This is a problem from Lee smooth manifolds, 17-13: Let $\mathbb{T}^2 = \mathbb{S}^1 \times \mathbb{S}^1$ be the 2-torus. Consider the two maps $:f,g:\mathbb{T}^2 \rightarrow \mathbb{T}^2$ given by $f(w,z)=(w,z)$ and $g(w,z)=(z,\bar{w})$. Show that $f$ and $g$ have the same degree but are not homotopic.
I want to prove this using induced map on the first de Rham cohomology group as suggested by Lee. So clearly $f$ induces identity map on $H^1_{dR}(\mathbb{T^2})$. I want to prove $g$ doesn't induce the identity map. How should I do this?
If you don't like the fundamental group and insist on using the suggested approach, I think the following works: Give $\mathbb{S}^1 \backslash \{i\} \times \mathbb{S}^1 \backslash \{i\}$ the coordinates $(s,t)$ via stereographic projection. We wish to find an exact $1$-form $\omega \in \Omega^1(\mathbb{S}^1\times \mathbb{S}^1)$ such that $g^*(\omega)$ is not cohomologous to $\omega$.
Consider $\omega = e^{-(st)^2}(sdt+tds)$. Then $d\omega =0$, so this is exact, and it represents a form on all of $\mathbb{S}^1 \times \mathbb{S}^1$ (recall that under the transition maps to the stereographic projections on $\mathbb{S}^1 \backslash \{-i\}$, $t \mapsto \frac{1}{t}$ and $s \mapsto \frac{1}{s}$, and note that $x^{-n}e^{-\frac{1}{x^2}}$ extends smoothly from $\mathbb{R} \backslash \{0\}$ to $\mathbb{R}$). Since $s \circ g =s$, $t \circ g = \frac{1}{t}$ (check this is what complex conjugation does under the stereographic projection relative to $i$), we have $g^*(\omega) = e^{-(\frac{s}{t})^2}(-\frac{s}{t^2}dt+\frac{1}{t}ds)$.
Now, note that $d(\int_0^{st}e^{-r^2}dr) = e^{-(st)^2}(sdt+tds)$ as forms on $\mathbb{R}^2$, but $\int_0^{st} e^{-r^2}dr$ does not extend to a smooth function on all of $\mathbb{S}^1 \times \mathbb{S}^1$, since when one fixes $t = \infty$ (i.e., the we consider the circle $\mathbb{S}^1 \times \{i\}$) the function is $\frac{\sqrt{\pi}}{2}$ everywhere except at $s=0$ (the point $(-i,i)$), where it is $0$. This shows $\omega$ is not closed. We can do a similar analysis with $\int_{0}^{\frac{s}{t}} e^{-r^2} dr$ to see that $g^*(\omega)$ is not closed (indeed, precisely the same computation applies after a change of coordinates to the stereographic projections on $\mathbb{S}^1 \backslash \{i\} \times \mathbb{S}^1 \backslash \{-i\}$), and more importantly that $g^*(\omega)-\omega$ is not closed, so $[g^*(\omega)] \neq [\omega]$ in $H^1(\mathbb{T}^2)$.