I am studying about an action $I^*$ on a de Rham cohomology group $H^1(S^1\times S^2)$ induced from an action $I\cdot (z,x)=(\overline{z},-x) $ where $S^1\times S^2\subset \mathbb{C}\times \mathbb{R}^3$. Note that, by Kunneth formula, $$H^1(S^1\times S^2)=\mathbb{R}.$$
Thus, I want to find a nonzero element in $H^1(S^1\times S^2)$ and want to see how $I^*$ acts to the element.
And my teacher taught me as below.
Let $d\theta \in \Omega^1(S^1)$ be a generator of $H^1(S^1)=\mathbb{R}.$ And let $\pi:S^1\times S^2\rightarrow S^1$ and let $\omega=\pi^*(d\theta)$. Then clearly, $d\omega=0$ so $[\omega]$ is nonzero element in $H^1(S^1\times S^2)$. If $\iota : S^1\times \{\text{north pole}\}\hookrightarrow S^1\times S^2$ is an embedding, observe that $$\pi\circ\iota =Id \implies \iota^*\pi^*=Id \implies \iota^*(\omega)=d\theta \implies \iota^*[\omega]=[d\theta] $$
Thus, $$I^*\omega=I^*\pi^*(d\theta)=\pi^*(I_1^*d\theta)=\pi^*(-d\theta)=-\omega.$$
So we get $I^*=-Id$.
But I got stuck at why $I_1^*d\theta=-d\theta$.
If we see carefully, $I_1^*d\theta(z)=d\theta(\overline{z})$. I don't know where is wrong and where I am missing.
I would very appreciate for any help and solution for this issue! Thank you!
Note that $I_1^*d\theta = d(\theta\circ I_1)$. As $(\theta\circ I_1)(z) = \theta(I_1(z)) = \theta(\bar{z}) = -\theta(z)$, so $\theta\circ I_1 = -\theta$ and hence $I_1^*d\theta = d(\theta\circ I_1) = d(-\theta) = -d\theta$.