Previously, I asked about how to prove that $dx + dy$ is a generator of the de Rham cohomology group of the torus.
Now it occurred to me that $dx$ and $dy$ are both also generators of $H^1(T)$. Please can you tell me if this is correct?
(1) To find a complete set of generators of $H^1(T)$ I have to find two differential $1$-forms that are closed but not exact and that do not differ by an exact $1$-form.
(2) Since $dx,dy$ is a pair of closed $1$-forms that are not exact they are promising candidates.
(3) Now I only need to prove that $dx , dy$ do not differ by an exact $1$-form.
How can I prove that $dx$ and $dy$ do not differ by an exact one form?
This is very similar to a part of John's answer to your previous question, so I will leave you to fill in the details.
Suppose $dx$ and $dy$ differed by an exact form. Then $dx - dy = df$ for some $f \in C^{\infty}(T)$. Now let $\gamma$ be a closed loop and consider $\int_{\gamma}dx - dy$; you should be able to conclude that this must be zero. Now find a particular loop $\gamma$ for which the integral isn't zero.