I am trying to understand the proof of the Theorem 2.5 here(Theorem of the Cube).
It is required to prove that an invertible sheaf on $X\times Y\times Z$ is trivial when the restrictions to $\left \{ x \right \}\times Y\times Z,X\times \left \{ y \right \}\times Z,X\times Y\times \left \{ z \right \}$ are trivial. $X,Y, Z$ are complete varieties(integral) over $k$ and $x\in X(k),y\in Y(k), z\in Z(k)$.[$Z$ may be taken as only a connected scheme]
The proof follows from David Mumford's book "Abelian Varieties". I understood most of the proof but not the last part where it is proved that $H^1(X\times Y,O_{X\times Y})=0$
The obstruction for lifting s to a global section of $L_2$ is an element $ξ ∈ H^1(X × Y,O_{X×Y} )$ We know that the restrictions of $L_2$ to ${x} × Y × Z_2$ and to $X × {y} × Z_2$ are trivial. Writing $i_1 = (id_X, y): X→ X × Y\ and\ i_2 = (x, id_Y ): Y→ X × Y$ , this means that ξ has trivial image under $i_1^{*}: H^1(X ×Y,O_{X×Y} ) → H^1(X, O_X) $ and under $i_2^{*}: H^1(X ×Y,O_{X×Y} ) → H^1(Y, O_Y)$.
I am missing something and cannot figure out how the images in $H^1(X, O_X),H^1(Y, O_Y)$ are $0$. Any and all help is appreciated!