0

Let $S$ and $T$ be two arbitrary sets and consider the free vector space $C(S)$ and $C(T)$ generated respectively by $S$ and $T$. Show that $C(S \times T)$ is isomorphic to $C(S) \otimes C(T)$.

I know that in this link: $C(S \times T)$ is isomorphic to $C(S) \otimes C(T).$ there is an answer to this question. But I want to solve the problem in a different way.

My approach is: I want to construct a bilinear map $\varphi:C(S) \times C(T) \rightarrow C(S) \otimes C(T)$ and then by the universal property exists a linear map $ f: C(S) \otimes C(T) \rightarrow C(S \times T) $ such that $\varphi = f \circ \otimes$ and i want $f$ to be bijection. So, i think that $\varphi(s,t) = (s,t)$ and $f(s \otimes t) = (s,t) $ but i think that these maps does not work.

Any Suggestion? Thanks for your help.

Mateo Soto Arango
  • 95
  • You must assume that in $C(S \times T)$, $\lambda \times (s,t) = (\lambda s,t) = (s,\lambda t)$ and $(s,t) + (s,t') = (s,t+t')$ because this is true in $C(S) \otimes C(T)$ ? Does this happen in $C(S \times T)$ ? I think not ? – Balaji sb May 15 '23 at 02:38
  • @Balajisb Hello! I don't know, if we assume that the problem is done, right? But I think we cannot assume this – Mateo Soto Arango May 15 '23 at 03:11
  • Then one way to argue to resolve this problem is that $(s,t+t')$ and $(\lambda s,t)$ and $(s,\lambda t)$ does not even belong to $C( S \times T)$. In otherwords in every equivalence class corresponding to $s \otimes t$ only one element is present in $C(S \times T)$. This must be the way to prove this mapping is a bijection. – Balaji sb May 15 '23 at 10:23

0 Answers0