Question on definition of Free Vector Space From typed notes given by Professor:
Let $F$ be a field and $S$ be a set.  $\prod_{s \in S}F$ is a two-sided $F$ vector space, whose elements are indexed sets    $(y)_{s \in S}$    The direct sum    $\oplus_{s \in S}F$    is a subspace of     $\prod_{s \in S}F$    whose elements are indexed sets   $(y)_{s \in S}$   of finite support.  The free $F$ vector space on the set $S$, denoted by $FS$ is defined by
$$
FS: = \oplus_{s \in S}F.
$$
My question regards the set $S$, it seems like there needs to be more requirements for the set $S$, otherwise the free $F$-vector space defined doesn't really depend on the set $S$ itself but rather the cardinality of the set $S$.  For example, if I let $S_1 = \{1,2,3\}$ or if I let $S_2 = \{4,5,6\}$ and consider a particular field $F$ then the free $F$-vector space defined would be the same, i.e. I would have $FS_1 = F \times F \times F = FS_2$.
Is this really the case, or am I missing something with regards to how the free $F$-vector space is defined?
 A: Yes and no.
What exactly $\prod_{s\in S}F$ is? I mean formally. What exactly are "indexed sets of finite support"? Again, formally.
The set-theoretic definition of $\prod_{s\in S}X_s$ is as follows:
$$\prod_{s\in S}X_s:=\big\{f:S\to \bigcup_{s\in S}X_s\ \big|\ f(s)\in X_s\text{ for all }s\in S\big\}$$
In other words it is the collection of all choice functions. When all $X_s$ are equal to say some fixed $X$ (like in our case) then this definition simplifies to
$$\prod_{s\in S}X:=\big\{f:S\to X\ \big|\ f\text{ is a function}\big\}$$
With that definition our $\prod_{s\in S}F$ becomes a vector space via pointwise addition and pointwise scalar multiplication. Then $FS$ is defined as
$$FS=\bigoplus_{s\in S}F=\big\{f\in\prod_{s\in S}F\ \big|\ f(s)=0\text{ for all but finitely many }s\in S\big\}$$
This shows that if $S\neq S'$ then $\prod_{s\in S}F\neq\prod_{s\in S'}F$ and $FS\neq FS'$ as sets, simply because functions inside have different domains. Also these are not Cartesian products (defined as tuples), at least formally, even though they are naturally equinumerous with the corresponding Cartesian products in finite case. That's why we often think about them as infinite Cartesian products.
Anyway, in your particular example we formally have
$$FS_1\neq F\times F\times F$$
$$FS_2\neq F\times F\times F$$
$$FS_1\neq FS_2$$
And so clearly the definition of $FS$ does depend on $S$, not only on its cardinality. But not up to isomorphism.
Indeed, if $g:S\to S'$ is any set-theoretic function, then this function induces a linear map
$$\Lambda_g:FS'\to FS$$
$$\Lambda_g(f)(s)=f(g(s))$$
Note that this definition makes sense, because $f:S'\to F$ by definition, $\Lambda_g(f)$ is supposed to be a function $S\to F$ and $s\in S$. It is not hard to see that if $g:S\to S'$ is a bijection then $\Lambda_g$ is a linear isomorphism with the inverse $\Lambda_{g^{-1}}$. In the language of category theory we say that $\Lambda$ (without the subscript) is a contravariant functor from the category of sets to the category of vector spaces.
Concluding: two equinumerous sets $S\sim S'$ induce isomorphic linear spaces $FS\simeq FS'$. But these are not literally equal.
A: There is nothing missing. Free vector spaces defined over sets with equal cardinality (same field) are isomorphic as vector spaces. The nature of the elements in $S$ is immaterial in the definition, you are just considering formal sums, which you can interpret as combinations of delta-functions, or subsets of $S$. If you have $S_1\sim S_2$, the corresponding free vector spaces are isomorphic by identifying corresponding subsets via the bijection $S_1\leftrightarrow S_2$.
