Let $\mathrm{P}_{\mathbb{C}}^{n}$ be the n-dimensional projective space, we pick the local charts \begin{equation} U_i=\lbrace [z]\in\mathrm{P}_{\mathbb{C}}^n\mid z_i\ne0\rbrace,\quad\phi_i([z_0:\dots:z_n])=(z_0/z_i,\dots,z_{i-1}/z_i,z_{i+1}/z_i,\dots,z_n/z_i) \end{equation} then the transition maps are $(i>j)$ \begin{equation} \phi_{ij}(u_1,\dots,u_n)=(u_1/u_i,\dots,u_{j-1}/u_i,1/u_i,u_j/u_i,\dots,u_{i-1}/u_i,u_{i+1}/u_i,\dots, u_n/u_i). \end{equation} If instead we consider \begin{equation} \tilde{\phi}_{ij}(u_1,\dots,u_n)=(u_1/u_i,\dots,u_{j-1}/u_i,u_j/u_i,\dots,u_{i-1}/u_i,1/u_i,u_{i+1}/u_i,\dots, u_n/u_i) \end{equation} then Huybrechts claims that
$\phi_{ij}$ is the composition of $\tilde{\phi}_{ij}$ with the permutation $(j+1,i)$ of parity $(-1)^{i-j-1}$.
Huybrechts, Daniel, Complex geometry. An introduction, Universitext. Berlin: Springer (ISBN 3-540-21290-6/pbk). xii, 309 p. (2005). ZBL1055.14001. See page 92, Remark and Proposition 2.4.3.
Is this claim true?
To obtain $\phi_{ij}$ from $\tilde{\phi}_{ij}$ I need to move $1/u_i$ from the $i$-th place to the $j$-th place, which is done by $i-j$ subsequent transpositions. E.g.: Moving $7$ to the 3rd place: \begin{equation} 123456789\to 123457689\to 123475689\to 123745689\to127345689 \end{equation}
is achieved in $7-3=4$ steps.
The previous fact is used to prove that
the determinant of the jacobian is \begin{equation} \det \operatorname{J}\phi_{ij}=(-1)^{i-j+1}\det \operatorname{J}\tilde{\phi}_{ij}=(-1)^{i-j}(1/u_i)^{n+1}, \end{equation} where $\det \operatorname{J}\tilde{\phi}_{ij}=-(1/u_i)^{n+1}$.
\begin{equation} \lvert \operatorname{J}\tilde{\phi}_{ij}\rvert=\begin{vmatrix} u_i^{-1} & & & & & &-u_1u_i^{-2} &\\ &\ddots & & & & &\vdots &\\ & & u_{i}^{-1} & & & &-u_{j-1}u_i^{-2 } &\\ & & & u_{i}^{-1} & & & -u_j u_{i}^{-2} &\\ & & & &\ddots & & \vdots &\\ & & & & & u_{i}^{-1} & - u_{i-1}u_{i}^{-2}\\ & & & 0 & & & \fbox{$-u_i^{-2}$} & &\\ & & & & & & -u_{i+1}u_{i}^{-2} & u_{i}^{-1}\\ & & & & & & \vdots& & &\ddots &\\ & & & & & & -u_nu_i^{-2} & & & &u_{i}^{-1}\\ \end{vmatrix}=-(u_i)^{-(n+1)}. \end{equation}
If instead I calculate $\det \operatorname{J}\phi_{ij}$
\begin{equation} \operatorname{J}\phi_{ij}=\begin{bmatrix} u_i^{-1} & & & & & &-u_1u_i^{-2} &\\ &\ddots & & & & &\vdots &\\ & & u_{i}^{-1} & & & &-u_{j-1}u_i^{-2 } &\\ & & & 0 & & & \fbox{$-u_i^{-2}$} & &\\ & & & u_{i}^{-1} & & & -u_j u_{i}^{-2} &\\ & & & &\ddots & & \vdots &\\ & & & & & u_{i}^{-1} & - u_{i-1}u_{i}^{-2}\\ & & & & & & -u_{i+1}u_{i}^{-2} & u_{i}^{-1}\\ & & & & & & \vdots& & &\ddots &\\ & & & & & & -u_nu_i^{-2} & & & &u_{i}^{-1}\\ \end{bmatrix} \end{equation} I expand with respect to the row of the $(j,i)$-element, which yields \begin{equation} \det \operatorname{J}\phi_{ij}=(-1)^{i+j}(-u_i^{-2})(u_i^{-1})^{n-1}=(-1)^{i-j+1}(u_i)^{-(n+1)}. \end{equation} This differs from Huybrechts' result by a factor $(-1)$. Moreover the jacobian of $\phi_{ij}$ should be related to that of $\tilde{\phi}_{ij}$ by moving the $j$-th row below the $i$-th one (i.e. below $-u_{i-1}u_i^{-2}$) which requires $i-j$ switches, consistently with what I said before.
So I am probably making some mistakes in both the computation of the parity of the permutation and that of the determinant, which should be embarrassingly easy, but nonetheless I can't see where. I'm so concerned about this $(-1)^{i-j}$ because it is interpreted as the Čech coboundary of $\lbrace U_i,(-1)^i\rbrace$ so that the cocycle of $K_{\mathrm{P}_{\mathbb{C}}^{n}}$ and that $\mathcal{O}_{\mathrm{P}_{\mathbb{C}}^{n}}(-n-1)$ are equal in Čech cohmology.
Can someone help?
If $i>j$ then \begin{equation} \begin{split} \phi_i\circ\phi_j^{-1}(u_1,\dots,u_n)&=\phi_i([u_1:\dots:1:u_j:\dots:u_n])\\ &=(u_1/u_{i-1},\dots,1/u_{i-1},\dots, u_{i-2}/u_{i-1},u_{i}/u_{i-1}\dots u_n/u_{i-1}) \end{split} \end{equation} where the insertion of $1$ in the $j$-th position causes the successive elements to shift by one place (therefore the $i$-th element is $u_{i-1}$).
\begin{equation} \operatorname{J}\phi_{ij}=\begin{bmatrix} u_{i-1}^{-1} & & & & & &-u_1u_{i-1}^{-2} &\\ &\ddots & & & & &\vdots &\\ & & u_{i-1}^{-1} & & & &-u_{j-1}u_{i-1}^{-2 } &\\ & & & 0 & & & \fbox{$-u_{i-1}^{-2}$} & &\\ & & & u_{i-1}^{-1} & & & -u_j u_{i-1}^{-2} &\\ & & & &\ddots & & \vdots &\\ & & & & & u_{i-1}^{-1} & - u_{i-2}u_{i-1}^{-2}\\ & & & & & & -u_{i}u_{i-1}^{-2} & u_{i-1}^{-1}\\ & & & & & & \vdots& & &\ddots &\\ & & & & & & -u_nu_{i-1}^{-2} & & & &u_{i-1}^{-1}\\ \end{bmatrix} \end{equation} The determinant is then \begin{equation} (-1)^{i-1+j}(-u_{i-1}^{-n-1})=(-1)^{i-j}(z_i/z_j)^{-n-1}, \end{equation} since I have expanded w.r.t. the $(j,i-1)$ element (and the indices of the homogeneous coordinates start from $0$). For the same reason the sign of the permutation is correct.