[proofplan] The assertion is local, so we work in holomorphic coordinate charts on $X$ and $Y$. In such coordinates, a form of type $(p,q)$ is a sum of coefficient functions multiplied by wedges of exactly $p$ holomorphic coordinate differentials and exactly $q$ antiholomorphic coordinate differentials. Holomorphicity of $F$ implies that the pullback of each holomorphic coordinate differential on $Y$ is a linear combination only of holomorphic coordinate differentials on $X$, and similarly for antiholomorphic differentials after conjugation. Linearity and compatibility of pullback with wedge products then preserve the number of holomorphic and antiholomorphic factors in every local term. [/proofplan]

[step:Express the form locally in holomorphic coordinates] Let $x_0 \in X$ be arbitrary, and set $y_0 := F(x_0) \in Y$. Choose a holomorphic coordinate chart $(U,\varphi)$ on $X$ with $x_0 \in U$, where \begin{align*} \varphi: U \to \varphi(U) \subset \mathbb{C}^m \end{align*} has coordinate functions $z_1,\dots,z_m: U \to \mathbb{C}$. Choose a holomorphic coordinate chart $(V,\psi)$ on $Y$ with $y_0 \in V$ and, after replacing $U$ by the open neighbourhood $U \cap F^{-1}(V)$ of $x_0$, assume $F(U) \subset V$. Write \begin{align*} \psi: V \to \psi(V) \subset \mathbb{C}^n \end{align*} with coordinate functions $w_1,\dots,w_n: V \to \mathbb{C}$. Since $\alpha \in A^{p,q}(Y)$, on $V$ it has the local form \begin{align*} \alpha|_V = \sum_{I,J} a_{I,J}\, dw_I \wedge d\bar w_J \end{align*} where $I=(i_1,\dots,i_p)$ ranges over strictly increasing $p$-tuples in $\{1,\dots,n\}$, $J=(j_1,\dots,j_q)$ ranges over strictly increasing $q$-tuples in $\{1,\dots,n\}$, each coefficient is a smooth function $a_{I,J}: V \to \mathbb{C}$, and \begin{align*} dw_I := dw_{i_1}\wedge \cdots \wedge dw_{i_p} \end{align*} while \begin{align*} d\bar w_J := d\bar w_{j_1}\wedge \cdots \wedge d\bar w_{j_q}. \end{align*} If $p=0$ or $q=0$, the corresponding wedge factor is interpreted as $1$. [/step]

[step:Compute the pullback of holomorphic coordinate differentials]For each $a \in \{1,\dots,n\}$, define the smooth coordinate component \begin{align*} f_a: U \to \mathbb{C}, \qquad f_a := w_a \circ F. \end{align*} Because $F$ is holomorphic and $w_a$ is a holomorphic coordinate function on $V$, the function $f_a$ is holomorphic on $U$. Hence, in the coordinates $z_1,\dots,z_m$, \begin{align*} \frac{\partial f_a}{\partial \bar z_r} = 0 \end{align*} for every $r \in \{1,\dots,m\}$. The differential of a smooth complex-valued function in complex coordinates is \begin{align*} df_a = \sum_{r=1}^m \frac{\partial f_a}{\partial z_r}\, dz_r + \sum_{r=1}^m \frac{\partial f_a}{\partial \bar z_r}\, d\bar z_r. \end{align*} Therefore \begin{align*} F^*(dw_a) = d(w_a \circ F) = df_a = \sum_{r=1}^m \frac{\partial f_a}{\partial z_r}\, dz_r. \end{align*} Thus $F^*(dw_a)$ is a linear combination of the holomorphic coordinate one-forms $dz_1,\dots,dz_m$ and has type $(1,0)$.[/step]

