 Solution

Use homogeneous coordinates $X_0,\dots ,X_n$. Multiplication by these degree-one sections defines the sheaf morphismOn $U_i=D_{+}(X_i)$, write $t_j^{(i)}=X_j/X_i$, with $t_i^{(i)}=1$, and let $E_j^{(i)}$ be the standard vector in component $j$ using the local line bundle frame $X_i^{-1}$ of $\mathcal{O}(-1)$. Then $ϵ(E_j^{(i)})=t_j^{(i)}$, so $ϵ$ is surjective since $ϵ(E_i^{(i)})=1$. Its kernel is a locally free sheaf with basisIndeed any vector in the kernel is uniquely a sum of these vectors, by solving for its $i$-th component.

The Kähler differential sheaf ${\Omega }_{{\mathbf{P}}^n/k}^1$ has basis $d{t}_j^{(i)}$ on $U_i$, since this chart is a polynomial affine scheme. Define the local map $d{t}_j^{(i)}\mapsto F_j^{(i)}$. To check that these isomorphisms glue, on $U_i\cap U_{\ell }$ put $u=t_i^{(\ell )}$ and $v=t_j^{(\ell )}$. ThenThe twisting sheaf on projective space frame changes by $E_j^{(i)}=u^{-1}{E}_j^{(\ell )}$, henceFor $j=\ell$, use $t_{\ell }^{(\ell )}=1$, $d{t}_{\ell }^{(\ell )}=0$ and $F_{\ell }^{(\ell )}=0$; the same formula applies. Thus the differential and kernel frames have identical transition matrices. We obtain the cotangent form of the Euler sequence:This construction works over any field, including positive characteristic.

The Kähler differential sheaf has rank $n$. Taking determinant line bundles in this locally split short exact sequence of sheaves givesThe left side is the tensor product of $n+1$ copies of $\mathcal{O}(-1)$, and $\det (\mathcal{O})=\mathcal{O}$. ConsequentlyIt is the canonical bundle of projective space. The determinant identity can also be seen directly by taking the wedge of the kernel basis followed by a lift of the quotient basis; changing that lift by a kernel vector leaves the wedge unchanged.