[proofplan] We define the maps on Dolbeault forms by applying the holomorphic bundle maps only to the coefficient vector in the [tensor product](/page/Tensor%20Product) of forms with bundle fibres. In holomorphic local frames, the bundle maps are represented by matrices of holomorphic functions, so the local formula for the Dolbeault operator shows that these induced maps commute with $\bar{\partial}$. Exactness in each degree is then proved at the level of smooth bundle-valued forms: injectivity is pointwise, the kernel is the smooth subbundle $E \subset F$, and surjectivity follows by smooth local lifting and a smooth [partition of unity](/page/Partition%20of%20Unity). [/proofplan]

[step:Define the induced maps on bundle-valued forms] Fix an integer $q \geq 0$. Let $\Lambda^{p_0,q}T^*X$ denote the complex vector bundle of smooth complex-valued forms of type $(p_0,q)$ on $X$. We use the standard identification \begin{align*} A^{p_0,q}(X,E) = \Gamma\left(X,\Lambda^{p_0,q}T^*X \otimes E\right), \end{align*} and similarly for $F$ and $G$. Define the smooth vector bundle morphism \begin{align*} \operatorname{id}_{\Lambda^{p_0,q}T^*X} \otimes i: \Lambda^{p_0,q}T^*X \otimes E \to \Lambda^{p_0,q}T^*X \otimes F \end{align*} over $X$. The induced map on smooth sections is \begin{align*} i_*^q: A^{p_0,q}(X,E) \to A^{p_0,q}(X,F). \end{align*} Likewise, define \begin{align*} \operatorname{id}_{\Lambda^{p_0,q}T^*X} \otimes \pi: \Lambda^{p_0,q}T^*X \otimes F \to \Lambda^{p_0,q}T^*X \otimes G \end{align*} and let \begin{align*} \pi_*^q: A^{p_0,q}(X,F) \to A^{p_0,q}(X,G) \end{align*} be the induced map on smooth sections. These maps are complex-linear because $i$ and $\pi$ are complex-linear on every fibre. [/step]

[step:Show that the induced maps commute with the Dolbeault differentials]Let $U \subset X$ be an open coordinate neighbourhood on which $E$, $F$, and $G$ admit holomorphic frames. Let $r_E := \operatorname{rank} E$, $r_F := \operatorname{rank} F$, and $r_G := \operatorname{rank} G$. Choose holomorphic frames $(e_a)_{a=1}^{r_E}$ for $E|_U$, $(f_b)_{b=1}^{r_F}$ for $F|_U$, and $(g_c)_{c=1}^{r_G}$ for $G|_U$. Since $i$ and $\pi$ are holomorphic vector bundle morphisms, there are holomorphic functions \begin{align*} I_{ba}: U \to \mathbb{C} \end{align*} and \begin{align*} P_{cb}: U \to \mathbb{C} \end{align*} such that \begin{align*} i(e_a)=\sum_{b=1}^{r_F} I_{ba} f_b \end{align*} and \begin{align*} \pi(f_b)=\sum_{c=1}^{r_G} P_{cb} g_c. \end{align*} Let $\omega \in A^{p_0,q}(X,E)$. On $U$, write \begin{align*} \omega|_U = \sum_{a=1}^{r_E} \alpha_a \otimes e_a, \end{align*} where each \begin{align*} \alpha_a \in A^{p_0,q}(U) \end{align*} is a smooth scalar-valued form. The local formula for the Dolbeault operator in a holomorphic frame gives \begin{align*} \bar{\partial}_E(\omega|_U)=\sum_{a=1}^{r_E} \bar{\partial}\alpha_a \otimes e_a. \end{align*} Therefore \begin{align*} i_*^{q+1}\left(\bar{\partial}_E\omega\right)|_U = \sum_{a=1}^{r_E}\sum_{b=1}^{r_F} I_{ba}\bar{\partial}\alpha_a \otimes f_b. \end{align*} On the other hand, \begin{align*} i_*^q(\omega)|_U = \sum_{a=1}^{r_E}\sum_{b=1}^{r_F} I_{ba}\alpha_a \otimes f_b. \end{align*} Since each $I_{ba}$ is holomorphic, $\bar{\partial}I_{ba}=0$. The scalar Leibniz rule for $\bar{\partial}$ gives \begin{align*} \bar{\partial}(I_{ba}\alpha_a)=I_{ba}\bar{\partial}\alpha_a. \end{align*} Hence \begin{align*} \bar{\partial}_F\left(i_*^q\omega\right)|_U = \sum_{a=1}^{r_E}\sum_{b=1}^{r_F} I_{ba}\bar{\partial}\alpha_a \otimes f_b = i_*^{q+1}\left(\bar{\partial}_E\omega\right)|_U. \end{align*} Since the equality holds on every such neighbourhood $U$, it holds globally: \begin{align*} \bar{\partial}_F \circ i_*^q = i_*^{q+1} \circ \bar{\partial}_E. \end{align*} The same computation with the holomorphic matrix entries $P_{cb}$ gives \begin{align*} \bar{\partial}_G \circ \pi_*^q = \pi_*^{q+1} \circ \bar{\partial}_F. \end{align*} Thus $i_*$ and $\pi_*$ are cochain maps.[/step]

