Let $X,Y$ be $S$-schemes, let $x\in X$ and $y\in Y$ be points over $s\in S$ and let $\mathcal O_{Y,y}\to\mathcal O_{X,x}$ be a local $\mathcal O_{S,s}$-homomorphism. If $Y$ is locally of finite presentation at $y$, then this homomorphism extends to an $S$-morphism $U\to Y$ for some neighbourhood $U$ of $x$.
Here a nice proof is given. In brief, everything is reduced to the following algebraic task:
Given $R$-algebras $A,B$ (with $A$ being of finite presentation over $R$) with prime ideals $\mathfrak p\subset A$ and $\mathfrak q\subset B$ that lie over the same prime ideal $\mathfrak r\subset R$ and given a local homomorphism $A_{\mathfrak p}\to B_\mathfrak{q}$ over $R_{\mathfrak r}$ (hence over $R$ as well), we need to find $b\in B\setminus \mathfrak q$ such that the composition $A\to A_{\mathfrak p}\to B_{\mathfrak q}$ factors over $B_b\to B_{\mathfrak q}$.
In the linked proof, an appropriate $b\in B$ is constructed such that the composition $R[T_1,\dots,T_n]\twoheadrightarrow A\to B_{\mathfrak q}$ factors over $B_b\to B_{\mathfrak q}$ in such a way that the resulting map $R[T_1,\dots,T_n]\to B_b$ factors through $A$. Here the fact that there are only finitely many relations comes into play. However, my initial proof circumvents this finiteness condition and I don't seem to find a logical error:
In my proof, I simply write $A=R[a_1,\dots,a_n]$ (i.e. I only assume $A$ to be of finite type over $R$) and I observe that the images $\frac{b_i}{s_i}$ of the $a_i$ along the map $A\to B_{\mathfrak q}$ all lie in $B_{b}$ for $b=\prod_i s_i$. Hence the map factors over $B_b\to B_{\mathfrak q}$, regardless of any finiteness condition with regard to the relations between the $a_i$.
Can anybody point out any error in my proof? Otherwise I don't see why the condition of $Y$ being locally of finite presentation over $S$ at $y$ is needed.
