I'm trying to solve the following exercise:
Let $k$ be a field and $f : \mathbb{P}_k^1 \longrightarrow \mathbb{P}_k^1$ a morphism of schemes defined by (in homogeneous coordinates) $$ [x:y] \mapsto [x^2:y^2]$$
- Show that $f_*\mathcal{O}_{\mathbb{P}_k^1}$ is locally free of rank 2
- Show that $f_*\mathcal{O}_{\mathbb{P}_k^1} \cong \mathcal{O}_{\mathbb{P}_k^1} \oplus \mathcal{O}_{\mathbb{P}_k^1}(-1)$
The idea I have in mind is to consider the morphism in the principal opens $D_+(x)$ and $D_+(y)$, because restricting there we have a morphism between affine spaces, and therefore can get the associated morphism of rings.
But in order to get $f_*\mathcal{O}_{\mathbb{P}_k^1}(D_+(x)) = \mathcal{O}_{\mathbb{P}_k^1}(f^{-1}(D_+(x)))$ I need to know how $f$ acts on all points of $\mathbb{P}_k^1$. I don't see how that follows from the definition of $f$ given above, because it only defines the image of the rational points of $\mathbb{P}_k^1$.
Let me change a little bit the notation, just to show you a more compact argument.
To not go in confusion, I denote by $Y:=\mathbb P^1$ the domain of $f$ with coordinates $x_0,x_1$, and by $X:=\mathbb P^1$ the codomain of $f$ with coordinates $z_0,z_1$, so that $f\colon Y\to X$ is sending $[x_0,x_1] \mapsto [x_0^2,x_1^2]$.
First of all we observe that this map is a double cover of $Y=\mathbb P^1$. In other words, given the following action of $\mathbb Z_2$ on $X=\mathbb P^1$
$$ \overline{0}=Id_{\mathbb P^1}, \qquad \overline{1}\colon \mathbb P^1\to \mathbb P^1, \quad [x_0,x_1]\mapsto [x_0,-x_1],$$
then the invariant functions with respect this action are $x_0^2$ and $x_1^2$, so that the quotient map is exactly our map $f\colon \mathbb P^1\to \mathbb P^1/\mathbb Z_2\cong \mathbb P^1,\quad [x_0,x_1]\mapsto [x_0^2,x_1^2]$.
Let us study now the sheaf $f_*\mathcal O_Y$ on $X$. Denote by $z_0, z_1$ the coordinates on $X$ and consider $U_i:=\lbrace z_i\neq 0 \rbrace$.
Then
$$(f_*\mathcal O_{Y})_{|U_i}=\lbrace \text{functions on } \ Y \ \text{defined on} \ f^{-1}(U_i)=\{x_i\neq 0 \} \rbrace=\mathbb C[\frac{x_j}{x_i}], \quad j\neq i.$$
Now we have to understand which is our ring as $O_X-module$. In order to do this, let us take a polynomial $g\in (f_*\mathcal O_{Y})_{|U_i}=\mathbb C[\frac{x_j}{x_i}]$, $g=a_0+a_1\frac{x_j}{x_i}+\dots a_d\left( \frac{x_j}{x_i}\right)^d$.
Which are the $O_X(U_i)$-functions on $X$? They are just the invariant functions on $Y$, since $Y$ is the quotient of $X$ with respect the action of $\mathbb Z_2$. But the invariant functions on $Y$ are exactly polynomials involving $1$ and $\left( \frac{x_j}{x_i}\right)^2$.
Therefore our polynomial $g$ can be written as
$$g=\left(a_0+a_2\left( \frac{x_j}{x_i}\right)^2+\dots +a_{d-1}\left( \frac{x_j}{x_i}\right)^{d-1}\right)\cdot 1 + \left(a_1+a_3\left( \frac{x_j}{x_i}\right)^2+\dots +a_{d}\left( \frac{x_j}{x_i}\right)^{d-1}\right)\cdot \frac{x_j}{x_i}, $$
where $a_0+a_2\left( \frac{x_j}{x_i}\right)^2+\dots +a_{d-1}\left( \frac{x_j}{x_i}\right)^{d-1}, a_1+a_3\left( \frac{x_j}{x_i}\right)^2+\dots +a_{d}\left( \frac{x_j}{x_i}\right)^{d-1}\in O_X(U_i)$ (here I am supposing that $d$ is odd, but it works is a similar way for $d$ even).
We have just proved that $(f_*\mathcal O_{Y})_{|U_i} =O_X(U_i)\cdot 1\oplus O_X(U_i)\cdot \frac{x_j}{x_i}\cong \bigoplus_{i=1}^2 O_X(U_i)$ as $O_X(U_i)$-moduli just by fixing the basis $\lbrace 1,\frac{x_j}{x_i}\rbrace $. Since $X=U_0\cup U_1$, then we have proved that $f_*\mathcal O_{Y}$ is a locally free sheaf of rank $2$.
Please observe that we should expect what we have obtained since the action of $\mathbb Z_2$ on $Y$ induces an action on $(f_*\mathcal O_{Y})_{|U_i}$, that it will be splitted in this case as a sum of invariant functions and anti-invariant functions on $Y$. Here the space of invariant functions is $O_X(U_i)\cdot 1$ whilst the space of anti-invariant functions is $O_X(U_i)\cdot \frac{x_j}{x_i}$.
It remains to establish which is exactly this locally free sheaf of rank $2$ on $X$. In order to do this, we have just to give a look to the $co-cycles$ of our sheaf. We have the following linear map of $O_X(U_i\cap U_j)$-moduli
$$\bigoplus_{i=1}^2 O_X(U_i\cap U_j)\to (f_*\mathcal O_{Y})_{|U_i\cap U_j} \to \bigoplus_{i=1}^2 O_X(U_j\cap U_i), \qquad \begin{pmatrix}h_1 \\ h_2\end{pmatrix}\mapsto h_1\cdot 1+h_2\cdot \frac{x_j}{x_i}\mapsto \begin{pmatrix}1 & 0 \\ 0 & \frac{z_i}{z_j} \end{pmatrix}\begin{pmatrix}h_1 \\ h_2\end{pmatrix}$$
Therefore the co-cycles are $g_{ji}=\begin{pmatrix}1 & 0 \\ 0 & \frac{z_i}{z_j} \end{pmatrix}$ which are exactly the co-cycles of $O_{\mathbb P^1}\oplus O_{\mathbb P^1}(-1)$ on $U_i\cap U_j$. We have therefore proved
$$f_*\mathcal O_{Y}=O_{\mathbb P^1}\oplus O_{\mathbb P^1}(-1).$$
Now I would like to say some comments. The theory behind what I have written is a lot. As you can expect, what I said holds in much more general situations. In particular, when we are considering abelian covers of algebraic varieties $\pi \colon Y\to Y/G\cong X$ given by the action on $Y$ of a finite abelian group $G$. In this sense, we would get the following splitting
$$\pi_*O_Y=\bigoplus_{\chi \in G^*}L_\chi^{-1},$$
where $L_\chi$ are suitable invertible sheaves on $X$ and $\chi \in G^*$ is a character of the group $G$. Moreover, $L_0=\mathcal O_X$ always. In our context, we have got that $L_0=\mathcal O_{\mathbb P^1}$, whilst $L_{\overline{1}}=\mathcal O_{\mathbb P^1}(1)$.
This is the starting point of the theory of abelian coverings of algebraic varieties, where the invertible sheaves $L_\chi$ play a crucial role to the existence of our covering $\pi$. I will not go into more detail, but if you are interested in, please see https://eudml.org/doc/153330 .