[proofplan] Let $L := \mathcal{O}_{\mathbb{CP}^n}(-1)$ be the tautological line subbundle of the product bundle $\mathbb{CP}^n \times \mathbb{C}^{n+1}$, so that $\mathcal{O}_{\mathbb{CP}^n}(1) = L^*$. The proof is the bundle-theoretic exact sequence obtained by dualizing the tautological line in the product vector bundle only in the first factor: tensor the quotient sequence $0 \to L \to \mathbb{CP}^n \times \mathbb{C}^{n+1} \to (\mathbb{CP}^n \times \mathbb{C}^{n+1})/L \to 0$ with $L^*$. The middle bundle becomes $\mathcal{O}(1)^{\oplus(n+1)}$, the left bundle becomes $\mathcal{O}$, and the right bundle is identified with $T\mathbb{CP}^n$ through the standard tangent-space description $T_{[\ell]}\mathbb{CP}^n \cong \operatorname{Hom}_{\mathbb{C}}(\ell,\mathbb{C}^{n+1}/\ell)$. We verify these identifications and the holomorphic exactness in affine charts. [/proofplan] [step:Define the tautological quotient sequence and tensor it by $L^*$] Let \begin{align*} L := \mathcal{O}_{\mathbb{CP}^n}(-1) \subset \mathbb{CP}^n \times \mathbb{C}^{n+1} \end{align*} denote the tautological holomorphic line bundle, whose fiber over a point $[\ell] \in \mathbb{CP}^n$ is the one-dimensional complex subspace $\ell \subset \mathbb{C}^{n+1}$. Define \begin{align*} Q := (\mathbb{CP}^n \times \mathbb{C}^{n+1})/L \end{align*} to be the holomorphic quotient bundle. The inclusion of $L$ into the product bundle gives a short exact sequence of holomorphic vector bundles \begin{align*} 0 \longrightarrow L \longrightarrow \mathbb{CP}^n \times \mathbb{C}^{n+1} \longrightarrow Q \longrightarrow 0. \end{align*} Since $L^* = \mathcal{O}_{\mathbb{CP}^n}(1)$ is a holomorphic line bundle, tensoring by $L^*$ is fiberwise an exact operation. Therefore we obtain a short exact sequence \begin{align*} 0 \longrightarrow L^* \otimes L \longrightarrow L^* \otimes (\mathbb{CP}^n \times \mathbb{C}^{n+1}) \longrightarrow L^* \otimes Q \longrightarrow 0. \end{align*} [/step] [step:Identify the first two tensor bundles with $\mathcal{O}$ and $\mathcal{O}(1)^{\oplus(n+1)}$] The bundle $L^* \otimes L$ is canonically identified with $\mathcal{O}_{\mathbb{CP}^n}$ by the trace-contraction map from $L^*_{[\ell]} \otimes L_{[\ell]}$ to $\mathbb{C}$ that sends $\lambda \otimes v$ to $\lambda(v)$, where $[\ell] \in \mathbb{CP}^n$, $\lambda \in L^*_{[\ell]} = \ell^*$, and $v \in L_{[\ell]} = \ell$. Let $(e_0,\dots,e_n)$ be the standard basis of $\mathbb{C}^{n+1}$. The product bundle decomposes as \begin{align*} \mathbb{CP}^n \times \mathbb{C}^{n+1} \cong \bigoplus_{i=0}^{n} \mathcal{O}_{\mathbb{CP}^n} e_i. \end{align*} Tensoring this decomposition by $L^*$ gives \begin{align*} L^* \otimes (\mathbb{CP}^n \times \mathbb{C}^{n+1}) \cong \bigoplus_{i=0}^{n} L^* \cong \mathcal{O}_{\mathbb{CP}^n}(1)^{\oplus(n+1)}. \end{align*} For each $0 \leq i \leq n$, let $Z_i \in H^0(\mathbb{CP}^n,L^*)$ denote the holomorphic section whose value at $[\ell]$ is the restriction to $\ell$ of the $i$-th coordinate functional on $\mathbb{C}^{n+1}$. Under these identifications, the map