[guided]We prove the commutation with $\bar{\partial}$ locally, because the Dolbeault operator with coefficients in a holomorphic vector bundle has its simplest expression in holomorphic frames. Choose an [open set](/page/Open%20Set) $U \subset X$ on which the three bundles have holomorphic frames. Let $r_E := \operatorname{rank} E$, $r_F := \operatorname{rank} F$, and $r_G := \operatorname{rank} G$, and write these frames as $(e_a)_{a=1}^{r_E}$, $(f_b)_{b=1}^{r_F}$, and $(g_c)_{c=1}^{r_G}$. Since $i: E \to F$ is holomorphic, its local matrix entries in these frames are holomorphic functions \begin{align*} I_{ba}: U \to \mathbb{C} \end{align*} defined by \begin{align*} i(e_a)=\sum_{b=1}^{r_F} I_{ba} f_b. \end{align*} Likewise, since $\pi: F \to G$ is holomorphic, there are holomorphic functions \begin{align*} P_{cb}: U \to \mathbb{C} \end{align*} defined by \begin{align*} \pi(f_b)=\sum_{c=1}^{r_G} P_{cb} g_c. \end{align*} Now take an $E$-valued form $\omega \in A^{p_0,q}(X,E)$. On $U$ it has a unique expression \begin{align*} \omega|_U = \sum_{a=1}^{r_E} \alpha_a \otimes e_a, \end{align*} where each coefficient \begin{align*} \alpha_a \in A^{p_0,q}(U) \end{align*} is a smooth scalar-valued form. In a holomorphic frame, the bundle Dolbeault operator acts only on the scalar form coefficients, so \begin{align*} \bar{\partial}_E(\omega|_U)=\sum_{a=1}^{r_E} \bar{\partial}\alpha_a \otimes e_a. \end{align*} Applying $i$ to the coefficient vector then gives \begin{align*} i_*^{q+1}\left(\bar{\partial}_E\omega\right)|_U = \sum_{a=1}^{r_E}\sum_{b=1}^{r_F} I_{ba}\bar{\partial}\alpha_a \otimes f_b. \end{align*} We compare this with first applying $i$ and then applying $\bar{\partial}_F$. First, \begin{align*} i_*^q(\omega)|_U = \sum_{a=1}^{r_E}\sum_{b=1}^{r_F} I_{ba}\alpha_a \otimes f_b. \end{align*} The only possible obstruction to commuting with $\bar{\partial}$ would be the derivative of the coefficient functions $I_{ba}$. But those functions are holomorphic, so $\bar{\partial}I_{ba}=0$. Applying the scalar Leibniz rule for the Dolbeault operator yields \begin{align*} \bar{\partial}(I_{ba}\alpha_a)=I_{ba}\bar{\partial}\alpha_a. \end{align*} Therefore \begin{align*} \bar{\partial}_F\left(i_*^q\omega\right)|_U = \sum_{a=1}^{r_E}\sum_{b=1}^{r_F} I_{ba}\bar{\partial}\alpha_a \otimes f_b. \end{align*} This is exactly the same local expression as $i_*^{q+1}(\bar{\partial}_E\omega)|_U$, so \begin{align*} \bar{\partial}_F \circ i_*^q = i_*^{q+1} \circ \bar{\partial}_E. \end{align*} The proof for $\pi$ is identical in structure, with the matrix entries $P_{cb}$ replacing $I_{ba}$. Since the $P_{cb}$ are also holomorphic, $\bar{\partial}P_{cb}=0$, and the same Leibniz-rule computation gives \begin{align*} \bar{\partial}_G \circ \pi_*^q = \pi_*^{q+1} \circ \bar{\partial}_F. \end{align*} Thus both induced maps are cochain maps. This is precisely where holomorphicity of the original bundle maps is used; for merely smooth bundle maps, the extra $\bar{\partial}$ of the matrix entries would not vanish.[/guided]

