Let $k$ be algebraically closed, and let $X$ be a closed subset in $k^n$ with corresponding radical ideal $I$. Let $\mathcal O$ be the standard sheaf associated with the space $k^n$, and let $\mathcal O_X$ be the sheaf associated with $X$. So e.g. $\mathcal O(k^n) \cong k[X_1,\dots, X_n]$, and $\mathcal O_X(X) \cong k[X_1,\dots, X_n]/I$. Is $\mathcal O_X$ the inverse image sheaf for the inclusion map $X \rightarrow k^n$?
I think I've constructed a morphism of sheaves $i: \mathcal F \rightarrow \mathcal O_X$, where $\mathcal F$ is the presheaf $U \mapsto \varprojlim\limits_{V \cap X = U} \mathcal O(V)$. If $f \in \mathcal O(V)$, then every point $x$ in $V$ has a neighborhood $V_x$ and polynomials $g_x,h_x$ such that $f=\frac{g_x}{h_x}$ on $V_x$ and $h_x \neq 0$ on $V_x$. Seeing as how $k[X_1,\dots, X_n]/I$ can be identified with the restriction of polynomial functions in $k[X_1,\dots, X_n]$ to the closed set $X$, it is easy to see that $f_{|V \cap X}$ lies in $\mathcal O_X(V \cap X)$, because every point $x$ has a neighborhood $V_x \cap X$ on which $f$ is equal to the quotient $\frac{\bar{g_x}}{\bar{h_x}}$, with $\bar{h_x}$ nonvanishing on $X \cap V_x$. We then get a map from the direct limit $\varinjlim\limits_{V \cap X = U} \mathcal O(V)$ into the ring $\mathcal O_X(V)$.
Now I just have to show that the pair $(i, \mathcal O_X)$ is actually a sheafification of $\mathcal F$. I would really appreciate a hint on how to do this.
I prefer work in the framework of (affine) schemes.
Let $X$ be a scheme and let $Y$ be a (non empty) closed subet of $X$ with immersion $i:Y\hookrightarrow X$; we would like define a(n affine) scheme structure over $Y$ as the scheme structure induced from $(X,\mathcal{O}_X)$.
A possible solution is check that $(Y,i^{-1}\mathcal{O}_X)$ is a scheme structure on $Y$ such that $(i,i^{\sharp}):(Y,i^{-1}\mathcal{O}_X)\to(X,\mathcal{O}_X)$ is a closed immersion of (affine) schemes, where $i^{-1}\mathcal{O}_X$ is the inverse image sheaf of $\mathcal{O}_X$ via $i$; but by definition $(i,i^{\sharp})$ is a closed immersion of schemes only if $Y$ is isomorphic to $i(Y)$ as scheme, that is we need of a scheme structure over $Y$ while we are defining a scheme structure over $Y$: all this is a "short circuit"!
We need change the "strategy"!
Considered the functor: \begin{equation*} U\subseteq X\,\text{open},\,\mathcal{I}_Y(U)=\{h\in\mathcal{O}_X(U)\mid\forall x\in Y\cap U,\,h(x)=0\in\kappa(x)\} \end{equation*} where $\kappa(\cdot)$ is the residue field (in $X$) of the point $\cdot$; one can prove that $\mathcal{I}_Y$ is a quasi-coherent ideal (sheaf) of $\mathcal{O}_X$, it is called the vanishing ideal of $\,Y$, and (because $Y$ is a closed subset of $X$) one can prove that: \begin{equation*} Y=V(\mathcal{I}_Y)=\left\{x\in X\mid\left(\mathcal{O}_{X\displaystyle/\mathcal{I}_Y}\right)_x\neq0\right\}\simeq\operatorname{Spec}\mathcal{O}_{X\displaystyle/\mathcal{I}_Y}; \end{equation*} in general the isomorphism $V(\mathcal{I}_Y)\simeq\operatorname{Spec}\mathcal{O}_{X\displaystyle/\mathcal{I}_Y}$ holds.
If $X$ is an affine scheme, let $R$ be the structure ring $\mathcal{O}_X(X)$; one can prove (easily) that $\mathcal{I}_Y$ is the sheafification $\widetilde{I}$ of an ideal $I$ of $R$; therefore one can prove (via definition of categorical limit) that: \begin{equation*} \forall x\in Y,\,\mathcal{O}_{Y,x}=\left(\mathcal{O}_{X\displaystyle/\mathcal{I}_Y}\right)_x=\mathcal{O}_{X,x\displaystyle/\mathcal{I}_{Y,x}}=R_{x\displaystyle/I_x}. \end{equation*} Let $p:R\to R_{\displaystyle/I}$ the canonical projcetion of $R$ onto the quotient $R_{\displaystyle/I}$, because $i$ is the morphism $p^{*}$ induced from $p$, $i^{-1}\mathcal{O}_X$ is not isomorphic to $\mathcal{O}_Y$.
In general, $Y$ is a closed (not necessarily affine) subscheme of a (not necessarily affine) scheme $X$ if and only if for all $x\in X,\,i^{\sharp}_x:\mathcal{O}_{X,x}\to\mathcal{O}_{y,x}$ is surjective.
For completeness, I suggest: