y/m_y^n ,$$
and properties of separated morphisms, we see $X_n \to \mathrm{Spec} \mathcal O_y/m_y^n$ is separated. But since $\mathrm{Spec} \mathcal O_y/m_y^n$ itself is separated, $X_n$ is separated. Consequently, we can use Cech cohomology to calculate $H^i(X_n, \mathcal F_n)$. So the natural morphisms $g: X_m \to X_n$ induced from $\mathcal O_y/m_y^n \to \mathcal O_y/m_y^m$, and $g^*(\mathcal F_n) \to \mathcal F_m$ give the map $H^i(X_n, \mathcal F_n) \to H^i(X_m, \mathcal F_m)$ we want.   Keep in mind all $X_n$ are homeomorphic as topological spaces.   
A: Note that $\mathfrak m^{n+1}_y \subset \mathfrak m^n_y$, so the natural quotient map goes $\mathcal O_y/\mathfrak m^{n+1} \rightarrow \mathcal O_y/\mathfrak m^n_y$, hence the map on the spectra goes the other way round. We end up with the following diagram:
$\require{AMScd}$
\begin{CD}
    X_n @>>> X_{n +1}\\
    @VVV @VVV\\
    \mathrm{Spec} \, \mathcal O_y/{\mathfrak m_y^n} @>>> \mathrm{Spec} \, \mathcal O_y/{\mathfrak m_y^{n+1}}
    \end{CD}
Now I would suggest to consider all your sheaves as sheaves on $\mathrm{Spec} \, \mathcal O_y/{\mathfrak m_y}$. The fact that they form inverse limits is then a consequence of Remark III.9.3.1 on page 255.