[step:Prove injectivity degree by degree] Fix $q \geq 0$ and let $\omega \in A^{p_0,q}(X,E)$ satisfy $i_*^q\omega=0$. For each $x \in X$, the value $\omega_x$ lies in \begin{align*} \Lambda^{p_0,q}T_x^*X \otimes E_x. \end{align*} The map on fibres is \begin{align*} \operatorname{id}_{\Lambda^{p_0,q}T_x^*X} \otimes i_x: \Lambda^{p_0,q}T_x^*X \otimes E_x \to \Lambda^{p_0,q}T_x^*X \otimes F_x. \end{align*} Since the original bundle sequence is exact, $i_x: E_x \to F_x$ is injective. Tensoring over the field $\mathbb{C}$ preserves injectivity, so $\operatorname{id}_{\Lambda^{p_0,q}T_x^*X} \otimes i_x$ is injective. Hence $\omega_x=0$ for every $x \in X$, and therefore $\omega=0$. Thus $i_*^q$ is injective. [/step]

[step:Identify the kernel of $\pi_*^q$ with the image of $i_*^q$] Fix $q \geq 0$. Since $\pi \circ i=0$, we have \begin{align*} \pi_*^q \circ i_*^q = 0, \end{align*} so \begin{align*} \operatorname{im} i_*^q \subset \ker \pi_*^q. \end{align*} Conversely, let $\omega \in A^{p_0,q}(X,F)$ satisfy $\pi_*^q\omega=0$. For each $x \in X$, the element $\omega_x$ lies in \begin{align*} \Lambda^{p_0,q}T_x^*X \otimes F_x, \end{align*} and its image under \begin{align*} \operatorname{id}_{\Lambda^{p_0,q}T_x^*X} \otimes \pi_x \end{align*} is zero. Since \begin{align*} 0 \to E_x \xrightarrow{i_x} F_x \xrightarrow{\pi_x} G_x \to 0 \end{align*} is exact, the fibrewise kernel of $\pi_x$ is $i_x(E_x)$. It remains to see that the fibrewise preimage is smooth. Let $U \subset X$ be an open set on which the smooth subbundle $\ker(\pi|_{F|_U}) \subset F|_U$ has a smooth frame. Such open sets exist by local triviality of smooth vector bundles. Since $i|_{E|_U}: E|_U \to \ker(\pi|_{F|_U})$ is a fibrewise linear isomorphism of smooth vector bundles, its inverse is smooth in local frames. Extending a smooth frame of $\ker(\pi|_{F|_U})$ to a smooth frame of $F|_U$ defines a smooth projection $F|_U \to \ker(\pi|_{F|_U})$. Composing this projection with $(i|_{E|_U})^{-1}$ gives a smooth vector bundle morphism \begin{align*} s_U: F|_U \to E|_U \end{align*} such that \begin{align*} i \circ s_U \end{align*} is the identity on $\ker(\pi|_{F|_U})$. Define \begin{align*} \eta_U: U \to \Lambda^{p_0,q}T^*U \otimes E|_U \end{align*} by \begin{align*} \eta_U = \left(\operatorname{id}_{\Lambda^{p_0,q}T^*U} \otimes s_U\right)(\omega|_U). \end{align*} Then $\eta_U$ is smooth and satisfies \begin{align*} i_*^q\eta_U=\omega|_U. \end{align*} On overlaps, these local sections agree because $i_*^q$ is injective. Therefore the $\eta_U$ glue to a global section \begin{align*} \eta \in A^{p_0,q}(X,E) \end{align*} with \begin{align*} i_*^q\eta=\omega. \end{align*} Hence \begin{align*} \ker \pi_*^q \subset \operatorname{im} i_*^q. \end{align*} Combining the two inclusions gives \begin{align*} \ker \pi_*^q = \operatorname{im} i_*^q. \end{align*} [/step]

