Extending holomorphic 1-forms I'm working on the following problem:
The holomorphic $1$-form $\frac{dz}{1+z^2}$,
defined on $\mathbb{C}\setminus\{\pm i\}$, can be extended to a
holomorphic $1$-form $\omega$ on $\mathbb{P}^1\setminus\{\pm i\}$.
My idea is to choose two coordinate neighborhoods of $\mathbb{P}^1\setminus\{ \pm i\}$ (with one of them being a subset of $\mathbb{C} \setminus \{\pm i\}$), write down a holomorphic 1-form in each local coordinate, and show that they agree on the overlap. In particular, consider the coordinate patches $(U_1,z)$ and $(U_2, \frac{1}{z})$ where $U_1 = \mathbb{C} \setminus \{ \pm i\}$ and
$U_2 = \mathbb{P}^1 \setminus \{0, \pm i\}$. Let $w = \frac{1}{z}$, then $f(w)dw = -\frac{1}{1 + w^2}dw$ is a holomorphic 1-form on $U_2$. On $U_1 \cap U_2$, one has
\begin{equation*}
    f(w)dw = f(\frac{1}{z})\frac{\partial w}{\partial z}dz = (-\frac{z^2}{z^2 + 1})\cdot (-\frac{1}{z^2})dz = \frac{dz}{z^2 + 1}
\end{equation*}
so one can patch $-\frac{dw}{1 + w^2}$ and $\frac{dz}{1+ z^2}$ together to obtain a holomorphic 1-form on $\mathbb{P}^1 \setminus \{ \pm i\}$.
I'm wondering whether this is correct? I'm not so sure because when I compute the pullback form $\tan^{*}(\omega)$ with respect to the map $\tan: \mathbb{C} \to \mathbb{P}^1 \setminus \{ \pm i\}$, I don't get a unique expression. Specifically, the pullback form on $U_1$ is $dz$, while the pullback form on $U_2$ is $-\frac{1}{1 + \tan(z)^2}\sec^2(z)dz = -dz$, so I made at least one conceptual mistake.
 A: I find your notation (using $\tan$ and $\sec$) slightly confusing, so let me just work it out in the way that I consider to be "clean". We take the standard open cover $\{U_0,U_1\}$ for $\mathbb{CP}^1$. Denote the coordinate on $U_0$ by $z$, and on $U_1$ by $w$. These are related via $w=1/z$. Suppose that the $1$-form $\omega=\frac{dz}{1+z^2}$ is defined on $U_0\setminus\{\pm i\}$. We want to see if this can be extended to $\mathbb{CP}^1\setminus\{\pm i\}$. The way to do this, is to first pullback via the map $\varphi:U_1\cap U_0\to U_0$ given by $w\mapsto 1/w$. This yields
$$\varphi^*\omega=\frac{d(1/w)}{1+(1/w)^2}=-\frac{dw}{1+w^2}$$
So far, this is in agreement with what you wrote. Now, the points $\pm i$ are mapped to $1/\pm i=\mp i$. As such, $\varphi^*\omega$ may be extended to all of $U_1\setminus\{\pm i\}$ in the obvious way, by the formula above across the origin of $\mathbb{C}\cong U_1$. By construction, the respective $1$-forms on $U_0\setminus\{\pm i\}$ and $U_1\setminus\{\pm i\}$ agree on $U_0\cap U_1\setminus\{\pm i\}$, so they glue to give a holomorphic $1$-form on $\mathbb{CP}^1\setminus\{\pm i\}$.
I do not know why you are using $\tan$ and $\sec$. I have had to calculate things on $\mathbb{CP}^1=S^2$ for multiple different courses and I have never seen these being used as charts maps. I think it leads to unnecessary confusion. I hope the explanation above is clear.
A: You also asked how one can find the pullback of the $1$-form $\omega$ (on $\mathbb{P}^1\setminus\{\pm i\}$) by the tangent function $\operatorname{tan}\colon \mathbb{C}\rightarrow \mathbb{P}^1\setminus\{\pm i\}$. This was omitted in Quaere Verum's answer.
As you indicate, when using the local representation $\frac{1}{1+z^2}dz$ for $\omega$ on $U_1$, we have
$${\operatorname{tan}}^*\omega=\Big(\operatorname{tan}^* \frac{1}{z^2+1}\Big)d\Big(\operatorname{tan}^*z\Big)=\frac{1}{\operatorname{tan}(z)^2+1}\cdot\frac{\partial}{\partial z}\Big(\operatorname{tan}(z)\Big)dz=\frac{\operatorname{cos}(z)^2}{\operatorname{cos}(z)^2}dz=dz.$$
As expected, when using the local representation $-\frac{1}{1+w^2}dw$ for $\omega$ on $U_2$, we also find
$$\operatorname{tan}^*\omega=\Big(\operatorname{tan}^* -\frac{1}{1+w^2}\Big)d\Big(\operatorname{tan}^*w\Big)=\Big(-\frac{1}{1+\big(\frac{1}{\operatorname{tan}(z)}\big)^2}\Big)d\Big(\frac{1}{\operatorname{tan}(z)}\Big)=-\operatorname{sin}(z)^2\frac{\partial}{\partial z}\Big(\frac{1}{\operatorname{tan}(z)}\Big)dz=dz.$$
