Problem 3 pg. 12 from Greub's Multilinear Algebra.
Show that $C(S) \otimes C(T)$ is isomorphic to $C(S \times T)$ where S,T are sets, and C(T) is the free vector space on T.
My first instinct is to show that $C(S \times T)$ satisfies the universal property of the tensor product on $C(S) \times C(T)$ but I'm not sure where to begin.
The universal property of the tensor product says that the tensor product $C(S) \otimes C(T)$ along with the bilinear map $\phi: C(S)\times C(T) \rightarrow C(S)\otimes C(T)$ (given by $(v,w) \rightarrow v\otimes w$) satisfies the property that given any vector space $V$ and a bilinear map $F$ from $C(S) \times C(T)$ to $V$, there exists a unique linear map $f: C(S) \otimes C(T)\rightarrow V$, such that $F = f \circ \phi$.
Now consider the vector space $C(S\times T)$ along with the bilinear map $\psi:C(S)\times C(T)\rightarrow C(S\times T)$, given by $(e_s,e_t)\rightarrow e_{(s,t)}$, where $s\in S, t\in T$ and $e_x$ is the basis element corresponding the the element $x$ in the respective free vector spaces. Observe that since we want $\psi$ to be bilinear, for arbitrary vectors $v = \sum_{s\in S}a_se_s \in C(S)$ and $w = \sum_{t\in T}b_te_t \in C(T)$, $\psi(v,w)$ must be given by \begin{align} \psi(v,w) &= \psi\left(\sum_{s\in S}a_se_s,\sum_{t\in T}b_te_t\right) \\ &= \sum_{s\in S}a_s \psi\left(e_s,\sum_{t\in T}b_te_t\right) \\ &= \sum_{s\in S,t\in T}a_sb_s \psi\left(e_s,e_t\right) \\ &= \sum_{s\in S,t\in T}a_sb_s e_{(s,t)} \end{align} Here $a_s$ and $b_t$ are elements in your base field, being zero for all but finitely many $s\in S$ and $t\in T$. This makes sure that the sum at the end is also finite and hence is an element of $C(S\times T)$. Now this is well defined, since basis representations for $v$ and $w$ are unique, and can be checked to be bilinear.
Now to check for the universal property, consider a bilinear map $G: C(S)\times C(T) \rightarrow V$. We wish to show that there exists a unique linear map $g:C(S\times T)\rightarrow V$ such that $G = g\circ \psi$. Observe that the equality $G = g\circ\psi$ forces that $g(e_{(s,t)})=g(\psi(e_s,e_t))=G(e_s,e_t)$ for $s\in S$ and $t\in T$. Since elements of the form $e_{(s,t)}$ form a basis for $C(S\times T)$, this uniquely determines the linear map $g$. Now one must check that the linear map $g$ given by $g(e_{(s,t))} = G(e_s,e_t)$ satisfies $G = g\circ\psi$, which I leave to you as an exercise.