[step:Lift every $G$-valued form by smooth local lifts and a partition of unity] Fix $q \geq 0$ and let \begin{align*} \gamma \in A^{p_0,q}(X,G). \end{align*} Choose an open cover $(U_\lambda)_{\lambda \in \Lambda}$ of $X$ such that over each $U_\lambda$ the smooth vector bundle surjection \begin{align*} \pi|_{F|_{U_\lambda}}: F|_{U_\lambda} \to G|_{U_\lambda} \end{align*} has a smooth right inverse \begin{align*} r_\lambda: G|_{U_\lambda} \to F|_{U_\lambda} \end{align*} with \begin{align*} \pi \circ r_\lambda = \operatorname{id}_{G|_{U_\lambda}}. \end{align*} Indeed, after shrinking $U_\lambda$ if necessary, local trivializations identify $\pi|_{F|_{U_\lambda}}$ with a smooth surjective matrix-valued map of constant rank; choosing a nonvanishing maximal minor gives a smooth local right inverse by the usual matrix inverse formula on that minor. Since $X$ is a paracompact complex manifold, its underlying smooth manifold admits a smooth locally finite partition of unity subordinate to the open cover. Choose such a partition $(\rho_\lambda)_{\lambda \in \Lambda}$. For each $\lambda$, define the local smooth lift \begin{align*} \widetilde{\gamma}_\lambda: U_\lambda \to \Lambda^{p_0,q}T^*U_\lambda \otimes F|_{U_\lambda} \end{align*} by \begin{align*} \widetilde{\gamma}_\lambda = \left(\operatorname{id}_{\Lambda^{p_0,q}T^*U_\lambda} \otimes r_\lambda\right)(\gamma|_{U_\lambda}). \end{align*} Extend $\rho_\lambda\widetilde{\gamma}_\lambda$ by zero outside $U_\lambda$. The locally finite sum \begin{align*} \widetilde{\gamma}=\sum_{\lambda \in \Lambda}\rho_\lambda\widetilde{\gamma}_\lambda \end{align*} defines an element \begin{align*} \widetilde{\gamma} \in A^{p_0,q}(X,F). \end{align*} Using $\pi \circ r_\lambda=\operatorname{id}_{G|_{U_\lambda}}$ and $\sum_{\lambda \in \Lambda}\rho_\lambda=1$, we compute \begin{align*} \pi_*^q\widetilde{\gamma} = \sum_{\lambda \in \Lambda}\rho_\lambda\pi_*^q\widetilde{\gamma}_\lambda = \sum_{\lambda \in \Lambda}\rho_\lambda\gamma = \gamma. \end{align*} Thus $\pi_*^q$ is surjective. [/step]

[step:Assemble the degreewise exact maps into a short exact sequence of complexes] For every $q \geq 0$, we have proved that \begin{align*} 0 \to A^{p_0,q}(X,E) \xrightarrow{i_*^q} A^{p_0,q}(X,F) \xrightarrow{\pi_*^q} A^{p_0,q}(X,G) \to 0 \end{align*} is exact. We have also proved that $i_*$ and $\pi_*$ commute with the corresponding Dolbeault differentials. Therefore the degreewise maps assemble into a short exact sequence of cochain complexes \begin{align*} 0 \to A^{p_0,\bullet}(X,E) \xrightarrow{i_*} A^{p_0,\bullet}(X,F) \xrightarrow{\pi_*} A^{p_0,\bullet}(X,G) \to 0. \end{align*} This is the desired short exact sequence of Dolbeault complexes. [/step]