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On the standard affine charts $U_i=\{X_i\ne 0\}$, use the formal frame $e_i=X_i^m$ and glue rank-one free sheaves byThe transition functions are regular units on overlaps and satisfy the cocycle identity. This constructs the twisting sheaf on projective space $\mathcal{O}(m)$ as an invertible sheaf for every integer $m$, including negative $m$.

On a nonempty $U$, a section has local expressions $h_i{e}_i$, where $h_i\in \mathcal{O}(U\cap U_i)$. Pulling to the punctured affine cone gives $h_i(X/X_i)X_i^m$. Their agreement on overlaps makes them one homogeneous rational function $r$ of degree $m$, regular on ${\pi }^{-1}U$. To make its polynomial representation explicit, write one nonzero local coefficient as $p(y)/q(y)$ in the affine coordinates. Homogenize $p,q$ to degrees $a,b$; the resulting rational expression isIf the exponent of $X_i$ is negative, move its power to the denominator. Cancel common factors; the greatest common divisor of homogeneous polynomials can be chosen homogeneous. Thus $r=F/G$ with $F,G$ coprime homogeneous polynomials, $G\ne 0$, and $\mathrm{deg}F-\mathrm{deg}G=m$.

Conversely, for such a homogeneous rational function regular on ${\pi }^{-1}U$, its coefficient $r/X_i^m$ is regular on $U\cap U_i$: restrict to the slice $X_i=1$. These coefficients obey the displayed transition rule and hence define a section. This proves the homogeneous rational sections of a twisting sheaf description in both directions, for arbitrary open $U$ rather than only a standard chart.

Now lift a regular degree-zero rational function $f=P/Q$ to the punctured cone. The quotient rule givesThis is a rational function homogeneous of degree $-1$. It is independent of the chosen representation because differentiation is a derivation of the rational function field. It is regular wherever $f$ is regular: a derivation of a polynomial ring extends to each localization by the quotient rule, and regular functions are locally such fractions. The preceding homogeneous-section description therefore provesZeros of a particular displayed denominator do not invalidate this argument; regularity is a property of the rational function, which can have another local representation.

The map $f\mapsto ({\partial }_{0}f,\dots ,{\partial }_{n}f)$ is a $k$-linear sheaf derivation with values in $\mathcal{O}(-1{)}^{\oplus (n+1)}$. The universal property of Kähler differentials consequently definesMultiplication by the coordinate section $X_i\in \Gamma ({\mathbb{P}}^n,\mathcal{O}(1))$ defines the other mapFor a degree-$d$ homogeneous polynomial, differentiating each monomial gives ${\sum }_i{X}_i{\partial }_{i}P=dP$. Applying this to the equal-degree numerator and denominator gives ${\sum }_i{X}_i{\partial }_{i}f=0$ for degree-zero $f$, so $\sigma D=0$. This identity is valid in every characteristic.

To prove all the exactness assertions, fix $U_i$ and put $y_j=X_j/X_i$ for $j\ne i$. The Kähler differential sheaf is free there on $d{y}_j$, because a derivation of the polynomial ring is determined freely by its values on the coordinates. Trivialize each $\mathcal{O}(-1)$ with frame $X_i^{-1}$. In this frame,where $e_j$ now denotes the $j$th coordinate vector of the direct sum. The map $\sigma$ is surjective, since its $i$th coefficient is one. Its kernel consists exactly of tuples with $b_i=-{\sum }_{j\ne i}{y}_j{b}_j$, so the $n$ displayed vectors $e_j-y_j{e}_i$ form a free basis of that kernel. The map $D$ sends the differential basis bijectively to this kernel basis and is therefore injective. Exactness on these charts proves the cotangent Euler sequence in homogeneous coordinatesThe proof uses no division by an integer and hence no characteristic-zero assumption.

For $n\ge 1$, the cohomology of twisting sheaves on projective space givesIn particular, $H^0(\mathcal{O})=k$, $H^{q>0}(\mathcal{O})=0$, and $H^q(\mathcal{O}(-1))=0$ for every $q$. The long exact sequence in sheaf cohomology of the displayed Euler sequence beginsand gives zero in the remaining degrees. ThusThe nonzero group is generated by the connecting image of the section $1$ of $\mathcal{O}$. If $n=0$ is admitted, ${\mathbb{P}}^0$ is a point and its cotangent sheaf is zero, so the sole requested group $H^0$ is zero.