from $\mathcal{O}_{\mathbb{CP}^n}$ to $\mathcal{O}_{\mathbb{CP}^n}(1)^{\oplus(n+1)}$ is the one induced by the homogeneous coordinate sections $Z_0,\dots,Z_n$. Indeed, at $[\ell] \in \mathbb{CP}^n$, choose a nonzero vector $z = (z_0,\dots,z_n) \in \ell$ and choose $\lambda \in \ell^*$ satisfying $\lambda(z)=1$. The identity element of $\operatorname{End}_{\mathbb{C}}(\ell) \cong \ell^* \otimes \ell$ is $\lambda \otimes z$, and its image in $\ell^* \otimes \mathbb{C}^{n+1}$ is \begin{align*} \lambda \otimes z = \sum_{i=0}^{n} z_i(\lambda \otimes e_i). \end{align*} Thus its components are precisely the restrictions of the coordinate functionals $Z_i$ to $\ell$. [guided] The tensor sequence becomes the Euler sequence only after we identify its first two terms carefully. First, for each point $[\ell] \in \mathbb{CP}^n$, the fiber $L_{[\ell]}$ is the line $\ell$, and the fiber $L^*_{[\ell]}$ is its dual [vector space](/page/Vector%20Space) $\ell^*$. The [tensor product](/page/Tensor%20Product) $\ell^* \otimes \ell$ is naturally the vector space $\operatorname{End}_{\mathbb{C}}(\ell)$ by the map $\lambda \otimes v \mapsto (w \mapsto \lambda(w)v)$. Since $\ell$ is one-dimensional, $\operatorname{End}_{\mathbb{C}}(\ell)$ is canonically identified with $\mathbb{C}$ by sending an endomorphism to the scalar by which it acts. Equivalently, the trace-contraction map from $\ell^* \otimes \ell$ to $\mathbb{C}$ that sends $\lambda \otimes v$ to $\lambda(v)$ identifies $L^* \otimes L$ with the holomorphic line bundle $\mathcal{O}_{\mathbb{CP}^n}$. Next, the middle term is simple because the product bundle has a fixed global basis. Let $(e_0,\dots,e_n)$ be the standard basis of $\mathbb{C}^{n+1}$. Then $\mathbb{CP}^n \times \mathbb{C}^{n+1} \cong \bigoplus_{i=0}^{n} \mathcal{O}_{\mathbb{CP}^n} e_i$, and tensoring with $L^*$ gives $L^* \otimes (\mathbb{CP}^n \times \mathbb{C}^{n+1}) \cong \bigoplus_{i=0}^{n} L^* \cong \mathcal{O}_{\mathbb{CP}^n}(1)^{\oplus(n+1)}$. It remains to understand the actual injection. Fix $[\ell] \in \mathbb{CP}^n$, choose a nonzero vector $z = (z_0,\dots,z_n) \in \ell$, and choose $\lambda \in \ell^*$ with $\lambda(z)=1$. The identity map on $\ell$ is represented by $\lambda \otimes z \in \ell^* \otimes \ell$, because for $w = a z \in \ell$ one has $\lambda(w)z = a z = w$. After including $\ell$ into $\mathbb{C}^{n+1}$, this element becomes $\lambda \otimes z = \sum_{i=0}^{n} z_i(\lambda \otimes e_i)$ in $\ell^* \otimes \mathbb{C}^{n+1}$. Here $Z_i \in H^0(\mathbb{CP}^n,L^*)$ is the section whose value at $[\ell]$ is the restriction to $\ell$ of the $i$-th coordinate functional on $\mathbb{C}^{n+1}$. Thus the $i$-th component of the injection is multiplication by the homogeneous coordinate section $Z_i$. This is exactly the coordinate-section map $\mathcal{O} \to \mathcal{O}(1)^{\oplus(n+1)}$ appearing in the theorem. [/guided] [/step] [step:Identify $L^* \otimes Q$ with the holomorphic tangent bundle] For $[\ell] \in \mathbb{CP}^n$, the fiber of $L^* \otimes Q$ is \begin{align*} (L^* \otimes