I am new to differential geometry and am learning it. My question here is just a technical one, but I am unable to figure it out and become confused after trying for quite a while, so I hope someone can help me out.
I am checking the statement $df^*\omega=f^*d\omega$ for a particular example, where $\omega$ is a differential form on a manifold $N$, and $f$ is a map from another manifold $M$ to $N$. From my explicit calculation, I find $df^*\omega\neq f^*d\omega$ in my case.
More precisely, consider a 2-manifold $N$ with coordinates $\theta$ and $\phi$, and consider a 2-form $\omega=\sin\theta d\theta\wedge d\phi$ on $N$. So $N$ can be viewed as $S^2$, and $\omega$ can be viewed as the canonical volume form on $S^2$. In particular, $d\omega=0$.
Now consider a 3-manifold $M=N\times [0, 1]$, so the coordinates of $M$ can be taken as $\theta$, $\phi$, and $t\in [0, 1]$. Define a map $f: M\rightarrow N$ as: \begin{equation} \left( \begin{array}{c} \theta\\ \phi\\ t \end{array} \right)\in M \stackrel{f}\longrightarrow \left( \begin{array}{c} \frac{\pi}{2}t+(1-t)\theta\\ (1-t)\phi \end{array} \right)\in N \end{equation}
My goal is to check if $df^*\omega=f^*d\omega$, where the RHS vanishes because $d\omega=0$. However, I found that \begin{equation} f^*\omega=\sin\theta(1-t)^2d\theta\wedge d\phi+\phi\sin\theta(1-t)dt\wedge d\theta+\left(\frac{\pi}{2}-\theta\right)\sin\theta(1-t)dt\wedge d\phi \end{equation} and \begin{equation} df^*\omega=\left(\frac{\pi}{2}-\theta\right)\cos\theta(1-t)d\theta\wedge dt\wedge d\phi\neq 0=f^*d\omega \end{equation}
On the other hand, one can prove that $df^*\omega=f^*d\omega$, so I think I have made a mistake. I appreciate if someone can let me know where my mistake is.