[guided]Fix $a \in \{1,\dots,n\}$. The coordinate function $w_a: V \to \mathbb{C}$ selects the $a$-th holomorphic coordinate on $Y$. Pulling this coordinate function back along $F$ gives the smooth function \begin{align*} f_a: U \to \mathbb{C}, \qquad f_a := w_a \circ F. \end{align*} Since $F$ is holomorphic and $w_a$ is holomorphic in the chosen chart on $Y$, the composite $f_a$ is holomorphic in the chosen chart on $X$. In complex coordinates, holomorphicity is exactly the vanishing of the antiholomorphic derivatives: \begin{align*} \frac{\partial f_a}{\partial \bar z_r} = 0 \end{align*} for every $r \in \{1,\dots,m\}$. Now compute the pullback of the coordinate one-form $dw_a$. Pullback commutes with exterior differentiation on functions, so \begin{align*} F^*(dw_a) = d(w_a \circ F) = df_a. \end{align*} For a smooth complex-valued function, the [exterior derivative](/theorems/1525) decomposes into holomorphic and antiholomorphic coordinate parts: \begin{align*} df_a = \sum_{r=1}^m \frac{\partial f_a}{\partial z_r}\, dz_r + \sum_{r=1}^m \frac{\partial f_a}{\partial \bar z_r}\, d\bar z_r. \end{align*} The second sum vanishes because $f_a$ is holomorphic. Hence \begin{align*} F^*(dw_a) = \sum_{r=1}^m \frac{\partial f_a}{\partial z_r}\, dz_r. \end{align*} This is the key point: no $d\bar z_r$ term appears. Therefore the pullback of a holomorphic coordinate differential is again a form of type $(1,0)$ on $X$.[/guided]

[step:Compute the pullback of antiholomorphic coordinate differentials] For each $a \in \{1,\dots,n\}$, the antiholomorphic coordinate function on $V$ is \begin{align*} \bar w_a: V \to \mathbb{C}, \qquad y \mapsto \overline{w_a(y)}. \end{align*} Using $f_a = w_a \circ F$ as above, we have \begin{align*} \bar w_a \circ F = \overline{f_a}. \end{align*} Since $f_a$ is holomorphic, $\overline{f_a}$ is antiholomorphic in the coordinates $z_1,\dots,z_m$, so \begin{align*} \frac{\partial \overline{f_a}}{\partial z_r} = 0 \end{align*} for every $r \in \{1,\dots,m\}$. Therefore \begin{align*} F^*(d\bar w_a) = d(\bar w_a \circ F) = d\overline{f_a} = \sum_{r=1}^m \frac{\partial \overline{f_a}}{\partial \bar z_r}\, d\bar z_r. \end{align*} Thus $F^*(d\bar w_a)$ is a linear combination of the antiholomorphic coordinate one-forms $d\bar z_1,\dots,d\bar z_m$ and has type $(0,1)$. [/step]

[step:Apply multiplicativity and linearity of pullback to each local summand] Pullback is linear over sums, pulls back coefficient functions by composition, and is compatible with wedge products. Hence on $U$, \begin{align*} F^*(\alpha|_V) = \sum_{I,J} (a_{I,J}\circ F)\, F^*(dw_I) \wedge F^*(d\bar w_J). \end{align*} For $I=(i_1,\dots,i_p)$, \begin{align*} F^*(dw_I) = F^*(dw_{i_1})\wedge \cdots \wedge F^*(dw_{i_p}), \end{align*} and each factor $F^*(dw_{i_s})$ is a linear combination of $dz_1,\dots,dz_m$. Therefore $F^*(dw_I)$ is a sum of wedge products containing exactly $p$ holomorphic coordinate differentials. Similarly, for $J=(j_1,\dots,j_q)$, \begin{align*} F^*(d\bar w_J) = F^*(d\bar w_{j_1})\wedge \cdots \wedge F^*(d\bar w_{j_q}), \end{align*} and each factor is a linear combination of $d\bar z_1,\dots,d\bar z_m$, so $F^*(d\bar w_J)$ is a sum of wedge products containing exactly $q$ antiholomorphic coordinate differentials. Multiplying by the smooth coefficient function $a_{I,J}\circ F: U \to \mathbb{C}$ does not change type. Hence every local summand of $F^*\alpha$ has type $(p,q)$ on $U$, and therefore \begin{align*} F^*\alpha|_U \in A^{p,q}(U). \end{align*} [/step]

[step:Conclude the type condition globally] The point $x_0 \in X$ was arbitrary. For every point of $X$, we have found a holomorphic coordinate neighbourhood on which $F^*\alpha$ is represented as a smooth sum of terms with exactly $p$ holomorphic coordinate differentials and exactly $q$ antiholomorphic coordinate differentials. Since being of type $(p,q)$ is a local condition on a complex manifold, it follows that \begin{align*} F^*\alpha \in A^{p,q}(X). \end{align*} This proves that holomorphic pullback preserves type $(p,q)$. [/step]