Q)_{[\ell]} = \ell^* \otimes (\mathbb{C}^{n+1}/\ell) \cong \operatorname{Hom}_{\mathbb{C}}(\ell,\mathbb{C}^{n+1}/\ell). \end{align*} We identify this vector space with $T_{[\ell]}\mathbb{CP}^n$. Let $U_j := \{[Z_0:\cdots:Z_n] \in \mathbb{CP}^n : Z_j \ne 0\}$ be the $j$-th affine chart. Define the chart map $\varphi_j: U_j \to \mathbb{C}^n$ by $\varphi_j([Z_0:\cdots:Z_n]) = (w_0,\dots,w_{j-1},w_{j+1},\dots,w_n)$, where $w_i := Z_i/Z_j$ for $i \ne j$. On $U_j$, every line $\ell$ has a unique representative $s_j(w) := (w_0,\dots,w_{j-1},1,w_{j+1},\dots,w_n) \in \mathbb{C}^{n+1}$, where $w = \varphi_j([\ell]) \in \mathbb{C}^n$. For a tangent vector $a = (a_i)_{i \ne j} \in T_w\mathbb{C}^n \cong \mathbb{C}^n$, define $A_a: \ell \to \mathbb{C}^{n+1}/\ell$ by sending $c\,s_j(w)$ to $c\,[a_0,\dots,a_{j-1},0,a_{j+1},\dots,a_n] \bmod \ell$. This gives a complex-linear isomorphism \begin{align*} T_{[\ell]}\mathbb{CP}^n \cong \operatorname{Hom}_{\mathbb{C}}(\ell,\mathbb{C}^{n+1}/\ell). \end{align*} The formula depends holomorphically on $w$. On an overlap $U_j \cap U_k$, write $u = \varphi_k \circ \varphi_j^{-1}(w)$. The normalized representatives satisfy $s_k(u) = s_j(w)/w_k$ when $k \ne j$, and differentiating this relation in the tangent direction $a \in T_w\mathbb{C}^n$ gives \begin{align*} (d(\varphi_k \circ \varphi_j^{-1})_w a)_i = \frac{a_i w_k - w_i a_k}{w_k^2} \end{align*} for indices $i \ne k,j$, while the component corresponding to $j$ is $-a_k/w_k^2$. Substituting this derivative into the definition of $A_a$ gives the same element of $\operatorname{Hom}_{\mathbb{C}}(\ell,\mathbb{C}^{n+1}/\ell)$, because the difference between the two lifted vectors in $\mathbb{C}^{n+1}$ is a scalar multiple of $s_j(w)$ and therefore vanishes modulo $\ell$. Thus the local identifications transform by the differential of the holomorphic coordinate change $\varphi_k \circ \varphi_j^{-1}$. Hence these fiberwise isomorphisms glue to a holomorphic vector bundle isomorphism \begin{align*} T\mathbb{CP}^n \cong L^* \otimes Q. \end{align*} [/step] [step:Conclude exactness of the Euler sequence] Substituting the identifications $L^* \otimes L \cong \mathcal{O}_{\mathbb{CP}^n}$, $L^* \otimes (\mathbb{CP}^n \times \mathbb{C}^{n+1}) \cong \mathcal{O}_{\mathbb{CP}^n}(1)^{\oplus(n+1)}$, and $L^* \otimes Q \cong T\mathbb{CP}^n$ into the tensor sequence gives $0 \longrightarrow \mathcal{O}_{\mathbb{CP}^n} \longrightarrow \mathcal{O}_{\mathbb{CP}^n}(1)^{\oplus(n+1)} \longrightarrow T\mathbb{CP}^n \longrightarrow 0$. The first map is the homogeneous-coordinate map described above, and the second map is induced fiberwise by the quotient map $\mathbb{C}^{n+1} \to \mathbb{C}^{n+1}/\ell$ followed by the tangent-space identification $\ell^* \otimes (\mathbb{C}^{n+1}/\ell) \cong T_{[\ell]}\mathbb{CP}^n$. Exactness holds because it was obtained from the exact tautological quotient sequence by tensoring with the line bundle $L^*$ and then applying holomorphic vector bundle isomorphisms. This proves the Euler sequence for $\mathbb{CP}^n$. [/step]