$f:S\to Y$ of $S$ into any module $Y$ over $R$ extends to a uniquely determined homomorphism $h:X\to Y$ of $X$ into $Y$ 
Let $S$ be an arbitrary basis of a module $X$ over $R$. Show that any
  function $f:S\to Y$ of $S$ into any module $Y$ over $R$ extends to a
  uniquely determined homomorphism $h:X\to Y$ of the module $X$ into the
  module $Y$

By what I know about basis and modules, I know that any element of $X$ can be written as a linear combination of elements of $S$. I know that $Y$ has a basis, so any element there can be written as a linear combination of this basis. Since the basis representation is unique, I should found a bijection $f$ from basis elements $S$ of $X$ to the basis elements of $Y$, right? Then, I should just extend like this:
$$\overline{f} = h:X\to Y; h\left(\sum_{i\in M}\alpha_i s_i \right) = \sum_{i\in M}\alpha_i f(s_i)$$
 A: $\quad$If a (unitary) $R$-module $X$ has a (nonempty) basis $S$, then yes, any element of $X$ can be written as a $\mathit{unique}$ $R$-linear combination of elements of $S$. 
$\quad$Given an $R$-module $Y$ and a function $f:S\rightarrow Y$, it does $\mathit{not}$ follow that $Y$ also has a basis. It does follow that there exists a unique $R$-module homomorphism $h:X\rightarrow Y$ such that $hi=f$, where $i:S\rightarrow X$ is the inclusion function. 
$\quad$This is part of a theorem which characterizes $\mathit{free}$ modules, which says that a free (unitary) $R$-module (i.e. one which has a (nonempty) basis) is a free object in the category of (unitary) $R$-modules. 
$\quad$To prove it:
(1) define $h$ in the way you described; 
(2) verify $hi=f$; 
(3) show that $h$ is an $R$-module homomorphism 
(4) show that $h$ is unique with the properties in (2) and (3) (suppose there is another $R$-module homomorphism $g:X\rightarrow Y$ such that $gi=f$ and show $g=h$). 
$\quad$For a great reference, see Hungerford's $\mathit{Algebra}$, chapter IV, section 2, theorem 2.1.
