Direct calculation of the canonical bundle of the complex projective space 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?
 A: 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.
